Physics 1st Year Quantum_.pdf

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Physics (Sem-} & 2)
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9
1-2 A (Sem-1 & 2) Relativistic Mechanics
'ft~l;, .(J(;tlik(u~
.' it;'l'Q$tultites
Relativistic Mechanics CONCEPT OUTLINE: PART-l
Frame of Reference: It is that coordinate system which is used to
identify the position or motion of an object.
Types ofFrame of Reference:
P art-l (1-2A to 1-9A) a. Inertial frame ofrefeJ:ence, and
b. Non-inertial frame of reference.
!. Frame ofReference';
inertial andlllQll-ine'
l
.. (;'alilean Tra:nsfo'rm
~".'., ~
" /lv/ichelson~MorleyEXl'''''~''~'.'' '" ; .. ~,e;,(~lestions I
Postulates of Specidl·'Phe<jjjfidfRela1Jif)#Y~ l1,?»'3ifmr:wt@,'M;tlf"'''~M]m;H!il?Wl'1t#f¥'''f''''u:i5>'·m;ff'@t':Irw'''>n: ·'.-x·~,,,,~'1£'Zn;:!~{W%""~~~~(",, -, , :. _
,,-;,; ,', . .";,,,.,.. ,;.
.---_.
PAl'"Show that the frame ofreference lDovtng with constant
(,'oncept Outline: Part-l 1-3A
'>. Long and Medium Answer Type Questions 1-2A
velocity v is an inertial frame of reference.
Part-2 (1-9A to 1-23A)
....
1..., . "
Let 8 is a frame of reference which is in rest to an obser'('r and s' i~
another frame of reference moving with constant velocit:v ' in till'
~ Lorentz Transformatiiin.s;~>;·
., Length Contraction
positive x direction with respect to the same observer.
Time Dilation
2. o and 0' are origin offrame 8 and 8' respectively.
". Felocity Additio(L Theg1;11lrtj
y,8 y' +s' ~ V
I
x , ..
.t. Concept Outline :. Part-2 l-9A
I
B. Long and Medium Answer Type Questions 1-9A I
:- x' - p
vt~
P art-3 (1-23A to 1-84A)
I
I ,
;" Variation ofMas~t1{;:j/i;lW;Pi:.'W:ri.'j,· <j!~~"f;:;:,ij:/·: 1 0 ' .
~ ,)----------.
.. Einstein's Mass Ene, ,
,
,
, RelatilJisticRelati()n ,
,
z
~ lV1assless Particles' z' ,
II
A. Concept Outline: Part-3 1-24A
~~ltl?~;IjI$I~.
iI. Long a.nd Medium Answer Type Questions 1-24A 3. Initially 0 and 0' coincide with each other at time t =t' =0,
where t and t' =time measured in 8 and s' frames respectively.
1-1 A (Sem·l & 2)
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Physics 1-3A (Sem-l & 2)
4. Suppose P be a point in the space.
5. Now from Fig. 1.1.1, x = x' + vt ...(1.1.1)
y = y' 1 •••(1.1.2)
z = Z ~ No relative motion ...(1.1.3)
t = t' J •••(1.1.4)
6. Eq. (1.1.1) to eq. (1.1.4) are position and time transformation equations
in sand s' frame.
7. Differentiating eq. (1.1.1) w.r.t. t on both sides,
dx dx' vdt'
- = -+-- ...(1.1.5)
dt dt dt
dx dx'. vdt'
- = -+-- ( .: t =t' dt = dt')
dt dt' dt'
=> u = u' + v ...(1.1.6)
8, Differentiating eq. (1.1.2) ~.r.t. t,
dy dy'
...(1.1.7)
dt = &
dy dy'
(.: t=t')
dt = dt'
=> u = u' ...(1.1.8)
9. Similarly, tl = u,Y ...(1.1.9)
10. Now differentiating ~q. (l~1.6) w.r.t. t, we get,
dux du: dv
-- = --+
dt dt dt
du du: ( .: v = constant)
d/ = dt
dux _ du:
(.: t =t')
dt - dt'
=> a =a' ...(1.1.10)
11. Similarly on differeritiati~geq. (1.1.8) and eq. (1.1.9), we get
ay =a'y . ...(1.1.11)
az= a'z ...(1.1.12)
12. Eq. (1.1.10), eq. (1.1.11) and eq. (1.1.12) shows that the acceleration is
invariant in both frames.
1:~. So a frame of reference moving with constant velocity is an inertial
frame.
Que 1:2:: '1 Derive the Galilean transfonnation equations and show
that its acceleration components are invariant.
Answer I
1. Suppose we are in an inertial frame of reference s and the coordinates
of some event that occurs at the time t are x, y, z as shown in Fig. 1.2.1.
I~A(Sem·l&2) Relativistic Mechanics
2. An observer located in a different inertial frame s' which is moving with
-+
respect to s at the constant velocity v, will find that the same event.
occurs at time t' and has the position coordinates x', y' and z'.
y y!
s s ' _
..v
o x 0' x'
z z'
...
3. Assume that v is in positive x direction.
4. When origins ofs and s' coincide, measurements in the x direction made
in s is greater than those ofs' by v
-+
t (distance).
5. Hence, x'=x-vt (1.2.1)
y' = y (1.2.2)
z'= z (1.2.3)
t'= t (1.2.4)
These set ofequations are known as Galilean transformations.
6. Differentiating eq. (1.2.1) with respect to t, we get
dx' = dx _ v dt} no relative motion
dt dt dt
dx' dx
--= --v (.: t = t' dt' = dt)
dt' dt
dy' dy
7. Similarly
dt' = dt
dz' dz
and
dt' = dt
8. Since, dx'/dt' = ur', the x-component of the velocity measured in s', and
dx/dt = ur' etc., then,
u'=u -v
r r ...(1.2.5)
u' = u
y y ...(1.2.6)
9.
u/ = Uz ...( 1.2.7)
Eq. (1.2.5), eq. (1.2.6) and eq. (1.2.7) can be written collectively in the
vector form as
-+
u' = u - v ...(1.2.8)
10. To obtain the acceleration transformation, we differentiate the eq. (1.2.5).
eq. (1.2.6) and eq. (1.2.7) with respect to time such that
du; = .!!:....(u -v)~ dUr
dt' dt r dt
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__
l'hysics 1-5 A (Sem-l & 2)
duy' = duy and dU; = duz
Similarly,
dt' dt dt' dt
dux' ,duy' ,du; ,
Since, --= a '--=a '--=a
dt' x 'dt' y' dt' z
du du du
-_X_=a . --y-=a ; __z_=a
dt x' dt y dt z
Then we get a'= a ...(1.2.9)
.• x
ay
'= ay
...(1.2.10)
a'= a ...(1.2.11)
z z
or writing these equations collectively, ;;, = ~
U, The measured components,ofacceleration ofa particle are independent
of the uniform relative velocity ofthe reference frames.
,i Tn other words, acceleration remains invariant when passing from one
inertial frame to another that is in uniform relative translational motion.
'~ue 1.3. IShow that the distance between points is invariant under
";,,iilean transformations.
Answer·'· :1
Let (xl'.Y1' Zl) and (x2' Y2, Z2) be the coordinate oftwo points P and Q in
l'est frame 5. Then the distance between them will be
d = J(x2·- XI )2 + (Y2 - YI)2 + (Z2 - ZI)2
.Now the distance between them measured in moving frame 5' is
d' = J(x2 '- Xl ')2 + (Y2 '- YI ')2 + (Z2 '- Zl ')2
U sing Galilean transformation
x2' = x2 - v/,Y2' =Y2 -v/ and Z2' = Z2 -v,t
Xl' = Xl -v/,YI ' =YI -vi and Zl' = Zl -v,t
1, Hence
,,i' J[(x~ - V
x
t)-(~xtW-+[(Y2 -vyt)-(YI _vy t)]2 + [(Z2 -vzt)-(Zl -vzt)]2
'" J(x~ - XI )2 + (Y2 .,.- YI)2 + (Z2 - ZI)2
d'= d
Que 1.4;,/.,1 Discuss the objective and outcome ofMichelson-Morley
experiment.
An~~~~~!i~
A. Objective of Michelson-Morley Experiment:
The main objective of conducting the Michelson-Morley experiment
was to confirm the existence ofstationary ether.
According to Morley ifthere exist some imaginary medium like 'ether' in
the earth atmosphere, there should be some tilue difference between
relative motion ofbody with respect to earth and against the motion of
earth.
I~A (Sem-l & 2) Relativistic Mechanics
3. Due to this time difference there exist some path difference and if such
path difference occurs, Huygens concept is correct and if it does not
occur then Huygens concept is wrong.
B. Michelson-Morley Experiment:
1. In Michelson-Morley experiment there is a semi-silvered glass-plate P
aild two plane mirrorsMI andM2
which are mutually perpendicular and
equidistant from plate P.
2. There is a monochromatic light source in front of glass plate P,
3. The whole arrangement is fixed on a wooden stand and that wooden
stand is dipped in a mercury pond. So it becomes easy t.o rotate,
4. Let vbe the speed of imaginary medium (ether) w.r.t. earth and (' is the
velocity oflight, so time taken to move the light ray from plate P to M,
and reflected back,
C
T = ~ + ~ = d[ .2 l
I
c+v c-v (C2_V~)J
2dc 2d
T1
=
2
c [ 1-;:J=F ~;J
5. Expanding binomially,
TI
= 2d[I+~J ,.. (1.4.1)
2
C c
[Neglecting higher power term]
6. Time taken to move a light ray from plate P to M, and to reflect back,
2d 2d 2d [ v~ J~
I
T2 = .Jc2 _ v 2 = ~ = --;;- 1 - c2
~ "--2
C
M 1 M'l
I I ' I ':./r~ ':.1
':.
I ,
1 I ,
' ,
I
,,, M 2 M'2
, I I
, ;' ....
I'
I III ! ~'" , ; ' , -II
S > > 2 N ,. -....-- I I
,./ ,./ P' I I
.... ;'
I I
I- d" "I I..!
L
T
_.iht~~~~i~*l#Ji1~~nt.
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1-7 A (Sem-l & 2)
Physics
2
7. Expanding binomially, T = 2d [1 + V 2 ] •••(1.4.2)
2
c 2c
[Neglecting higher power terml
8. So, time difference,
2
2d [ v ] 2d [ v ]
2
M =T - T = - 1+- - - 1+
1 2 C c2 C 2c2
= -
2d [ v
1+
2
-1--
~] =-
2d ['v
- 2
2
] =-s-
d v
2
...(1.4.3)
2 2
c c 2c c2c c
9. Now, the apparatus is rotated by 90' so that the position ofmirror M 1
and M gets interchange. So time taken from P to M 1 is,
2
T' = 2d[1+~]
1 c 2c2
and time taken from P to M 2 is,
T '= 2d [1+ ';2]
2 C c2
10. Time difference, ""'t' = T]' - T 2'
2 2
= _
2d[1+----1
v
2
v
2
] =---=--
dv
S
2dv ...(1.4.4)
2
c 2c2
c 2c
s
C
11. So, total time difference,
AT _ A# At' _ dv
2
( dv"' 2dv
2
L.> - L.>£ - L.> - - -I - - ) = - 3
3 3
c . c c
12. Since, path difference =speed of light x !'>.T
2dv2
2dv2
...(1.4.5)
!'>. = c x !'>.T = ex - -
S = - 2
C c
If A is the wavelength oflight used then the path difference in terms of
13.
the number of fringes is given by,
2
n = !'>./ A= 2dv
15. Taking. d = 11 m
C
2
A
Velocity of earth, v = 3 X 104
mls
Velocity of light, c = 3 X 108
mls
[1 A= 10-10
ml
A= 6000 A
2x.11x9xlOa 22 ><102
22 11
.=-=
n=
9 x 1016
X 6000 X 10-10
6000 60 30
n = 0.36
Ifsuch type of medium like 'ether' exists in the atmosphere, there must
16.
be a fringe shift of 0.36.
Michelson-Morley performed that experiment several times in different
17.
situations, in different weather conditions but no fringe shift was obtained
hence Huygens concept of'ether drag' is wrong. This is known as negative
result.
1-8A (Sem-l & 2) Relativistic Mechanics
_ What will be the expected fringe shift on the basis of
stationary ether hypothesis in Michelson-Morley experiment ? If
the effective length of each part is 8 m and wavelength used is
8000 A? ..
-
1. We know that fringe shift is given by
n = 2dv
2
= 2x8x(3x10
4
)2 =0.2
C
2
A. (3 x 108
)2 X 8 X 10-7
[.: v =3 )( 104
mls and c =3 X 108
m/.~'
_ State Einstein's postulates ofspecial theory ofrelativity.
Explain why Galilean relativity failed to explain actual results of
Michelson-Morley eitperiment. _ ' i i i
A. Einstein's Postulates: There are two postulates ofthe special theory
-ofrelativity proposed by Einstein;
i. Postulate 1 :
The laws of physics are the same in all inertial frames of reference
moving with a constant velocity with respect to one another.
Explanation:
1. Ifthe laws ofphysics had different forms for observers in different
frames in relative motion, one could determine from these
differences which objects are stationary in space and which are
moving.
2. As there is no universal frame ofreference, therefore this distinction'
between objects cannot be made. Hence universal frame of
reference is absent.
ii. Postulate 2 :
The speed oflight in free space has the same value in all inertial frames
of reference. This speed is 2.998 )( 108 mls.
Explanation:
1. This postulate is directly followed from the result of Michelson-
Morley experiment.
B. Reason for Failure Galilean transfonnation to Explain Actual
Results of Michelson Morley Experiment:
1. In Galilean transformations the speed of light was not taken to be
constant in all inertial frames.
2. These equations were based on absolute time and absolute space.
3. The above two assumptions contradict the Einstein postulates.
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t'hysics 1-9A (Sem·l &2)
So the Galilean transformation failed to explain the actual results of
Michelson-Morley experiment.
CONCEPT OUTLINE: PART-2
Lorentz Transformations: The equation relating the coordinates
of a particle in the two inertial frame on the basis of special theory of
relativity are called Lorentz transformations.
(~ue f.'f~'l What are Lorentz transformation equations? Derive
1he expression for it.
'nswer I
The equations in special theory of relativity, which relate to the space
and time coordinates of an event in two inertial frames of reference
moving with a uniform velocity relative to one another, are called Lorentz
transformations.
Let us consider two frames of reference sand s' in which frame s' is
moving with velocity v along x-axis. The coordinatel1! of frame s are
(x, y, z, t) while the coordinates of frame s' are (X',y', Z', t ' ).
According to first postulates ofspecial theory ofrelativity in frame s',
x'ex(x-vt) :::> x'=k(x-vt) ...(1.7.1)
where, k = Proportionality constant.
In frame s, x ex (x'.+ vt')
X = k (x' +vt') ...(1.7.2)
Y +y'
5 : 5' _v
I
(x, y, z, tf : (x', y', z', t')
J · · · x'
x ,I' 0'
,
,
,
'z·
,
,
I-lOA (Sem·l &2) Relativistic Mechanics
5. Putting x' in eq. (1.7.2),
x = k [k (x - vt) + vt'] = [k2 (x - vt) + v kt'J
x = k 2
x - k2vt + kvt'
kvt' = (1 - k 2 )x + k2vt
t'= 0-k
2
)x+k
2
vt = (1-k~:::4 kt ...( I. 7 .:~ I
kv kv
6. According to second postulate of special theory of relativity, speed of
light is a constant quantity.
In frame s, x = ct
In frame 5', x' = ct'
7. Putting the value ofx' and t' from eq. (1. 7.1) and eq
k(x-vt)= ~0-k2)+ckt
vk
c
x[k- O-k
2
)] =lck+vklt
V k
[ck + vkJt
x= I-~-,--~
lk - vh0- k- )J
9. On comparingeq. (1.7.6) with eq. (1.7.4),
(ck + vk)
c
2
.- [k -~ck (1- k ) J
.,
c- 0
ck + vk = ck --(1-k-)
vk
2
vk = _~_(1-k2)
vk
v2 k 2 =_c" + c2k?
k2 (v2 _ c2 ) + e2 = 0
k 2 (e 2 _ v2 ) = c2
~[c2_V2] =1
2
c
2 v2l
k [1-
71J=1
k2 - _1_
- v 2
1---
2
e
1
or k= -
C?
yl- ~i-
10. Putting the value of k in eq. (1. 7.ll,
_
..( 1.7.41
...r1.751
(1.7.3) in eq. (1.71)),
...( 1.7.())
x' =-J 1 ., [x - v t1 (First Lorentz tran!'form;l1 iCl!1t>f]uatinll J
I v·
i 1- ..
~)1t_
v C'
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1-11 A (Sem-l & 2)
Physics
y' =y (Second Lorentz transformation equation)
z' = 2' (Third Lorentz transformation equation)
xO-k
2
) x [1 ]
t' = --- + kt =- - - k + kt
11. Now, kv v k
~lI -1 1+t
v 1 v2 1 v2
~,_;; J,-" J'-"
x[ ~ 1] 2
t
;- ~1-~-H2 +J .v
1-- 1--
2
c2
c
1---1 --
2
V2 + t = ~[ V21
~ c2
1 c + t
v2
v U -U V 1 _ -U
~.l-~ V~-~ V
1
c2 "1-~
l 2
_~[ v ]+ t =_-V=X+=C
t
2
U -U C2P
C .
~1-? ~1-C2 ~1- C2
vx
t-
C;
z
t'=
V.l-?
12. Similarly putting the value of k in eq. (1.7.2), inverse Lorentz
transformation equations,
H
1
2
x = [x' + vt'l
1--
2
c
Y =y'
g:2
z == z
vx'
t' +'--2
t=
1--
2
c
Qti~bfJ. '.. Deduce the Lorentz transformation equations from
Einstein's postulates. Also show that at low velocities, the Lorentz
transfonnations reduce to Galilean transformations.
-
1-12.A(Sem-l &2) Relativistic Mechanic>
A Lorentz Transformation Equations from Einstein's Postulates;
-Refer Q. 1.7, Page 1-9A, Unit-I.
B. Condition at which Lorentz Transformations Reduce to
Galilean Transformations :
1. Lorentz transformation equation,
...(1.8.1)
x=g2
1-
c
2. At low velocities means v«< c
2
v
Thus, 1- 2= 1
C
So eq. (1.8.1) reduces to; x = x' + vt' ...(1.8.21
Eq. (1.8.2) is a Galilean transformation equation.
3. It means atlow velocities, the Lorentz transformation reduces to Galilean
transformation.
_ Show that the space time interval:r + :r + %2 - c 2 t 2 = 0 is
invariant under Lorentz transfonnation.
1. The Lorentz transformation are
- vx'
x'+vt' " t'+-2
x= H ,y=y,z=z andt= gC
2 2
v v
1--
2 1--
2
c c
2. Using eq. (1.9.1) in given equation
x2 + y2 + Z2 = c2 t2
I ,2 t'+--
2
x + v t ] ,2 12 2 c
VX'J2
=> 2 +Y +z =c 2
v
g v
[ 1-- [ 1-
2 2
c c
H 2
y I2 +Z'2 = __
1 [ c2 t'2+ v X'2 X'2_ v 2t '2 J
=> v2
c2
1---
2
c
=> y'2+ Z'2 =~[(C2tI2_X12{1_ ;:)J
1-~
2
c
2 2 2
12(1 v ) '2(1 V ) '2(1 V ) ., '2(1 VZJ
=> x -?" + y - "?") + Z l -"?") = c"t l - c2)
...0.9.11
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Physics 1-13A (Sem·l & 2)
=> X'2+ y,Z+ Z'2 = c2 t'Z
3. Hence x2 + y2 + Z2 =c2 t2 is invariant under Lorentz transformation.
QuC:],p;w~JIAs measured by 0 a bulb goes offat x =100 kIn, y =10 kIn,
z =1 km and t = 5 x 1~ sec. What are the coordinates x', y',:t' and t' of
this event as determined by a second observer 0' moving relative to
o at - 0.8 c along the common x-x' axis ?
Answer:1
Gi*~:I1:x'= 1(1,(
v = --':'O~8x;3x1.
To Find i Cddr~
x-vt
L Since, xl
=
-~
~~--~2
100-(-0.8 x 3 x 10
5
)x 5x 10-4 = 366.66 km
x
l
= J1-(0.8)2
[.,. C = 3 X 105 km/s]
-4 (-0.8 x 3 x 105
) X 100
t - .;- x 5 x 10 - -----'(3 X 105)2
:l,.
t' = Jl~~;u = )1- <0.8)2
= 12.77 X 10-4
s
", y' =y = 10 km
"
z' = z =1 km
Quel,ll:<1 Show that a moving circle will appeared to be an ellip,'-'''
if it is seen from a frame which is at rest.
Answer I
l The equation of circle is x2 + y2 = a2.
:0: Putting the values from Lorentz transformation,
. + V ,2 2
t' ]2
---- +Y =a
J1-~:'
l
'
c "
X'"+V
2
t'2+2x'.vt' '2 0
_____ + y = a.~
( ,,21
·1-, I
. c- I
') a'2 v:a
y'"! + v-':'1''2 + 2x' Fr + y'2 - .V'2 V": 0.- ---,,
c- C'
1-14A (Sem.l & 2) Relativistic Mechanics
'Z ,2 (VZj 2( VZJ 2 '2 "
x +Y ll-~) =a 1- c2- -v t -2x vt
X'2 y'2 2 VZt'z+ 2x'vt'
...0,1111
V
2
-(~J+l = a _ ( )
1-"?- ll-?
'l v'l t''2+ 2x'vt '
3. Let, p2 = 1-~ and QZ= a - ----- '- ----
2
c ( v' I
~ 1 - ," J
Hence, eq. (1.11.1) becomes,
X'2 Y'z
_+_ =Qz
p2 12
x.2 y. 2 , •
(QP)2+Q2 =1 ..,(11121
4. Eq. (1.11.2) represents an equation of ellipse.
.~!;~a~~ What is length contraction? Find out its equation using
Lorentz transformation.
i::f3
1. The appeared decrease in the length of a body in the direction of mot ion
-is called length contraction.
2. Let us consider two frame of reference sand s' in which frame s' is
moving with velocity v along x-axis.
3. A rod oflength 'La' is moving horizontally in frame s'.
Y IY'
I --..v
: p2?2Zj
I I I
sl S'I I I ,
/ ; -;; - - - ;:27 - . x
~
~
~
f4 Lo-I
z',~
z
li'ig~.1,,12;1.
3. According to Lorentz transformation,
x '= Xl - vt
1 ~
1- '.2
c
x ' = gX2-V~
z
1-.Y."
C
x2
' -Xl' = x~
1- c
('
~
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Physics 1-15A(Sem-l&2) 1-16A (8em-l & 2)
Relativistic Mechanics
4. x == L (apparent length)
x2 - '. _ L 4. Length of moving.rod,
X I-Xl - 0
2 L _
L =: ~ L = JL}+ L/ = [<o.7Lo)2 + (~oJT2
= 0.86L
o y1-<'i
o
- 5. Percentage contraction in length = Lo - L x 100
f V2
4>
L =: L o 1 - -c2
= 4 - 0.86Lo x 100 =14 %
<W:~·'j;:if3'J···1 What do you mean by proper length ? Derive the £0
expression for relativistic length. Calculate the percentage What will be the apparent length of rod of.length 5 ,
contraction of a rod moving with a velocity of 0.6 c in a direction
and inclined at an angle 600
to horizontal. This rod is moving with
inclined at 300
to its own length. _ a speed of3 )( 107 mls.
Answer I
A Proper Length: The length of a body measured by an observer in the
frame in which the body is at rest clilled proper length or actual length.
B. Expression for Relativistic Length: Refer Q. 1.12, Page 1-14A,
Unit-I. 1. Since, frame of reference is moving along x-direction. So length of rod
C. Numerical: appears to change inx-direction only.
2. So, new length in x-direction,
1. Suppose the proper lenboth of rod isLo' g 2 ,oJ (3x107
12
L = L 1-- = 5cos60 1-1- - )
2. Observed length, along the direction of motion, o ~ 3xl~
Lx =Lo cos 300
)( J1- <0.6)2 = 0.7Lo Lx =2.487m
3. Observed length, perpendicular to the direction of motion, 3. New apparent length ofrod,
L = L sin 300
= L o
o =~sin600)2
.Y 2
y
../(2.487)2 + (5sin600)2 = 4.99 m
z
.xl07lD!s
-.'''~'''-«-''-'''"'""'"''~'t·;'-~·'':'<~·;:.~'::'<';''Y:;i~'
{.,~-:",':,>.~:/~:i' ,',"" ."
Given: v =O;<?C,incUi
To Find: PElrcel'l;tage'c
:- - - --- -£
o' U
<':II
8
I'l I
';) I
S:I
,I' v )-- , .. x
Lo cOs 30°
I" .. ,
z
,·;Fl't1{;,li.~,lt! y
.".,< ..M'"~i)m~"Al:i
z' a.
, v = 3 x 10
7
m/s
I
: s'
I
,
I
I
,
,
I
'5 sin60°
,
I
,
,
,
,, ---------...
,
,
,
,
,
,
Jly ' ,.i_1i
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I
Physics 1-17A <Sem-l &; 2)
1-18A <Sem-I &; 2) Relativistic Mechanics
Que:l~[*wl How much time does a metre stick moving at 0.1 c
relative to an observertake to pass the observer? The metre stick is
parallel to its motion.
Answer"·' I
GiVe~=:t-,Q~.~'i1
To Fifubmime:
. " " ., ,.:~;.~,;c::," ... ';~,_~,- "
1. Since, L= Lo~l- ;: = Lo.J1-0.01 =1 x .J0.99 =0.994 m
L 0.994
2 Time = - = 8 =3.31 x 1Q-l'sec
v 0.1 x 3 x 10
Quelfl'6.~;J What is time dilation? Find out its equation using
Lorentz transformation and give an example to show that time
dilation is a real effect.
~$~~~;i)ti'l~fl
A. Time Dilation:
1. In the special theory of relativity, the moving clock is found to run
slower than a clock at rest does. This effect is known as time dilation.
2. Suppose sand s' are two frames ofreferences. Frame s' is moving with
constant velocity v in the positive x direction w.r.t. frame s.
:l. !f(tl
', t2
') be the times ofoccurrence oftwo events measured by the clock
in frame s' and to be the corresponding time interval, then we have
to = t'2- t ...(1.16.1)
'-I. If (tl' t2
) be the times ofoccurrence of the same events measured by the
another clock in the stationary frame sand t be the corresponding time
interval, then we have
t = t2 - t1
lJ sing Lorentz transformation equation,
I VX' I VX'
t 2 +-2 ·- t 1 +-2
t= ~
ct
j.l--~2- ~.l-?,
t' +VX' -t' _ vx'
t=" g " 1- V
1-Z z
c
"Fi
to ...(1.16.2)
t=g2
C
1--2
C
6. From eq. (1.16.1) and eq. (1.16.2), we get, t > to
So the relativistic interval oftime is more than proper interval oftime.
s , s' --+ v
y. y •
t2
p [2'
t 1 :
•
I t 1'
I
I
I
I
I
I
I
: x'
'):0,....--------,-07' - - - - - - - - ....
, x
,
,
, "
,
z','
1J1!~i~~.!,)~~
7. Therefore a moving clock appears to go slow (i.e., take more time to
complete a rotation compared to a rest clock).
8. Ifv =c, then, t =00
9. If v is very less than c then t = to
B. Time Dilation is a Real Effect:
1. The cosmic ray particles called Il-rneson are created at high altitude
about 10 km above earth's atmosphere and are projected towards earth
surface with velocity 2.994 x 108 mls.
2. It decays into e+ (Positron) with an average life time of about 2 X 10-(; sec
3. Therefore, in its life time Il-meson can travel a distance,
d = vt = 2.994 x 108
x 2 X 10-6
= 600 m
4. But it is found at earth surface alstJ. It is possible because oftime dilation
effect.
to 2 x 10""
t=
)l-Y
; = 1 _ ( ~:.9_~4x !~8 J"
c 3 X lOR
[.,' t" = 2 X 10-6
sec.1
= 3.17 x 10- 5 sec
5. In this time a wmeson can travel a distance
d =2.994 X 108
x 3.17 X 10-6
=9490.98 m '" 10 km
6. This shows that time dilation is a real effect.
_ The proper mean life time of + 11 meson is 2.5 )( 1(}-8 sec.
Calculate:
I. Mean life time of + 1.1 meson travelling with the velocity
2.4)( IOSm/s.
2. The distance travelled by this + 1.1 meson during one mean life
time.
3. The distance travelled without relativistic effect.
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Physics 1-19A (Sem-l &2)
Answer I
• . . . '-'Iv ,;-, J':,
GIven: to = 2.5 x 10 II. ¥,.==.~.~~,
To Find: i Me@ IU'~,t~
ii. .n~tan#i;!tl'
iii.Distlm~ tro"'.
2.5 x 10-$ 2.5 x 10-$
1. Mean life time, t = to =
8 J1- (0.8)2
Jl-~:- 1_(2.4 X10 )2
3 X 108
8
_
2_.5....,x_1_0_-_ 4 17 1"-"
=. Xu-sec
0.6
2. The distance travelled = (2.4 x 108
) x (4.166 X10-8
) = 10 m
:3. The distance travelled without relativistic effect
= (2.4 x 108
) x (2.5 x 1Q-B) = 6 m.
Que 1.18. IThe half life of a particular particle as Dleasured in the
laboratory comes out to be 4.0 x lo-a see, when its speed is 0.8 c and
3.0 l( 10-" sec, when its speed is 0.6 c. Explain this.
Answer I
1. The time interval in motion is given by
to
t = H 2
(where. to = proper time interval)
v
1--
2
c
2. The proper half life of the given particle is
to = t ~1- ;:
3. In the first case t =4.0 l( 10-$ sec and v =0.8 c
. to = 4.0 X 10-8
)1 _ (0.: c)2 = 2.4 xlO-S sec
4. As proper half life is independent of velocity, therefore half life of the
particle when speed is 0.6 c mustbe given by
t = __
t_
o _ = 2.4 x 10-$ = 2.4 X10-
8
= 3 x lO-S sec
Jl-~:- ~1 _ (O':-~:r 0.8
which is actual observation.
5. Thus the variation ofhalf life of given particle is due to relativistic time
dilation.
Qu-ei;tt:~;;'l At what speed should a clock be Dloved so that it may
appear to lose 1 Dlinute in each hour ?
1-20A (Sem-l & 2) Relativistic Mechanics
...
1. Since. rest clock takes 60 minutes for Ttime interval.
:. Rest clock takes 1 minute for T/60 time interval.
2. Now. moving clock takes 59 minutes for same T time interval.
:. Moving clock takes 1 minute for T/59 time interval.
T T
3. Here, to = 60 and t = 59
4. From time dnation fo=ula, t: g2
v
1-2
c
T
2
T= 60 ::::;. J1- v _ 59
59 ~ c2 -60
2
c
v = 5.45 X 107 mls
_ Deduce the relativistic velocity addition theorem. Sho'
that it is consistent with Einstein's second postulate.
_1illl·:::;;.-·=·'i""il;;:r.J::::.:J::C'47'F~-=-l:5::-
•. '.-:
.•...+VI~_=-a-r::-k-s~05i
-'
_1.~J;lrkS 071
A Relativistic Velocity Addition TheoreDl :
1. Let sand s' be two frame of references in which s' is moving with"
constant velocity v in the x direction w.r.t. frame s.
2. LetPbe a point having coordinate (x,y,z. t) and (x',y', z', t') in frame:
and s' at any instant oftime.
3. In these two frames the components of the velocities of that partic1(
along x, y and z axis will be given by
dx ,dx'
u = - and u , =
"dt "dt'
dy ,dy'
u = -andu =
Y Y
dt dt'
dz ,dz'
u = -andu =
Z Z
dt dt'
4. Now using the Lorentz transformation equation,
x'+vt'
X=g2
1--
2
c
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I'hvsics 1-21 A (Sem-l & 2) 1-22 A (Sem-l & 2) Relativistic Mechanics
y =y' 2
V , u' 1-
y c 2
t'+2 X 10
uy = .,,0.20,6)
c
z=z'andt= 102
1 v •
+2 u,.
c
c
s s' ~v
1- 2
U' 1--
• • z c2
y y' • H2
9. SlOrilarly, u. = v ",0,20.7)
I
I 1+-u'
I
p (x, y, z, t) c2 x
•
I
I (x', y'. z', t') 10. Ifmotion ofobject is only in x direction then u =ux
and equation becomes
I
I
u'+v
u= - -
I
vu'
I
1+
I
. c2
I
B. Consistency with Einstein's Second Postulate :
o Case-I: If u' = c, then
c+v (c+v),c
U=---=>u= =c
z 1+~c (c+v)
2
~~~S,{i~~;j c
~~~f Case-2 : If v = c then
.>. Differentiating above equations,
u'+c
dx = _d_x;='+=v=d=t_' ...(1.20.1) u=---,=c
r;7
C.u
1+-2
c
'J~-~
Case-3 : If v = c and u' = c then
dy =dy' ..,(1.20.2) c+c
dz = dz' ...(1.20.3) U=---=-c
c,c
1 +-
dt'+~dx' c2
2
dt = c ... (1.20.4) 1. So the velocity ofany object cannot be greater than 'e', whatever be thE'
velocity of moving frame or velocity ofobject in that frame,
R
'J~- ~2 2. Therefore, the relativistic velocity addition theorem is consistent with
Dividing eq, (1.20.1) by eq. (1.20.4), the Einstein's second postulate ofspecial theory ofrelativity.
"
dx dx'+vdt'
.~ Rocket A travels towards the right and rocket B travels
dt = dt'+~dx'
c2 towards the left with velocity O.Se and 0.6c respectively relative to
the earth. What is the velocity of rocket:
IVlultiplyand divide by ~ on R.H.S. a. A, measured from B, and
dt' b. B, measured from A?
dx dt' +v It '+v
~'----,,.......,. => U = _.&x_ _ ...(1.20.5)
dx'
::::::
.
dt" v dx' x v
dt
-+--- 1+-u ..'r~i!"':fP,8ctl:lJldJ).6cfElspectively.
c2 2
dt' dt' c x
)C~~t:A,~~a~f.iredfrom
B,
Similarly dividing eq. (1.20.2) by eq. (1.20.4), (~~~~.a,.l'ile@~r~dfrom A.
1 1. The velocity addition formula,
dY)1_y2
dy c' x dt' u'+v
u= - -
dt 1 u'v
dt'+~dx' 1+-·
c2
dt' c2
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_ _
7
1-23 A (Sem-1 & 2)
Physics
y
0.8c 0.6 c
A
---
B
x
Earth frame
Velocity of h z I ' ' ' ' ' ' '
"
.....
• eart .. '
velocity ofA w.r.t.B,v=06 ..". ,'!'$if, .
2. wrt . c
3. elocity ofA .. , earth, u' = 0 8
w.r.t. B, . c
4. V
u'+v O.8c+O.6c
u=
1 + ~v = 1 + (0.8c)(O.6_~
2
c2 c
l.4c 4
U= --=0.9 6c
=:
1.48
5. Velocity of earth w.r.t. A,
v = -0.8c
1-24A(El~m-1&2) Relativistic Mechanico:
Variation ofMass with Velocity: Mass is a function ofthe velocity
of the body. It increases with increasing velocity represented by thp
relation:
m= gmO
v2
1- 2
c
Mass Energy Equivalence: The variation ofmass with velocity has
modified the idea ofenergy, so that, a relationship can be established
between mass and energy.
J
Deduce expression for variation of mass with velocity.
Velocity ofB w.r.t. earth,
6. OR
u' = -0.6c Show that the relativistic invariance of the law of conservation of
Velocity of B w.r.t. A, momentum leads to the concept of variation of mass with velocit)
u'+ v (-O.6c) + (-0.8c)
l.l = --,-= _'1&1J§j:,~rks10J
u V (-O.8)(-O.6)c
2
1 +--
2 1+ 2
c c
u::: -0.946c
8. Negative sign indicates that velocity ofB w.r.t. A is towards left.. 1. Suppose s and s' are two frame ofreferences in which s' is moving with
a constant velocity 'v' w.r.t. observer o.
Qlifil 1.:2,2,.1 Show that no particle can attain a velocity larger than 2. Two identical bodies A and B having same mass m are moving with
velocity u' but in opposite direction in s' frame.
velocity of light. 3. After some time both collides and sticktogether and momentarily come:,
to rest in s' frame.
An~~~'" ,I 4. Now from velocity addition theorem,
u'+v
1. Let vx'=candv=c
_x__
v'+v ...(1.23.11
U 1 = -----u'V
2. we know, V'C = v'v 1+
1+-" 2
c
-u'+v
c'
u2 = -----u'V ...(1.23.2)
1-2
c
Where, u l and u 2 =velocity ofA and B in s frame before collision
and
m l
and m 2
= their masses in s fraIIle.
5. From the law of conservation of linear momentum,
miul + m 2u2 =(m + m 2
) v ...(1.23.3)
1
C + C 2c
vx = ~=2=c
1+
c2
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Physics 1-25 A (Bem.l & 2)
8 .8' -+ V
I
I
I A B
I
I
@
I
..
I
eY -,
I u' -u
I
-' ,,1 _ _ _ _ _ x'
",,"" 0' -----~
0' I
~~~ x
z ~
z' ¥"" '~
~f7t%~
Putting the value of u and u from eq. (1.23.1) and eq. (1.23.2), in
l 2
eq. (1.23.3) we get
( u'+v ) (-u'+v'l
=> ~ l 'j+ m 2 l--'-j = (m l + m 2)v
l+~ 1-~
2 2
c c
=> m [ u' + v ] ~[v- -u'+V]
1 u'-v 1- u'v
1+-
2
c
u'v27
U'V2] m v-~+u,-v]
m UI+V-V-~
=> 1 u'v 2
1- u'v
1+- [
[
c2
c2
[1-::] mu,[1- ~;]
-.,,:> ~u' u'v 2
1- u'v
1+2 2
c , c
1+~
=> = _~C2 ...(1.23.4)
~
m 2 1- u'v
2
c
Now, squaring eq. (1.23.1), we get
o ( u' + V J2
u1- = 1 + u' vi c2
U'V)2 (U '+V)2
_ 1-~ ( u'+ v 'I _ c c
2
C
1- U~
- "C2l1+ uI V j
2
-
( 1+-
(u
2
----
'v)2
2
c2 1+ 2
C
2 2
U'2 v 2u'v (U'2 v 2U'VJ
1+--+--- -+-+--
2
c4
c2 2 2
c c c
1+ U'V)2
( 7
,
1-26A (Bem-l & 2) Relativistic Mechanics
V2J U'2 (
2
J
v
(1-?, -7l1-?""
1- u~
2
c U'V)2
( 1+?
V2)( U'2)
(1-?"")ll-?-)
1+ u'v
2 ...( 1.23.5)
c2 = u
l----L
2
c
7. Similarly, we can take eq. (1.23.2) and proceed in the same manner,
V2)( 12
)
U
(1~?"" 1- ~2-
1- u'v
2
...( 1.23.6)
C
2 =
1-~~
2
c
8. Putting eq. 0.23.5) and eq. (1.23.6) in eq. (1.23.4),
~
~
1--~
c2
n~2 ~
1-
c2
~ ~
mJ j1 - --j- = m2j1 - --j-= ma = constan t
9. Ifbody B is at rest in stationary frame 8 that is U o =0 before cojjision and
m = m in frame 8.
2 o
10. As bodies A and B are identical and have same mass in s', So. /II, = 1?1
(relativistic mass) for u l
=v.
a
Therefore,
m
m=g2
1- 2
c
:IJ~:£~K.~I Derive Einstein mass enerfD' relation E =me· and discuss
:~lf,~>,{:~})';~~_'><.~,
it. Give some evidence showing its validity.
~~~l~V;f:~
A Einstein Mass Energy Relation:
1. Suppose a force 'F is acting on a body of mass '/II' in the sallle direction
as its velocity 'v'.
2. The gain in K.E. in the body is in the form of work done.
3. If a force 'F displaces the particle through a small distance 'ds', then
work done,
dW= dK=F.ds ... (1.24.1>
4. According to Newton's Law of motion,
F = dp = fj( rn_v,!
dl 011
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a. Derivation for ES =p·c· + m
1. Total energy ofa particle is, E =me2
2. The relativistic mass, m
3.
:::::;. F= mdv +v dm
dt dt
~---~!_---~. F
1l1li_
5. Multiplying 'ds' on both sides, we get
ds ds
=> Fd = m -.dv+ v-.dm
. S
dt dt
6. From eq. (1.24.0, dK =m v dv + v2
dm ..,(1,24.2) (-: :=v)
m
7. But we know tnat, Tn = J (mo = rest mass of particle)
U
2
v
1- --2
c :
On differentiating, we get
(
1) ( v2Y~ (-2V) ."Zo vdv
dm == mo - 2" ll-;Z) 7" d v
2
( 1-~)~
mvdv
dm = ( 2) [,
c
m=g]
c2
1-~
2
c
dm (e2
- v2
) == mv dv ...(1.24.3)
8. Now putting the value ofeq. (1.24.3) in eq. (1.24.2), we get
dK = (e 2
- v2
) dm + v2
dm = C Z dm
9. Ifthe change in kinetic energy ofa particle be KWhen it's mass changes
from rest mass m o to relativistic mass m, then
K In ~
2
fdK = f e dm
o m"
K == c· (m - m o) =c2 (&In)
to. Total energy of particle,
E = Relativistic K.E. + Rest mass energy
2
E =("~ - m o) C
Z
+ m oc =mcz
This is Einstein's mass energy-relation, which states mass energy
equivalence.
B. Evidence of its Validity:
1. In nuclear reaction such as fission and fusion. These reactions take
place in nuclear' reactor and during the explosion of atom bo~b. The
cause of production ofenergy in stars arid some other processes becomes
known toda:v only due to the discovery of this important mass energy
n~lati{)n.
1-28A (Sem-l & 2)
Relativistic Mechanics
2. In process ofannihilation ofmatter, an electron and a positron give up
all its mass into two photons. The entire mass is converted into energy
This verifies mass-energy relation.
Derive the relation
--~--
a. E8 := pac· + m ·c", and
o
b. p =J.r:.. +2l11o K where, Kis kinetic energy.
2C ':
o
,..(1.25.1 i
= g"Zo ...(1.25.21
VZ
1-Z
C
Putting the value ofeq. (1.25.2) in eq. (1.25.1),
z Z
E = Tnoc = "Zoe "Zo C
g
z
2 HZvz e
1-- 1--- 1_,n2
v 2 2
CZ m ZCZ rn 2?--
"'·cz
=> E= ""'!l
I pzcz
1--
m2
c'
=:> mJc' m;c'
E2 = PZez =~z [ '.' E =m.e2
)
1--- 1--
mZc' E 2
( p2C2,
=:> EZr1---j == m zc"
'E2 0
=:> E2 - p2c2
=m "
o2C
=:> 2e•
EZ = p2cz + m o
b. Derivation fOr P. J
.r:.. + 2m K :
o
1. Total energy, E =relativistic kinetic energy + rest mass energy
=> E== K+moc2
=> K=E-m c2
o
=:> K= JmJc4+PScZ-mocz (·.·E= J~e4+P2e2)
=:> K + moc· = JT":; c· :t- p2 c2
...(1.25:;
2. On sparing both side of eq. (1,25.3), we get
=:> f(l + fno2C' + 2Kmoc 2 = m 2C'I + ~C2
o
=:> 1(2 + ZKmoC2 = J1'2c2
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Physics 1-29A (Sem-l & 2) 1-30A (Sem·l & 2) Relatiyistic Mechanics
K2
Pl.=2Kma
=> + 2
c
IK2
p= V7+ 2maK
=>
Quel~~&~:,lShow that the relativistic form of Newton's second law,
when F is parallel to V is
_ dv ( v 2 y812
F = m o dt ll-~)
An's~~~;"'1
1. Newton's second law,
- dP d
F = -=-(mv)
dt dt
m"
But
m=g
dt
d(:.V ]
So,
F = J1-~;
F = m I 1 dV + v(-~)(1-~r'2(-~~)dv1
"g2dt (y2 J dt
1- -2 1--2
c c
j
y2
1 dv ~2 dv
m" ~ - y2 dt + ( Y2)312 dt
1--· 1-
c c
r 2 2
y2 J-312
_ dv( y2y·312 [( y2 y2] dv (
F = 171. -I 1 - ---;-) 1 1- -) + - = m - 1-
2 2 2 2
" dt c , c c " dt c
-fine 1.27. The mass of a moving electron is 11 times its rest mass.
I-'i nd its kinetic energy and momentum. [~::fJ;,t:~Oll';~~~~i'lc~,;;~t):'1
"Answer, I
~~v;:~~ = 1i~m~ronetic'~ll~~gy.·:,::"'";'·' '
ii. Momentum ' .
m
a
1. Since,
m=g
1-.~
ma
So
11 ma
= ~ y2
1-~
g 1 y2 __!..
1 ~ = -=>1- 2
- c2 11 c 121
~ = 0.995
c
y == 0.995 x 3 x 108
= 2.985 X 10" m/s
2. Kinetic Energy, K.E. = (m - m ) c2
a
:z (11 m a - m a) c2
=10 m c2
o
= 10 x 9.1 X 10-31 x (3 X 10")2 =8,19)( 10 I:lJ
3. Momentum, P= mv = 11 moy = 11 x 9.1 x 10·:1I >< 2.98fi x 10'
P = 2.987 X 10-21 N-s
~;l~;t~Ii.~t~The total energy of a moving meson is exactly twiee its
~iM~L_,d~~'~.1~?,
rest energy. Find the speed of meson. 1~tl111~Ol.2.:~.a,.I~:~~~ks
i)Ii
~~""-,.:>,·.,c.;1.~:,,§.~,$&&.sX-",,""ffi.,,A
1. As given E= 2Ea
mc2 = 2m c2
a
or m= 2ma
2. Since, InC.
m=g
1-~;'
, c·
m"
2m
a
= J1 y.2
- 2
C
1
2 = F----'" ::-.:> Y =0.866c
V"
1-~"
= 0.866 x 3 x 108
=2.59 x 10" m/s
.'.11Show that the relativistic K.E. will convert into classical
K.E. if v «c.
L
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Physics 1-31 A (Sem-1 & 2) 1-32 A (Sem-1 & 2)
Relativistic Mechan,'
I = 2.08 X 105
x 1.6 X 10-19 J
Answer
['.' 1 eV = 1.6 x 10 "',
1. The expression for relativistic KE. is = 3.328 X 10-14J
. [( 2-112 ]
A charged particle shows an acceleration OY
K.'m-m,,'e" [g m,Je' =m,e' l'-:') -1
4.2)( 1011 cm/s
l
under an electric field at low speed. COUlpute til;
acceleration ofthe particle under the same field when the speed h,,'
2. Expanding using binomial theorem, reached a value 2.88)( 1010 cm/s. The speed oflight is 3 x 1010 clll!s.
1 v
2
3 v
4
]
K= m oc2
1+-2 +- 4 + ... -1
[ 2 c 8 c
3. Since. v« c i.e., vic « I, so, higher terms may be neglected.
Thus, K =moc" [1 +.! v; -lJ"'.!mo v2
.... (Classical KE.) 1. At low speed v« c, the effective mass m = m
2 c 2 o
, 2. Acceleration at low speed,
4. Therefore, ifv« c then relativistic KE. will convert into classical KE.
F
Que 1.30. ICalculate the workdone to increase speed ofan electron a o = - =4.2 X 1012
cmf S2
mo
of rest energy 0.5 MeV from 0.6e to 0.8c. 1_.,),_
'->",f!' ",' ',I'. ',;,' '" -"".~.";' '.. -', " .' 'ok"'" .,'''('' ,',",,' " . 3. Now, mo = m
m= ~ o
~
l 0.28
Answer I 1 - (2.88 x 1010) 2
V
1
-C2 1
3 X 1010
Given: Initial velocity=o.;6c,~~ldt~~~~~t~ 4. Now, acceleration at high speed,
To Find: Amount ofworkdone. · " i
F F 0.28F
a= _= = _
2 m m 10.28 m
1. K.E. of electron, o
K= m o = v 2 0
o
-Inc [g-m ]c2
1- -- a. = 0.28 x 4.2 x 1012 =1.176 X 10'2 em/s2
c q
2
:lfltl If the kinetic energy of a body is twice its rest mas,.
energy, find its velocity.
~IiII,{
,',",';:',,2:Qlt5:~16<'Maioks05,1
K= m.e' [H~lT; -1] P'. "<~'. ',·1·,"..;, "...".,-.. :!.~., '. ""', ;. -"!' .' . _.I
2. Now initial kinetic energy,
K, .m.e'[Ho.:ert-1] -025moe' )~i~~:~~~;1~~~~?''''':'
K, = 0.25 x 0.5 x 106
eV = 1.25 x 105
eV 1. Kinetic energy, K.E. =0 2 x rest mass energy
.! mv2
= 2x n~ c 2
2 0
:3. Final K.E., K,.moe'[Ho:ert -1] 2. We know that, m= m o
K, = 0.666 x 0.5 x 106 eV = 3.33 x 105 eV v2
..j Amount of work =K: - K, g
1--
c2
= :3.33 x 10" - 1.25 X 105
= 2.08 X 105
eV
'"
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jlhvsies 1-33 A (Sem.l & 2)
1-34 A (Sem·l & 2) Relativistic Mechanics
,5 SO, ! rno 2 2
2 10.v = 2 x rn c
1-~ 0
2
C
,
V
2
= 4c2
1- ~
2
10
v4
= 16 c4
( 1 - ;: J 1. As we know rno
rn=
c
g2
v 4 = 16 c4 - 16 c2y2 ' 1- 2
v,' + 16 c 2 v 2 - 16 c4 = 0
c
- 16 c2
± J256 c4
+ 4 x 16 c4 1.5 rno= ~
v 2
= ~
2 V1-~
2 -16 c2
± 17.89 c2
2
v =
~1- v2
2
= ~
v = 0.972 c [Taking positive sign] c 3
2. According to length contraction formula
Que 1.33.1 Show that the rest mass of photon is zero.
g 2
L=Lo 1-
Answer 2
I c
A photon travels with the velocity oflight. Its momentum is given by, 2 2
L = 1 x -' = - =0.667 m
p = rnb v 3 3
~
~~ -?
©©©
But the momentum of a photon of radiation of wavelength A is
p = !!:.. (h is Planck's constant)
A
h rnov
A
10
2
1-
c2 '
h102
1
or rn
o
= -;;
VA
But for photon v = c,
So, rno = 0
Therefore, the rest mass of a photon is zero.
'Que 1.34. ·1 Calculate the length of one meter rod moving parallel
; ;> its length when its mass is 1.5 times of its rest mass.
1_
:.~ ..,. '_~,.",*~Pi'Oil",' rrJN'Cr'v== ~.
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.
r,
2-2 A (Sem-! & 2) Electromagnetic Field Them'.
2
:":";'/:'>! PART~l I_
.·qtj:q}liqrQ~~~~~,i:!~:nsi~y., Displacement Curren i
:1J~~f9!llor tlJ-~C.'rf~l:.<!#}J!{J;gnetic Field toSatish'
Electromagnetic '.',.~ytl1~~t~q;£~Y:,~q#-Q,tiOti;.1iI(/iliJe~t"~~q,i;;ationin Vacuum
LJ N IT
Field Theory '." !Sf;;;;t·::i;;;.Lb;.{'!<i<{::·:~Wt.~~!N<g!J:p~;iJ..~~~E~'t3:~:/M~¢i~m.
CONCEPT OUTLINE: PART-'
Part-! (2-2A to 2-l6A) Electromagnetic: It is the study ofthe mutual interactions betvec,'
electric charges. Charges may be stationary, they may move wi: i
y constant velocity or they may be in accelerated motion.
- C;ntinuityEquation for. curre
.. ··ft:..t Dens..J ....•........... ./,,,,./,,
:...•:.t ;/>: :<, /) Equation of Continuity: It states, "The net change in the tota'.
• Dzsplacement Current .;'.:'"
• Modifying Equation for the Curl ofMggn.~·". amount (of the conserved quantity) inside any region is equal to the
Equation '.. ",.•{ ..;':C: resultant of the amount that flows in and out of the region through
[ the boundary".
• Maxwell's Equation in Vacuumd~d,in'Nl
. . ; .".. ~
Maxwell's Electromagnetic Theory: Maxwell proposed the theor."
oflight to explain the different phenomenon (e.g., reflection, refraction.
t.otal internal reflection, polarization, etc.). This theory is known a,-'
A. (.'oll,epl Olltline : Part-l .. 2-2A
n. LOIIE! and Medium Answer Type Questions .. 2-3A
Maxwell's electromagnetic theory.
Maxwell's Equations in Free Space or in Non Conductin,,~
Part-2 (2-17A to 2-26A)
Medium:
• Energy in an Electromcuglte.ti,
------1
• Poynting Vector a.nd POYIi f''f'l~:".
~
f
~ ~
• Plane ~lectromagnetic'V'f.f.t,; 1. div D = 0 D·dS=O
• Relation Between J£le9~f!'J,b
Waves
~ ~
2. div B
~
= 0
fB·dS = 0
B. Long and Medium Answer ~ype Questions 2-l7A I I -i
I - cB I f~~ d(f-"'-» I
~. curl E = - fit c E . dl = - dt l s B . dS) i
Part-3 (2-26A to.2-30A)
• Energy and MomentU11'1i(JQ,rri/!i
~ aD ~ ~ aD ~
• Resultant Pressure' '. ..' !4. curl H =- ~! f. H 'dl = f -.
~ dS ,
. Ot c sOt j:
• Skin Depth
Displacement Current: It is defined as EO times to the rate of
A. Concept Oil tline : Part-3 2-26A change ofelectric flux through the surface.
B. Long and Medium Ansr.per Type Questions 2-26A
oq,
I d = 9>-
Ot
~
2-1 A (Sem-l & 2)
I
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Physics 2-3 A (Sem-l & 2)
Q:Uei21~~I~~Derive the equation of continuity. Also write its physical
significance.
"''''''''''''''''1
An .'. '..."!i:i"",','';
s'w'~""::;11'
A. Equation of Continuity:
! . The equation ofcontinuity represents the conservation law ofcharge in
electromagnetism.
2. If I is the corresponding current due to decrease of charge inside the
given region, then we have
1=- dq ...(2.1.1)
dt
...
If p is the volume charge density ofthat region and J the corresponding
surface current density, then eq. (2.1.1) can be expressed as
...... d
fJ .ds =: - - fpdV ...(2.1.2)
s dt v
,(, U sing Gauss' divergence theorem
fA.d; = fdiv AdV
s v
On left hand side of eq. (2.1.2), we get
-d
fdivJdV dt fpdV
v v
or fdivJdV
v
= J(ZJdV
. ... ap )
or J
(dlV J + at dV = 0 ...(2.1.3)
'J. Since Vis an arbitrary volume, eq. (2.1.3) holds onlyifits integrand will
be zero, i.e.,
... ap
divJ+-=O
iJt
This expression is known as equation of continuity.
B. Physical Significance:
The equation ofcontinuity is expressed as,
,W,
....
2-4 A (Sem-l & 2) Electromagnetic Field Theory
-+ a
div J + ---.e. = 0 .:.(2.1.4)
at
2. According to the Gauss's theorem in electrostatic
-+ -+-+
div D = V.D=p
-+
where D is the displacement vector.
3. Thus, eq. (2.1.2), becomes
-+ a· -+
div J + -(divD) =0
' at
( -+1
or div J+ div la:j = 0' (.,' Divergence is independent of time)
. (-+ aD"
or dlV lJ + -j = 0 ...(2.1.5)
, at
4. The solenoidal vector is very important e.g., we consider the discharge
ofa capacitor through resistance R .
5. Ifwe consider a closed surface 8 1 on a part of the circuit Fig, 2.1.1, then
the current leaving 8 1 is equal to the current entering it. Same cannot
be said to the closed sUrface 8 2 , until one admits a displacement current
-+ -+
<a D fat) as equivalent to the conduction current J .
6. But with this equivalence we have a closed circuit, which takes the form
of conduction current in the wire and a displacement current in the
capacitor dielectric between the plates.
7. Eq. (2.1.5) signifies the treatment ofthe capacitor as a circuit element.
8 2
_ Derive the modified equation for the curl of magnetic
field to satisfy continuity equation.
...
1. Ampere's law is given as
VxH=J ...(2.2.1)
Taking the divergence of both sides of eq. (2.2.1), we get
~&1'j)l:'ir"
~~;~~
L
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2
2-5 A (Sem-l & 2)
Physics
.
V' (V'xHl = V'.J
~
But according to vector identity, divergence of a curl is zero, therefore,
v (V'xH) =0
! .l' V·J = 0
This result.ant is not consistent with the continuity equation ('V'J = : )
:·L
'.
i.e., c"tnkment. of Ampere's law is inconsistent and some modification is
required in it.
SuppoO'p we add an unknown term G to eq. (2.2.1), then
...(2.2.2)
V;v;B = J-rG
,J.
" Takmg divergence of bot.h sides of eq. (2.2.2), we have
V . (V x Bl = 'V. (J + G) = 0
i .<' Y·}+V'·G =0
or V.G = -V·J
V' . G = _ (_ a~<. J= a~" [.,' 7.J =_ a~" ] ...(2.2.3)
0. We know t.he point form of Gauss's law as, 7 ·15 = Pu
Taking difTerentiation of both sides V . aD = ap"
at at
Putting this value in eq. (2.2.3), we get
- aD
7·G=v·
at
aD ...(2.2.4)
or G =
et
7. Using eq. (2.2.2) and eq. (2.2.4) we get,
--+
--+ --+ aD
V'xH=J+
at
Que 2.3. IDerive the Maxwell's equation in differential form.
Answ~J:" I
A. Derivation of Maxwell's First Relation:
1. According to Gauss' law ofelectrostatics:
"The nel electric flux (4)E) passing through a closed surface is equal to
(11£0) times the total charge q contained in the surface".
Thus, 4>E = !L ...(2.3.1)
EO
2-6 A (Sem-l & 2) Electromagnetic Field Theon
2. The electric flux can also be expressed as
4>E = IE .ds l': :.i
s
where E is the strength ofthe electric field and It. d s is the electnc
flux passing through a surface elements ds ofa closed surface S
3. From eq. (2.3.1) and (2.3.2), we have
IE .ds = !L ..,(2.:,;:
s EO
4. rfVbe the volume enclosed by the surface Sand p be the volume c!1~U',:,
density, then the total charge enclosed in the closed surface can hi
expressed as
d'S E
_~1i"'~lf~~'t~il!~1~h~~eq.
q = IpdV ...(2.3.4)
v
5. From eq. (2.3.3) and eq. (2.3.4), we get
J 1
E
- .ds
- ='- IpdV
s EO v
or EO IE.d s = Ip dV
s v
or JEoE .ds = JpdV
v
or ID .ds = IpdV ...(2.3.51
s V
where D=EO E is the electric displacement vector in the presence of
free space.
6. Using Gauss' divergence theorem on the left hand side ofeq. (2.3.5), we'
get,
(J-+ -+ J 'I
-+
Idiv D dV = Ip dV l," sA'dS= vdivAdVj
V V
l
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1
j
1
2-8 A (Sem-l & 2) Electromagnetic Field Theory 1
IJhysics 2-7 A (Sem-l & 2)
-+
J
div B = 0
or f(div D dV - pdV) =0 ...(2.3.6) This is the Maxwell's second relation or equation.
v 6. This relation states that there are no magnetic monopoles in the world.
C. Derivation of Maxwell's Third Relation:
Since Vis an arbitrary volume, eq. (2.3.6) holds only ifits integrand will
1. According to Faraday's law ofelectromagnetic induction:
be zero, i.e.,
"The induced emf (electromotive force) produced in a current carrying
(div D
~
- p) =0 coil is equal to the negative time-rate of magnetic flux <DM associated
. with the coil".
i.e., div D
~
= P
This is the Maxwell's first relation or equation. Thus, e == - dt
d (C1>M) ...(2.3.11)
,'j In free space, volume charge density, p =0 ~
2. If E is the strength of the electric field corresponding to the induced
So, div D
~
= 0
emf e, then the induced emf can be expressed as line-integral of E
B. Derivation of Maxwell's Seco:ud Relation:
around the coil Fig. 2.3.3, i.e.,
We know that an isolated magnetic pole does not exist in the nature.
Hence, the net magnetic flux M passing through a closed surface is e = fE ·d[ .. (2..3.12)
zero. c
Thus, M = 0 ...(2.3.7)
---~-
- ....
" , - -- ,
", ..........::-- --~:::~,
~~~JT
N "s ~_~c .
~
,,,....::----.--=,~
,.
!tllftj~~~ii!;r:9~u.~?fi~~f~;li~u:t'tep;t.carrYing coil.
, ...: :::::~::::. ' ",
3. Comparing eq. (2.3.11) and eq. (2.3.12), we have
,:1~lJ~~;.1X:V~;. 'Y" ' ' •
d
2. The magnetic flux can be expressed as fE .d[ = - -(<DM) ...(2.3.13)
dt
c
M = fB. ds ...(2.3.8)
4.. The magnetic flux can be expressed as
s
M = fIi ·ds ...(2.:3.14)
where Ii is the strength of the magnetic field and B. ds the net
s
5. Substituting the value ofM in eq. (2.3.13), we get
magnetic flux passing through a surface elements d sofa closed surface
dl
S-
fE . -~ rfIi 'dS')
From eq. (2.3.7) and eq. (2.3.8), we get c ,C
fB.ds = 0 ...(2.3.9) ...(2.3.15)
or fE .d l = - f~ (Ii .d s)
s c sat i
·i
Using Gauss' divergence theorem on the left hand side ofeq. (2.3.9) we 6. Using the Stokes curl theorem on the left hand side of eq. (2.3.15), we
get, get
i
.,' ,A.dl= curlA'ds
.,' sA. ds = !div AdV ...(2.3.10)
fdiv]j dV =0
( f
-+ -> • -> )
fcurlE .ds f~(Ii .ds) [ f
.... .... f ....~. ) 1.
- at
V . c· s .
s s
Since V is an arbitrary volume, holds only if its integrand will be zero,
,.e.,
~
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._~----------_.-,,
2-9A (Sem-1 & 2) 2-10 A (Sem-1 & 2)
Physics Electromagnetic Field Theory
o. Substituting the value ofI from eq. (2.3.19) in eq. (2.3.18), we have
f Icurl E .d; + ~ (B. d;)] = 0
s c a t fH. dl = fJ· d'S ...(2.3.20:
C 's
.f [curl If + l~~l(ds) = 0 ...(2.3.16) Using Stokes curl theorem on left hand side of eq. (2.3.20), we get
6.
7. Since, S is an arbitrary surface, eq. (2.3.16) holds only ifits integrand is
JcurlH.ds = IJ.ds 'Or.: f.... --> fcurl A
........1
A. dl "" . ds
zero, i.e.,
s s c S !
curl If + alJ
at =0 or curl ii = J ...(2.3.21)
--> ali 7. Using.eq. (2.3.21) in the equation of continuity, we get
or curl E
at --> a [ --> cp )'
This is the Maxwell's third relation. div (curl H ) + ---.e. = 0 '.' div J + - = 0
at at
D. Derivation of Maxwell's Fourth Relation:
. ap .....
1. According to Ampere's circuit law: l.e., at = 0 r·.· dlV (curl H) = 0] ...(2.3.22)
"The line-integral of magnetic 'induction vector B
-->
ar01111d a closed 8. Eq. (2.3.22) is applicable for the steady-state conditions in which the
current carrying loop is equal to !lo times the current flowing in the charge density is not changing with time. This shows that for time
loop". varying fields, Ampere's law should be modified. For this; Maxwell
suggested that eq. (2.3.21) should be modified as follows:
2. Thus, fB. d! = !lol ...(2.3.17)
.... .... ....
('
curl H = J + J D ...(2.3.23)
;) Since. B = Po H , we have
-->
where J D is known as the displacement current density.
fPo H .dl =!lal 9. Taking d~vergenceon both sides of eq. (2.3.23), we get
('
-+ -) ~
div(curIH)=div(J+ J D
)
or fH. dl =1 ...(2.3.18)
.... ....
c or -.
o= div ( J + J D) [.: div (curl H ) = 0 I
.... .,
4. Let us consider a small surface elements ds of the surface S bounded or div J = - div J D
--> a -+
by the closed loop C. If J be the surface current density of the loop, or ---.e. =div J [ --> ('P
D .,' div (J) + -:- = 0
at ot
then the current flowing in the closed loop can be expressed as
I ....
1 = fJ.d; ...(2.3.19) o(div D) d' J
or = IV D
s
,
f
at
~
[anJ .,
or div ~ =div J D
r alf ....
i
or - ::: J D ... (2.32,
at
B curl ii = J + aD
Fig. 2.3.4. A ClR~~9'~"'~~*7~~llIi)~il~ at
~-~
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,
Physics 2-11 A <Sem-l & 2)
I
!
This is the Maxwell's fourth relation or equation.
11. From this relation, it is clear that the displacement current density
relates the electric field vector E as (D =e E)to the magnetic field
~
vector H.
....
12. In free space, surface current density J =0
....
~ aD
So, curl ( H ) =
at
Qllej~~~t:~J1{1 Derive the Maxwell's equation in integral form..
Ait~~¥~;f~ft
A Maxwell's First Relation in Integral Form. :
1. The Maxwell's first relation is
div D
~
= p ...(2.4.1)
2. Integrating this over an arbitrary volume V bounded by a closed surface
S, we have
f (div D) dV = f p dV ...(2.4.2)
v v
3. U sing Gauss' divergence theorem on the left hand side ofeq. (2.4;2), we
get
f D.ds =- f pdV [.: !A. ~= !div AdVJ
S V
where S is the surface which bounds the volume V.
4. In free space, volume charge density, p = o.
So, fD.d; =0
s
This is the integral form ofthe Maxwell's first relation.
B. Maxwell's Second Relation in Integral Form. :
1. The Maxwell's second relation is
div B
~
= 0 ...(2.4.3)
2. Integrating this over an.arbitraryvolume V bounded by a closed surface
S, we have
fdiv BdV = 0 ...(2.4.4)
v
3. Using Gauss' divergence theorem on the left hand side ofeq. (2.4.4), we
get
2-12 A <Sem-l & 2) Electromagnetic Field Theory
~ (~ ~ -> J
B d s = 0 .: fA. ds";' f div AdV ...(2.4.5)
f~
S S v
Where S is the surface which bounds the volume V.
4. This is the integral form ofMaxwell's second relation.
c. Maxwell's Third Relation in Integral Fonn :
1. The third Maxwell's relation is
~ aB
~
curl E = _ - ...(2.4.6)
at
2. Integrating this over an arbitrary surface S bounded by a closed loop C,
we have
-- fa-- ...(2.4.7)
curlE .ds = - atCB. ds)
f
s s
3. Using Stoke's curl theorem on left hand side of eq. (2.4.7), we get
fE .d! = - f ~(Jj.ds) [.: fA. dl = fcurl A'd~ ]
c S at c s
i.e., 1E .d r = -d: l
(
fa .d sJ
')
Here C is the closed loop which bounds surface S.
4. This is the integral form ofMaxwell's third relation.
D. Maxwell's Fourth Relation in Integral Form :
1. The Maxwell's fourth relation is
~ an
curl ( H ) = J + - ...(2.4.8)
at
2. Integrating this over an arbitrary surface S bounded by a closed loop C,
we have
curl -
H . ds
-= f (-
J + -
an) .d s
- ..(2.4.9)
f
S S at
3. Using Stoke's curl theorem on left hand side ofeq. (2.4.9). we get
aD J [f ., ., f , ..
= J+----;;t ·ds .. A.d/= cudA-ds
fH . d! f (
c S (' .';
or fH.d1 = f(J +JD)·ds
c S
Here, C is the closed loop which bounds surface S.
4. This is the integral form ofMaxwell's fourth relation.
....
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Physics 2-13 A (Sem-l & 2) 2-14 A (Sem-l & 2)
Electromagnetic Field ThPP'
5, In free space, surface current density, J = 0
_ Using Maxwell equation Curl B = ~o [ J + ~~J'prow'
So, IH.d 7 = IJD d ; that div D '" p.
c s
I ~ ~ oD'" 1.
01' H.dl = f-.ds The Maxwell's equation is given by :
c s Dt
...... --> aD
-->}
Que ~.5~ IGiven the physical significance of Maxwell's equations. Curl B = ~o {J + iii
.... ......
Answer "I But we know, B = I-lo H
A. Physical Significance of Maxwell's First Equation:
1. It signifies that the total flux of electric displacement through a closed So, Curl H = {; + ~~} ... (2.G.l i
2. Taking the divergence ofeq. (2.6.1), we obtain
surface enclosing a volume iS"equal to the net charge q [=!pdV J
div(Curl H) = div {; + aaf} == 0
contained within that volume.
B. Physical Significance of Maxwell's Second Equation: 3. Since the div ofany vector quantity is zero i.e.,
1. It signifles that the net outward flux of magnetic induction through a div (Curl H) =0
:;ud'uce enclosing a volume is equal to zero.
:Z. This shows't.he non-existence of monopoles in nature. , Therefore, div {; + ~f} == 0
C. Physical Significance of Maxwell's Third Equation:
div ; + div { a:} = 0
1. It signifies t.hat. the emf [ '" JE.d 7) induc~d around a closed path is
I
~
mv; + {Od:;~} = 0 ...(2.Ei.2
I
r
equal to the negat.ive rate of change of magnetic flux (=-fB.d -;J
4. From continllity equation, we have div ; + (:', ,
=0
linked wit.h t.hat. closed path.
D. Physical Significance of Maxwell's Fourth Equation: l.e.,
. d'
IV J.... = op
_
at .,.(2.6.3)
,
,
i 5. Using eq. (2.6.3) in eq. (2.6.2), we "btain
[ -->
1. It. signifies that. t.he mmf [ = JH.d/')Jaround a closed path is equal to
I: _ ap + adiv D = 0
f at at
I
[
the sum of the conduction current and displacement current linked
with that closed path. op odiv D
at ot
-->
1 div D = p
"i
1
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S
1 ~
2-16 A (Sem-1 & 2) Electromagnetic Field Theory
Physics 2-15 A (Sem-l &.2)
_ aD
J D- ....(2.7.5)
Ot
Que;~~7,.;MilllExplain the concept of displacement current and show l:
~
11. From eq. (2.7.4) and eq. (2.7.5), we have
how it led the modification of Ampere's law.
~ I = A aD
Answ:~;~;t(;W;f D
Ot
1. Let us consider an electric circuit consisting ofa batteryB, resistance R, f I =- A a(eoE )
or I': D =1:(lEI
key K and a capacitor C in series as shown in Fig. 2.7.1. i
~ D
at
I =Aeo a(E)
or D ... (2.7.6)
at
12. If cr is the surface charge density of the plates of capacitor and q is the
~~K
~ charge on each plate, then we know that
. cr q
E=-=- [.: cr=(qIA))
~,,~ eo Aeo
13. Using this relation in eq. (2.7.6), we get
Fig. 2.7"l.;]ill~~ti~1~~.I.lliilL, ..,Ji;
When we close the circuit by pressing the key K, the charging of the I D = Aeo ~(-.5L)1
capacitor starts. at Aeo
:i Consider a circular surface 8 1 and a semispherical surface 8 2 such that
i.e., I D= aq = I
both are bound by the same closed path. During charging, there is no at
actual flow of charge between the plates ofthe capacitor.
14. This shows that the displacement current in the space between the
4. Hence, the current flows through surface 8 1 but not through 8 2 , plates ofcapacitor during charging is equal to the conduction current.
;). Applying Ampere's law for the surface 81' we have 15. Thus, the concept of displacement current needs a modification to
Ampere's law.
IB.dl =!-lJ ...(2.7.1)
_ Explain the characteristics of displacement current.
Applying Ampere's law for the surface 8 2, we have
=0 ~: .,' ' ' , ' , ' , ,,:
IB.dl ...(2.7.2)
. ,,~ . ~
-",...
G. Since the results of eq. (2.7.1) and eq. (2.7.2) contradict each other, 1. The displacement current is the current only in the sense that it produces
these equations cannot be corrected. a magnetic field. It has none ofthe other properties ofcurrent since it is
not linked with the motion ofcharges.
'7 Maxwell tried to improve the contradiction between the two equations
by adding an additional term !-lolD on the right-hand side ofeq. (2.7.1). 2. The magnitude ofdisplacement current ·is equal to the rate ofchange of
magnitude ofelectric displacement vector, i:e., J D = <aDIcrt).
..~. Thus, the modified Amperes law can be expressed as
3. It serves the purpose to make the total current continuous across the
f1Ldl = !-lo (l + ID) ...(2.7.3)
discontinuity in conduction current. AB an example, a battery charging
where I D
is known as di~placementcurrent. a capacitor produces a closed current loop in term's of total current
:J. The displacement current is given by l total =1 + I D.
I =- A J ...(2.7.4) 4. The displacement current in a good conductor is negligible compared to
D D
where A is the area ofthe plates of capacitor and J D is the displacement the conduction current at any frequency lessthan the optical frequencies
current density. ("" 1015 Hz).
j 0, According to Maxwell's fourth relation, the displacement current density
is given by
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r
Physics 2-17 A (Sem-l & 2) [
t 2-18 A (Sem·l & 2) Electromagnetic Field T1Ie""
/'
-;"
/PART-':2J > .'i"""'> : .
A.
Energy in an ElectromagneticField,RoY1,Lttni1:'MBJJ.tor<and·.J:!(iy;nti~·.
Theorem, Plane ElectromagneticWdtJ~8<l~';t?'?iP#uirf:al~dt.heir:.,:<.
Transverse Nature, Relation. BetU)e~'rt'8te,~ff:i¢'an#:lft,4s.~~i<;·
,,'
Fields ofan Electr'()m4jfn:!i#c;W'~pes
•..,.'••... ';Y>;."f/:;'<i·.
Poynting Vector : The magnitude and the direction offlow ofenergy
per unit area per unit time in an EM-wave travelling in free space (or
->
vacuum) can be expressed by a vector known as Poynting vector S.
Electromagnetic Waves: Electromagnetic waves are coupled electric
and magnetic oscillations that move with speed of light and exhibit
typical wave behaviour,
Energy Density of Electromagnetic Wave The total
...lectromagnetic energy density
tl = coW
l,__. ,.__,,, ....._"
Que 2.9. IWhat is Poynting vector?
Answer' I
A. Poynting Vector :
1. The magnitude and the direction offlow ofenergy per unit area per unit
time in an EM- wave travelling in free space (or vacuum) can be expressed
..
by a vector known as Poynting vector S .
B. Expression for Poynting Vector:
1. Let us consider a small volume element dV = A dx, where A is its cross
sectional area and dx its length along X-axis,
2, if U IS the ~M-energydensity of the EM-waves, then
1 2 1 2
u = 2" EO E + 2" ~OH ...(2.9.1)
I .X
-I dx ~
Z
;"'i:iii~ifiifGl,"{~~i~~B~~l~3~UMI~~'~~£.!¥')tor•.·..
~·<J~:?~~~~:I"~l:'f.'~~,,';:»?':-ti.$'1t~~~~~;t;ft_,,fC0l,~M){r,;> .~~" :,.J:.::;,>
3. Hence, the energy associated with volume element dV is given by
U 2 1 2)
= u dV = ( "2
1
EO E + "2 ~OH A dx ",(2,9 ...
4. The relation between the magnitudes of field vectors E and H i"
~ _ ) ~o _ C- _1_ ,'<.
H- EO - ~o - EOC ",("..
5. Using eq. (2.9.3) and eq. (2.9,2), we have
(1 H I E '
u= l-EO E-+-~oH-)Adx
2 EOC 2 ~oC
or u = (...!..+ ...!..) ERA dx
2C 2C
i"
,
f or u = ERA dx [ .., C c. ,(,
(dx Idt) <II
i
or U = EHAdt ,..(2,~).j
I
6. Hence, by definition, the EM-energy passing through per unit area ]JP],
unit time with EM-wave, i.e., magnitude ofPoynting vector is
s';' ..!!....... = EHAdt =EH
Adt A dt
f 7. In vector form, this expression can be expressed as
-> -> ->
t S=ExH
Derive Poynting theorem and explain its physical
significance.
I
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, 2-20 A (Sem-l &2) Electromagnetic Field Theory
Physics 2-19 A (Bem-l & 2)
"
Answe,r, ,:'1
A. Poynting Theorem:
According to this theorem, "The time rate ofEM-energy within a certain
volume plus the time rate of EM-energy flowing out through the
boundary surface is equal to the power transferred into the EM-field".
B. Derivation of Poynting Theorem:
Suppose we have some charges and current, which at time t produces
fields E
--+
and B .
Then, the work done on an element charge dq contained in a volume'
element dV due to its displacement d l in time dt under the influence of
Lorentz force It' is
It' .d l = dq (it + vx B) .dl
It' .d l = dq (it + vx B) .vdt [.o' v=~~J
dq (it . v) dt + dq (v x B). vI dt
(it. v)dq dt
[since (v x B). v=[v B v] =0] ...(2.10.1)
l. If p and J
--+
be the volume charge density and current density
respectively, then we have
....
dq = p dV and p v = J ...(2.10.2)
Using eq. (2.10.2) in eq. (2.10.1), we have
It'.dl = [E '~lpdVdt = (it·J)dVdt
Hence, the total work done per unit time on all the charges in some
volume V is given by
dW = f(E . J) dV ...(2.10.3)
dt v
Maxwell's fourth relation is
curlH = J'I- aD
at
- - aE ........
or curl H =J + 60 - [.o' D = 60 E) ...(2.10.4)
at
7. Taking the scalar product of E on both sides ofthe above e4·(2.10.4),
we have
E.curlH = E . [J + 60 a!]
aE
= E .J + ~ E-
or E .curlH QO .
at
or E·J = E . curlH - 60 E . ait
tv
- - - - .... (,j!;
H . curl E - div (E x H) - 1;0 E·
or E·J f't
...(2.10.5)
~ ~ --t ~ » •
[.o' div (A x B) = B. curl A - A. curl B]
8. Maxwell's third relation is
~ alI
curl E = -at
~ aH ...(2.10.6)
or curl E = -!-to
at
9. Using eq. (2.10.6) in eq. (2.10.5), we have
- - - [ aH J . - - - (lE
E .J = H -!-to - - dlV (E x H ) - So E ,.
. at ~
-aH - a"E --
or E.J - !-toH--EoE·-- div(E x H) I
J
at at
_ ~ (!-to HZ + So E z) _ div (it x H ) j
at 2 2
or E.J
j
_ ~ (!-to HZ + So E Z ) _ div (8) ...(2.10.7)
or E.J at 2 2
--+ --+--+
where S = (E x H) is the Poynting vector.
10 Using eq. (2.10.7) in eq. (2.10.3), we have
d: = ![_~(!-t;HZ+S~EZJ-diV(8)]dV
or dW = -d f(!-tOHZ+60EZ)dV-fdiV(S)dV
dt dt v 2 2 v
...(2.10.8)
Since, dWldt represents the power density that is transferred into EM
11.
field and increases the mechanical energy (kinetic, potential or whatever)
of the charges, therefore, if uM denotes the mechanical energy, then we
have
dW =!!:.- fUM dV ...(2.10.9)
dt dt v
I
l
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I
r
2-22 A(SeDi·l & 2) Electromagnetic Field Theorv
2-21 A (Sem·l & 2)
Physics
12. Also, the EM-energy density is given by E2 = 500 x 377 = 15000.36
4n
...(2.10.10)
U =: I-lo H 2 +~E2 or E =122.475 Vim
I
em 2 2
13. Using the eq. (2,10.9) and eq. (2.10.10) in eq. (2.10.8), we obtain
dt
dJ = .-dJUem dV
UM dV ill J--'
div (8) dV
V V V
~ . J
d
Of div(8)dV = - dt (uM +uem)dV
J
V
. Ja
V
or div (8) dV = at (UM + uem ) dV
J
v v
-'> a
div S = - - (UM + uem )
or at
c. Physical Significance:
-'>
The Poynting vector 8 describes the flow ofenergy in the same way as
-'>
the current density vector J describes the flow of charge. Since the
equation ofcontinuity expresses the conservation ofcharge, the poynting
theorem represents the conservation of energy.
.I .
A 500 watt lamp radiates power uniformly in all
directions. Calculate the electric and magnetic field intensities at 1
m distance from the lamp. . .
AnsWer.1
Given: Energy ofth~lalQ.p;
To Find: i. Electri~:
ii. .i1:·fiW1~.
2
1. Area illuminated =41tr2
=41t (1)2 =41t m
Therefore, Energy radiated per unit area per second = 500
2. 41t
:3. Hence, from Poynting theorem
-> -'> -> 500 ...(2.11.1)
I S I = I E x HI=' EH = 41t
...(2.11.2)
and E = r& =377 n
H V~
4. Multiplying eq. (2.11.1) and eq. (2.11.2), we get
and H = ~ =0.325 Aim
377
. . , Calculate the magnitude ~f Poynting vector at the
surface of the sun. Given that power radiated by the sun
=3.8 )( 1()26 Watts and radius of the sun =7 x 108
m.
_ill
i:i:i'-'4~ " '"'. ,iilii;
.fi:~6ftnesun.
1. The Poynting vector is given by :
8 _ Power _ 3.8 x 1026
3.8 x 1026
- 4nr2 - 4n(7 x 108 )2 - 1.96n x 1018
=6.171 x 107 W/m2
. _ Define EM waves. State a few properties of
electromagnetic waves.
..
~
1. EM waves are coupled electric and magnetic oscillations that move with
speed oflight and exhibit typical wave behaviour.
2. The properties of electromagnetic waves are as follows:
i. In free space or vacuum, the EM wave travel with speed of light.
ii. The electrostatic energy density is equal to the magnetic energy
density.
iii. These waves carry both energy and momentum, which can be
delivered to a surface.
iv. EM waves are transverse in nature.
v. Electromagnetic waves ofdifferent frequencies can exist.
_ _ _ Derive electromagnetic wave equation in free space.
OR
Using Maxwell's equations, derive electromagnetic wave equations
in va~~um. and prove that wa..·c propagate with speed of light.
-
1. Maxwell's relations are given by,
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2-23 A (Sem·l &2)
Physics
divD =p
div E =0
- aE ~
curl E = - - I ... (2.14.1)
at
- - aD
curIH=J+
. at.
For the propagation of EM-waves in free space (or vacuum), we have
a =0, p =0, sr =I, IJr =1
i. c., J '= rrE =0, E =EO Er = EO and IJ = lJo IJr = lJo
Thus, the Maxwell's relations for the propagation of EM-wave in free
space (or vacuum) are given by
div 15= 0
div E =0
- aE
curIE=-
at
- aD
curIH=
at
...(2.14.2)
I. For free space, l5 = soE and Ii = 1J0H ,so we have
div E =0
divH =0
curl E= _ 1J0 aH l ...(2.14.3)
at
- aE
curlH =&0
at
Taking curl on both sides ofthe third relation in eq. (2.14.3), we have
: ~
- (aHJ
curl (cur} E) = curl -!-to fit
. - (aHJ
2
grad(d1vE)-V E = -!-to curl fit
["~ grad (div .if )- v2.if = curl (curl .if )]
~ (aJ. - ->
- V2 E = .:.;. 1J0 fJt· (curl H) [.: div E = 0]
a ( aE] [.. =IH.~ a:]
- V2 E=- I-lo at So fit
"'I
Electromagnetic Field Theory
2-24 A (Sem-l & 2)
~ (a2
E J
_V2 E = EO 1J0 at 2 .. (2.14.4)
6. Now taking curl on both sides of fourth relation in eq. (2.14.3), we can
show that
~ (a2 Hl
V2H = sOlJo l7) ...(2.14.5)
7. Comparing eq. (2.14.4) and eq. (2.14.5) with the general wave eq.,
-> 1 (a2j1 1 .
v2 jI = 2l-2-j ,v is the wave velocity
v at
1
we get EO lJo = 2
v
1 1
i.e., v-
- .J &0 1J0 -. I(4re &0) ( ~~)
1 1
v=
(9x 1109J00-7
) = J(9-~i'Of6 J
(.,' 114 reE =9 x 109 N_m2/C2 and ~IJ4TC = 10-7
Wb/A-m)
o
v = ~ 9 x 1016 =3 x 108 mls =c (speed of light)
This shows that EM-wave travel with the speed of light in free space (or
vacuum).
'.'1'1'1 Show that E, H and direction of propagation form a set
of orthogonal vectors.
!
I
1. Considering EM-waves as plane waves, the electric and magnetic vectors I
I
E and if can be expressed as I
I
E = E0 exp [i( K.; - rot)]
if = if 0 exp [i( K .r - rot l] I
2. These two equations represent the plane-wave solutions of Maxwell's
. ~
'I
relations for the propagation of EM-waves in free space (or vaCUUOlI. In Ii
_. i
this condition, the del operator (V) and partial time derivative operator 1
I'
(~) can be expressed as ~;
I
~
~
l
f:
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PART-3
r
II ,1.
V' == iK
and (:e) = -ioo
...(2.15.1)
where K is the propagation constant and 00 the angular frequency of
EM-wave.
•• --+
K· E = 0
--+ .~
K·H =0
--+ --+ --+
i K x E = - Jlo (- ioo H )
and
_.
i K x
-t
H
-
= eo (-ioo E )
or
--+ .~
K·E
-. ~
K·H
=0
= 0
K x E =
--+
Jlooo H
K x H
~
= -eooo E
From the set of eq. (2.15.2), it is clear that E , K and Hvectors are
4.
perpendicular to each other. This indicates that EM-waves propagating
in free space (or vacuum) are transverse in nature. This fact is shown in
Fig. 2.15.1.
y
) ' )'"~_. ~ .. }{
o K
Z
Fi • 2.15.1. Transversenatn-fe~£;~lK~W~'V~~,·i~.~;~~~I;·~K~~l~;0
g . "." ''', .~ .,"'_,'? )k;-,~,~;,,;,.i(;;-: .:..: ;,l~:;"B·}·Nltt~~~J;~R~.~r",~r.,~",,-;.;
Que 2.16. IDiscuss the energy density of electromagnetic wave.
...(2.15.2)
--+ ~ ~

2-26 A (Sem-l & 2) Electromagnetic Field Theon
_.
1. When the electromagnetic waves travel in free space, the electrostatic
energy L.ensity u and magnetostatic energy density u are given by
e m
1 . 2
U = -f'ftE
e 2 ."
1 2
U = -JloH
m 2
2. Total energy density ofEM-wave is given by
1 2 1 2
u=u +U = -aoE +-lloH
e m 2 2
3. .But for a plane electromagnetic wave in free space
E = ) Ilo or H = ) 60 E
Jl 60 Jlo
1 2 1 602
4. S0, U = - 60 E + - Ilo . - E
2 2 Jlo
U = eaJtl
This is the total electromagnetic energy density.
/~~:~~~~~~;~~gneticWaves,
Skin Depth: It is defined as the depth ofa conductor from the surface
at which the amplitude ofthe wave is lJe ofthe amplitude of surface.
_....~. . ·!1;'.~1 Derive the relation between energy and mOlllenturn
4'" "%.''''''"....):0
carried by electromagnetic waves.
1. The momentum ofa particle ofmass m moving with velocity v is given
by
p=mv ...(2.17.1 i
,
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Physics 2-27 A (Bem-! & 2)
According to Einstein's mass-energy relation
U
Energy, U =mc2 or m =2
c
U->
P -v ...(2.17.2)
2
c
1. The energy density in plane electromagnetic wave in free space is given
by
u = EJt2 ...(2.17.3)
where E is the magnitude of electric field.
-1. Thus, the momentum density or momentum per unit volume associated
with an electromagnetic wave is
u -->
P = 2V ...(2.17.4)
c
5. If the electromagnetic waves are propagating along X-axis, then
v = ci
2-28 A (Bem-! & 2) Electromagnetic Field Theory
IIIIIIIIDefine radiation pressure. Derive the relation between
radiation pressure and energy density.
._.
, . -.. .~'.: - ~
A. Radiation Pressure :
1. When electromagnetic wave strikes a surface, its mOriwntum changes.
The rate ofchange of momentum is equal to the force, This force acting
on the unit area ofthe surface exerts a pressure, called radiation pressure.
B. Relation between Radiation Pressure and Energy Density:
1. Let a plane electromagnetic wave incident normally on a perfpd Iy
absorbing surface of area A for a time t.
2. Ifenergy U is absorbed during this time, the momentum p delivered to
the surface is given, according to Maxwell's prediction, by
U
P=
c
3. IfS is the energy passing per unit area per unit time, then
U=SAt
--> u~
p = -I ...(2.17.5) f SAt
c P= -
t). The Poynting vector c
where S is the magnitude of Poynting vector.
...... 1 - --+
S = -(ExB)
~o
, 4. But ~ = u (energy density)
E2 I c
-> ~ [--> -> E2~] lI;
S = - 1 .,' Ex B =- 1 ...(2.17.6) i .. P =uAt
~oc . c ~
5. From Newton's law average force F acting on the surl:ICe is equal to the
7. Substituting the value ofE2 from eq. (2.17.3) in eq. (2.17.6), we get
average rate at which momentum is delivered to th.. :illI'face. Thel'efore.
--> _U_ i = uc i F=!!.=uA
s= l·,'c=~J
EoC~lo t
6. The radiation pressure Prad exerted on the surface.
or ui = S ...(2.17.7)
-->
F
c Prad = A =u
S. Putting the value of u i from eq. (2.17.7) ineq. (2.17.5), we get Hence, the radiation pressure exerted by a normally incident plane
electromagnetic wave on a perfect absorber is equal to the energy density
S _ _
1_(ExB) in the wave.
P - - 2
2
c ~oc 7. For a perfect reflector or for a perfectly reflecting surface, the radiation
after reflection has a momentum equal in magnitude but opposite in !
-> --> .. _1 -~ J
• 2 - ~o direction to the incident radiation. The momentum imparted to the I·',
1·,
or P = Eo(E x B) .,,(2.17.8)
( ~oc
surface will therefore be twice as on perfect absorber. That is,
~). Eq. (2.17.8) represents momentum per unit volume for' an
Prad = 2u
electromagnetic wave. The value of this momentum is,
_ Discuss depth of penetration or skin depth.
P = -
u
or u =pc
c
i.e" Energy density = wave momentum x wave velocity.
t
f
I
[
l ...
1
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i{i'
Answer I
1. Thl' depth of penetration is defmed as the depth in which the strength of
ell'ctric field associated with the electromagnetic wave reduces to 1/e
times of its initial value.
2. Depth of penetration or skin depth is a very important parameter in
describing conductor behaviour in electromagnetic field and in radio
communication.
l.v
E
o= penetration depth
-------~ . -
I I
- i s • • X
Distance-
Fig. 2.19.1.
:1. TIll' rl'cipl"()('nl of attenuation constant is called skin depth or depth of
!-,t'!w(l'a( ion i.""
1
6=
attenuation constant
4. For good conductors, penetration depth decreases with increase in
frequency and is given as
is = U)~cr
~
I), FOJ' poor conductors, skin depth is independd.t of frequency.
0= ~H
Que 2.20.1 Show that for a poor conductor, the skin depth can be
expr('ssed as
o=! ~
cryj;
Answer I
1. We know that fur a conducting medium, the propagation constant can'
lw cxpn'".~('d n,.;
K = (1 + if3
!fl']"c. (1 == co fi [
1 + C~J2 -1
]
112
2-30A(Sem-l & 2) Electromagnetic Field Thl'o!'
U2
and
f3 = (0 J¥ J1 + ( e:r+ 1
[ ]
and skin depth is given by
0= .!.
a
2. For a .poor conductor cr « &(0, so we can approximate the first term II'
square root bracket of right hand side of expression of fJ. using I I"
binomial theorem as
cr 2 ( 0'2 2 2 cr
).! 0' 4
1+ ( - ) = 1+ =1+---.---+ 1 +-~
£w 8 2m2 21:2(02 884(JJ4
2c:.!",
i.e., J1 + ( cr)2 _ 1 = ~cr2
lew 2 &2 (02
3. The expression of u therefore reduces to
l]/2
ql cr2
a == (OJ¥[ 2 &,,2 J
{EI; cr cr r;;
or a == (0 '12 ../2 &(0 Or a = "2 V-;
4.. Hence, the skin depth for a poor conductor can be expressed as
0= ..!.=~ ~
a cr V;
For silver, Il =Ilo and cr =3 x 107 mhos/nl. Calculate the
skin depth at 108
Hz frequency. Given, Il = 47t x 107 N/A2.
I
1
o
I 8
",;::;~.j;~~f~~~h~~:~~~2~2::,cr~?;XJ~I~o$!~' Hz
f=.10
1. Since silv@or is a good conductor, therefore, the skin depth is given 11',
,
e- /2 _ C-2
1
f V~ V~f)j.!oO'
!)
F
0= ~ 2
l
!
~
(2n:xl08
)x4n:xl0- 7
x3xl07
o= 9.19 x 10 - 6 m
I
~I; ©©©
i
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I
- Stefan/sL.aw
- Wien'sLaw
- Rayleigh-JeansLav
- Wave ParticlelJuizlf
- Matter Waves
r. rOf/cept Outline: Part-! 3-2A
H. Long and Medium Answer Type Questions 3-2A
Part-2 (3-IIA to 3-25A)
- Time Depende~t:4Jt
EquatiQ.,! .....'
• Born lnterpretq,tr~ii
- Solution to s'tl " "
..'
discuss enel'gy
II: .....••.
' '..':;' ' :
..'.
.;";: ...•
":.::....'
UNIT
'. """",W£i',CXiC;< Quantum Mechanics
Part-I ....••.•..••..•.•.•...••••.••••••••.••••••••••••••••••••.••••••••••••••••••• (3-2A to 3-IIA)
I· Black BodYRi:iILi'a~i~
" :11)
3-2 A (Sem-l & 2) Quantum Mechanics
.' . I
":'PART-l ....,••
i~i~4tan·s',l4ffJti:W.~n's Law, Rayleigh-Jeans
¢;i.W~·:W'(;fl.J~e.(j,tftt.ql'(j,J:)rtality,
Matter Waves.
CONCEPT OUTLINE PART-l
I
Wave Particle Duality:
According to Einstein, the energy of light is concentrated in small
bundles called photon. Hence, light behaves as a wave on one hand
and as a particle on the other hand. This nature of light is known as
dual nature, while this property of light is known as wave particle
duality.
de-Broglie Wave or Matter Waves:
de-Broglie wavelength is given by
A= !:..
p
de-Broglie wavelength in terms of temperature is given by
h h
A- -----===
- P - )3mkT
- SchrodingeY,s
too Compton EffeQt
A. Concept Outline: Part-2 ..,....................................................... 3-IIA
B. Long and Medium Answer Type Questions 3-llA
3-1 A (Sem-l & 2)
{
_ . " '
j}
A. Black Body:
1. A body which absorbs completely all the radiations incident upon it,
o reflecting none and transmitting none, is called a black body.
2. Absorptivity ofablack body is unity for all wavelengths.
3. It appears black whatever the wavelength of incident radiation is.
4. When a black body is heated to a suitable high temperature it emits total
radiation which is known as black body radiations,
5. From the energy point ofview, black body radiation is equivalent to the
radiation of an infinitely large number of non-interacting harmonic
oscillations, the so called radiation oscillations. .
!
l ,
.1
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1
r
!
3-3 A (Sem-l & 2)
Physics
ii. No actual body is a perfect black body, it is only an ideal conception.
7. Lan,p black is the nearest approach to black b9dy which absorbs nearly
99 % of the incident radiation.
R. Platinum black is another example ofa black body.
B. Energy Distribution:
1. Results of the studies of black body radiation spectra are shown in
Fig. :3.1.1 in which variation of intensity with wavelength for various
tt.'mperatures are shown.
t
.~
co
I:l
Q)
...,
I::l
.....
1646 K
(A.T) = A function of the product A.T and is giveL '.,
I itA.T) = e-cll/l.kT = e-<,/;'T
! u;. dA. = AA.-s e-<,/I·T d"•
2. For A. = 0::', it;. = °and for A. =0, 11;. =O.
Wavelength ! :3. Thus eq. (3.3.2) shows that no energy is emitted by a wave ot tnfu. i' :
Fiji. 3,1.1
wavelength as well as by a wave of zero wavelength.
(
~. The energy distribution in the radiation spectrum of black body is not
<I. For T = 00, eq. (3.3.2) reduces to: u;. dA. =AA.-s , which is finite quantit·,
uni forlll. As the temperature ofthe body 'rises, the intensity ofradiation
and is in open contradiction with the Stefan's fourth power law (c/T" :
fOI' each wavelength increases.
5. Wien's law of energy distribution, however, explains the enc;·;'.
i
:1. At a g-ivcn temperature, the intensity ofradiation increases with increase
distribution at short wavelengths at higher temperature and fails io'
in wavelength and becomes maximum at a particular wavelength. With
i
! longwavelengths.
further increase in wavelength, the intensity of radiation decreases.
B. Displacement Law:
4. The point of maximum energy shifts towards the shorter wavelengths
1. As the temperature ofthe body is raised the maximum energy tends"
as the temperature increases,
be associated with shorter wavelength, i.e.,
5, For a given temperature the total energy ofradiation is represented by f AmT= constant ...(3.~L;j
the area between the curve and the horizontal axis and the area increases
with increase of temperature, being directly proportional to the fourth
power of absolute temperatures.
Que 3.2. IDiscuss Stefan's law.
Answer I
1. The tbtal amount of heat (E) radiated by a perfectly black body per
second per unit area is directly proportional to the fourth power of its
absolute temperature (T), i.e.,
E oc '['4 or E = a'['4
where. a = Universal constant and is called Stefan's constant.
2. This law is also called as Stefan's fourth power law.
3, Ifa black body at absolute temperature T is surrounded by another
black body at absolute temperature To then the net amount ofheat (E)
lost by the former per second per cm2
is given by
3-4 A (Sem-l & 2) Quantum Iledwlli
E = cr(T4-To4)
4. This law is also known as Stefan-Boltzmann law.
. ._t...·Explain Wien's laws of energy distribution.
".l?NY,li-.hr'S.•. .,~.'1J,.Ji
~~lWi!~i:~<
A. Fifth Power Law:
1. The total amount ofenergy emitted by a black body per unit voj un:,
an absolute temperatw'e T and contained in the spectral region!: ,," ".
within the wavelength A. and A. + dA. is given as,
u/-dA.= ~f(A.T)dA. ...(3:3'
where, A = Constant, and
where, A.m
=wavelength at which the energy is maximum, and
T = absolute temperature.
2. Thus, ifradiation: ofa particular wavelength at a certain temperature Ie
adiabatically altered to another wavelengths then temperature change'.
in the inverse ratio.
_aDiscuss Rayleigh·Jean's Law.
>' ' . ~ ",
;;::i:
1. The total amount ofenergy emitted by a black body per Wlit volume at
-an absolute temperature T in the wavelength range A. and Ie + die is given
as,
u dA. = 81tkT dA.
/- A.4
where, k =Boltzmann's constant =1.381 x 10-23
J IK.
l
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2
Physics 3--3 A (Sem-l & 2)
2. The energy radiated in a given wavelength range A. and A. +dA. increases
rapidly as A. decreases and approaches infinity for very short wavelengths
which however can't be true.
3. This law, thus, explains the energy distribution at longer wavelengths
at all temperatures and fails totally for shorter wavelengths.
4. The energy distribution curves of black body show a peak while going
towards the ultra-violet wavelength (shorter A.) and then fall while
Rayleigh-Jean's law indicate continuous rise only.
G. This is the failure ofclassical physics and this failure was often called as
the "Intra-violet catastrophe" ofclassical physics.
Que 3.5. :I Derive Planck's radiation law and show how this law
successfully explained observed spectrum of black body radiation.
AnSt~~~(~tf~1
A. Planck's Radiation Law:
1. According to Planck's quantum hypothesis the exchange of energy by
radiations with matters do not take place continuously but
discontinuously and discretely as an integral multiple ofan elementary
quantum of energy represented by the relation
E= hv
where, v =frequency of radiations, and
h =Planck's constant.
Thus, the resonators can oscillate only with integral energy values hv,
2hv, 3hv, ... , nhv or in general En = n hv (n = 1, 2, 3,...).
3. Hence, emission and absorption ofenergy by the particles ofa radiating
body interchanging energy with the radiation oscillation occur discretely,
not in a continuous sequence.
4. In relation En = n hv, n is called a quantum number and the energies of
the radiators are said to be quantised and allowed energy states are
called quantum states.
5. On the basis of his assumptions Planck derived a relation for energy
density (u) of resonators emitting radiation offrequency lying between
v and v + dv which is given as follows:
8nhv3
dv
u, dv = ~ eh"kT -1 ...(3.5.1)
8n he dA.
or ul. dA. = --5- hclMT ...(3.5.2)
"- e -1
6. The eq. (3.5.1) and eq. (3.5.2) are known as Planck's radiation law.
B. Experimental Verification ofPlanck's Radiation Law: According
to Planck's radiation law expression for energy density is given as
8n he d,,
ul.d"-= ~ • ".
3-6 A (Sem·l & 2) Quantum Me.;hanics
a. Wien's Law from Planck's Radiation Law :
1. For shorter wavelengths "-Twill be small and hence e'''likT» 1
2. Hence, for small values of ,,-T Planck's formula reduces to
8n he d,,- .
=8n he A-5 e-hchkTdA:
U d,,- = -- --
l. "-5 ehc/MT
or Ul. d,,- = A,,--5 e-<Ul.T d,,- ...(3.5.3)
where, A = Constant (= 8n he), and
a = Constant (= he/k).
3. Eq. (3.5.3) is Wien's law.
4. This result shows that at shorter wavelengths Planck's law approaches
Wien's law and hence·at shorter wavelengths Planck's law and Wien's
law agrees (Fig. 3.5.1).
Planck's law
, Rayleigh-Jean's law
Ul.tl : ~
I ,
I ,
I "
I ; ....
I .... ~
I Wein's law ~ -
A ---
lIil~.~~l
b. Rayleigh-Jean's Law:
1. For longer wavelengths ehclUTis small and can be expanded as follows:
he' he
ehcll.kT = 1 + - - "'-
"- kT "- kT
2. Hence, for longer wavelengths Planck's formula reduces to
dA = 8n he AkT d,,-
Ul. ,,-5 he
dA = 8n kT d,,- ...(3.5.4)
or ul. "-4
3. Eq. (3.5.4) shows that for longer wavelengths Planck's law approaches
to Rayleigh-Jean's law and thus at longer wavelengt.hs Planck's law and
Rayleigh-Jean's law agree (Fig. 3.5.1).
4. Thus, it is concluded that the Planck's radiation law successful Iy explained ~"."
the entire shape of the curves giving the energy distribution in black
body radiation.
_ _ Discuss the wave particle duality.
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Answer I
ACl'ordillg to the Planck's theory of thermal radiation; Einstein's
"xplanCltlof! of photoelectric effect; emission and absorption of radiation
b,i s.I[,>;tal(;('; black body radiation etc., the electromagnetic radiation
cOl~i,;t ofdiscl'l'te indivisible packets ofenergy (ltv) called photons which
man ifes( partide character of radiation. On the other hand, macroscopic
[,ptical phenomena like interference, diffraction and polarisation reveal
and firrllly confirm the wave character of electromagnetic radiation.
Th,.,,·...[o,·('. we conclude that the electromagnetic radiation has dual
thar'(H:tel', in certain situation it exhibits wave properties and in other it
acts like a particle, .
:0; The pnrticle and wave properties of radiation can never be observed
sim.ultaneollsly, To study the path of a beam ofmonochromatic radiation,
IV" us(' tIll' W:W(' theory, while to calculate the amount of energy
t ransLlctions of the same bea.l1l, we have to recourse to the photon or
pnrt iele tlWOl')',
,T It hmi be('n found impossible to separate the particle and wave aspects of
electromagnetic radiation,
Que 3.7. IWhat are de-Broglie's waves or matter waves?
t}'~·U:Mil1f;til~i
-f";~'t:4~,::,<,.,·"~,,":''''i,.f!"",,:,:;:
Answer I
1. W11en a material particle moves in a medium, a group of waves is
associated with it due to which it shows the wave particle duality, These
waves are known as matter waves or de-Broglie waves.
2. AccOl"ding to de-Broglie's concept, each material particle in motion
behaves as waves, having wavelength '')..' associated with moving particle
of momentum p.
A= !!.. => AOC..!:
p P
Wa"e nature oc
1
Particle nature
Que 3.8. IDeduce expression for wavelength of de·Broglie wave.
Ab:s"W'cr' I
1. Let a photon having energy,
E = hv = he ...(3.8.1)
')..
2. Ifa photon possesses mass, it is converted into energy.
3. Now according to Einstein's law,
E= mc2
...(3.8.2)
3-8 A (Sem-l & 2)
Quantum MechanH':"
4. From eq. (3.8.1) and eq. (3.8.2),
he he
me2 = _ => A=-
').. me2
h h
A= - => A =,.- ['.' me :~, 1-' I
me p
5,
In place of photon, we take material particle having mass 'm' movm.'
with velocity 'v'. The momentum,
p=mv
6. The wavelength of wave associated with particle is,
A = --!!:.....=!!:..
mv p
This is de-Broglie's wavelength.
7.
If Ek is kinetic energy of material particle of mass 'm' moving ',', ':,
velQcity 'v' then,
Ek = -
1
mv2
2
2 2
m v
Ek = -
2m
(mv)2 p2
or E - --=_
k- 2m 2m
or p = )2mEk
8. The de-Broglie's wavelength, A == ) h
2mE"
9.
According to kinetic theory ofgases, the average kinetic energy (E ,) of
the material particle is given as r
3
Ek = -KT
2
10. The de-Broglie's wavelength,
h h
t..=
J2m x .?~ J3mKT
where, K = 1.38 X 10-23 JIK
T == temperature (K).
11.
SUppose material particle is accelerated by potential difference ofV volt
then,
Ek==qV
where, q == charge ofparticle.
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l
Physics 3-9A (Sem-I & 2)
12. The de-Broglie's wavelength,
h
")..- -
- .j2mqV
Que3~!lit~tJ Derive an expression for de-Broglie wavelength of
helium. atom having energy at temperature T K.
AnSV~l~W~
1. According to kinetic theory ofgases, the average kinetic energy (Ek ) of
the material particle is given as
3
Ek = -KT
2
<.:. The de-Broglie's wavelength,
").._ h _ h _ h
- .J2mEk - )2 3KT - .J3mKT
mx-
2
where, K = 1.38 x 10-23 JIK, and
T = temperature (K).
Que.S.l0:d The kinetic energy ofan electron is 4.55 Ie 1()-Z5J. Calculate
the velocity, mom.entum and wavelength of the electron.
Answ';~!,l~l
GiVen.~~*;="
..
ToFind:
cV~l~l!~'
If m is the rest mass ofelectron, v is the velocity ofthe electron, then its
a
kinetic energy (Ek
) is given by
2
E k ="2
1
m ov
26 "
2Ek 2 x 4.55 x 10- =1 x 103 mls
v = V =
m 9.1 x 10 31
o
[ oo m = 9.1 x 10- 3 kg]
• 0
Momentum of electron,
p = mov = 9.1 x IQ-31 x 103 = 9.1)( 10-28 kg rnls
3. Wavelength of electron,
").. = hlp = (6.63 x 10-34)/(9.1 x 10-28) = 7.29 x 10-7m
·1
..!
I
"I :
3-10A (Sem-l & 2) Quantum Mechanics
" F i n d the de-Broglie wavelength of neutron of energy
12.8 MeV (given that h = 6.625 x 10-34J-s, mass of neutron
(m ) • 1.675 x 10-27 kg and I eV = 1.6 Ie 10-19 Joule).
n
I
~ !
- V:~:~;~"6zY325 x 10-34 J-s, m = 1.675 x 10- 27 kg,
n
<>::',;~~::i'>'2;:" ,
)gth~"
1. Rest mass energy of neutron is given as
rna c2 = 1.675 X 10-27 X (3 X 10R )2
= 1.5075 X 10-10 J
10
= 1.507 X 10- = 942.18 MeV
1.6 x 10- 19
2. The given energy 12.8 MeV is very less compared to the rest mass
energy ofneutron, therefore relativistic consideration in this case is not
applicable.
3. Now de-Broglie wavelength of the neutron is given as
")..- -~
- .j2mEk
Ek
= 12.8 x 106 x (1.6 x 10-19 ) J
").. = 6.625 x 10 :H
J2 x 1.675 x 10' 27 x 12.8 X 10" x 1.6 x 10. 10
6.625 X 10-34
8.28 X 10. 20
= 8 X 10-15 m
=8xlO-5 A
••11 Calculate the de-Broglie's wavelength associated with
1
a proton moving with a velocity equal to 20 th of light velocity.
_I
" i, . '"
108 mls = 1.5 x 107 rnls
1. Formula for de-Broglie's wavelength:
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Physics 3-11 A (Sem·l & 2)
h
A=
III V
A = 6.63 x 10- 34 =2.646 x 10- 14 m
1.67 x 10- 27
X 1.5 X 107
f·,' In =1.67 x 10- 27 kg and h =6.63 x 1O-34J-sl
=2.646 x 10 - 4 A
I PART-2[
Time Dependent and Time Indepen,d(!ntSchr~.clinger'p,.
Equation, Born Interpretationofw,d,rAeFunct~trl!$"?~i"'"
Stationary State, Schrodinger Wave Eg/J,lqJion(or,l?f1/e,ri;
Particle in a Box, Compton Effei:l.
CONCEPT OUTLINE PART·2
----~-----
Wave Function and its Significance:
The wave function IjI is described as mathematical function whose
variation builds up matter waves. 11jI!2 dermes the probability density
of finding the particle within the given confined limits. IjI is defined as
probability amplitude and IIjIj2 is dermed as probability density.
Schrodinger's Wave Equation:
This wave equation is a fundamental equation in quantum mechanics
and describes the variation of wave function IjI in space and time.
Compton Effect: The phenomenon in which the wavelength of the
incident X-rays increases and hence the energy decreases due to
scattering from an atom is known as Compton effect.
I Questions-An$w·~"s
f----·
I Long Answer Type and Medium Answer TYpe Questions
...__._----- -"- -_.. - -~ ~
Que 3.13.1 What is Schrodillgel' wave equation 'l Derive time
in<!(c'!wnl!Pllt and time dependent Schrodingel' wave equations.
IA:f{'rtJ 20~5"lQ~..~~1~~~·1:.~
Answer I
1. Sehrudinger's equation, which is the fundamental equation of quantum
I11l'chanics, is a wave equation in the variable 1jI.
I
f 3-12 A (Sem-l & 2)
Quantum lVl"ch'lIl:
A Time Independent Schrodinger Wave Equation:
I. Consider a system ofstationary wave to be associated with particle an'
the position coordinate of the particle (x, y, z) and IjI is the period!,
displacement at any instant time 't'.
2. The general wave equation in 3-D in differential form is given as
2
....2 _ 1 c. 1j1 '" '.
Y 1jI- --2- ... I.i. J
v
2
Dt
where, v = velocity of wave, and
iP c""j 2 (}2
V
2
= --2 + -2- + ~ =Laplacian operator.
ax c!y OZ
3. The wave function may be written as
IjI = ljIoe-iw, ,.,(:3,]:~.:i
4. Differentiate eq. (3.13.2) w.r.t. time, we get,
(~1jI . ,
...:- =- i «) IjI e-/l'J
,r:i}:) ,
r't 0
5. Again differentiating eq. (3.13.3), we get
(}2", '2 2 . t
--:!.. = l (t) IjI e-/l"
8t2 0
(121j1
__ = _ «)21j1
.. '(3.1:3.,
fit 2
6. Putting these value in eq. (3.13.1),
_0)2
2
V 1j1 = -~1jI ... (;:L j ~
v
2n v co 2n
7. But. «) =2n:I = -- ~ _ .. 1;:3.1:3(;,
A' v f.
. 4n2
V2 1 = - -2-1j1
' , .':3 1.
A
9. From de-Broglie's wavelength, t. = h
nJ.v
'J -4n:2
m 2
v 2
then, 9- ~I = ;,__ '1 . .,1 ::3,1:)..'",
h- '
10. IfE and V are the total energy and potential energy ofa partiele and E •
is kinetic energy, then,
2
Eli = E- V or 1. m.v = E - V
2
m 2V2 =2m (E - V) ... (:.3. 1;~L~' .
1
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Physics S-13 A (Sem.l &2)
v2 _ -47t2
2m[E - V)jI
jI- h2
2 2m[E-V]jI [ h ]
V jI + 2 =0 where, n=- ...(3.13.10)
!l 2ft
This is required time-independent Schrodinger wave equation.
12. For free particle, V = 0
V2 jI + 2m EjI = 0
n2
B. Time Dependent Schrodinger Wave Equation:
!. We know that wave function is jI =jI oe-irot
!-. On differentiatingw.r.t. time, we get,
ojl = _ iro jI e-irot
ot 0
or ojl = _ i (27tV) jI ...(3.13.11)
ot
:3.
l.
E =hv
But
So, eq. (3.13.11) becomes,
~ v == .!E.
h
ojl =_i2ft (E) jI
ot h
Ojl
ot
= _ ~
Ii
Ej1
[-: n= ~]
2ft
or
Ii a'll
EjI=- Tat
:l.
6.
or E ." Ojl
jI=ln
ot
Now time independent Schrodinger wave equation is,
2 2m
V jI + - (E V) jI = 0
li2
2m
v2 jI + (EjI- VjI) = 0
li2
Using eq. (3.13.12), we g.et,
...(3.13.12)
V2 jI + 2m
-
112
[. Ojl ]
lh
- VjI
at
= 0
2
V jI -
2m
-
112
VjI = -
2m.
- l h
112
ojl
-
ot
.~!'1lI
.. ,
3-14A (Sem-l &2) Quantum Mechcnics
2m ) 2m . cljI
( V 2 _-V 1jI=--lfl-
112 n2
&
(,,2 ') . <"1jI
or l--v2
+ V) IjI =In
2m at
This is required time dependent Schrodinger wave equation.
2 .
7. - !'!.-v2
+ V =H ~ is known as Hamiltonian operator.
2m
Ojl
and, i n -~- =EjI ~ enel'gy operator.
et
Then, HjI =E>¥
Q:qe~~·~~·~; •.lDiscuss the Born's interpretation of wave function.
~~~'~l~ij.::'1
1. Max Born interpreted the relation between the wave function '.'.. II
and the location of the particle by drawing an analogy between (Iw
intensity ofIight or photon beam and the intensity of electron be" Ill.
2. Consider a beam of light (EM-wave) incident normally on a scn~(>11 Th(,
magnitude of electric field vector E ofthe beam is given by
E =Eo sin lkx - eM)
where, Eo = Amplitude of the electric field.
3. For an EM-wave, the intensity I at a point due to a monochmmatic lwam
of frequency v is given by
1= CEp < E2 > ...(3.] 4 ])
Here, <.E2> = Average of the squarC' of tIll' il1~tantHTH'(lll.~
magnitudes of the eh'c(ric field v('dor 01 t IH'
wave over a complete c~·clp.
C = Velocity oflight in free .<pace. ,11lll
£0 =Electric permittivity offl'E'E' space.
4. The intensity may also be interpreted as the number N of photon" ('d('1l
of energy hv crossing unit area in unit time at the point uncleI'
consideration normal to the direction of the photons.
Thus, 1= N hv ...(3.] 4.2)
5. Comparing eq. (3.14.1) and eq. (3.14.2). we get,
N hv = ceo < E2>
N = CEo < E Z >
or
hv
or N'>'.. <E2> ...(:3.]-1.:11
This relation is valid only when a large number of ph()ton.~ :ll'e invo! (·eI.
i.e., the beam has the large intensity.
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r
6. If we consider the scattering ofonly a single phQton by a crystal or the
passage ofonly a single photon through a narrow slit, then it is impossible
to observe the usual pattern ofintensity variation or diffraction.
7. In this situation, we can only say that the probability ofphoton striking
the screen is highest at places where the wave theory predicts a maximwn
and lowest at places where the wave theory predicts a minimum.
8. Eq. (3.14.3) shows that < E2 > is a measure ofthe probability of photon
crossing unit area per second at the point under consideration. Hence,
in one dimension, <E2> is a measure ofthe probability per unit length of
finding the photon at the position x at time t.
Que 3.15. IThe wave function of a particle confined to a box of
length Lis
(2.7tX
IjI (x) = VL SID L 0 < x < L
and ljI(x) = 0 everywhere else.
Calculate probability of finding the particle in region
o<x< L
2
Answer I
1. The probability of finding the particle in interval dx at distance x is
p(x)dx = !1jI1 2
dx = -! sin2
(rrx) dx
L L'
2. The probability in region 0 < x < L is
2
JLl2 ILl22 . 2(rrx)
p= 0 p(x)dx = 0 Lsm L dx
1 fLl2( 21tX)
L 0 I-COST dx
3-16 A (Sem·l & 2)
Quantwn Mechamc,.;
( )
~
IjI x, y, z, t = "
..::...a"f."(x, y, Z )e-iE"tlh
...(3.161
,,~l
The complex conjugate of eq. (3.16.1) is,
1jI*(x, y, z, t) = ~ am * fm *(x, y, z) ei~",flh ... tJ.16. __
m",.l
The product of'I' and 1jI* or probability distribution function jJ lit'" is give"
by
.. - [~a f. (x y z) e-iE.'lh] [~.a' f." (x y z) eiE",tt:, II
#I'" - 4...J n n , , 4..J mill' "
J
n=! m=!
= ~a"a: f.,(x, y, z) f.; (x, y, z)
" " • f. ( ) f.' ( ) "E", -E., 111 i' ("j 1 ('
+ 4..J4..Jan am n X,Y,Z In x,y,z e ... ,_ ..t " .
nl "
3. The probability distribution function that is, '1''1'* will be independent 01't i[,.
onlywhen an' s are zero for all values except for one value ofE". In such '!.o, .
the wave function contains only a single term and expressed as
'l'n(x, y, Z, t) =f,,(x, y, z) e-iE,,'h
4. Since '1''1'* = f.lr.: is independent of time, the solution repreSf'r,'. " ;. ,'
eq. (3.16.4) is stationarystate solution.
;Q~~'il~:t't~:i Write the Schrodinger wave equation
, for l 1, (, l".lrtkh~ in
"'l":'~:,)1'"L'!(l",,;~ ."::1,1.',, •
a box and solve it to obtain the eigen value and eig'('" "dH~dGC'.
~111'it)i3~~I~i~20i-:-i.i8i
: -...~._, -.....,----_...:__._-----,~._--'
:~~
1. Let a particle is conimed in one-dimensional bOA oflength 'L'. 'I'he particlp
is free i.e., no external force, so potential energy insid.e box is zero
(V= 0>.
I
I
v" 0 I
I
1 L 1
£'2:=2
I
Que 3.16. IDiscuss the stationary state solutions in brief.
=I~ )1v=oo •
Answer I
x=O x=L +x
1. A state of the system in which probability distribution function 1jI1jI* is
independent oftime is called stationary state ofthe system.
.~~~lii
where, IjI = wave function, and
r V =0 for 0 < x < L
1jI* = complex conjugate of wave function.
V =00 for x < 0 and x > L
2. Let the probability distribution function 1jI1jI* for a system in the state is I i.e., outside the box the potential energy is infinite.
given by the wave function
I
2. The particle cannot exist outside the box.
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Physics 3-17 A (Sem-1 & 2)
jI =0 for x =L and x =0
:', Schrodinger time independent equation for free particle (V =0),
2
a 1' 2m E 0 (3 17 1)
--" + -2 jI = ... . .
ax- tt
or ~~ + k 2
jI = 0 [Where. k 2 == 2rnEJ ...(3.17.2)
ax h2
4, Solution of eq. (3.17.2) is
jI(x) = A sin kx + B cos kx ...(3.17.3)
U sing boundary condition. jI = 0 at x = 0
o= A sin 0 + B cos 0
or B = 0
~d jI=O~x=L
o=A sin kL + B cos kL
or A sin kL =0 or sin kL =sin n1t [.,' B =0]
kL = n1t, n = 1, 2. 3.... But n '* 0
k = n1t
L
•-; Now eq. (3.17.3) becomes,
'I' (x) == A sin n1tX [Eigen function]
II L
n.2 ;r;2 2mE 2 2 2
= E == n rr. tt
and
£2 ~ 2mL2
Then. E = n h tEigen energy]
"
2 2
[tl=;rr.]
8mL2
2
Fen' 1/ == 1. E] = -- h 2 ' it is ground state energy of particle.
8mL
"~~ue 3~ A partil'le is in motion along a line between x == 0 and
, ,~ L with zero potential energy. At points for which x < 0 and
''- > L. the potential energy is infinite. The wave function for the
partick ;'I nth statp is given by:
. nrr.x
I, =A Sln-
.,." L
Find the expression for the' normalized wave function.
OR
del'iva normalization wave function.
:nswer I
The eigen funct ion is,
'I'" (x) == A sin nrr.x ...(3.18.1)
L
3-18 A (Sem-1 & 2) Quant.um Mechanics
2. Now applying normalization condition to find constantA,
L
2
fl jI n(X) 1 dx = 1
o
rA 2
sin2
(n1tX)dx 1
o L
A 2nrr.
-
2fL(1-cos--X) dx = 1
2 0 L
- . 2nrr.x]L
A 2 x sm-L 1
2 2nrr.
[
L 0
A 2 '
_L= 1
2
A==
Ji
3. So, eq. (3.18.1) becomes, jIn (x) = sin (n2x).
Ji
This is normalization function.
n=3
n=2
n==l
x=L x=O x=L
x=O
"~iik~';ll:i£",'r'
~Iif~clll:'/
An electron is bound in one dimensional potential box
which has width 2.5 )( 10-10 m. Assuming the height of the box to be
infinite, calculate the lowest two permitted energy values of the
electron., _lllJJI'F._i&1
=:;Jili,-::;'+:,:t
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1
We know that,
2
h 2
n
E = -
" 8mL2
n 2 (6.63 x 10-34)2
En = 8 x 9.1 X 10-31 X (2.5 X 10-10)2
[.: h =6.63 X 10-34 J-s, m =9.1 X 10- 31 kg]
= 9.66 X 10-19 n2 J
9.66 X 10- 19
= eV
1.6 X 10-19
== 6.037 n 2 eV
2. For n = 1, E 1 == 6.037 eV
n=2, E 2 =24.15eV
Qu,eS.20. ICompute the energy ditTerence between the ground state
and first excited state of an electron in one-dimensional box oflength
10-8 m.
Answe~ I
~~v;~~~;~:::~ft~~n~~'~j;
stat,e.
2
h2
n
1. We know that eigen energy, E" == --2
8mL
E == n 2 (6.63 x 10-34)2 == 0.60 X 10-21 n 2 J
" 8 x 9.1 x 10-31 x <10-8 )2
2
0.6 x 10-
21
n eV
1.6 x 10 19
E" == 3.75 X 10-3 n 2eV
2. For ground state (n == 1), E 1
= 3.75 X 10-3 eV
3. First excited state (n = 2), E 2 = 0.015 eV
4. Difference between first excited and ground state,
E 2
-E1 = (15 - 3.75) 10-3 eV =11.25 meV
• m.a~I';1 A particle confined to move along x·axis has the wave
function IJf =ax between x = 0 and x == 1.0, and 'If = 0 elsewhere. Find
the probability that the particle can be found between x =0.35 to
x =0.45. Also, find the expectation value < x > of particle's position.
3-20 A (Sem·l & 2) Quantum Mechani('~
....
' .....
..
1.
2
p == j I'"n 1dx
XJ.
2. Here, Xl == 0.35 and x 2 == 0.45
== f 0.45 ax 2dx == 2 f 0.45 x 2dx
ThE-refore, P 0.35 () a 0.35
a 2 045 a 2
p== s[x3135 =3 [(0.45)3-(0.35)3]
2
= ~ [0.091125 - 0.042875] =0.0161 a 2
3
3. The expectation value ofthe position ofparticle is given by
<X> = j X 1",,,(x)1
2
dx
4. Since, the particle is confined in a box having its limit X == 0 tox = 1 then.
1 1
<x> = fx.(ax)2dx =a2 fx3dx
o 0
2
a
<x> = - ==0.25a2
. 4
_ _ Determine the' probabilities offinding a particle trapped
in a box oflength L in the region from 0.45 L to 0.55 L for the ground
state. LmtTtr 2017-]81
~IU
........., ..... "." .. ,;(~#.a~.~abox .
1. The eigen function of particle trapped in a box of length L is
n7tX
IJf X = - Sln-'
() Ji .
" L L
2. Probability, p= JIIJfIl(x)1
2
dx = ~.Jsin2
n;:dx
Xl Xl
The probability ofimdingthe particle between Xl a:n.dx2when it is in //11,
state is,
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I
X
2 x..' 1 ( 21t1lX) 1 [ L. 2nnx J'
p =- - 1 - cos - - dx = - x - -- sin -
s
L '" 2 L . L 27tn L XI
Since, xl = 0.45 Land x 2 =0.55 L, for ground state, n =1
O55L
1 [ L, 2nx J.
P= - X--Sln-
L 27t L O.45L
i [(0.55L - ::. sin 1.l7t) -(0.45L - ~ sin 0.97t) ]
1 1
" [( 0.55 - 27t sin 198
0
) - ( 0.45 - 27t sin 162
0
)]
P =0.198362 =19.8 %
(~ue 3.23. IDiscuss Compton effect and derive an expression for
'ompton shift.
OR
Der'ive an expression for Compton shift showing dependency on
:lng-Ie of scattering.
Answer: I
When a monochromatic beam of high frequency radiation is scattered
by a substance, the scattered radiation contain two components-one
having a lower frequency or greater wavelength and the other having
the same frequency or wavelength.
The radiation of unchanged frequency in the scattered beam is known
<I" 'unmodified radiation' while the radiation oflower frequency or slightly
higher wavelength is called as 'modified radiation'.
This phenomenon is known as 'Compton effect'.
Let a photon of energy hv collides with an electron at rest.
During the collision it gives a fraction ofenergy to the free electron. The
electron gains kinetic energy and recoil as shown in Fig. 3.23.1.
Scattered
X-rays
Modified
Source _",,-4,"" .II n X-rays
.:F;~"~;;~,~~lW~~~~;~'
3-22 A (Sem·l & 2) Quantum Mechanics
._--
e~ E h "
_e-<; nergy 2 V ' S i n
!." ~ v e hv'
'?::J<:,,'b'~o""O Momentum hv'
Electron C
. at rest v'cos e
Incident photon ·····r·?~::~~~·~~~~_::~·gle : "J~"""" .......--...
Energy hv •••.•/ <V hv •••• 4>
Momentum hv ••..•• c·..... In' cos <j>
C Recoil ."-. Fe
electron
mv sin 4> mv
(a) Geometry of Compton scattering (b) Components of momen tum
before and after collision
tifA:~4~~;$.
a. Before collision:
i. Energy of incident photon = hv
ii. Momentum of incident photon = h v
c
2
iii. Rest energy of electron = m oc
iv. Momentum of rest electron = 0
b. After collision:
i. Energy of scattered photon =h v'
ii. Momentum of scattered photon = h v'
c
2
iii. Energy of electron = mc
iv. Momentum of recoil electron = mv
6. Accordingto the principle ofconservation of energy,
hv + m c2 = hv' + mc2 ... (3.2:3.11
o
.Again using the principle of conservation of momentum along and
perpendicular to the direction ofinCident, we get,
Momentum before collision = Momentum after collision
hv hv'
7.
...(3.2:1.2)
- + 0 = -- cos 8 + mv cos <V
c c
hv'. 8 . '" ...(3.23.3)
O + 0 = -- sin - m v sin 't'
c
8. From eq. (3.23.2), we get,
...(3.23.4)
mvc cos ep = hv - hv' cos 8
9. From eq:(3.23.3), we get,
mvc sin ep = hv' sin 8 ...(3.23.5)
Squaring eq. (3.23.4) and eq. (3.23.5) and then adding, we get,
10.
m2v2c2=(hv - hv' cos 8)2 +(hv' sin 8)2
= h2v2 _ 2h2 vv' cos 8 + h 2 v·2 cos2 e + h 2v'2 sin2 e
= h 2 [v2 + v'2 - 2vv' cos 81 ...(3.23.6)
11. From eq. (3.23.1), we get,
2
mc2 =h(v - v') + m oc
-:~,
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r
3-23 A (SeDl-l & 2)
Physics
m 2C4 = h2( v2 _ 2vv' + v'2) + 2h(v - v') m oc2
+ m o2C4
Squaring,
...(3.23.7)
Subtracting eq. (3.23.6) from eq. (3.23.7), we have,
12. 4
2 2
rn2 4 _ m 2v2c 2= - 2h2vv'( 1 - cos 6) + 2h( v - v') m oc + m o c
c
2 4
m2c2 (c2 _ v2) = _ 2h2vv'(I- cos 6) + 2h(v - v') m oc
2
+ mo c
or
2 2 4
ln
2
c 2
_0-2-(c2 _ v2) = _ 2h2vv'(I- cos 6) + 2h(v -v') m oc + m o c
or
v
1- ~2
[ .. m= g]
2 2 4
mo2 c4 = - 2h2yv'( 1 - cos 6) + 2h(v - v') m oc + m o c
or
2h( v - v') l7l C2 = 2h 2 yV '( 1 - cos 6) ...(3.23.8)
0
v- v' h
or -,- = --2(1- COs 6)
Vy mac
1 h ...(3.23.9)
--;-- = --,
2
(1-cos6)
' ' m oc
Eq. (3.23.9) shows that v' < y as m ' c, h are the constants with positive
I:L o
values and the maximum value ofcos 6 = 1. This shows that the scattered
frequency is always smaller than the incident frequency.
14. From eq. (3.23.9), we have,
c c h
--- = --(I-cose)
v' v TrlQc
or 'Ie' -'Ie = _h_ (I-cos 6)
moc
2h . 26 ...(3.23.10)
or 6.'A= --sIn
moc 2
From eq. (3.23.10), it is noted that Compton shift depends on angle of
IS.
scattering.
Qu~ S~~4. IExplain the experiDlental verification ofCompton effect•
_ III
AA$~~I' I
1. The apparatus used by Compton for experimental verification of
Compton effect is shown in Fig. 3.24.1.
3-24 A (Sem-l & 2) Quantum Mechamc;
Scattered X-ray "(: BS
•.•'
........~.. ;
@
T
.. -----~:~aterial (T) ./
.'
.......................
~.'1!Il_~~~·;~~t.
2. Monochromatic X-rays of wavelength A. from a Coolidge tube CT an'
allowed to fall on a target material T such as a small block of carbon.
3. The scattered X-rays of wavelength A.' are received by a Bragg
spectrometer BS, which can move along the arc of a circle.
4. The wavelength ofthe scattered X-rays is measured for different valwo,
of the scattering angle.
cjl = O· cjl = 45° cjl = 90· .p = 1350
6.A. LA.
t A I - f
I
A. __
~_j,~_~~~t~;,
5. The plots ofintensity of scatteredX-rays against their wavelength are
shown in Fig. 3.24.2 for Ijl =0°, 45°, 90° and 135°.
6. These plots show the two peaks for non-zero values of Ijl, which is thp
indication of the presence of two distinct lines in the scattered X-mv
radiation.
7. One of these lines is known as the unmodified line which has the samp
wavelength as the incident radiation and the other line is known as the
modified line which has the comparatively longer wavelength. TIll'
Compton shift 6.A is found to vary with the angle of scattering.
_ Why Compton shift is not observed with visible light?
t_
"!:';- - 't ,"
.~
1. The energy of a visible light photon say of wavelength 'Ie = 6000 A
(= 6 X 10- 7 m) is given by
E =hv = hc
A.
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-------------------
! "l1ysics 3-25 A (Sem.l & 2)
= (6.6 x 10- 34 x 3 x 108 ) J
(6 x 10- 7 )
(6.6 x 10- 34 x 3 x 108 ) eV
(1.6 x 10 19 x 6 x 10- 7 )
=2.06eV"'2eV
Vhereas the energy ofX-ray photon, say of wavelength
i. 'c 1 A = 10- 10
m will be more than 1000 times the above value.
The binding energy of the electron in the atoms is of the order of the
10 eV. For example, the binding energy ofthe electron in the hydrogen
atom is,
E -
b -
21t2
k 2
Z 2m e4
0
h2
Where, k:: 1/41t&o = 9 x 109 N-m2/C2,
rna = 9.1 x 10- 31 kg,
h:: 6.6 x 10- 34 J-s,
e = 1.6 x 10- 19 C,
Z= 1.
Thus, E :: 21t
2
(9 x 109
)2 (9.1 x 10- 31 )(1.6 x 10-19)4 J
b (6.6 x 10- 34)2
21t2
(81 x 1018)(9.1 x 10- 31 )(1.6 x 10- 19)4 V
(6.6 x 10- 34)2 (1.6 x 10- 19) e .
= 13.68 eV
Hence, this electron can be treated as free when X-rays are incident but
this electron cannot be treated as free for visible light. So, the Compton
effect cannot be observed for visible light.
©©©
Wave. Optics
.C~~~rentSources. . '. .•.
• Int~'rrei;ii1(t.ceiji·JJnifonnand'wedgeBhaped Thin Films
~-Net:f!sS,i,.~Mr~{E~te1J:4ed
Sql:fr~es ,.
• Nii6Jt.jj;ti;;ll~/rj.I1~;qnd its Applicati"rJns
A. Concept Outline: Part-l 4-2A
B. Long and Medium Answer Type Questions 4-2A
Part-2 ; (4-24A to 4-42A)
• Frau~h~ferjjiffracti()nqtSingleSlit and at Double Slit
.Absehl"$pect,.~ '. .'. . . .
• ']Jt'ffrywfipn'0f,fJ,ting .
~....Speq~r:q':wttlJ,!f:tating
~l)i8p~is~i/e·Power ;
-:R;di.!w.g1t!s (Jf.it:eriortofResolution
•.Rei61vingPower ofGrating
, .. -------
4-1 A (Sem-1 & 21
i
i
J
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4-2 A (Sem-l & 2) Wave Optics
Coherent ~q(t'rc~$,
7?JHn 1Pj1:
Neli{li
Interference: The non-uniform distribution ofthe lightintensity due
to th(! superposition of two waves is called interference.
Necessary Conditions for Interference:
1. Light sources must be coherent in nature.
2. Light waves should be of same frequency.
:l. The sources of light must be very close to each other.
4. Light sources should be manochromatic in nature.
fl. The light waves must propagate along the same direction.
Types of Interfel'ence :
1. Constructive Interference: At certain points the resultant
intt'nsity ([) is greater than the sum ofindividual intensity oftwo
waves, The interference produced at this point is known as
constructive interference, it results into bright fringe. At
constructive interference,
I> I, + [2 '
l .
2. Destructive Interfel'ence : At certain points the resultant
intensity (lJ is less than the sum of individual intensity of two
waves. The interference produced at this point is known as
destructive interference and it results into dark fringe. At
destructive interference,
I < II + 12
CONCEPT OUTLINE: PART-l
Que 4.1. IWhat do you understand by coherent sources? How are
these obtained in practice?
Answer I
A Coherent Sources:
1. Two sources are said to be coherent if they emit continuous light waves
of the same frequency or wavelength, nearly the same amplitude and
Physics 4-3 A (Sem-l & ~ I
having sharply defined phase difference that remains constant w'
time.
B. Production of Coherent Sources:
1. Iftwo sources are derived from a s~gle source by some device, then
phase-change in one is simultaneously accompanied by the same phu.-,
change in the other. Thus the phase difference between the two sour." ..
remains constant.
2. Following are the devices for creating coherent sources oflight:
a. Young's Double Slit:
1. In this device, two narrow slits 8 1
and 8 2
receive light fi'ol"
same narrow slit 8.
2. Hence 8, and 8 2 act as coherent sources, as shown
Fig. 4.1.1.
s;w
c"ill~,~~~'~'
b. Lloyd's Mirror:
1. In this device, a slit 8 and its virtual image S' formed by reflection" I
a mirror are the coherent sources, as shown in Fig. 4.1.2.
s~
......~ a'
....::----~~.~:~.-.j:~~r
S··o: - - • •
~'1_zt'lJ
c. Fresnel's Double Mirror:
1. In this device t:wo virtual images 8 1
and 8 2
ofa single slit S. form""
by reflection at two plane mirrors M and M inclined at a S!1Lli,
1 2
angle to each other, are the coherent sources as shown II,
Fig. 4.1.3.
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/~illt!;t~£
d. Fresnel's Biprism :
1. In this device, 8 , and 8 2
, which are the images ofa slit 8 formed by
refraction t.hrough a biprism, act as coherent sources, as shown in
Fig. 4.1.4.
S'~
:,~~
Fig. 4.1<4.;
Michelson's Interferometer:
1. In this device, a single beam is broken into two light waves
perpendicular to each other, one by reflection and the other by
refraction.
2. The two beams, when reunite produce interference fringes. Here
these two beams act as coherent ·sources.
Q.ue 4.2. .1 Explain theory of interference by two waves.
OR
':xplain Young's double slit experiment.
Hlswer I
Let ll.~ considpl' tVO superimposed waves t.ravelling with same fi'equency
: -""-, E1J1d having constant phase difference in the same region.
. 2IT i
Ifa. ;md a., 3re amplitude oftwo waves, the displacement oftwo waves
:;It a'IlY instant t is given by
Y I = a I sin cot ...<4.2.1)
y, = a., sin ( rot + 6) ...(4.2.2)
WIIl'rC'. (~ = initial phase difference
,~
",IT x Path difference
A
x
p
..~~
8 ---..
~--8
-2
..
_'.'
'*",X
,"
y
~ > _ ~- ,ttf;- ,,:.'j
3. According to principle of superposition, the resultant displacement at
point Pis
Y =Y, +Y2
Y = a , sin rot + a sin (rot + 8)
2
= a , sin rot + a [sin rot cos 8 + cos rot sin iii
2
Y =sin rot (a , + a 2
cos 8) + (u2
sin 0) cos wt ...i4. 23)
or
4. Let us take,
A cos <p = a , + a2 cos 0 ...(4.2.4)
A sin <p = a2 sin 0 ...(4.2.5)
Then, eq. (4.2.3) becomes,
Y = A cos cjI sin rot + A sin cjI cos (l)t
or y = A sin (rot + cjI) .. (4.2.6)
5. Since, eq. (4.2.6) the resultant wave equation h,"liing amplitude A this
can be obtained by squaring eq. (4.2.4) and eq. 'A ::l.5) and adding,
, A2 = a/ + a2
2 cos2
8 + 2a1"c (as 8 + a/ sinz ()
A2 = a ,2 + a2
2 + 2a1a2 cos 8
6. By the definition, intensity is directly proportional to the square of
amplitude,
IaA2
or 1= KA2 [K = I, in arbitrary unit]
1= A2:; a,2 + a 2
2 + 2a,a 2 cos 8
Maximum Intensity (Constructive Interference,
.A. Condition 1.<"
1 ) :
1. If"" cos 8 = 1 i.e., 8 = 2n1t
where, n = 0, 1,2,3'00'
2. I max
=a ,2
+ a2
2
+ 2alu 2
I =(u + U 2)2
max l
I > II + 12
3. So, max
Path difference = ~ x Phase difference
4. 21t
A A
- x 2n1t = 2n
21t 2
= even multiple of ').../2.
.L·,
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~A (Sem-l & 2) Wave Optics
B. Condition for Minimum Intensity (Destructive Interference,
I min
) :
1. If cos 0 = -1 i.e., 0 = (2n + 1)n
where, n = 0, 1,2,3...
2. Im,n = a l
2
+ a/ - 2al a2
= (a - a )2
l 2
3. Hence, Im;n < II + 12
4, Path difference = ~ x Phase difference
21t
A A
- x (2n + 1)n = (2n + 1)
21t 2
= odd multiple 00J2.
QUe 4.3. IDiscuss the interference in thin film due to reflected
light. What happens when film is excess thin?
OR -
Explain the phenomenon ofinterference in thin films due to reflected
light and transmitted light. .
Answer I
1. Consider a parallel sided transparent thin film ofthickness t and refractive
index ~ > 1.
2. Let SA a monochromatic light ofwavelength A be incident on the upper
surface of the film at an angle i. This ray gets partially reflected along
AB and partially refracted along ACdirection.
3. Now at point C it again gets reflected along CD and transmitted along
DE
Q
:It'~Ir~b~(~~ri
Physi~s
4-7A (Sem·l &: 2)
A Inteli'erence in a ThfnFilm by Reflected Light:
I. According to Fig. 4.3.2; the path difference between AB and DE rays,
~ = pathACD in film - pathAL in air
~ = "dAC + CD) -AL .. ,(4,3,1)
2. Now in ~ANCand WCD,
CN CN
cosr= --=_
AC CD
t
AC= CD=
cosr
3. Nowin~ALD,
" AL ALAD"
sm £ = -- ~ == SIn £
AD
AL = (AN +ND) sin i
4. But, from ~ANC and WCD,
AN = t tan r and ND = t tan r
. So, AL == 2t tan r sin i
5. Putting the values'ofAC, CD, and AL in eq. (4.3.1),
A (2tl 2tt "
• U == /-I l--j - anr.SInz
. cos r
21Jt 2 sin r . .
= - - - t--.slnz
cos r cos r
~ = 2,.u [1-si02 r] sin 'iJ
','/-1=-
cosr [ sin r
~ = 2,.u cos r
6. Since, the ray AB is reflected at the surface of a denser medium.
A
Therefore, it undergoes a phase change of 1t or path difference of 2 .
The effective path difference between AB and DE is
A.
~ = 2/-1t cos r + 2' ...(4.3,2)
a. Condition for Maxima:
I. If A.
~= 2n
2
where, n = 0, 1,2,3...
2. Then, A. A.
2flt cos r + - == 2n _
2 2
2/-1t cos r == (2n - I) '2
A.
..,(4.3.:3)
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--
4-8 A (Sem-I &; 2) Wave Optics
b. Condition for Minima:
A.
1. <1 =(2n + I)
2
where, n = 0, 1,2,3...
A. A.
2. Then. 2J.!l cos r + 2" = (2n + 1) 2"
2J.!l cos r = nA ...(4.3.4)
B. Interference in a Thin Film by Transmitted Light:
1. From Fig. 4.3.1, the path difference between two transmitted rays, CP
andFQ,
<1 = Path CDF in film - path CR in air
<1 =~ (CD + DF) - CR ...(4.3.5)
2. Now in <1CDM, cosr= DM ~CD=_t_ and CM=ttanr
CD cosr
DM t
3. In illMF, cos r = -- ~ DF =-- and MF =ttanr
DF cosr
4. In t>.CRF, sin i = CR ~ CR =CFsin i
CF
or CR = (CM +MF> sin i
CR = 2t tan r sin i
5. On putting the value ofCD, DF' and CR in eq. (4.3.5),
2J.!l 2 ..
t>. = --- ttan rsm t
cos r
2J.!l sin r . sin i]
---2t--~slnr .: ~=-.-
[
cos r cos r sin r
2J.!l [1 . 2 ]
<1= - - -sIn r
cos r
<1 =2J.!l cos r ...(4.3.6)
a. Condition for Maxima:
A.
1. If t>.=2n2" [where, n = 0, I, 2, 3...]
')..
2. Then, 2J.!l cos r = 2n
" 2
or 2J.!l cos r = n'A. ...(4.3.7)
b. Condition for Minima:
1. If t>. = (2n + 1)
A.
[where, n =0, 1, 2, 3...]
2
2. Then, 2J.!l cos r =(2n + 1)
A.
...(4.3.8)
2
C. Condition for Excess Thin Film :
1. When the film is excessively thin such that its thickness t is very small
as compared to the wavelength of light, then 2~t cos r is almost zero.
Hence effective path difference becomes ~.
2
2. Thus every wavelength will be absent and film will appear black in
reflected light.
_ Discuss the formation of interference fringes due to a
wedge-shaped thin film seen by normally reflected monochromatic
light and obtain an expression for the fringe width.
[_==
""".
'. """
..,""'.
4""".""",I=I1=:(~=~,-:=';_=·J=~.·t"k=·s~···O=-=-71
A Formation of Interferences Fringes:
1. A wedge shaped thin film is one whose plane surfaces OA and OB are
slightly inclined to each other at a small angle e and encloses a film of
transparent material of refractive index 11 as shown in
Fig. 4.4.1.
2. The thickness ofthe film increases gradually from 0 to A. At the point
ofcontact thickness is zero.
3. When the upper surface OB ofthe film is illuminated by a parallel beam
ofmonochromatic light and the surface is viewed by reflected light, then
the interference between the two rays; one PQ reflected from the upper
surface ofthe film (glass to mm boundary) and the other FG obtained by
internal reflection (film to glass .boundary) at the back surface and
consequent transmissions at the film surface AB.
Q
B
·OC"""'=: IV ( .:... .~ i
Glass to film
boundary
••~.:~'I~I+l'¥
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Wave Optics
4-10 A (Sem-l & 2)
Since both the rays PQ and FG (or PEFG) are derived from the same
4.
incident ray SP (by the division ofamplitude) they are coherent and on
overlapping produce a system of equidistant bright and dark fringes.
The fringes are straight and parallel to the contact edge of the wedge.
With white light coloured fri?ges are observed. .
When a beam of monochromatic light is incident normally at point P on
5.
the upper surface of the film, the path difference between the rays
reflected from the upper and lower surfaces ofthe film is 2J.l t, where t
is the thickness ofthe film at P.
At point P, reflection occurs from the interface between the optically
6.
denser medium and optically rarer medium, therefore, there occurs an
additional path difference of1J2 or phase change of1t. Thus an additional
path difference of 'A./2 is introduced in the ray reflected from the upper
surface.
Hence the effective path difference between the two rays
7.
". 'A.
= 2J.lt +
2
8. The condition for bright fringe or maximum intensity is
2J.lt + IJ2 =2nAJ2
or 2J.lt = (211. - 1)1J2,
where, 11. = 1,2,3..... ...(4.4.1)
Similarly" the condition for dark fringe or minimum intensity is
9.
'A.
2J.lt + - =(211. +1) 1J2
2
or 2J.lt = nA,
where, 11. = 0, 1,2, .... . ...(4.4.2)
Expression for Fringe Width : Fringe width 00 or the separation
B.
between the two successive bright fringes or between two successive
dark fringes may be obtained as follows:
Let x be the distance of nth dark fringe from the edge 0 of the film as
1.
shown
n
in Fig. 4 . 4 . 1 . '
~B
o x ------.0 A
14--- n---'
I. xn+l~
....
tan a =.!.L or t =x tan a
n
Then, x 1 n
2. Putting this value oftl
in eq. (4.4.2), we get,
...(4.4.3)
21JX tan 9 = n~
1l
3. Similarly, ifxn
+
1is the distance of(11. + l)th dark fringe, then
2J.lXn+1 tan a=(11. + 1)1.. ...(4.4 .
4. Subtracting eq. (4.4.3) from eq. (4.4.4), we get,
2J.l (xn
+1
- xn
) tan a = A
A
or x -x = ... (4,4
n+1 n 2J.l tan a
5. For very small value of a, tan a '"" a
... (4.4.) ,
.. Fringe width, 00 = Xn+l - xn = 2~ 9
where, ais measured in radian.
6. Similarly, we can obtain same formula for the fringe width of bri~rJ.
fringes, that is the fringe width ofbright fringe is expressed as,
'A.
00= ...(4.-!
2J.la
7. It is clear from eq. (4.4.6) and eq. (4.4.7) that for a given wedge angle (I.
the fringe width of dark or bright fringes is constant (as 'A. and I-l I.'.
constant). It means that the interference fringes are equidistant fro l1 '
one another.
_'IIIWhite light falls normally on a film of soapy water whos.!
thickness is 1.5 ~ 10-5 cm and refractive index is 1.33. Whi<'!
wavelength in the visible region will be reflected most strongly ..
. ..,
"egion.
(211. -1);A..
1. Since, 211t cos r = -'--------'--'
... . 2
2 x 1.33 x 1.5 x 10-5 x 1 = (211. - 1) 1J2
, 4 x 1.33 x 1.5 x 10-5
7.98 X 10-6
~= cm
211.-1 211.-1
7.98 X 10-7 .
1..= m
(211.-1)
2. For 11. = 1, A, = 7980 X 10-10
= 7980 A(visible region).
;A.. = 7.98 X 10-
7
= 7.98 X 10-
7
=2660x 10-10
3. For 11. = 2,
2 (211.-1) 3
= 2660 A(not in visible region).
4. Hence, 7980 Ais most strongly reflected wavelength in visible region.
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4-12 A (SelD-I & 2) Wave Optics
Que ~.(J~~1~1A man whose eyes are 150 cm above the oil film on water
surface observes greenish colour at a distance of 100 em from his
feet. Find the thickness ofthe film.
(~oll =1.4, ~water=1.38, A_a'" 5000 A)
Answer ,'I
Given:~hil='~'~
To Ffud: Tblc'
I, The condition for maxima,
A
2,.u cos r = (2n - 1) [where, n = 1, 2, 3....]
2
or t = (2n -1)A
4~cos r
From Fig. 4.6.1,
tan i = 100 =~
150 3
.. 2
sin ~ = J13
., S' sini
.l. Ince, ~= -.
sin r
2
sin r = sin i = J13 = 0.3962
I-' 1.4
and cosr= ,h-sin2r=,,/[l-(0.3962)2]=0.9182
(2n -1M (2n -1)5x 10-7
!. Therefiore, t = = --:'---:-'_--=--:-=
4~cos r 4x 1.4 x 0.9182
=(2n-l) x 9.724 x lQ-8 m'
Putting n = 1,2,3...... value oft is calculated.
Eyes
Air
..... ::~_:::::::1
W~ie~~ :: :ilia~~ ~
::::::: : ~
r
~
~
....
-
il:'
Physics 4-18 A (SeID-1 & 2)
_ Light ofwavelength 5893 Ais reflected at nearly nonnal
incidence from a soap film of I-' .. 1.42. What is the least thickness of
this film that will appear :
a. dark
b. bright
-
i. Least Thickness of Dark Film :
1. Since, the condition for the dark film in reflected system is
2,.u cos r =nA
2. For normal incidence, r =0 and cos r =1
:. 2,.u = nA or t =nIJ21-'
3. For least thickness of the film, n = 1
A
t=
2~
8
t = 5893 X 10- = 2.075 X lO-fi cm
2 x 1.42
ii. Least Thickness of Bright FillD :
1. The condition for bright film
A
2~t cos r = (2n - 1) -
2
2. For normal incidence, r = 0 and cos r = 1
.. 2,.u =(2n - 1) '2
A
3. For least thickness, n = 1
A A
2,.u = (2 x 1 - 1) - or 21-'t = -
2 2
8
t = ~ = 5893 X 10- = 1.0375 X 10-5 cm
and
4~ 4 x 1.42
_ ,'White light is
,
incident on a soap film at an angle sin -1 i5
'
and the reflected light is observed with spectroscope. It is found
that two consecutive dark bands correspond to wavelength
6.1 x 10-" em and 6.0 x 10-" em. If the I-' of the film be 413, calculate its
thickness.
,I,
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Wave Optics
4-14 A (Sern-l & 2)
Answ~r'l
Given: i == siri+l
To Find,: TJ:Up
Since, the condition for dark band is
1. 2~t cos r = nA •••(4.8.1)
2.
If n and (n + 1) are the orders for dark bands for wavelengths A.1and A.2
respectively, then,
2~t cos r = nA. ...(4.8.2)
J
and 2~t cos r =(n + 1) 1..2
...(4.8.3)
or 2~t cos r =nA.1 =(n + 1)1..2
n=~
or 1..1 - 1..2
On putting the value of n i,n eq. (4.8.2),
:3,
'AI.. A. A.
2~t cos r = _1_2_ or t = 1 2 ,
1..1- 1.. (A.1- A.2)2~cos r
2
( sin i
(
. .)2
4. But, cos r = J1-sin2r =,/1- Sl:~ l'; ~ = sin r)
(4/52 ~ 4
cos r = 1-l4/3) = V
1- 25 = 5
6.1 X 10--6 x 6.0 X 10-5
5. Now, t = _ 4 4
(6.1 X 10-5 -6.0 x 10 5)X 2 x -x
3 5
= 0.0017 cm.
Cfj.l:tiJ(~.~·1 Two plane glass surface,s in contact along one edge are
separated at the opposite edge by a thin wire. If 20 interference
fringes are observed between these edges, in sodium light of
wavelength, A. = 5890 Aof normal incidence, find the diameter of the
wire.
~~.~i::,;&l
GiVen:N=
rtb tina.':.I)io,-.,p:.r;7'C~'T"""WF'i:"5Pi"''''_''''''''"'''~''''''''''"~'=''~'>-'-'''''··-',.,
Let the diameter of the wire be 'd' and the length of the wedge be 'Z'.
1.
2. The wedge angle is given as
d
tan 9 =
Z
tan 9 ~ 9 (As 9 is small)
Physics 4-15 A (Sern-l ~,
d
9= T !
~;
. ~I
~Ji~~
~~;1<'"
3. Now, fringe-width in air wedge is
or
A. A.
00= --=
2~" 29
ZA.
00=
2d
(.:
c()-
~ = 1 for air well"
4. IfN fringes are seen,
Z=Noo
Thus,
Woo
00= -
2d
d= NA.
2
d = 20x5890
2
d =58900A
d =5.8~ X 10-4 cm
_ Discuss the necessity of an extended source?
1. When a thin film is illuminated with monochromatic light from a IJ:,'
source and is viewed with a lens of small aperture, the light reflect ('
from all corresponding point on the film does not reach the."
simultaneously as shown in Fig. 4.10.1(a). Thus only a small portion",
the film will be visible.
2. To see the whole film, the eye will have to be moved from one position
to the other. Hence, with a point source the entire film cannot be view,,"
at a glance.
3. Ifwe employ an extended source, the light reflected by every poin t
the film reaches the eye (as shown in Fig. 4.10.1(b». Hence, the enw
film can be viewed simultaneously by keeping the eye at one place onl
4. Hence an extended source of light is necessary to view a film
simultaneously.
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-1
(a)

Eye
~(,0
~'t>O-O; 8 2
, '
:<;>~
:¢'0 ,
~81 "",
(b)
QU'~}1t'I~!~What are Newton's rings? Explain with diagram.
An~~~~::i!r;;j
1. When a plano-convex lens of large radius of curvature is placed on a
plane glass plate with convex surface in contact, a thin film between the
lower surface ofthe lens and the upper surface ofglass plate is formed.
2. The thickness ofthis film is very small (or zero) at the contact point and :1
gradually increases from contact point to outward.
3. When a monochromatic light falls on the film, we get dark and bright
concentric circular fringes having uniform thickness.
-t. These rings are first investigated by Newton and are called Newton's
rings.
A. Experimental Arrangement:
1. According to Fig. 4.11.1, S is a monochromatic source oflight placed at
the focus of lens L 1
•
2. A horizontal beam of light fallon the glass plate G placed at 45° to the
incident beam.
3. This beam is partly reflected from glass plate G.
4. This reflected beam fall normally on the lensL, placed onglass plateE.
5. Hence, the interferen«!! occurs between the rays reflected from the
upper and lower surface ofthe film. 'I.
6. Interference rings are seen with the help oflower power microscope M.
7. The fringes are circular because the air film is symmetrical about the
point ofcontact with lens and glass plate.
j
.f
'/F I'" /lII >S
Dark ring
~'tg~:;~~~l.l.
B. Explanation:
1. According to Fig. 4.11.2, rays (1) and (2) are reflected interfering rays
corresponding to incident ray SP.
S
1
air film
""-.00::
~
'
" ,
77
41
l11ItAt!ll·.~t
Now the effective path difference between (1) and (2) rays is given as:
2.
A
d =2!!t cos r + '2
where, !! = Refractive index, and
t = thickness.
3. For normal incidence, r =0, therefore
cos9=1
4. Hence, d = 21-lt + '2
A
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~.
4-18 A (Sem-I & 2) Wave Optics
5. At point of contact (0) of the lens,
t = 0
A.
=> 6 =
2
6. This is condition of minunum intensity hence the central spot ofthe ring
is dark. .
a. Condition for Maximum. Intensity (Bright Rings) :
1. If path difference, 6 =2n ~ [where, n = 0, 1, 2, 3... ] .
A. A. A
2. Hence, 21lt + "2 = 2n"2 or 2l-'t = (2n - 1) "2
3. For air, Il = 1 => 2t = (2n - 1) ~
b. Condition for Minimum Intensity (Dark Rings):
1. If path difference, 6 = (2n + I) ~ [where, n = 0, 1, 2, 3...]
A. A.
2. Hence, 21lt + "2 = (2n + 1) "2
2l-'t =nA.
3. For air Il =1 => 2t =nA
Que40t~•. 1Describe and explain the formation ofNewton's rings in
reflected monochromatic light. Prove that in reflected light
diameters of the dark rings are proportional to the square root of
natural numbers.
OR
-
What are Newton's rings? Prove that in reflected light diameters of
the bright rings are proportional to the square root of odd natural
number.
Answer I
A Newton's Rings: Refer Q. 4.11, Page 4-16A, Unit-4.
B. Diameter of Rings:
1. Let R is radius of curvature of lens 'L' and 't' is thickness of air film at
point 'p'.
2. From the geometrical properties ofcircle as shown in Fig. 4.12.1,
AP x AB = AO x AF
r x r = t x (2R - t)
r2
= 2Rt - t 2
Physics. ~19A (Sem-I & 2)
,""."'''--t -.......,
.
/ ...,
I .... '
f I
~
, f'
B /
~
~~~
3. In actual, R is quite large and t is very small. So, t2
is neglected.
Hence, r2
=2Rt
2
r
t= ...(4.1~.1.
2R
Be For Bright Rings:
1. Since, we know that,
A
2t =(2n -1)
. 2
On putting the value oft from eq. (4.12.1),
r2
A AR
2 - = (2n -1)- or r2 =(2n - 1)
2R 2 2
2. IfradiuB ofnth bright ring is rn'
Then, r 2 = (2n -1)AR ...(4.12.21
n 2
3. D n is diameter of nth bright ring,
2 _ (Dn J2
rn - 2
iW
Now, from eq. (4.12.2),
(2n-l)Il.R
(~nr = 2
or Dn
2= 2(2n-l)AR
or Dn = .j2(2n-l)AR
4. Let, K= .J2AR
D,. = K.J2n-l [where, n = 0,1,2. :3.
D oc .J2n -1
5. Diameter of bright ri~gis proportional to the square root ofodd natural
number;;.
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i
~20 A (Sem-l & 2) Wave Optics
b. For Dark Rings :
1. Since, we know, 2t = M #
Substituting the value of't' from eq. (4.12.1),
2
2 -
r
= nA.orr=nAR
2R
.,
... If radius of nth dark ring is rn'
Then, r 2 =n:>.R
.l. D" is diameter ofnthndark ring,
rn = Dn /2
then, (~nr = nAR or Dn2 =4nA.R
Dn = .J4n:iR
'!. Let, K = .J4:iR
Therefore, Dn = KJn or D" oc In
;-), Diameter of dark ring is proportional to the square root of natural
number. .
Que 4.1~;~i,1 Explain the formation of Newton's ring? Ifin a Newton's
ring experiment, the air in the interspaces is replaced by a liquid of
refractive index 1.33, in what proportion would the diameter of the
rings changed? _
AnSWf3r <:]
A. Formation of Newton's Ring: Refer Q. 4.11, Page 4-16A, Unit-4.
B. Numerical:
!.
Diameter of a ring in liquid film
Diameter of the same ring in air film =
1
~
= 1
,..-;;;;
,,1.33
= 0.867
,) So, the diameter of rings .decreased by the portion of 0.867 of natural
diameter.
Que,~~.~~~1 Show that the diameter D,. of the nth Newton's ring,
when two surface of ramus R i
and R a are placed in contact is given
. 1 1 4nA.
by the relation: - : 1 ; - =--2 •
. RI Ra D"
.;~.
Newton's Rings formed by two Curved Surfaces:
-
a. Case I:
1. When a planoconvex lens of radius of curvature R 1 is placed on the
planoconcave lens of radius R z.
2. Let at point A the thickness of air fihn is 't' and nth dark ring is passing
through A and its radius is r,,'
3. According to Fig. 4.14.1. .
i rll
A
r----~
C
.' Fig. 4.14.1.
t=t -t [','AC=t,andBC= f 1
l 2 2
r
2
r
Z
r z ( 1 1 )
= 21t - 2R or t =2 R] - R
z z
2 (1 1"
or ...(4,14.1)
2t = r" l---j
R 1 R?
4. For dark ring,
21-lt = nJ..
2t = nJ.. [.,' ~l = 1 for airl ... 14.1421
5. From eq. (4.14.1) and eq. (4.14.2),
J 1 1"
rn lR] - R ) = nJ..
2
1 1 nA
or - - - =
~ R2 r;~
6. But, Dn =2r"
1 1 4nA
., ~ - R = D:
2
b. Case II :
1. Let both the lenses are pl:lnoconvex and their curved surf'acl' is in
contact.
2. Let 'I' is thickness of air film at point A and R 1 and R e an' racliLl,~ of'
curvature oflenses respectively as shown in Fig 4.14.2,
.,
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['
Wave Optics
4-22 A (Sem-l & 2)
'~'":W~
G.~~~;:¥.~
t = t
a. Since, l
+ t z
z
r r,~ r,~[1 IJ
n
= -+--=:>- -+
2~ 2~ 2 RI Rz
z( 1 1 '1
or 2t = r 1-+-)
'." "RI R2
4. For dark ring, 2t = nA
"( 1 1) _,
r 1 - - + -) = rIA
""R, R"
1 1 n..t
-+- = -z'
R, Rl r"
fl. r = D /2
But,
" "
1 1 4n,,
-+
D"z
1 1 4n..t
~ Rl
6. Hence, - ± - = -z-
R, Rz D"
IDescribe how Newton~sring experiment can be used to
determine the refractive index of a liquid?
Answer I
1. The transparent liquid whose refractive index is to be determined is
introduced between lens and glass plate.
2. Since, diameter of nth dark ring is given by
...(4.15.1)
D z = 4n"-R
" I-l
:3. Similarly, for (n + p)th dark ring,
D Z = 4(n+p)"-R ...(4.15.2)
".p
j.l
Physics 4-23 A (SeIll.. l &:t
4. On subtracting eq. (4.15.1) from eq. (4.15.2),
D:+p _ D 2 = 4p"-R (For liquid) ...(4.15.:
n j.l
5. . For air, j.l = 1
D 2
_ D 2 =ho/-.R ...(4.154:
f.' n+p. n ~
6. On dividing eq. (4.15.4) byeq. (4.15.3),
(D 2 _D2)
_ n+p II. Rir
j.l- (D,~+p - D;)UQUid
_ Newton's rings are observed by keeping a spherk>r'
surface of 100 em radius on a plane glass plate. Ifthe diameter of th('
16th bright ring is 0.690 em and the diameter of the 5 th ring is
0.338 em, what is the wavelength of light used?
i I I f l j " " .
"""":~"";=-.··:::'"0~C:C<""_.Jli"":I4;""';""".~!i"" .•. =--a-r=-ks 05!
••. ,--'{:M"-:
1:t,
-,!':;::;'~~.<'<-"~".flk~U,-~~~t:m:.-&~h~
~~~'Y~~~W~
-5 =lO,R= 100 em.
~~~@~Ij _.... ...
1. IfD + and D be the diameter of(n + p)th and nth bright ring,
Il p Il
D
2
,,+p - D,~
then, "-= 4pR
"- = (0.590)2 - (0.336)2
4 x lOx 100
=5.88 x 10-5 cm = 5880 A
~,_ Newton·s rings are observed normally in reflected light
of wavelength 8000 A. The diameter of the 10th dark ring j s
0.50 em. Find the radius of curvature of lens and thickness of the
corresponding air film. lIIIIIW!19t~l.f~.1Marks
O~J
-:' ,.', " ' ,:,;
~
.
1. The diameter ofnth dark ring is given by
D2
D 2 = 4nAR or R= -"
" 4nA
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---
R==' O.50xO.50 ==104.17cm
4 x 10 x 6.0 x 10-6
~. If t is the thickness of the film corresponding to sring ofD" diameter, jJ!;
then,
. 2 2 ' .
2t == D" .' or t = D" =0.50 x 0.50 =2.99 x 10-4 em
4R 8R 8 x 104.17 "
Diffraction: Diffraction oflight is a phenomenon of bending oflight
and spreading out towards the geometrical shadow when passed
t Iwough an obstruction.
Rayleigh's Criteria : The spectral lines ofequal intensity are said to
he resolved, if the position ofthe principal maXima ofone spectral line
coincide with first minima ofthe other spectral line.
Long Answer TYP~~~M:EJdi~D1'~$wer
I
x x
Screen
B
Opaque
obstacle Y
Screen
~~:lli41.••~
The departure oflight path from true rectilinear path or the bending of
['>..;U around corners ofan obstacle is called diffraction oflight.
Que 4.18.1 What is meant by diffraction oflight ? Write name of the
two classes of diffraction and explain it.
Answer I
2. When light passes through smallapertures or by the side ofsmall obstacles
it does not follow rectilinear path strictly, but bends at round corners of
the obstacles.
3. Diffraction phenomena can be divided into two types:
a. Fresnel Diffraction:
1. In Fresnel diffraction, the source of light and the screen, both are
placed at finite distance from the diffraction element (obstacle or
apertures) in which incident wavefront is either sphericn.l Ol'
cylindrical and no lens are used,
b. Fraunhofer Diffraction:
1. In Fraunhofer diffraction, source of light and the screen both arl'
placed at infinite distance from diffraction element in which incident
wavefront is often plane.
2. Here convex lenses are used for focusing diffracted light.
~'1.'4 Describe Fraunhofer diffraction due to a single slit and
t"li~>,'·<'(.f;~::<.,,~.>,;;:l;.(';;"_->,
";<': •
show that relative intensities of successive maxima are nearly
1 1 1
1 : 22: 61: 121
OR
Discuss the phenomenon of diffraction at a single slit and show
that the relative intensities of successive maxima are nearly
4 . 4 4
1 :--2:--2:--2 . !;~g~:2Ql,.~tli,Marks
051
9n 25n 49n
OR
Explain the diffraction pattern obtained with diffraction at single
slit. By what fraction the intensity of second maximum reduced
from principal maximum? ~t••!t~·J.::I;;14~Marks 05 1
A. Fraunhofer Diffraction due to a Single Slit:
1. The light from a monochromatic source S is converted into parallel
beam oflight by convex lens L 1,
2. Now this beam is incident normally on a slitAB of width 'e'.
3. Now according to Huygen's wave theory, every point in AB sends out
secondary waves which are superimposed to give diffraction pattern on
screenXY.
4. In this diffraction pattern, a central bright band is obtained because the
rays from AB reach at C in same phase and here the intensity is
maximum.
The rays which are directed through an angle eare focused at point 'P'.
5.
,I
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--
!---
Wave Optics
4-26 A (Sem-l & 2)
6. To find intensity at P let us draw normalAK.
7. Path difference of rays meeting at Pis
BK= e sin 9
and, phase difference = (~)esin 9
8. Let AB be divided into large number of equal parts. The secondary
waves originating from these parts will be ofequal amplitude 'a' (say).
9. Then phase difference between two successive waves will be
0= ~(2;)esin 9
x
L2
;';'~~~-=1_
S ...... t I
~ ----..¥:
B
y
10. Now, according to n simple harmonic waves,
. nfl . (7teSin9)
aSln- asm
R= 2 = A.
sino/2 .
SIn(7teSin9)
nA.
11. Let no = 2a
and, a = 2:esin9
Then, R=
or R=
12. Intensity at point 'P,
1= R2 = Ao sin a ...(4.19.1)
2
a
a. Position of Maxima:
sin a = 1 when a. ~ 0
1.
a
A.
asina
--,-
[.,' ~ is very small. So, sin~ =~J
n n n
sinl;)
sin a
na--~R=....
.4'
sma [.,' na =A)
a a
2 2
4-27 A (Sem-l & 2)
Physics
. sino. . 1 ( as 0.
5
1
hm - - = 11m - 0.--+-- ...)
a->O a .......0 a ~!§
2. 1 = 1 (1)2 ~ 1 =10
0
3. 0.= 2!.esin 9
..t
"iesin9 = 0 ~ sin9 = 0
9=0
4. Point C is central maxima or principal maxima.
b. POliition of Minima:
1. If sino. =0
a
:::) sino. = 0
0.*0
= 1
a=:i: m7t (where, m = 1. 2, 3....1
e7tsin9
2. Hence, =:i: m7t
..t
e sin 9 = :i: (mA.) (where, m =1, 2. 3 1 ...(4.19.2)
3. Eq. (4.19.2) gives the direction offirst, second, third minima and thi:-;
equation is called diffraction equation.
Co Secondary Masbna :
1. The condition ofsecondary maxima may be obtained by differentiating
eq. (4.19.1) w.r.t. a and equating it to zero.
dl = .!!:...-[A 2(sina)2] =
do. do. 0 a
· h sino. 0 . 0
2. EIt er - - = :::) SIn a =
a
or acosa-sina=O
0.= tan a
A22 sina [aeosa -sina]
0 a 0.2
3. sin a = 0, gives position ofprincipal minima and position of secondary
maxima is given by
a = tan a ...(4.19.3)
4. Eq. (4.19.3) can be solved graphically by plotting the curves
y=a and y=tana
5. According to curves, the point ofintersection ofthese two curves gives
the value ofa satisfyingthe equation a = tan a. These points correspond
to the value of
3n ±5n 7n
0.= 0, :i: 2'2.±2....
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1
4-28 A (Sem-1 & 2) Wave Optics
ifil~i~~t~
G At a = 0, the position ofprincipal maxima,
I =1 (Si: a)2 =1
0 0
7. At a = 3re , the intensity offirst secondary maxima,
2
( . 3re '2
1=1 (Sina)2 ~I lSIn 2j =~
l O a 0 3re 91t2
2
8. At a = 5re , the intensity of second secondary maxima,
2
( , 51t '2
41
SInTj
1=1 -
2 0 51t = 25;'
l 2
Intensityr
4
Secondary
maximum
....
9. At a = 7Tt 1 = 410
2 ' 3
49lt2
:J.
Physics 4-29 A (Sem-I & 2)
. fit'" 4 4 4
1.
O The ratio 0 re a lve IntenSIty as: I: --2 :'--2 : --2 : .....
9re 25re 49re
1 1 1
or, 1: 22 : 61 : 121 : .....
11. Direction of secondary maxima is given by
e sin 8 = ± (2m + 1) ~
12. But, a = Tee sin 0 or a = ~.(2m + 1)-~
,1. ), 2
re
a = (2m + 1)"2 m = I, 2, 3, 4 .....
Hence a = 31t 5re Ire 9re
, 2 ' 2 ' 2' 2 .....
_ _ A light of wavelength 6000Afalls normally on a straight
slit of width 0.10 mm. Calculate the total angular width of the central
maxima and also the linear width as observed on a screen placed
1 meter away.
- "'~Qrii~;:{O;1 cm, D =I m =100 em
"?l~ntr~~axnna.
'1;11 m8X1ma..
1. Since, e sin 8 = nA
2. For n = 1 and for small 8,
A 6xlO-5
8 = - =--- =6 X 10-3
rad.
e 0.01
3. Total angular width, 28 =2 x 6 X 10-3
= 1.2 X 10-2 rad or 0.688°
4, Linear half width = 8D = 6 x 10-3
X 100 = 0.6 em
5. Total linear width =2eD =2 x 0.6 =1.2 em
I
~ Im--
_D_
_t~'~
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Wave Optics
4-30 A (Sem-l & 2)
Q-i.,i"l~1Discuss Fraunhofer diffraction at a double slit.
AD,j'~j;":1
1. Consider a paranel beam ofmonochromatic light having wavelength i..
incident normally on two parallel slits AB and CD as shown in
Fig. 4.21.1.
2. Width ofeach slit is 'e' and are separated by distance 'cr. Distsnee between
8 1
and 8 2
point is 'e + d'.
3. N ow each slit diffracts the light at an angle a to incident direction.
L 2
:£B "n-=;:;;sr
s ~ 0
y
~~
~~~~Uf
From the theory of diffraction due to single slit we know that, resultant
5.
amplitude is
sin a 1 t .
R=A - - and a=-esma
a "
Let 8 and 8 are two coherent sources, each sending wavelets of
6. 1 2
amplitude A sin a in the direction ofa.
a
7. Therefore, the resultant amplitude due to interference of these two
waves at point P can be calculated as: .
i. Draw perpendiculars 8 1K on 8,j(.
ii. Path difference, 8,j(= (e + d) sin a ...(4.21.1)
and, phas~ difference,
0= ~1t(e+d)Sin9 ...(4.21.2)
iii. IfR' is resultant amplitude at pointPthen, according to Fig. 4.21.2.
OB2 =OA2 + AB2 + 2AB.OA cos l)
R'2 = R2 + R2 + 2RR cos 0
= 2R2 + 2R2
cos 0
R'2 = 2R2
(1 + cos 0) =2R2(2COS2~)
Physics 4-31 A (Sem·l &:2;
R'2 = 4R2 cos 2 ~
2
z1
B
O R A
·'i~8i.r
~~.~J
iv. But, R =A sina
a
• 2 "
R'2 _ 4 A 2
sIn a 2 u
then, - 2 cos
a 2
v. Let, /3 = ~ = .! (e + d) sin e
2 "
• 2
th R'2 - 4A2 sm a 2/3
en, - a 2 •COS
8. Now, by definition,
• 2 2A
1= R,2 = 4A2
sIn a.cos ,.,
2
a
9. Hence, the resultant intensity depends upon following two factors:
. A2
sin2
a . .
I. 2 due to diffractIon, and
a
ii. cos2
/3, due to interference.
I
-1t 1t
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1~_
Conditions for Maxima:
If, cos2
f3 = 1 or /3 =± n1t
4-32 A (Sem·l & 2) Wave Optics
ff/M[('~'P
- 21t -1t 0 1t 21t
(b)
,
I " 4A2 «sin2 0.)/0.2) cos2 /3
,~
'"
.......
,-."".- .....
[where, n = 0, 1, 2, 3,...]
But, 13 = ~(e+d)sine
±n1t= ~(e+d)sine
(e + d) sin e= ± n').. [where, n =0, 1, 2, 3,....]
H. Conditions for Minima:
[f, cos2
/3 = 0 or cos /3 =0
1t
13 = ± (2n + 1) "2 [where, n = 0, 1, 2, 3,...]
Then, (e + d) sin e=± (2n + I)!::
. 2
Que4;22.iV
IWhat do you mean by a diffraction grating ? Derive
"'xpression of Fraunhofer diffraction due to N slits.
OR
,~ive the construction and theory of plane transmission grating.
<;xplain the formation of spe~traby it. _
Answer.:1
The diffraction grating consists of a large number (N) of parallel slits
having equal width and separated by an equal opaque space.
Physics 4-33 A (Sem-! & 2)
2. It is constructed by rolling a large number of parallel and equidistant
lines on a glass plate with a diamond point.
A Explanation:
1. Let a parallel beam of monochromatic light ofwavelength ''A' be incident
on 'NsUts.
2. This light diffracted at an angle eis focused at point P on the screen by
lens L 2
having same amplitude.
R == A sino.
a
3. Let e be the width of each slit and 'd' be the opaque space between two
slits, then (e + d) is called grating element.
4. Path difference = (e .+ d) sirl e
and, phase difference,
2f3 = 21t (e + d) sin e.
'A
'5. Therefore, as we pass from one vibration to another, the path goes on
increasing by an amount (e + d) sin eand phase goes on increasing by an
amount 2')..1t (e + d) sin e. Thus, phase increases in arithmetic progression.
6. Now, the resultant amplitude and intensity at point P due to N slits can
be obtained by vector polygon method,
. Nd
SIn - .
R'=R 4
Sln
2
).
L~
/'
L 1
s ('
y
7. But, d = 2f3
·li'ig:4,..22.1.
x
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Wave Optics
4-34 A (Sem-l & 2)
Hence.
or
where,
8. Intensity,
9. Hence, intensity
R sin 2NI}
R'= ~
sin 2j3
2
R' = R sinNJ3
sin p
R = A sina (due to single slit)
a
1= R'2 =A2 (sina)2(si~NPJ2
a smp
distributed is product of two
...(4.22.1)
...(4.22.2)
terms lot term
A 2 ( s i: (l J2 represents diffraction pattern due to single slit and lInd term
2
(sinNp1 • fi t d N l'
1-.--J I'epresents mter erence pat ern ue to SitS.
. sm P .
a. Condition and Intensity of Principal Maxima:
1. ",en, sin J3 = 0
J3 = ± n1t [where. n = 0, 1, 2, 3...) ...(4.22.3)
2. Then, sinN13 = 0
sin NJ3 0.. d . fi
Hence, --- = - IS In etermlnate orm.
sin J3 0
It is solved by L-Hospital rule.
.~(sinNJ3)
lim sinN13 lim _d--'-13--.---- _ NcosNp =N
lim A
Ii ~:tlln-
. '
smJ3 11·-.,,," d ( . A) ~_t,,1t COS...,
-_.. sin ...
d13 •
.~~,¥&~.,£~
3. Putting the value of si~ NI} = N in eq. (4.22.2), we get,
sm J3 .
• 2
= A2 sin a N2 ...(4.22.4)
1 2 •
a
Physics 4-35 A (SeID- l .:
4. The direction of principal maxima is given by, sin p = O. P = ± !U
5. We know, J3 = i(e+d)sin 9
So, . ±nlt= i(e+d.~8in9=>(e+d)Sin9=±nA. ...(4.:~
6. For n =0 we get, 0 =oand get zero order principal maxima.
n = I, 2,3, correspond to 1st, lInd, IIlrd order principal maxI E'·
h Condition for Minima:
I. The intensity is minimum when, sin Np = 0
2. But, sin J3 "" 0
Hence, NJ3 =± mit or p =± mit
N
or, NIt(e+d)sin9=±mlt or ·N(e+d)sin9=±mA. ... (4 ..
A.
where, m can take all integral values except 0, N. 2N, 3N .....
m =0 gives maxima and m =1, 2,3, ..... (N -1) give minima.
B. Secondary Maxima:
1. There are (N - 1) minima between two consecutive principal m;l"
therefore, there are (N - 2) other maxima coming alternatively w'!
minima between two successive principal maxima.
2. Position ofsecondary maxima is obtained by differentiating eq. l+_
w.r.t. /3 and equating it to zero.
dI =A2 sin
2
a 2[sinNP][NcosNI} sinp -sin NJ3cos J3 ] = 0
dJ3 a 2 sinp sin2
J3
or, N cos Np sin I} - sin Np cos P = 0
and, tan Np = N tan I}
3. From Fig. 4.22.3,
N tan l3
1
;rl!i~~."~:
sin Np = N tan p
~l + N 2
tan2
/3
Squaring both sides and dividing by sin2p,
sin2Np _ N 2
tan2
p
sin2
J3 - [(I+N2
tan 2p)sin2
p)
N 2
[1+(N2
-l)sin" Iii
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-,-a6 A (Sem-l & 2) Wave Optics
1
H d" da
4. ence, IsperSlve power, dA =
. . sm 13
Putting the value of si~2;v13 in eq. (4.22.2),
~(e:dr _
A
2
2
l' = A 2 (sina)2 N ...(4.22.8)
o a [1 + (N2
-1)sin2/3J _ What do you understand by missing order spectrum?
Intensityofsecondary maxima l' 1 What particular spectra would be absent if the width of
Intensity of principal maxima = I = [1 + (N2 -1)sin2/3J transparencies and opacities of the grating are equal? Show that
only first order spectra is possible ifthe width of the grating element
Zero order is more than wavelength of light and less than twice the wavelength
principal
of light.
maxima II!'
OR
What are the conditions absent spectra in the grating?
1. Sometime for a particular angle ofdiffraction '8' satisfying the rdn.tioll
Second (e + d) sin a=nA, there is no visible spectrum obtained. This phenomellon
principal is known as missing order spectrum.
maxima 2. We know that condition for a minima in a single slit is givPll by
e sin e = m ) . . 1 4 . 2 4 . 1 J
and the condition for the principal maxima in the n'h onil", spec:tr'um is
Secondary minima Secondary minima given by
(e + d) sin e =nA ...(4.24.2)
Fig;4.2Z~4.,
3. Ifboth conditions are simultaneously satisfied, the diffracted rays from
;Jue 4.23./ What is diffraction grating? Show that its dispersive all transparencies are superimposed upon each other but the resultant
intensity is zero, i.e. the spectrum is absent.
1 - (4 d h e + d n
i'O'Ner can be expressed as where all terms have 4. From eq. .24.1) an eq. (4.24.2), we ave, ---- =
e In
2
J( e:c!r_1. This is the condition for absent spectra.
5. If e =d then, n =2 m
So that 2nd , .... order ofthe spectra will be mif'sing COlTc'spDnding
, 4th , 6th
.,,- usual meanings. I""",·.~=-·_=·'J'U='=
~~=,.J=;~."""'/~~=:;:=;'.=.!~;;=~Jt""";'s"""'.O=6J
.• ••
to m = 1, 2, 3, ...
6. When d =2e then, n =3 m
;~~~wer I 6th • 9th
Hence, 3rd , ... spectra will be missing correspolHjing t.o
J}iffraction Grating: Refer Q. 4.22, Page 4-32A, Unit-4.
m =1, 2, 3_
Proof:
7. Themaxh:"-'1 number of spectra available with a diffraction grating in
Since. (e + d) sin a = nA :::::> sin a= ----!!:!:::.. ...(4.23.1) the visible region can be evaluatpd by using the grating equation for
(e + d) normal incidence as
Differentiating eq. (4.23.1) w.r.t. ;>_, we get, _ (e+d)sin8"
(e + d) sin an =nA or n -
da A
(e + d) cos e - = n
dA 8. The maximum possible value of8 is 90°.
. n'dA ndA (e + d)sin 90 (e + d)
n =--
da- ---~= ...(4.23.2) max
- (e + d) cos a - (e + d)JI= sin2 a A A
9. If the grating element (e + d) lies between Aand 2A or grating element
I 'utting the valve of sin afrom eq. (4.23.1) in eq. (4.23.2), we get
(e + d) < 2A
ndA ndA dA
dH =
2A
2
A
2
( d) /1- ---.!!:.2A
2_ = ,j(e+ d)2 - n = J(e + d)2 _ 2 Then, n < -<2
max A
e + . (e + d)2 n A
1
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10. Therefore, for normal incidence only first order will be obtained.
11. Hence, if the width ofgrating element is less than twice the wavelength
of light, then only first. order is possible.
_e44~,ldWhat is dispersive power ofgratingand resolving power
of an optical instrument? Explain Rayleigh's criterion ofresolution.
'""",.' ...'. "J
~swer
A. Dispersive Power of a Diffraction Grating:
1. The dispersive power of a diffraction grating is defined as the rate of
change of the diffraction angle with the wavelength. It is expressed as
de
dA
2. For a grating, (e + d) sin e =nA
Differentiating w.r.t. A, we get,
. de
(e + d) cos 8 dA = n
de n
or - = --
dA (e + d) cos 8
B. Resolving Power of an Optical Instrument:
1. The' ability of an optical instrument to produce the separate images of
two objects placed very close to each other is known as resolving power.
C. Rayleigh's Criteria of Resolution:
1. According to Rayleigh's criterion, the spectral lines ofequal intensity
are said to be resolved, if the position of the principal maxima of one
spec! raj line coincide with first minima ofthe other spectral line.
A. I )'2
Principal maxima
Al
/-,
Fig. ~.25.b'
Physics 4-39 A (Sern-l &. l.
What do you mean by resolving power ofgrating? Dcrl'
the necessary expression.
" , . ' , . ,.
" ' . ' .
·.·.~I.~
' ..
A. Resolving Power of Grating:
1. It is defined as the ratio of wavelength (A) of any spectral line to t I)'
smallest difference oftwo wavelengths (dA), for which the spectrallin<
can be resolved at the wavelength A.
· . f . "
ResoIvmg power 0 gratmg = d"
B. Expression:
1. Let a light consisting of two wavelengths AI and "-2 is incident norm:d i
on a grating element (e + d) and the spectral lines corresponding t (j •
and "-2 are formed on screen P and P .
1 z
2. These spectral lines,just resolve ifthey satisfy the Rayleigh's critenoJ I
the direction ofnth principal maxima for wavelength (1'1) is given h'.
(e + d) sin 8 =nA
I
N(e + d) sin a=Nn"-I ...(4 :.!'.
3. And I" minima in direction (8 + d8) is
N(e + d) sin (e + de) = mAL ...(4.:'::(,
except (m = 0, N, 2N ..... or 1, 2, 3..., N - 1)
I
-----..
~
..I
- I
4.
5.
6.
7.
A N(e+d)sina
dA A.
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Qll~~~~,7[~'1 Find the angular separation of 15048 A and 5016 A
wavelengths in second order spectrum obtained by a plane
diffraction grating having 115000 lines per em.
~i~f:;;.~~
1. Since, dA =Ai - A2
= 5048 - 5016 = 32 A= 32 x 1O-1
°m
A= Ai + A2 =5048 + 5016 =5032 A =5032 X 10-10 m
2.
2
3. Now, angular separation is given by,
32 X 10-10
de = dA
re·696 X
,[(e:qr_A2
] 2 10-0 r-(5032 x 10'
1
0)2]
32x 10-10
32x 10-10
-.171.65 X 10-14
- 25.32x 10-14
,,110-14
(71.65 - 25.32)
32 X 10-10
4.63 X 10-6 = 6.91 x 1O~ rad
Que 4.28. IA diffraction grating used at nonnal incidence gives a
yellow line (A = 6000 A) in a certain spectral order superimposed on
a blue line (A = 4800 A) ofnext higher order. Ifthe angle of diffraction
is sin-1 (3/4). calculate the grating element.
ffi=~"""'.~=. ..=·'Irn;;::;JiI:~'e;·I="'1
·""""'·O"""""'··d=~=
tE~~?~11;1~I';4,~~1Ri;~
An$~~r··,;l
Given: Ai =6000!A'=~g()~xl()t~~m;.~~·.i!:48M,~{¥'~i ,10-:'Sl<:n1,
~:;~~~~:~;:~~~llI:~t ....."I,,:;;,;;f~~,,'i ;;:;i~;0i:'
1. Ai - A2
= (6000 - 4800) x 10 --8
= l?OO X 10- 8
em
2. Since grating element,
A1A2
e+d=
(Ai - A2 ) sin e
6000 x 10-8
x 4800 X 10-8
e+d=
1200 X 10-8
x 0.75
= 3.2 x 10- 4 cm.
• •1A diffraction grating used at nonnal incidence gives a
yellow line (A = 6000 A) in a certain spectral order superimposed on
a blue line (A = 4800 A> of~ext higher order. Ifthe angle of diffraction
is 60°. calculate the grating element. W81U20:ra-;l4, Mal-ks05 J
Same as Q. 4.28, Page 4-40A, Unit-4. (Ans. 2.77 x 10 - 4 em)
-
_ Find out if a diffraction grating will resolve the lines
8037.20 A and 8037.50 Ain the second order given that the grating is
just able to resolve two lines of wavelengths 5140.34 A and 5140.85 A
.in the first order. li~.~0~,~~~~(i.¥~~l!!0I5I
.. fi.rllt()rder = 5140.34 A and
~condorder= B037.20A and
1. The resolving power of So grating is given by ~ = nN
dA
Therefore, N= .!(~)
n dA
A = 5140.34 + 5140.85 = 5140.595 A
where,
2
dA =5140.85 - 5140.34 =0.51 A and n =1
N = .! (5140.595) = 10080
1 0.51
2. Hence, the resolving power of a grating in second order
')JdA == nN =2 x 10080 = 20160.
~
3. The resolving power required to resolve the lines 8037.20 A and
8037.50 Ain the second order is A/ dA
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Wave Optics
4-42 A (Sem-l & 2)
A' = 8037.20 + 8037.50 =8037.35 A
In this case
2
dA' = 8037.50 - 8037.20 = 0.30
· A' 8037.35 267 17
ResoIvlng power, - = = 91.
dA' 0.30
4. Thus, the grating will not be able to resolve the lines 8037.20 A and
8037.50 A in the second order because tho required resolving power
(26791.17) is greater than the actual resolving power (20160).
Que 4.31.1 A diffraction grating used at normal incidence gives a
green line (5400 A) in a certain order n superimposed on the violet
line (4050 A) of the next higher order. lIthe angle ofdiffraction is 30",
find the value of n, also find how many lines per cm there in the
grating.
Answer I
Given: A, = 5400 A = 5400 x 10-8cm• 1..2 =4()50 x lIro"m.
10008
).'j - 1..2
= 1350 X em
To Find: i. Order of spectrum., n.
ii. Grating line~p.el'em,
1. The direction of principal maxima for normal incidence is given by
(e + d) sin e =nA
~. Let Il 'h order maxima of A" coincide with (n i 1)Ih order maxima oft..2• we
have.
(e + d) sin e= nAl = (n + I)A2
or nAt = (n + 1)1..2 or n = ~
Al - t..2
4050 X lO-
B
= 3
n=
1350 X 10-8
. AA At..
Now (e +d)sln8= __
1_2_ or e+d=
:I. 12
, Al - A (AI - t..2 )sin 8
2
e + d = 5400 X 10-B
x 4050 X 10-
B
1350 X 10-8 x 8in30·
4. N umber of.lines per cm,
8
N = _1_ _ 1350 x 10 = 3086
e+d 5400x4050x 2
@@@
Fiber Optics and
Laser
Part-1 ••••••••••••••••••.•••••••••••••••••••••••••••••••••••••••••••••••••••.••••. (5-2A to 5-16A)
A. Concept Outline: Part-l 5-2A
Part-2 ••••••.•••••••••••.••••••••••••••••••••.••••••••••••••••••••••••••••••••. (5-16A to 5-28A)
15-1 A (Se..-l 4: 2)
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5-2 A (Sem-l & 2) Fiber Optics and Laser
Optical Fibre: Optical fibre consists ofa core surrounded by a cladding
and a sheath. It is a thin, transparent and flexible strand. It is made up
of glass or plastic. It works on the principle oftotal internal reflection.
Acceptance Angle: It is defined as the maximum angle that a light
ray can have relative to the axis ofthe fibre and propagates down the
fibre.
Numerical Aperture:It is a dimensionless number that characterizes
the range of angles over which the fibre can accept or emit light.
Dispersion: The amplitude of the optical signal propagating in an
optical fibre attenuates due to losses in fibres as well as it spreads
during its propagation. Thus, the output signal received at the end
becomes wider compared to the input signal. This type of distortion
arises due to dispersion effect in optical fibres.
Que 5.1•.'11 What is optical fibre?
Answe~!~.:,)l
1. Optical fibre is a long, thin transparent dielectric material made up of
glass or plastic, which carries electromagnetic waves of optical
frequencies (visible to infrared) from one end ofthe fibre to the other by
means ofmultiple total internal reflection.
2. Optical fibres work as wave guides in optical communication systems.
:J An optical fibre consists of.an inner cylindrical material made up ofglass
or plastic called core.
'l The core is surrounded by a cylindrical shell ofglass or plastic called the
cladding.
;) The refractive index of core (n) is slightly larger than the refractive
index of cladding (nz)' (i.e., n l
> n2 ).
5-3 A (Sern-} & 2)
Physics
6. The cladding is enclosed in a polyurethane jacket as shown in Fig. 5.1.1.
This layer protects the fibre from the surrounding atmosphere.
7. Many fibres are grouped to form a cable. A cable may contain one to
several hundred such fibres.
Polyurethane
protective jacket
:,~i~;5:~;it~;
1••_ F.xplain the principle of optical fibre.
~~-
1. The working of optical fibre is based on the principle of total internal
reflection.
2. Total internal reflection is the phenomenon in which a light ray reflects
completely in the first medium, when it is incident on the boundary of
two different mediums.
3. When a light ray is incident on a high to low refractive index interface,
then from Snell's law, Fig. 5.2.1(a).
sin al = _
nz
__ ...(5.2.1)
sin az nl .
n and n = Refractive indices ofdenser and rarer mediums
where, z
l
respectively.
I
nl I
I
I
I
I 82 = _90°
~ ((")
~) ~)
l&lI~l~i~qi~'~~:oPtical
fibre.
4. Since n l
> nz, so from eq. (5.2.1), we have
sin al < 1
sin ez
i.e., sin a1< sin az
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5-4 A (Sem-1 & 2) Fiber Optics and Laser
'..e., 9, < 92 ...(5.2.2)
5. When the refracted light ray emerges along the interface, i.e., when the
angle ofrefraction becomes 90', i.e. 92
= 90', the correspoqdingval~e
of
the angle of incidence is called the critical angle and denoted by 9e'
Fig. 5.2.1(6).
6. Thus for 8, =8,,82
=90', eq. (5.2.1) becomes
sin 8 = n2
...(5.2.3)
, n,
7. If OJ =0 > 0.., the incident light ray is totally reflected back into the
originating medium with high reflection efficiency 99.9 %. This event is
known as total internal reflection, Fig. 5.2.l(c).
8. The necessary conditions for total internal reflection ofa light ray in an
optical fibre are therefore as follows:
l. The refractive index of fibre core should be higher than that of the
cladding.
il, The light ray should be incident between the core-cladding interface
flnd the normal to the core-cladding interface at an angle greater
than the critical angle,
ili. The respective refractive indices n 1
and n2
of core and cladding
materials of the fibre should be related to the critical angle by the
relation given in eq. (5.2.3).
~Cladding
~core
Fi~.,~~2.2.
Q*~6.3. IWhat do you mean by acceptance angle and numerical
aperture? Obtain the expression for acceptance angle and. num~rical
aperture.
.AItsWer I
A Acceptance Angle: It is defined as the maximwn angle ofinciqence at
the end face of an optical fibre for which the ray can be propa/!t~j;~.din
the optical fibre. This angle is also called acceptance cone half-angle.
B. Numerical Aperture :
1. It represents the light-gathering capacity ofan optical fibre.
2. Light-gathering capacity is proportional to the acceptlmCe angle, ~o'
3. So, numerical aperture can be the sine of acceptance angle ofthe fibre
i.e., sin 90
,
Physics
c. Expression for Acceptance Angle:
1. Applying Snell's law at points B in Fig. 5.3.1.
n1
sin '(90° - 91
) = n2
sin 90°
cos 9
n1 1
=n2
cos 9 = n2
1
n1
or sin 9 = Jr-1---C-O-S""'2-9-1
1
~1- n: ...(5.3.1 !
n;
2. Snell's law at 0,
no sin 90 = n 1
sin 91
or sin 9 = n1
sin 9 ".(G.:L
0 1
no
3. On substituting eq. (5.3.1) in eq. (5.3.2),
sin 9 = n1 ~1 _ n: _ Jr-n-=~---n""'; ... (5. ~ ,
o no n~ - no
90'
~ ~;.~:_*#:~.::
4. As the fibre is in air. So, the refractive index no = 1
The eq. (5.3.3) becomes, .
• 9 - 1~2--=-2 (5" ,
sm 0 - "n1
- n2 ... <. ,J .
This is the equation for acceptance angle.
D. Expression for Numerical Aperture (NA) :
1. According to the definition for numerical aperture (NA),
NA= sin 00 = Jn~-n; ...(5.3,r;;
2. Let the fractional ch8l1ge i.~ the refractive index (~) be the ratio between
the difference in refractive indices ofcore and cladding to the refractiv,'
index of core.
Core
Cladding
8 1
,
~ 90°-81
~---------------
n2
Light ray
c
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7
5-6 A (SeJD-l &2)
Fiber Optics and Laser
i.e., ..1.= n1 -n2
n1
...(5.3.6)
or n 1 - n2 = ..1. n 1
...(5.3.7)
3. Eq. (5.3.5) can be written as :
NA = In; - n; =J<n1 - n2
) <n,. + n)
...(5.3.8)
2
4. Substituting eq. (5.3.7) in eq. (5.3.8), we get
NA = J..1. n1 (n1
+ n2
)
.'). Since,
n 1 = n2 ; so, n1
+ n2
"" 2n
NA = J2..1.n; = n1 J2i
1
...(5.3.9)
(;. Numerical aperture can be increased by increasing '..1.' and thus enhances
the light-gathering capacity ofthe fibre.
We cannot increase..1. to a very large value because it leads to intermodal
dispersion, which cause signal distortion.
Que 5,,,,;,/'1 Write the classification of optical fibres.
OR
Discuss the different tyPes ofoptical fibre. Why graded index fibre is
iwtter than muItimode step index fibre.
AnswerJ
A. Classification of Optical Fibres:
K. Classification of Optical Fibre Depending on Material used:
a. Glass Fibres:
1. These fibres consist ofglass as the core and also glass as the cladding.
2. These are the most widely used fibres.
a. To reduce the refractive index of cladding, impurities such as
Germanium, Boron, Phosphorous or Fluoride are added to the
pure glass.
b. Plastic Clad Silica or P.C.S. Fibres:
1. By replacing the cladding with a plastic coating of the refractive
index lower than that ofcore, a plastic clad fibre is achieved.
2. Its advantage is only that the replacement of the glass cladding
with plastic offers the saving in cost.
;,. The limitations are:
i. .Losses are more than the glass fibres.
ii. Refractive index varies with temperature.
iii. Fibre life is small, mainly in humid environment.
Physics
Co Plastic Fibres :
1. These fibres consist ofboth core and cladding of the plastic material.
2. These fibres are cheaper in comparison to the above fibres.
3. But these fibres have high losses arid low bandwidth.
4. Also life of these fibres is small and refractive index varies with
temperature.
5. These fibres don't need protective coating and they are more flexible.
6. Attenuation of plastic fibres is more than glass or silica fibres but
even then they are frequently used for short distance computer
I applications.
ii. Classification of Optical Fibres Depending on Number of Modes:
f a. Monomode or Single Mode Fibre:
1. In this, fibre is capable of transmitting only one mode.
2. Suppose we make the core of the fibre for any small ray of order of
2 to 8 I-!m, then only one ray of light can enter the core and get
guided by total internal reflection.
3. Major advantage of single mode fibre is that it exhibits minimul11
dispersion loss and hence, the highest transmission bandwidth
4. Only high-quality laser sources that produce a vel''y focused bealll
of nearly monochromatic light can be used fol' single-mode
operation.
5. Because of the superior transmission characteristics, such fibres
are extensively used for long-distance applications.
b. MultiJDode Fibres:
1. In this, the fibre is capable of transmitting more than one mode, so
the name multimode fibre.
2. The multimode fibre has the core diameter of the ordel' (If
50 I-!m i.e., larger than the monomode fibre.
3. As the core radius is large enough, it accommodates many dim'n'nt
rays of light or modes, each entering the core at different angles
4. Since the different mode have different gl'OUpS velocities, t.!WI'('
exists considerable broadening of transmitted light pulf'Ps.
5. Hence, dispersion losses are more and bandwidth length product I;;
small of order of 1 GHz-km.
6. These fibres are useful for moderate distances.
7. The loss ofinformation capacity however is compensated by certai n
benefits of multimode fibres over monomode fibre such as:
i. Incoherent optical source can be used in multi mode fibre due
to large core diameter and large acceptance angle.
ii. Ease ofsplicing or joining.
iii. Lower tolerance requirements on fibre connectors.
• I
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-----------
----------------
iii. Classification of Optical Fibres Dependingon the Index Profile:
a. Multimode Step Index Fibre (MMSIF) :
1. It consists of a core material surrounded by a concentric layer of
cladding material with a uniform index ofrefraction n2 that is only
slightly less than that of core of refractive index n 1
• .
2. Ifthe refractive index is plotted against the radial distance from the
core, the refractive index abruptly changes at the core-cladding
surface creating a step, hence the name step index.
I.'~:/I
I~I
I I I
1 I I I
I I n, I I
+11 ~ :
t2I ~
III > 112 > no ..... +,.
Fig.5.4.~;
3. The name step index is due to this index profile and the term
multimode is due to its feature of propagating a number of modes.
4. Its manufacturing is such that its core radius is large enough to
accommodate many different rays of light or mode each entering
the core at different angles.
,ol'e~;-:'--:'-:'-:'=:~~-~l
F;bre ax;'
.....-- - ---------- . m
S
-- ~--~}--~~ ---~
z
Fig. 5.4.2. Propagation in mU,ltJmqde st~Rbde~fll1t~t
." '-." '".,.-.:", -. ~"'.-
1>. Multimode Graded Index Fibre (MMGIF) :
1 In this, the material in the core is modified so that the refractive
i nc!ex .profile does not exhibit step index change but a parabolic
H'frBctive index profile which is maximj,lm at the fibre axis.
2. In this fihl"(O', index of refraction has a mroomum value n, at the axis
and Jesser values falling off gradually and hence the name graded
index is given to this fibre.
:3. Since the light travels faster in a medium with lower refractive
index, the light ray, which is farther from the fibre axis travels
faster than the ray which is nearer to the axis.
4. As the refractive index is continuously changing across the fibre
axis, the light ray is bent towards the fibre axis in almost sinusoidal
fashion.
Pi. Light rays are curved towards the fibre axis by refraction.
Physics 5-9A(Sem
:;~ii~~!li~?~l~~~~~:'mm_tm'(jd~,grad~dindex
fibre.
c. Single Mode Step Index Fibre (SMSIF) :
1. In this fibre, the core ofafibre is made so small that onl~' U!l<
light can enter the core and get guided by the total internal )('1'
hence the name single mode.
2. This will be the only ray oflight or mode that can enter thl'
such a shallow angle.
-----------~
m~lt::::::=5:-?s:--
Fig. 5.4,.5. Propagati0llW.a.~ingle mode step index fibre:.
3. Major advantage of this fibre is that modal dispersion I." ;,
eliminated and because ofthis, such fibres are extensivelY II
long distance communication.
4. Different fibre designs have a specific wavelength called ";,
wavelength above which it carries only one mode.
5. Single mode step index fibre haa a auperior transmis,;ioll <,
over other fibre types ofthe above because ofthe absenCe' 0 I
dispersion.
Fibre axis
;liflA~.i·!t~
6. Light rays periodically diverge and converge along the length (,j ..
fibre.
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B. Graded Index Fibre is Better than Multimode Step Index Fibre :
In graded index fibre, the index of refraction in the core decreases
continuously while in multimode step index fibres the refractive index
of a core has a constant value. Therefore graded index fibre is better
than multimode step index fibre.
Que 5.5.;1 What are advantages of optical fibre over copper wire?
Answer,:
1. The information carrying capacity of a fibre is much greater than the
microwave radio system.
Attenuation in optical fibre is much lower than that of coaxial cable or
twisted pair.
Smaller in size and lighter in weight.
The life of fibre is longer than corresponding copper wire.
Fibre communication system is more reliable as it can better withstand
environmental conditions.
). The cost per channel is lower than that of metal counterpart.
7. H~mdling and installation cost ofoptical fibre system is very nominal.
Que 5.6. IDifferentiate between single mode fibres and multimode
r1bres.
Answer I
-_.
Single MocJe Fibr~s '.' ..... ri';·. ::l:t~'~odeF'ihres
. No.
1. In single mode fibres there is In multimode fibres, large
only one path for ray numbers of paths are available
propagation. for light ray propagation.
? Single mode step index fibres Multimode step indexfibres have
have less core diameter larger core diameter (50 10
I « 10 !-1m) and the difference 200 /-lm) and the difference
Ibetween the refractive indices between the refractive indices of
1
of we and dadding i, very core and cladding is large.
small.
:;. In single mode fibres, there is There is signal distortion and
no dispersion. dispersion takes place in
~~~Signal transmissioncapacity is
multimodefibres.
Signal transmission capacity is
Iless but the single mode fibres more in multimode fibres.
are suitable for long distance Because of large dispersion and
I communication. attenuation, they are less suitable
for long distance transmission.
';----t-Launching of light into single Launching of light into multi
mode fibres is difficult. mode fibres is easy.
---
6. Fabrication cost is very high. Fabrication cost is less.
s
5-11 A <Sem·l & 2)
Physics
q:ue 5,.7. i If refractive indices of core and cladding of an optical
fibre are 1.50 and 1.45 respectively determine the values of nUJDericaJ
aperture, acceptance angle and critical angle of the fibre.
IA,;l('fU 2014·15, Marks 05 I
i.~~~t:~;:~,:gi
Given :'h1
= 1.50, n 2 = 1.45
To Find: i. Numerical aperture.
.ii. Acceptance angle.
iii. Critical angle of fibre.
1. . , 'rieal aperture, -1(21) .
N"me NA ~n, 1 45 () "3.J
1.50 - _ :__ .. = .
III -_~J2 = -- -5'-0
where, :'1=---
n 1.
.
l
So, NA = 1.50 .j(~ 0.033) = 1.50 x 0.257 = lIl85
2. Acceptance angle, Ell) = sin-1 (NA) = sin-1
(0.385) = 22.64°
3. According to Snell's la'w, .
. 1 ( n,) . 1 I 1.45)
sin 8 = 2 or 8 = SlD'- ~ =8m i, l'.50
c
c
nJ
= 75.16°
Que 5.8. l A step index fibre has core and cladding refrad i .(' i nd icc.;
1.466 and 1.460 respectively. If the wavelength of light 0.H5 flm i",
propagated through the fibre of core diametel' 50 flm, find lht'
normalized fl'equency and the number of mode suppOI·tN by the
fibre.
Answer '1
d 50 =25 pm
Given: n =1.466, /1 2
= 1.460, }" =0.85 ~lm, a = 2=2
l
To Find: i Normalized fn~quency.
ii. Number of mode,
1. Normalized frequency ·i.., given by,
27W C-·,-,:'
. = -.-Vll'- n~
A
2 x ;r ,25 ~·-.c-,
--_. - .".; -", '" .
.__ .... .j!lAG!) , (1.-H)O I - --'. I Hz
0.85
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2. Number of guided modes,
y2
(24.48)2 = 299.635 "" 300
N=2 2
Q1!i~''&~'f)~ IDescribe the basic principle of communication of wave
in optical fibre. A step index fibre has core refractive index 1.468,
cladding refractive index 1.462. Compute the maximum radius
allowed for a fibre, if it supported only one mode at a wavelength
1300nm. ~ . ~
Answer I
A. Basic principle of clOommunication of wave in optical fibre:
Refer Q. 5.2, Page 5-3A; Unit-5.
B. Numerical:
~:v;::~~~~~=u~~:=~;w::~~V~f;:j~~~l:"·Y
..
y2
1. Number of modes supported, N =
2
y2
1=
2
y = 1.414
2. Let, a is radius allowed for a fibre.
y = 21ta 0 In 2 _ n2
).. "1 2
1.414 = 21t x ct J{l.468)2 _ (l.46W
1300 x 10"
0292.56 x 10-9
=a JU.468)2 - {1.462)2
a = 2.2 x 10-6 m = 2.2/-lm
Que 5.10. IDiscuss the different types oflosses In optical fibre.
Ansvver I
A. Absorption Losses:
1. Absorption is the most prominent factor causing the attenuati~n in
optical fibre.
2. The absorption of light is caused by the following three different
mechanisms:
i. Intrinsic Absorption:
1. It is the absorption oflight by the material ofthe core itself.
Physics
5-13 A (Sem-l & 2)
2. The intrinsic absorption is a material property ofglass itS(' i I
3. There is a tendency of the fibre material to absorb a sn I
amount of light energy.
ii. Extrinsic Absorption:
1. The presence of impurities in the fibre material is a m;!i'.
source of loss in practical fibres. This is known as extrilJ.... ,
absorption.
iii. Absorption byAtomic Defects:
1. Atomic defects in the fibre material are also responsible t,,,
the loss of light energy.
2. The atomic defects are created in the manufacture ofthe lih
3. These defects are also created when the fibre is exposed t,.
rays, y-rays, neutrons and electron beams.
B. Scattering:
1. It is the loss ofoptical energy due to imperfections in the fibre.
2. Due to this phenomenon, the light is scattered in all directions wh;
causes the loss ofthe optical power in the forward direction.
3. This loss is known as Rayleigh scattering loss.
4. Rayleigh scattering loss is found to be inversely proportional to II:.
fourth power of the light wavelength.
5
4
3
2
1
I I , I I I I I i Vavelength(un;;
0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 16
liilla'ii~1il'<,;I¥;
t'lltA1fu~ll!t'~J
C. Bending Losses:
1. The bending losses occur due to bends present in the fibre structUJ"l
2. These are of two types:
i. Micro Bending:
1. Micro bending losses are caused either dUring th'
manufacturing or during the cabling process.
2. Microbends may not be visible with the naked eyes.
3. During the manufacturing the microscopic bending of tlw (', " .
of the fibre occurs due to thermal contraction betwel'l1 1 I,.
core and cladding.
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;>-14 A (Sem-l & 2) Fiber Optics and Laser
~~,...-><;-~
'ltli..~~; .'

I. Macro Bending:
1. Excessive bendingofthe cable or fibre may result in loss known
as macrobend loss.
2. The fibre is sharply bent so that light traveling down the fibre
can't make turn and is lost in the cladding.
~MR"Obend
Fig"5.lOi'l~
(~ue 5.11.1 Define attenuation. Explain attenuation constant.
·nswer I
.... Attenuation:
It is defined as the reduction in amplitude (or power) and intensity of
signal as it is g'uided through optical fibre.
It i.'-; mainly due to absorption and scattering.
.Ut'Jluation Constant:
i! i' i.- 1'Ie optical power launched at t he input end ofthe fibre, then the
i'''"''C'J' 1'" at a distance L down the fibre is given by
p
"
=P
, e-" L ...(fj.11.1)
]H:l·l'. [X = Attenuation constant.
T" k; ,';u'i thm" "n both the sides of eq. (5,11.11,
u = 1. In P, ...(5.11.2)
L·· Po
In units ofdB/km. (j is defined through the equation
[J,dBikm = L
10
log P.
P
o
in ca,se of an ideal fibre, Po = Pi and the attenuation would be zero.
<,rue 5.12. IWrite a short note on dispersion.
Physics 5-15 A (Sern-l & 2)
i_
''.....~+.",.·.,.l.~,$.i.~~
"" "'-'.,,,,,,. ~~;P.
1. Dispersion is the time distortion of an optical signal that results from t.he
time offlight differences ofdifferent component of that signal, t.ypically
resulting in the pulse broadening.
2. 'In digital transmission, dispersion limits the maximum data rate, the
maximum distance, or the information carrying capacity of a single
mode fibre link.
3. In analog transmissioIl, dispersion can cause a waveform to become
significantly distorted and can result in unacceptable levels of compoo;ite
second order distortion (CSO).
4. When no overlapping of light pulses takes place, the digital bit rat.!' BT
must be less than the reciprocal ofthe broadened (through dispersion)
pulse duration (2,). .
5 Dispersion in optical fibres can be classified into three main lVIK'o; ,
i. Material dispersion,
ii. Waveguide dispersion, and
iii. Modal dispersion.
~1~)~kI:1l1I A communication system uses a 10 kIn fibre having a
loss of 2.5 dB/km. Compute the output power if the input power is
500I-lW.
i~.~~~~;·l
"':'/~:o···:··~~~· .... :..··';,;i:.......I1~:..:..... 50'O' 1"-" W
~~.;;?;.::::·.·:~Ja.~~~~~,U:·~iaull,.r.= ," x v -
;;g;#~~!<1~J)q~~~;:; •..... ,
1. We know that loss in fibre,
10 P
= -log 10 ----'
<XdBlkm
L Po
10 I 500 10-6
2 5
. - 10 og 10 Po x
6
(10)2.5 = 500 X 10
Po
P _ 500 X 10--6
=> 0 - ,. ~,.. 1.58 I-lW
. .Explain dispersion and attenuation in optical fibre. The
optical power, after propagating through a 500 mlong fibre, reduced
to 26 % of its original value. Calculate fibre loss in dBIkm.
_ . .~1;IJii.ijlC'1!f071
N:WYU;"'; ,e,,/<j"i'f:,:><;',:-ii'i:L,<,x;;:~~::'- ."
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An~.~~;/,;il
A. Dispersion: Refer Q. 5.12, Page 5-14A, Unit-5.
B. Attenuation: Refer Q. 5.11, Page 5-14A, Unit-5.
C. NUlDerical:
Given: L =56~
To Find: Fibr~f
1. The attenuation is given as,
10 (P)
a = -loglO --...!...
L Po
10 [P]
a- -10 --'
- 0.5 g10 0.75 P;
a = 20 x 0.1249
a = 2.498 dBIkm '" 2.5 dB/km
Laser: AOiJQTbJ
Emission ofRq,~i,~
Various LetJelsofL
Laser: It is an acronym for Light Amplification by Stimulated Emission
of Radiation,
Spontaneous Emission: It takes place when excited 'atoms make
transition to lower energy level without any external stimulation.
Stimulated Emission: It takes place when a photon of energy
(h v = E 2 - E 1) stimulates an excited atom to make transition to lower
energy level.
s
E · ., C ffi' ~l 8nhv
Insteln s oe IClents: --=--s
B21 c
Population Inversion: The phenomenon is which the number of
atoms in the higher energy state becomes comparatively greater than
the number of atoms in the lower energy state is known as population
inversion.
_ Explain LASER and different types of process of
radiations.
A LASER:
1. LASER stands for "Light Amplificat~9n by Stimulated Emission of
Radiation".
2. It is a device used to produce a strong, monochromatic, collimated and
highly coherent beam of light and it depends on the phenomenon oj
"stimulated emission".
B. Processes ofRadiation:
a. Absorption ofRadiation :
1. When an atom is,in its ground state and a photon of energy hI' is
incident over it, it comes to its excited state after absorbing that
photon. This process is known as absorption ofradiation.
E, Before 2
, ]
~hv
E 1 • 1 -------E
l
2. The probability ofabsorption ofradiation is given by :
P = B
-p(v)
I2 12
where, =Einstein's coefficient ofabsorption ofradiation ,
B I2
and
p(v) = Energy density.
b. Spontaneous Emission of Radiation:
1. When an atom is in its excited state, it can remain there only for
10-8 sec. After that it comes to its ground state and releases a photon
ofenergy hv. This process is called spontaneous emission ofradiation.
E2 • Before 2 ] After E
z
~hv
E 1 1 E]
.
2. The probability ofspontaneous emission ofradiation is given by :
(P21).PODIlIDeoWl = A21 .
where, = Einstein's coefficient of spontaneous emission
A 21
ofradiation.
c. Stimulated Emission (Induced Emission) ofRadiation :
1. When an atom is in its excited state and a photon of energy hi j:;
incident over it, atom comes to its ground state.
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2. But now instead ofoile, two photons ofenergy h~' each are released.
3. When these two photons ofenergy hv each are incident on another
two excited state atoms, four photons ofenergy hv each are released.
E
2
• IE2=E I - - llli§Ehv
~ : ~hvl I ~hV
I ~hvi I !Iv
E1-------IE1 . - _ _ " ~hV
!
_LASER
.,--.Beam
-
-
(Ii'fj~i~~~~f
4. This process goes on continuously and as a result, a monochromatic,
unidirectional beam of photon is released, which is known as
stimulated emission ofradiation.
5. The probability ofstimulated emission ofradiation is given by:
(PZI )stimulatcd = BzIP(v)
where, =Einstein's coefficient ofstimulated emission of
B21
radiation, and
p(v) :: Energy density.
Que 5:16.1 Differentiate between spontaneous emission and
stimulated emission.
Answer I
I S.No. Sponta.neo~!!.~~~Ij!~~P:", .; ".~~~~u~~te.~~1Di~si~p ".
1. It is a natural transition in
which an atom is de-excited
after the end of its life-time in
the higher energy level.
It is an artificial transition which
occurs due to de-excitation of an
atom before the end ofits life-time
in the higher energy level.
The photon emitted due to
2. The photon emitted due to
! spontaneous emission can
move in any direction.
stimulated emission can move only
in the direction of the incident
photon.
The probability of stimulated
emission depends on the
properties ofthe two energy levels
involved in the transition as well
as on the energy density of
incident radiation.
3. The probability ofspontaneous
emission depends only on the
properties of the two energy
levels between' which the
transition occurs.
Que 5~.17.1 Discuss necessary condition to achieve laser action.
5-19A(Sem·1&2l
Physics
;j~#~:~~1:~t"1,:,1
1. There are three conditions to achieve laser action as follows:
i. The number of atoms in higher energy state must be greater thil n
that.in lower energy state so that the rate of emission becomes
greater than the rate of absorption.
ii. The radiation must be coherent so the probability of spontaneous
emission should be negligible in comparison to the probability of
stimulated emission.
iii. The coherent beam oflight must be sufficiently amplified
;'iuei"'~~'i'{,:1 What are Einstein's coefficients A and B ? Establish a
relation between them.
AnsW'ef I
A. Einstein's Coefficients A and B :
1. The probability that an absorption transition occurs i,; given hy
Plz.=B p(v) ..(fi WI;
1Z
BIZ =Constant of proportionality known ;IS the
where,
Einstein's coefficient for inducl'd ahsol'pt 1011.
2. The probability that a spontaneous transition occurs is gi yen by
(P ).pont.ne"u, , .. ( 1),11',:2. i
= A ZI
21
where, A =Constant known as the Einstein's coeffIcient
ZI
for spontaneous emission
3. The probability that a stimulated transition occurs is given by
(P ).timulated = BzIP(u) ...(5.18.:3)
21
B =Constant of proportionality known as the
where, ZI
Einstein's coefficient for stimulated emission.
B. Relation Between Einstein's Coefficients A and B :
1. Under thermal equilibrium, the mean population N] and N" in the lower
and upper energy levels respectively must remain constant.
2. This condition requires that ·the number of transitions from E, to E]
must be equal to the number of transitions from E] to E,.
Thus,
The number of atoms absorbing J = lThe numbel' of atoms emitting I
( photons per second per unit volume photons per unit volume :
3. The number of atoms absorbing photons per second per unit vol ume
= B P(v)N I
I2
4. The number of atoms emitting photons per second per unit volume
= A ZI
N 2
+ BZ1P(v)Nz
5. As the number of transitions from E I
to E 2 must equal the number of
transitions from E 2 to El' we have .. ,(5.18.4)
B
I2
p(v)NI
=AZIN 2 + BZ1P(v)Nz
p(v) (B1zN"1 - B 21N 21 = A21N 2
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I·
~'r-
p(v) = ~IN2 ...(5.18.5)
-
[B12N 1 B21 N 2 ]
6. By dividing both the numerator and denominator on the right hand side
of the eq. (5.18.5) with B 1
"N'2' we obtain,
p(v) = ~1 / B12 ...(5.18.6)
N 1 _ B21 ]
[ N 2
B12
7. But, according to Boltzmann distribution law,
N 2 -tE,-E,l/kT
-- = e
N1
As E 2
-E1
= hv,
N2 e'hvlkT or N 1 = ehvlkT
N I N2
p(v) ~1 [ 1 ] ...(5.18.7)
B ehvlkT - B / B
'2 21 12
8. To maintain thermal equilibnum, the system must release energy in
the form ofelectromagnetic radiation.
9. It is required that the radiation be identical with black body radiation
and be consistent with Planck's radiation law for any v8.J.ue ofT.
10. According to Planck's law,
8nhv3 J[ 1 ] ...(5.18.8)
p(v) = ( ~ ehvlkT -1
where, c = Velocity oflight in free' space.
11. Energy density p(v) given byeq. (5.18.7) will be consistent with Planck's
law given by eq. (5.18.8), only if .
~1 87thv3
...(5.18.9)
B = ~
12
B
and, --.ll - 1 or B - B ...(5.18.10)
B - 12 - 21
12
12. The eq. (5.18.9) and eq. (5.18.10) are known as the Einstein's relations.
The coefficients B ,2, and A are known as Einstein's coefficients.
B 21 21
13. It follows that the coefficients are related through
c'
= - - a ~1 ••• (5.18.11)
B I2 = B 21
8nhv
14. The relation (5.18.11) shows that the ratio ofcoefficientB ofspontaneous
versus stimulated emission is proportional to the third power offrequency
ofthe radiation. This is why it is difficult to achieve laser action in higher
frequency ranges such as X-rays.
.".,Jllt:tWhat is population inversion ?
Physics 5-21 A (Sem-l & 2;
-
1. The phenomenon is which the number of atoms in the higher energy
state becomes comparatively greater than the number of atoms in th,
lower energy state is known as population inversion.
.2. According to Boltzmann's equation, if N 1
and N 2
are the number
atoms in the ground and excited states,
then, N 2 == e-IE,-E,l/kT
N 1
or N 2 == e - 6ElkT
N 1
where, i!!.E == Energy difference between the ground etat i'
and excited state,
k == Boltzmann's constant, and
T == Absolute temperature.
3. But for atomic radiation t:>E is much greater than kT. Therefon
thermal equilibrium the population ofhigher state is very much smailc.
than the ground state i.e., N 1
>N2
•
4. As a result the numbers of stimulated emissions are very little a,.
compared to absorption. Therefore laser action will not take place.
5. Ifsomehow the number ofatoms in excited state are made larger than
in the ground state i.e., N 2 > Nt' the process of stimulated emission
dominates and the laser action can be achieved. .
E r:::=:::;::J E3 excited state
t
1-: ,k";;'.",. . I E 2 metastable
state
I·· ) }~l ground state !>..',S'i'.;'i .. ·1 E 1 ground state
o No. of atoms""'" 0 No. of atoms ___
(0) In thermal equilibrium (b) After population inversion
tl.~~tJf~~ii:
_ Explain the concept of 3 and 4 level laser.
OR
Discuss the principal pumping schemes.
A. Three Level Pumping Scheme:
-
1. A typical three level pumping scheme is shown in Fig. 5.20.1.
2. The state E 1 is the ground level; E3
is the pump level and E2
is tho
metastable upper lasing level.
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,.
S-22 A (Sem-l & 2)
3.
When the medium is exposed to pump frequency radiation, a large
number of atoms will be excited to E level.
4. a
They do not stay at that level but rapidly undergo downward transitions
to the metastable level Eo through transitions.
Pumping level •
~JIIOU
p 00 d 0" Q E
3 o 0 0 Q 0 E
Upper a
lasing level
• L>
E2 a ROaR Du 0 <t; q, R E
2
E! 0 t> Yo E!
Ground state Lower lasing level
r'
Ca ).'tKar;'!?i2q;~; Cb)
D.
The atoms are trapped at. this level as spontaneous transition from the
level E2 to the level E is forbidden.
1
6.
The pumping continues and after a short time there will be a large
accumulation of atoms at the level E •
7. 2
When more than halfoftheground levelatoms accumulate atE , the population
inversion condition is achieved between the two levelsE! and
2
E •
'"!"
Nowa photon can tri/{/{erstimulated emission. 2
B. Four Level Pumping Scheme:
l.
A typical tour-level pumping scheme is shown in Fig. 5.20.2.
2.
The level E 1 is the ground level, E is the pumping level, E is the
4
a
metastable upper lasing level and E2
is the lower lasing level. E , E and
E" are the excited levels. 2 a
Pumping level
E 4 °'l1l8 'Nl ~ ~,O
Rapid E4 g Q Q P Upper lasing
level
decay
L.A.. E Q8%1P Cd ad °ii~E3
Metastable a
state ~
vI'
Q a E 2
Lower lasing
level
1 Q0 d u Q U a
E %" 8~~.:RfJb%%8 E 1 0 0 U 0 0'9
Ground state
Ca)
Cb)
3.
-lIB
E2
When light of pump frequency vp is incident on the lasing medium, the
active centers are readily excited from the ground level to the pumping
level E,.
';;;:
4. The atoms stay a~ the E4
level for only about 10-" sec, and quickly drop to
the metastable level Ea'
5. As spontaneous transitions from the level E j
to level E 2
cannot take
place, the atoms get trapped at the level E3
. The population at the level
E grows rapidly.
a
6.. The level E 2 is well above the ground level such that (E1
- E,) > kT.
Therefore, at normal temperature atoms cannot jump to level E., on the
strength of thermal energy,
7. As a result, the levelE2
is virtually empty. Therefore. popuJation inversion
is attained between the levels E 3
and E 2
•
8. .A photon of energy hv =(E1
- E z) emitted spontaneously ("an "tart a
chain of stimulated emissions, bringing the atoms to til(' ]0"('1' laser
level E2
.
9. From the level E 2
the atoms subsequently und('rgo nOIl-r;Hli;ll iv('
transitions to the ground level E, and will be once ag'lin "";nlahl" lilr
excitation.
Qti.e5.21.'1 Show that two level laser system has no pl'a<'t ical
significance for lasing. Explain the principle of three level lasers.
. IAKTU 2013-14, Marks 05 I
Answer I
A. Two Level Laser System has no Practical Significance for
Lasing :"
1. The two level laser system has no practical significanet' 1)('(';lu",' ill t"o
level system pumping is not suitable for obtaining pOpU!;It inll iIl', 'r"i"n.
2. The time span M. for which atoms have to stay at till' UpP"1" Ir·"{,j F.).
must be longer for achieving population inversion cOllditi<ln
3. Bencl', in t"o len'l sy"tem condttion of populati<ln ill"ll ·,,'I.ill 11<.'
achieve because (Iv'! = N 2 1.
4. Thus,.;timulall'demissiollwill nottake p!m'l' ,lilt! !. ..~"I":ln" :1, .. 1."1, ' ill
not occur.
B. Principle of Three Level Laser: Refer q. iJ.:W. Page:; :.11 . l '1111·;,
Que 5.22./ Explain Ruby laser with its construdion and wol'ldng.
Also explain its draw backs.
Answer I
A. Ruby Laser :.
1. Rub~' is basically A1 0 (silica) crystal containing ab,'ut n.o:)·; 111' "('ighll
2 3
of chromium atoms.
2. The Ala. ions in the crystallat.tice are substit uted b.· Cr" ions.
3. Cr:1+ ions constitute the active centres whereas the aluminium and
oxygen atoms are inert.
4. The chromium ions give the transparent AlzO" crystal a pink O!' n'd
colour depending upon its concentration.
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5-24 A (Sern-l & 2) Fiber Optics and Laser
B. Construction:
1. The construction for generating Ruby laser is shown in Fig. 5.22.1.
2. Active material is a small cylinder ofpink synthetic ruby, about 0.5 cm in
diameter and few centimetres long.
3. Two parallel mirrors are used, one mirror M is fully silvered and the
1
other mirror M 2 is partly silvered so as to enable the coherent light
radiation to be emitted through that end.
4. The mirrors must be separated by a distance that is an exact number of
halfwavelengths apart.
Ruby crystal
I ..
M,
I Power '. I
supply
Fig.G.~~u;,~pii.~~.ltt
,J.
f"'
Cooling is used to keep the ruby at a constant temperature.
6. SilkC quite a lot of the energy pumped into it, is dissipated into heat,
pumping process is carried with the help ofaxenon flash tube.
C. Working:
1. ChromiLlIl1 ions are excited by the optical pumping, which is achieved by
the xenon nash tube and raised to higher energy states H.
2. The excited atoms return to the lower state L from higher state H in two
steps as shown in Fig. 5.22.2.
3 First they return to meta-stable state M.
4. This transition is radiationless transition and energy ofthis transition is
passed to the crystal lattice as heat loss due to collisions.
5. The chromium ions that returned toM level can remain in this state f(Jr
several milli-seconds.
~ ~ H
i l' ~ i ~ ...: ~ ...... Radiationless transition
S 1 1.......................
: : .... , ............
5600 Ai! 1
;
, M
*" ,
i i E I I I
f .....- i ! i I I I Laser transition
_v· , ,; I
' I
f-~ i i i I I I (6943 A)
···11 _V : i, I I I
~~ ~ ~ t ttL
Fig.5.22;2.E~llf~fl~~~:~~~'i~_I~~J~1
6. Thus, the accumulation of the coming excited atoms at Mlevel fromH
level increases its population.
Physics 5-26 A (Sem-l &,'
7. When a chromium ion falls to the lower level L from the meta-stat
level M by spontaneous emission, it emits a photon of 6943 A.
8. This photon travels along the axis ofruby rod and is reflected baCK a."
forth by the silvered mirrors as shOwn in Fig. 5.22.3.
9. The photon travelling parallel to the axis of the tube will start phot:>,.
multiplication by stimulated emission ofother chromium ions ofM lev; .
10. When the photon beam becomes sufficiently intense, it emerges throl!~'
.•
the partially silvered end of the ruby rod in the form oflaser pulse".
11. The laser beam is red in colour and corresponds to a wavelength
(6943 A).
l~~~m:~~:::::~~~
_::-----:~===="~=.=
:~~:=ti:~1=
...._h..··...~-"M"'"
------------~------ ... '",'~', ~_JlIf.J'
" . ,:
dHi
100 % reflective 99 % reflective
mirror mirror
~1YRl~f~t!fI~A~~~I~I~t~J,t;'i#.'~UbY·laser;
D. Drawbacks:
1. The output ofthe laser is not continuous but occurs in the form of p,;,
of microsecond's duration.
2. The efficiency of ruby lasers is very less.
3. It requires greater excitation in order to achieve population inven;;·
~.J~lli.Discuss the He·Ne laser with necessary diagrams. G!" ,
its superiority over ruby laser. r,i.~~J~;i;l5;Markso~u
~~~.~'~I~;~~::
A. He·Ne Laser:
1. Helium-neon (He-Ne) laser is a gaseous laser.
2. The laser action of this laser is based on a four level pumping 8(1" ,
3. ThE) population inversion is achieved through inelastic atom-at
collisions. This is the basic principle ofHe-Ne laser.
B. Construction:
1. The construction for generating He-Ne laser is shown in Fig. 5.2:3.
2. It consists of a long discharge tube oflength about 50 cm and dianwi
1cm. .
3. The tube is filled with a mixture ofHe and Ne gases in the ratio 80
4. Helium is the pumping medium and Neon is the la5ing medium.
5. Electrodes are provided to produce a discharge in the gas and til,,··
connected to a high voltage power supply.
6. On the axis ofthe tube, two reflectorsM1
(fully silvered) and M 2
(partt":.
silvered) are fixed.
7. The distance between the mirrors is adjusted such that it equals !III. ~
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El
•
M1
:)-26 A (Sem-l & 2)
11. In this way, M 2 state of Ne can become more highly populated than
levelE.
,
12. The laser transition occurs when Ne atoms fall from level M" to E
through stimulated emission.
e~ 13. A red laser light ofwavelength 6328 A is obtained in He-Ne laser.
G
"
:M2 ; E. Superiority of He-Ne laser over Ruby laser:
.,' lir
}.
1. He-Ne laser produces continuous laser beam while ruhy l:oI>;er produces
light in the form of pulses.
,I
, 2. He-Ne laser employs a four level pumping scheme while ruby laser
employs a three level pumping scheme.
Fig.5.23,~:·.·Co.nstr~Ctio~:Pf,i{~.~N~·:~!~~.f~·?
c. Working:
_ _ What is the advantage of four level laser systems over
The encrgy level diagram is shown in Fig. 5.23.2.
,; three level laser systems? Describe the construction and working'
~l
of ruby laser. I_TV 2017·18, Marks OiJ
-l
-f
C ~ 2066eV
lB~~~;Jl)tl
20.61 eV
A. Advantages of Four Level Laser System over ThreE:' Level
E
Systems:
18.70 eV
1. It is easy to achieve population inversion with four level systenJ t h;1I1
.'14 Radiationless
with a three level system. '
a transition
2. In the four level laser, the transition does not terminate <It the g,'",md
1;1_ 02 I
state, the pumping power needed for the excitation of nt.oms ;," much
He
lower than in a three level laser.
,Fig. 5.23.2. Energy lc:>vel diagram ofHe-Ne laser.
Ne
3. The efficiency offour level laser is much better than that of <I t hrep level
laser.
1'lln~~ing is achieved by using electrical discharge in the helium-neon
111 ixtm·e. 4. In order to get population inversion in three level laser syst.ems mOI'e
than 50 % of the atoms in the ground level E j
must he lifted to level Hi
Electrons and ions in this discharge collide with He atoms raising them
i () ale'el !If" which is meta-stable. while it is not necessary in case of four level lasers.
5. The threshold pump power required for popuhtiT' ;nver'sion in t.hn>('
The He atoms are marc readily excitable than Nc:> atom because they are
level lasers is larger than in four levellasel'
fighter. Excitation level M 1 =20.61 eV of He is vcry close to excitation
B. Construction and Working of Ruby k. ,WI' : Refer Q, :'),2:2.
Page 5-23A, Unit-5.
i,>vel M. = 2C 66eV ofNe
:';ome of' the excited He atoms transfer their energy to Ne atOms of
~,lound statls by collisions between helium and neon atoms. ~~f~~t~~t,lWhat are various applications of '.,ASER beam '?
ill us. the e~, __ited He atoms return to ground state by transferring their
Iii 'rgYI () N,· atoms through collisions.
~~~~~:,.I
"hI' kilwti(' energy of helium atoms provides the additional 0.05 '.; I'D!'
::"jtillf~' 1111' neon atom. 1. The laser bean1 is used for drilling, welding and melting of hard m<lkri:tis
i'ilis is 1he main pumping sC'hen'le ofHe-Ne system. like dianlonds, iron, steel, etc.
Thus, neon atoms are active centres. 2. It is used in heat treatments for hardening or annealing in metallurgy,
3. The laser beam is used in delicate surgery like cornea grafting and in
Thp Iwliulll gas in the laser tube provides the pumping medium to attain
the treatment of kidney stone, cancer and tumor.
IlL' lle('essary population inversion for laser action. This population
!' 'C']'"ion is n1aintained because: 4. Laser is used in holography, fibre optics and nonlinear optics.
5. During war-time, lasers are used to detect and destroy enemy missiles,
Tlw meta-stability oflevel E a ensures a ready supply ofNe atoms in
level E". 6. Now, laser-pistols, laser-rifles and laser bombs are also being made,
which can be aimed at the enemy in the night.
Th,> Ne-Cltoms from level E decay rapidly to the neon ground stare
' ,
-' "
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--
7. Laser is very useful in science and research areas.
8. Laser is used for communications and measuring large distances.
9. Semiconductor laser is used for recording and erasing ofdata on compact
disks.
10. Semiconductor lasers and helium-neon lasers are used to scan the
unin'n;al barcodes to identify products in supermarket scanners.
Que 5.26. IIn a Ruby laser, total number of Cr+3 is 2.8 )( 1019• If the
laser emits radiation of wavelength 7000 A, calculate the energy of
the laser pulse. IA1tTtl'72~~~~'1'.~~;~~(1
Answer I
Given: 11 = 2.8 X lOlli, A. = 7000 A = 7000 X 10- 10 m
To .Find : Energy of the laser pulse.
1. We know that,
nhc
Enen.TV =
... I.
2.8 x 1019
x 6.63 X 10- 34
x 3 X 108
E nergy = 7000 x 10- 10
I'.' h = 6.63 X 10- 34
J-s and c = 3 X 108 mls)
= 7.956 J
OJ' Enp!<Tv = ~~ eV
ro. 1.67xI0'l!I
=4.764 x 10w eV
©©©
Physics (2 Marks Questions) SQ-1 A (Sem-l '" ~
._------------------..,.-----_. -_.._-_.
Relativistic MechaIlle:.·:"
(2 Marks Questionsf'
1.1. What is frame of reference '!
Ans~ A coordinate system with respect t.o which I"(' measun' t Ill' po>
of a point object of an event is cFlll~:d a franw of reh'J'l'!lct,.
1.2. Define inertial frame ofrefcl'ence.
ADj.· Inertial frame of reference is defined as the frame in which a!",,;
is at rest or moving with uniform velocity and is not under :11"
force.
1.3. What are non-inertial frames?
~ The frames of reference with respect to which an unaCCell'I<l1
body appears accelerated are called non-inertial frames.
1.4. What was the aim of Michelson-Morley experiment '!
~§. The aim ofthis experiment was to prove the existence ufthe el 11 I'
and to test whether the ether is fully or partially dragged WI!:
bodies moving in it.
1.5. What are the conclusions ofMichelson-Morley expcrimelJ [
:AJnl, a. There is no existence of hypothetical medium ether.
b. The velocity oflight in all inertial frames ofreferencl' rl'lll ,'II,
constant.
1.6. What are the Einstein postulates of special theory'.;
relativity?
81& Postulate I: The principle of equivalence.
Postulate II : The principle ofconstancy of the speed of light,
1.7. What do you understand by variant and invariant un~k
the Galilean transformation ?
Afi's; Variant means the physical quantities which change from (0;"
frame of reference to another frame of reference, e.g, ,·(,Ioci:·,
Invariant means the physical quantities which do llvl (; j, ",c;l.' ':
one frame of reference to another frame of l'eferencl' , EXUlllpi
distance between two points is invariant in true inert ial f!'lUlll"
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SQ-2 A (Sem-l & 2) Relativistic Mechanics
1.8. What do you mean by Lorentz transformation ?
~ll; The equations in special theory of relativity, which relate to the.
space and time coordinates of an event in two inertial frames of
reference moving with a uniform velocity relative to one-another,
;
are called Lorentz transformations.
1.9. What is the conclusion of Lorentz transformation?
Ans, The conclusion of Lorentz transformation is that it limits the
maximum velocity ofthe material bodies.
1.10. Write down the inverse Lorentz transformation equations.
Ans; Lorentz inverse transformation equations are:
x' + v t' , . t' + v x' / c2
X = , y = y , z = z and t = .j
.jl-v2 /c2 ' I_v2
/c 2
1.11. Define length contraction.
Ans; The length ofa moving rod will appear to be contracted if it is seen
from a frame of reference which is at rest. This decrease in length
in the direction of motion is called length contraction.
1.12. What do you understand by time dilation?
l¥=_=>;.=!t'a=.;.~=;!=~~=.~!l"""~~='1=:{lIGr)=t:~~"".
>""";'9~)~1
Ans. In the special theory ofrelativity, the moving clock is found to run
slower than a clock at rest does. This effect is known as time dilation.
1.13. Give the Einstein's mass-energy relation.
Aris. E = mc2
This relation is known as Einstein's mass·energy relation.
1.14. Give some examples of mass-energy equivalence.
Ans. Some important examples of the mass-energy equivalence are as
follows:
a. Pair production phenomenon,
b. Annihilation phenomenon (production ofy-rays photon),
c. Nuclear fusion, and
d. Nuclear fission.
F_'6~11111~·r~·'1
),,5. What are massless particles? l~;;" ,~~.; _,,~ ,:,,' <, "i.,:"",-"" ,ttJtw·"".~,'f>l; , " -, *~~";:~>:;<,.:"",,~
Am;. A particle which has zero rest mass (mo) is called a massless particle.
The velocity ofthe massless particle is same as that of light in free
space.
1.16. Which frames are known as accelerated frames?
i"ns. Non-inertinl frames are known as accelerated frames.
Physics (2 Marks Questions) SQ-3 A <Sem-l & 2)
1.1'7. Find relativistic relation between energy and momentum.
It~TlJ.20l5~1f)c,.Marks
02
.... Ai> the relativistic total energy E of a particle of ref,t mass (m{ll in
terms of its momentum p may be expressed as
2
E = Jm~ c· + p2 c
.: mo "" 0
E=pc
1.18. Why are Galilean transformations used ?
. . Galilean transformations are used to convert the laws of mechanics
from one frame ofreference to another frame, moving with constant
velocity with respect to the first frame.
1.19. Is earth an inertial frame of reference or not?
.... According to Newton's assumption earth is an inertial frame bec;1use
for the study ofany particle or body on earth, we can take earth as
inertial frame of reference.
But on other hand earth rotates about its axis as well as revolves
around the sun in its orbit so it can also be treated as non-inertial
frame of reference.
1.20. How the negative results of Michelson-Morley experiment
interpreted? EII.:IIt1fi~l.hM~'1'ks02J
ilIft1IW Following explanations were offered to interpret the negative
results ofMichelson-Morley experiment:
a. Ether drag hypothesis,
b. Fitzgerald-Lorentz contraction hypothesis, and
c. Constancy of speed oflighthypothesis.
1.21. What is proper length of a rod?
[j--'t_--':--':'l"U~.,-.::-.29--'J-:~-.1-7-,-
MIl-rk-s~~J
... The length of the rod measured by an observer in the frame in
which the rod is at rest is called proper length or actual length of
rod.
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SQ-4 A (SPJTl-l & 2) Electromagnetic Feild Th~ory
V
2
II
Ii Electromagnetic
I:
Field Theory
UNIT
(2 Marks Questions)
~.L. What is displacement current?
Ans. Tile rhang-ing electric field in vacuum or dielectric is equivalent to a
ClIITl'nt which produces the same magnetic effect as an ordinary
ClilTent in a conductor. This equivalent current is known as
dl,;placement current:.
2.2. Expl~in mathematically displacement current.
c' ...,. ..-+
AilS. ld = -- J. D . d s
Of; 'Yo<
(I', oD. d ;
I" =
Y. at
~,J: .d;
-,
·4
Wilere, aD . d' I d .
.Jr/ == -- IS ISP acement current enslty.
ot
2.3. State Gauss' law in electrostatics.
Ans. (;auss' law states that the electric flllX passing through any closed
hypothetical surface of any shape drawn in an electl'ic field is equal
to liEu times the total charge enclosed by the surface.
--~ ---io
1.('., <p== ~E.ds =
q
Eo
Whl'I'C', q = Total charge.
2.4. Define Gauss' law in magnetostatics.
l.ns; Gau;<s' law states that the magnetic flux around a closed surface is
()ljuallo zero.
$= ~B.d;=O
2.5. State the Ampere's circuital law.
Physics (2 Marks Questions) SQ-5 A (Sem-l & 2}
~ Ampere's circuital law states that line integral ofmagnetic inducti'.:'
Bfor a closed path is numerically equals to 1-10 times the currenj , .
through the area bounded by the path,
i,e., ~B, d i =J.1ol
Where, 1-10 == Permeability offree space.
2.6. State Gauss' divergence theorem.
·4
Jm;§~ It states that the flux ofvector field F over any closed surface'S
equal to volume integral ofthe divergence ofthe vector field C'neins," 1
by the surface's'.
H
F.d; == HIdiv FdV
• v
It is used to transform surface integral into volume integraL'
2.7. Give the physical interpretation of the Maxwell's equation
0;; a. First equation represents the Gauss' law in electrostatics fo'
the static charge;
b. Second equation represents Gauss' law for magnetism.
c. Third equation represents Faraday's law in electromagnet".
induction. .
d. Fourth equation represents the generalized form of Amp",
law.
2.8. What do you understand by equation of continuity?
.... The equation of continuity expresses the fact that the electnc
charges can neither be created nor be destroyed in macroscopIC
quantities. .
It is given as
-+ - ap
"V.J+at' =0
For static field, 0:; =0
V.J = 0
2.9. Write the physical significance of equation of continuity.
~ The current diverging from a small volume element must be equal
to the rate of decrease of charge within the volume.
2.10. Write Ma~ell'sequation in free space.
JI:IRlO For free space, p =0 and conductivity, 0' =0 therefore J = 0, Rene!'
the Maxwell's equations in differential form become:
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SQ--6A (Sem-I&2) Electromagnetic Feild Theory
-> -> -> ->
a. 'V.D =0 or 'V.E=O
-> ->
b. ~.R = 0 or 'V.H =0
~ -->
--> -> aB oH
c. 'V x E = - at = - flo --at
d. ~ x H = oD 8E
-=Eo-
Ot Ot
2.11. What do you mean by electromagnetic waves?
Aiig~ Electromagnetic waves are the coupled electric and magnetic
oscillations that move with speed oflight and exhibit wave behaviour.
'2.12. State Poynting theorem.
An8. According to this theorem, the time rate of EM-energy within a
certain volume plus the time rate ofEM-energy flowing out through
the boundary surface is equal to the power transferred into the
EM-field.
2.13. Write wave equation in free space.
~ The wave equations in free space are as follows:
....
'V2
H 02H
f.1oEo 7
....
02
E
'V 2
E Ioto Eo ot2
2.14. Write the mathematical expression of skin depth.
Anii'. The reciprocal ofattenuation constant is called skin depth or depth
ofpenetration.
1
0=
Attenuation constant
For good conductor 0= ~ 2
wiota
For poor conductor o=! ~
aV;
2.15. The electric field intensity j - 210 sin 1010, Vim for a field
propagating in the medium whose a -8.0 81m and E r • 1.0.
Calculate the displacement current density J d'
SQ--7 A (SeITl-l & 2)
Physics (2 Marks Questions)
ilIIIII:
·.i~o:",,"'50.Slm,e '" 1.0
:~r~f~ .,' ',~,. ' r
""'~:ij~'a~~siW,Jd'
1. The displacement curreIrt density J d is given by
J = dD = E dE ('.' E == E EO == 1 x 8.85 x 10-12
)
d dt dt r
J == 8.85 x 10-12
;t [250 sin10
10t]
d
== 8.85 x 10-12 x 250 x 1010
cos 1010
t
= ,22.125 cos 1010 t AJm2
2.16. Ifa plane electromagnetic wave in free space has ITlagnitude
of H as 1 AIm. What is the magnitude of E ?
~
1. For free space, the characteristics impedance,
Zo= ~: =~
41t X 10.7
1
Eo= Ho ~ = 1 x 8.854 x 10-12 '" 376.72 Vrn
2.17. Calculate the skin depth for a frequency of 10
10
Hz for silver.
Given a =2 )( 107 Slm and Jl =4n )( 10-
7
HIm.
:&'ill:
"f':tjn.n:.~·i'ii41tx .10- 7 Him
1. We know that, 0 = J2
fll<J(o
2 [.,' w '" 21tv)
= 4n x 10-7 x 2 X 107
X 2 x 3.14 X 1010
= 1.125 p.m
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SQ-8 A (Sem-I & 2)
3
UN~T
Quantum Mechanics
Quantum Mechanics
(2 Marks Questions)
3.1.
Anlil.
What is a black body?
A body which absorbs completely all the radiations incident upon it,
reflecting none and transmitting none, is called a black body.
3.2.
A;n:s.
Define black body radiations.
When a black body is heated to a suitable high temperature it emits
tot.al radiations which are known as black body radiations.
3.3.
An§~
Which body is assumed to be perfectly black?
Lamp black is the nearest approach to black body which absorbs
nearly 99 % of the incident radiation.
3.4. Define Wien's law.
ADS.
() ,.
Wien showed that the maximum energy, E ofthe emitted radiation
from black body is proportional to fi1'th power of absolute
temperature (T 5).
E U T5
En; = Constant x T 5
'"
",.
.;
3.5. What is Rayleigh-Jean's law 'I
Atls; Rayleigh-Jean's law states that the total amount ofenergy emitted
. by a black body per unit volume at an absolute temperature Tin
the wavelength range A and A. + dA is given as
u dA= Bit kT dA
). A4
~,. 3.6. Define Planck's law.
AtliL Planck's derived an equation for the energy per unit volume of
black body in the entire spectrum ofblack body radiation. It is given
by
Ii
!l),dA = B/t hc dA
AS e"cI).kT _ 1
Physics (2 Marks Questions) SQ-9A (Sem-l & 2:
3.7. What do you mean by wave particle duality?
:Al(j; According to Einstein, the energy of light is concentrated in SI,'.
bundles called photons. Hence, light behave as a wave on ntV' h,,,,.
and as a particle on the other hand. This hatw'e oflight is iUlO' •..·",
dual nature, while this property oflight is known as wave pari::
duality.
3.8. What are the properties of matter waves?
Dii'Z Following are the properties of matter waves:
a. Each wave of the group travel with a velocity known as pha'."
velocity.
b. These waves cannot be observed.
h
c. The wavelength of these waves, A =
p
3.9. What is matter or de-Broglie waves?
Dr. According to de-Broglie, a particle ofmass m, moving with velocit.
is associated with a wave called matter wave or de-Broglie wave
h h
A= - =
p mv
3.10. If uncertainty in the position of a particle is equal to d •.
Broglie wavelength, what' will be uncertainty in th-.
measurement of velocity ? 1_.I~~~ks o~:
MIl: Uncertainty in the position ofa particle =de-Broglie wavelength
~x = A = !!:... ...(3.10.1)
p
Where, h = Planck's constant, and
p = momentum of particle.
According to Heisenberg's uncertainty principle,
h
~ D.p?: ... (;).1O.:! ;
27t
Let ~v be the uncertainty in velocity,
D.p = m.~v ...(3.10.:3)
Putting eq. (3.10.1) and eq. (3.10.3) in eq. (3.10.2)
It h
-m~v =
p 2/t
So, uncertainty in velocity.
V
~v = [':p=mv!
27t
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SQ-10A (Sem-1 & 2) Quantum Mechanics
3.11. Define wave function.
~ The quantity whose variation builds matter wave is called wave
function ('1').
3.12. What are eigen values and eigen functions?
A;lilf; The values ofenergy for which Schrodinger's steady state equation
can be solved are called eigen values and the corresponding wave
functions are called eigen functions.
3.13. Write the characteristics of wave function.
Ami:
-
a. The wave function 'l' contains all the measurable information about
the particle.
b. It can interfere with itself. This property explains the phenomenon
ofelectron diffraction.
c. The wave function 'l' permits the calculationofmost probable value
ofa given variable.
3.14. How can we obtain a perfect black surface?
Amt,; An ideal model ofa perfectly black surface is obtained ifa small hole
is made in the opaque walls of a closed hollow cavity.
3.15. What is the physical significance of wave function?
Afilt The wave function 'l' itself has no physical significance but the
square ofits absolute magnitude I'l' 12
gives the probability offinding
the particle at that time.
3.16. Why Compton effect is not observable for visible light?
MiSli Compton effect is not observable for visible light because the·
maximum value ofCompton effect.1A =0.04852 A (when 9 =180°)
is very small (about 0.001%) as compared to the mean value of
wavelength ofvisible light (-5000 A). Compton effect is observable
only with X-rays and not with visible light.
©©©
SQ-ll A (Sem-l & 2)
Physics (2 Marks Questions)
Wave Optics
(2 Marks Questions)
4.1. Define interference.
:.;:mr; The modification in the intensity of light r0sulting from the'
superposition of two (or more) waves of light is called interfcrPllcP.
4.2. Explain interference in thin fHms.
gv: When a thin film oftransparent materiallil<;p oil clrop :,prcHd O'('!·
the surface of water is exposed to an extended S<lUITP of light. it
appears coloured. This phenomenon can be explailled ilS in tprfr'J'I.'IH'I'
ofthin films.
4.3. Explain the factor responsible for changing fl'inge width in
wedge shaped film. ~TU 2~1_6-~1~_Ma~~so~
Wedge angle '8' is the factor responsible for changing fringp widt h
in wedge shaped film because as the wedge angle is gradually
decrease, the fringe width increases and finally fringes disappenr
when 8 '" 0 or when the faces of the film become parallel.
4.4. Define Newton's rings.
~
When a monochromatic light falls on the film. we get dark ann
bright concentric fringes having uniform thickness, the"" rings an'
called Newton's rings.
Write down the conditions for bright and dark rings.
4.5.
DS; Condition for Bright Rings:
·2/-lt = (2m - 1) A./ 2
Condition for Dark Rings:
2/-lt =nV..
4.6. In Newton's ring experiment fringe width decreases with
the increase in the order of fringe. Explain why'?
IAlITU 2013-14, Mark-;Q2l
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_ _
__ __
r-
Wave Optics
SQ-12 A (Sem-l & 2)
ArtS.
4.7.
'Ans.
4.8.
, ns.
. '.1.
,,,.
4.10.
Ans.
a.
b.
4.11.
An,~.
4.12.
Ans.
[S. No.
~
i <1.
I
1---1
)
In Newton's ring experiment fringe width decreases with the
increase in the order of fringe because as the order of fringe
increases the thickness of the air film is also increased.
Why are fri.nges circular in Newton's ring experiment? Explain.
~!£,!,U:.,.~),tl~;l!~g;~~iP~;.1
In , Newton's ring set uRthe air film is enclosed below the convex
Ipns. The thickness of the film is constant over a circle having
cen t rc at the center of the lens. Hence the fringes are circular.
On which factor the condition of brightness or darkness
depends?
Thl' condition of brightness or darkness depends on the path
diff"I'l'IH;e between the two ref1ected rays.
Define diffraction.
Diflnlrtion of light is a phenomenon of bending of light and
::'pl'l'; ,ding out towards the geometrical shadow when passed through
un obstruction.
What are the types of diffraction?
Thl'n' me two types of diffraction :
Fn""J)e] diffraction, and
FraLlllhofer diffraction.
What happens to diffraction pattern when slit width of
single slit experiment increases?
l=i¥=<=.'fU=':"""'~~="~6=.·"=tr=~il1l=(~=,~'""'r.""".':,~::"::"".2/1
If WI' i11lT('aSe the width of the slit, diffraction pattern gets
narrower. Increasing the size of the opening reduces the spread
in the j)attern. .
Diffl'l'entiate between Fresnel and Fraunhofer diffractions.
Fre~~elDiffraction
Lateral distances are
important.
()h,,~-,;~:'r,d pattern is a projection
.FS5';tnh6t~~:nifi1+ij'<iti~ni:'
,""',."". .,,··.,UCC'·"· . ,;',,, " .,,'.,
The fl.ngular inclinations are
important.
Observed pattern is an imagE' ofthe
()f_~I~lifT:'~ld ing element. . I-,_so_u.:..::.r.:.c.:.e.:... _
(' n-h(, (,l'lllre of diffraction The centre of the diffraction
I pillll'rn may be bright or dark Ipattern is always bright for all paths
I d"j),',"Jing llIJUll the number 01 ['
parallel to the axis of the lens.
Fr('.~II,'l ZOlll'S.
_..1 _ _ - . - - - - - - - - - '
Physics (2 Marks Questions) SQ-13 A (Sem-l & 2)
. 4.13. Define diffraction grating•
. Ab1J; Diffraction grating is an arrangement consisting of a large numi),·,
ofclose parallel, straight, transparent and equidistant slits, c'ae I: .
equal widthe, with neighbouring slits being separated by an opaq;,
region of width d.
4.14. What do you mean by dispersive power of a plane diffracti",;·
grating?
'-.'. Dispersive power ofa grating is defmed as the ratio ofthe diffel'('IH'"
ftBlliio
in the angle ofdiffraction ofany two neighbouring spectral lim's t(.
the difference in wavelength between the two spectral ]jl1e,~.
4.15. How the dispersive power related to order ofthc sj)(>{'trulIl
Aft'Ii~ The di::;persive power.is directly proportional to lJw "I'll,,/,
spectrum, i.e., higher is the order greater is tIll' eli spc rsiq, i"" "
4.16. What is Rayleigh's criterion of resolution?
):'~TU~OI~~~~~~~~~~~;~
A.1Ui~ According to Rayleigh, the two point sources or two CqUel]].' int cnsc
spectral lines are just resolved by an optical instrument when t1...
central maximum ofthe diffraction pattern due to OIl(' ,~Ullrce fill!.,·,
exactly on the first minimum ofthe diffraction pattern oit he oj he I'
and vice-versa.
4.17.
Afi's;
.S;NO~
a.
b.
c.
Differentiate between the dispersive power and r('sojviHi~
power of grating.
--1
t>.ispef~~~·~w:~r·
•., •.•.. ,,' ·lte~()lvingPower i
It is defined as the rate of
change of angle of
diffraction with the
wavelength used.
Dispersive power is given
by dO = n
'd'A. (e+d)cos(l
Dispersive power depends
upon the grating element
It is defined as the ratio of the !
wavelength of any spectral line to
the smallest wavdength diffl' I'C lll'('
between neighbouring linl's {'or
which the spectral lines C;lll b" .ill.'l
resolved.
--_··------1
The resolving power ofa grating IS
.
gIven
b
y
I-..
dl-.. = nN !
~
Resolving power is indepl'l1dl'lll ,,'
grating element.
4.18. DifferenHate the i1::.terfen:'" y'l1d ;:,'1~ ~·~'·~:'l.ctj()n.
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SQ-!4A (Sem-! & 2) Wave Optics
Ans.
s. N~"l~;' -~··'~~~1K~~~!~t~<~:~1~~,~~
a.
b.
~
! c.
In this; the interference
occurs between t.he two
separate wavefront
emanating from two
coherent sources.
The interference fringes
are usually equally
spaced.
In this, the interference occurs
between the innumerable
secondary wavelets produced by the
unobstructed por.tion of the same
wavefront~
The diffraction fringes are never
equally spaced.
I -+1------------
In an interference pattern In diffraction pattern the intensity
all the bright fringes are of central maximum is n1a,'(lmum
ofequal intensity. and goes on decreasing as the order
of maxima increases on either side
. L of the central ~~~ima. ----.J
4.19. Why the centre of Newton's ring is dark?
If"""'.""".;!})"""' ..~=;.=,:~=)=.=;:=_.~"""'~;'9=·:")j
•.jl,.::.'·";:;'_'=
}'»8. At the point ofcontact oflens and glass plate, the path difference is
zero and phase change 'rt' takes place due to reflection on glass.
Hence dark spot will be formed at the centre ofring system.
4.20. What are the applications of thin film interference?
AU8. Applications of thin film interference are as follows:
a. ivleaslU'ement ofsmaU displacements,
tl. 'l'e:sting of surface finish,
c. Testing of a lens surface, and
d. Thickness of a thin film coating.
··1.:.~L Two independent sources could not produce interference.
Why?
Ans; Two independent sources could not produce interference because
. there will be phase difference development between the two waves
and hence sustained interference will not develop.
·L~2. What will be the effect on the intensity of principal maxima of
diffra<.-1;ion pattern when single slit is replaced by double slit?
An8. In single slit diffraction, intensity ofprincipal maxima,
" ...1 "sin" a
I = R"~, -...._--,,-
a-
By replacing single'slit by double slit, the resultant intensity at any
point on the screen is given by :
4A2 • 2
I = __s~n_~ cos2 13
a- '
So, the intensity ofprincipal maxima becomes 4 times.
©©©
SQ-15A (Sem·} '" 2)
Physks (2 Marks Questions)
~ ..._-
.... ~': ,>,~~~
: "". ..'f.
.'~._:, -. •.,-
.' ."."': ',' :".:' ~.:"'.:~.:.- .
','
. ' . ' , , "
"5"-i"""~':""""'"
Fibre Optics and Laser
UNIT
(2 Marks Questions)
5.1. Defin<.> fibre optics.
Fibre optics is a technology in which signnl" <1l' t'1l1'l'rt'r '"
Ans. I I I ~'
ele<'trie~,l intonpt.i('a) sil-'nal:-. tran,;mittpd thrllugh H tlllil ~I;l'"
and I'('on,t'rl ('0 into elt'clril':! sign:d:-,
!l.2. What is opticlll fihre '?
An opticul fibI'() is a cylindrical ','e Auid.. Ill<lt!(' lot [1:111"1'"
'!
. Ans. Ii
dielectric. which guides ligLl wa"l''': nlnng It,; Ic-ngth h 'Jt :t 1l1.
rei1£'ction.
What do you understand by totl-II intN'na) rel1t,etion ...
5.3.
Total internall'cJ1t'ction is the phenonwnon in which tlwre is con, ""1,,
Ails.
reflection oflight. within the In~'diumor the],£' is no refract.c'd ray. It (I•. ' .. ,',
when ,mgle of inc1dence is greater than critical angle.
Write down the advantages of optical fibres.
5.4.
The advantages ofoptica) fibres include:
Ai'il~;
High data transmission !'ates and bandwidth,
a.
b. Low losses,
Small cable size and weight; and
c.
d. Data security.
What are the functions of cladding?
5.5.
The cladding performs the following important functions:
A'.fi~
a. Protpcts the fibre from physical damage and absorbing ",urL" '
contaminants.
b. Prevents leakage of light f'ncrgy from the fi bl'!' t h]'[)u"h ,·vUTl·..·, ,. I, I
waves.
5.6. What is critical ray?
The ray incident, at the core·cladding bOlillt!rll')'. at t1" ,'it ical :.
I
AUs;
is call cd a cri tical ray.
5.7. Define acceptance angle.
Acceptance angle is t.he mnXilTIUm <Ingle th:1I :1 lii-(ht ,,,.'. can 1",',0
Ans~
relative to the axis of the fibre ancl propagait' clowll II" [ibn-.
5.8. Define modes.
The light ray paths along which t!H' waves al't' in ph:",' insid(' t 11<'
Alls;
fibre are known as modes.
,
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SQ-16 A (Sem-l & 2) Fibre Optics and Laser
5.9. What is attenuation?
AilS; The attenuation is defined as the reduction in amplitude or power
and intensity ofa signal as it is guided through optical fibre.
5.10. Define acceptance cone.
Affs; A cone obtained by rotating a ray at the end face ofan optical fibre,
around the fibre axis with acceptance angle is known as acceptance
cone.
5.11. What is dispersion?
AliS. The amount by which a pulse broadens as it passes through a
multimode fibre is commonly known as dispersion.
5.12. Give full form of LASER.
ADs. LASER is the acronym for Light Amplification by Stimulated
Emission ofRadiation.
5.13. Differentiate between the ordinary beam and laser beam.
Ans;
Laser beam is. produced by
stimulated beam.
It is monochromatic.
It is coherent.
Ordinary .~~~.,.
Ordinary light is produced
by spontaneous emission.
It is not monochromatic..
It is incoherent.
e.
a.
b.
S.No.
5.14. Define spontaneous emission and stimulated emission.
OR
What is stimulated emission of radiation in a laser?
_ _.liY!'~
Ans. Spontaneous Emission. : The process in which photon emission
occurs without any interaction with external radiation is called
spontaneous emission.
Stimulated Emission: The phenomena of forced emission of
photons are called induced emission or stimulated emission.
5.15. What do you mean by population inversion?
,
i _ 1 1 1
I ADs. The phenomenon in which the number of atoms in the higher
energy state becomes comparatively greater than the number of
atoms in the lower energy state is known as population inversion.
5.16. Define puinpin.g.:~
I,
Ans. The process of supplying energy to the medium to transfer it into
J: the state of population inversion is known as pumping.
5.17. Define metastable state. m
. . . .
t
_1
Ans. .1etastable state is particular excited state ofan atom, nucleus, or
other system that has a longer lifetime than ordinary excited states
It
~
I
Physics (2 Marks Questions) SQ-17A (Sem-l & ~.
and that generally has a shorter lifetime than the lowest eneq",'
state, called the ground state.
5.18. Give few important applications of optical fibre?
1
• •i~I'¢'Q46,Mark~-02
~ Following are the important applications ofoptical fibre:
a. In communication, '. b. In optical sensors,
c. In illumination applications, and d. In imaging optics.
5.19. Which fibres are generally used under sea water?
lIUDI: Single mode fibres are generally used under sea water,
5.20. Write down some applications of laser.
~ Following are the applications ofIaser :
a. In surgery, b. In holography,
c. In communications, d. In computer industry, and
e. Laser printing.
5.21. How can be modal dispersion minimized ?
~ Modal disp'ersion can be minimized by using single mode ''''.
monomode step index fibres or graded index multi mode fibn'.
5.22. Why does ruby laser emits red or pink colour?
&BIO Ruby laser emits red or pink colour due to the presence of eli; ,; ri, I
ions depending upon its concentration.
5.23. How is population inversion achieved in He-Ne las~'·"
D&l: In He-Ne laser, the population inversion is achieved throLl.!.!'!· .""':,,-"
atom-atom collisions.
5.24. What precautions are needed to minimize maLe,'j"
dispersion? lII.j~U::1~,;lJi7~Marks
02.1
Amr.: Material dispersion can be minimized either by choosing soun·!'·.
with narrow spectral range or by operating at longer waveleng'( J,
5.25. Differentiate between spontaneous w: .imulated emissi,,;
of radiation.
AWl:
s~:~~." ~'ij~ff;~~~~.lt~~~~';::'··· .~~#t~l:ated Emission
In this, light emitted is not MonochrOInatic light is emitted.
monochromatic.
b. INot controllable from I Controllable from outside.
outside.
c. I Incoherent photons are-' Coherent photons are emitted.
emitted.
Amplified beam is achieved.
No amplification oflight.
d.
©©©
•
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Ph.vsics SP-l A (Sem-l & 2)
B.Tech.
(SEM. I) ODD SEMESTER THEORY
E~fJN~TIO~,2013·14
E~GmEERI~GPHYSICS·I
Time: 3 Hours Max. Marks: 100
SECTION-A
Note: There are three Sections A, Band C in this paper. Questions are to
be done from aU three Sections.
1. Attempt all parts. Give answer ofeach part in short: (2 x 5 =10)
a. What do you understand by time dilation?
,os. Refer Q. 1.12, Page SQ-2A, 2 Marks Questions, Unit-I.
b. What are massless particles?
il,ns. Hl'fcr Q. 1.1fi. Page SQ-2A, 2 Marks Questions, Unit-I.
'C. In Newton·s ring experiment fringe width decreases with
the increase in the order of fringe. Explain why?
,ns. Het(~r Q. 4.6, Page SQ-llA, 2 Marks Questions, Unit-4.
d. How the unpolarized light and dn'ularly polarized light
distinguish'!
.. n .., { f1 Ull polarized light, t.he light. is passed 11
.Jl·ough a single plane while
i:1 circularly polarized light., the light. is passed through two planes.
e. "Vhat do you mean hy population inversion?
Ans. Refer Q. 5. 1~" Page SQ-16A, 2 Marks Questions, Unit-5.
SECTION-B
.~. .'' 1<'l1lpt :Ill." three part". All parts CUITY equal marks: (5)( 3 = 15)
;>,. (,;alculate t he length of one meter rod moving parallel to its
len/{th wh€Jl its mass is 1.5 times of it.s rest mass.
ns. Hel"r Q. 1.3-1, Pagel-33A, Unit-1.
b. The speed of an electron is measured to he 5.0 x 103
mls to an
accuracy of 0.003 %. Find the uncertainty in determining
the position of this electron (mass of electron is
9.1 x 10-:11 kg' and Planck's constant is 6.62 x 10-:l4 J-g).
Solved Palwr (:W I:1- 1~)
SP-2A (Sem-l & 2)
AKift.i
i~:~";;;{).q'~.'l{OmJs.accuracy = 0.083 %, me == 9.1 x lO-:n kg,
:62 X 10-34,J s· .
FJii.d1 Uncertainty in position ofelectron.
1. Uncertainty in velocity, /,>'; = Q.9~~ x 5.0 x 10
1
," 0.15 mho
100
h
h
=> tsx=
2. !'»: t>p = 2n 2IT ,. ,p
h 6.62 ~. 10- :4
!'»: - --~-_ .. ----- .--"- ...
- 2nm(6v) -2x:3.14x9.1·,1(l:lI
O.I'-;
!'»:' == 7.72 x 10 -. ,j m.
c. Newton's rings are observed in reflected light with
w3vdength 6000 A. It'the diameter of the 10
th
dark ring is
0.5 (~ln, find the radius of curvature of the lens and the
thickness of the corresponding air film.
~.; Refer Q. 4.17, Page 4-23A, Unit-4.
do A diffraction grating used at normal incidence gives a
yellow line (A. = 6000 A> in a certain spectral order
superimposed on a blue line (A. =4800 A> of next higher order.
Ifthe angle ofdiffraction is 60·, calculate the grating element.
:«if§; Refer Q. 4.29, Page 4-4lA, Unit-4.
eo The refractive indices of quartz for polarized ligh t f..le and f..lo
are 1.5508 and 1.5418reSJectively.Calculate the phase
retardation for A. =5000 A when the plate thickness is
0.082mm.
~
-,{~<,:~.,:,"F,~.,~>_>",'>';~-', ,'. _'" '" .~_ "_.' ': ': '. :',
~<l:.,~9J)$i'Q~:"');D41~~:t:b:<(jiOS2mm = 0.0032 cm,
····rl~~~~:. .
'
. 2n 2n
1. The phase retardatiOn == -A. 6 = -A. (II
r, - II
r" It
2 x 3.140.5508 -1.5418) x 0.0032
5000 x 1O. a
= 3.617rad.
SECTION-C
Note: Attempt all the question ofthis section. All questions carry equal
marks.
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light
Physics SP-3A (Sem-1 & 2)
3.
a.
Attempt anyone part of the following: (5 )( 1 = 5)
State Einstein's postulates of special theory of relativity.
Explain why Galilean relativity failed to explain actual
results of Michelson-Morley experiment.
AnS. Refer Q. 1.6, Page I-SA, Unit-I.
b. Show that the relativistic invariance of. the law of
Ans;
A
conservation of momentum leads to the concept of variation
of mass with velocity and equivalence of mass and energy.
Variation of Mass with Velocity: Refer Q. 1.23, Page I-24A,
Unit-I.
B. Equivalence of Mass and Energy: Refer Q. 1.24, Page I-26A,
Unit-I.
4.
a.
Attempt anyone part of the following: (5)( 1 = 5)
Deduce a relation between phase velocity and group velocity
in a medium where wave velocity is frequency dependent.
What happens if the phase velocity is independent of
frequency? .
~;
1.
2.
Ah 1
· dxro
s p ase ve OCIty vp
= - =
dt K
For the amplitude of wave packet to be constant
!::J.rot !::J.K
-- - --x = Constant
2 2
3. Hence, group velocity
!::J.ro
dx 2 !::J.ro
v = - = - - =
f: g dt !::J.K !::J.K
2,
. !::J.ro dro
v = hm - = -
g 6K ..... o!::J.K dK
l, _ d _
~::. vg - dK (Kvp ) -vp +K dK
V
I: [.,' ; =v p and ro=KVpJ
1.1
,
::
(21t) dvp
g p
v = v + T d(~1t)
dv
vg = vp - A. dA.
p
4. This is a relation between group velocity vg
and wave velocity vp
in
a dispersive medium in which wave velocity is frequency dependent.
dvp
SP-4A (Sem·1 & 2) Solved Paper (20 1.3-1~ •
dv
5. If = 0, then the phase velocity does not depends on frequeni
d:
and become independent
v =v
p g
b. A particle of mass m is confined to a one-dimensional box of
length L. Derive an expression for wave function and energy.
... Refer Q. 3.17, Page 3-16A, Unit-3.
5. Attempt anyone part ofthe following: (5 xl'" .
a. Discuss the interference in thin film due to reflected ligh;
What happens when film is excess thin ?
. . Refer Q. 4.3, Page 4-6A, Unit-4.
b. Explain the diffraction pattern obtained with diffraction a:
single slit. By what fraction the intensity of secon.!
maximlim reduced from principal maximum?
- Refer Q. 4.19, Page 4-25A, Unit-4.
6. Attempt anyone part of the following: (5)( 1 == .
a. What is diffraction grating? Show that its dispersive pow,"
1
can be expressed as where all terms hav,.
J( e:dr_1..
2
their usual meanings.
. . Refer Q. 4.23, Page 4-36A, Unit-4.
..b. What do you mean by double refraction 'l Explain tll('
working principle of Nicol Prism.
A Double refraction:
1. When a beam of unpolarized light is incident on the surface of an
anisotropic crystal such as calcite or quartz, it is found that it will
separate into two rays that travel in different directions. This
phenomenon is called birefringence or double refraction.
Unpolarized j •
Fig. 1.
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!'!l'"ics SP-5A (Sem.l & 2)
._----
2. The two rays are known as ordinary ray (O.ray) and extraordinary
ray (E-ray), which are linearly polarized in mutually perpendicular
directions.
B Working Principle ofNicol Prism:
1. When an unpolarized light ray8M parallel to the face DC' is incident
on the face A' D, it splits into O-ray and E-ray.
2. The O-ray is going from calcite to Canada Balsam travels from
optically denser medium (/-lo =1.66) to a rarer medium (/-leE =1.55).
;j. The refractive index ofO-ray W.r.t. Canada Balsam =I!cu'/-lo
4. The critical angle for O-ray, eo=sin'l (/-lcu'/-lJ =sin'l (1.55/1.66) =690
5. Since the length ofthe prism is sufficient, the angle ofincidence of
O-ray at Canada Balsam layer becomes greater than its critical
angle. Hence the O-ray is totally reflected from the Canada Balsam
layer.
6. On t.he other hand, the E-ray is going from calcite to Canada Balsam
travels frol11 an optically rarer medium (~lE =1.49) to a denser
I1lpdium (~lCR = 1.55).
'7. In this way, the light emerging from the Nicol prism is plane
polal'ized with vibrations parallel to the principal section.
7. Attempt anyone part of the following: (5 lC 1 =5)
a. Show that two level laser system has no practical
significance for lasing. Explain the principle of three level
lasers.
Ans. Refer Q. 5.21, Page 5-23A, Unit-5.
b. Discuss different types of optical fibre. Why graded index
fibre is better than multimode step index fibre?
Ans; Refer Q. 5.4, Page 5-6A, Unit-5.
©©©
SP-6A (Sem·1 & 2) Soh'pd f"lpP/' 120 14-1r; i
B.Tech.
EXAMINATION, 2014~15
ENGINEERING PHYSICS~I
Tilne : 3 Hours Max..Marks ; 100
SECTION-A
1. Attempt all parts ofthis question. Each part carries :2 mark..,.
(2)<5=101
a. What are inertial and non-inertial frames of reference?
~ Inertial Frame: Refer Q. 1.2, Page SQ-1A, 2 LVlarks (~lIPsti(m,;
Unit-l.
Non-inertial Frame : Refer Q. 1.3, Page SQ--lA. 2 Mm'k,; (,~ [('"~I iolt.-·
Unit-l.
b. What is double refraction ?
Aflit; The phenomenon in which we get two refract lI1g pl"jl(' po<" lzed
rays corresponding to one incident poladzec1light ray i,' C'Jlkd dnublt'
refraction.
c. Why are fringes circular in Newton's ring experiment?
Explain.
milr. Refer Q. 4.7, Page SQ-12A, 2 Marks Questions. Unit·A.
d. What is stimulated emission of radiation in a laser?
A'B1ft Refer Q. 5.14, Page SQ-16A, 2 Marks Questions, Unit-5.
e. What do you know about acceptance angle and cone in a
fiber?
~ Acce~tance Angle: Refer Q. 5.7, Page SQ-15A. 2 Marks
QuestIOns, Unit-5.
Acceptance Cone: Refer Q. 5.10, Page SQ-16A, 2 Marks
Questions, Unit-5.
SECTION-B
2. Attempt any three of this question. Each part carries Ii marks.
(5)<3=15)
a. Calculate the work done to increase speed of an electron of
rest energy 0.5 MeV from 0.6c to 0.8c.
~ Refer Q. 1.30, Page 1-31A, Unit-I.
b. An electron is bound in one dimensional potential box
which has width 2.5 lC 10-10
m. Assuming the height of the
box to be infinite, calculate the lowest two permitted energy
values of the electron.
a.-u; Refer Q. 3.19, Page 3-18A, Unit-3.
c. Newton's rings are observed by keeping a spherical surface
of 100 cm radius on a plane glass plate. If the diameter of
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Physics SP-7A (Sem-l & 2)
Afi:S~
d.
Arts.
e.
Ail's:;
3.
a.
Anlil'~
b.
AJ1&
4. a.
A'illi;;
A
B.
1.
the 15th bright ring is 0.590 cm and the diameter of the 5th
ring is 0.336 em, what is the wavelength of light used?
Refer Q. 4.16, Page 4-23A, Unit-4.
Find out if a diffraction grating will resolve the lines
8037.20 A and 8037.50 A in the second order given that the
grating is just able to resolve two lines of wavelengths
5140.34 Aand 5140.85 Ain the first order.
Refer Q. 4.30, Page 4-41A, Unit-4..
If refractive indices of core and cladding of an optical fibre
are 1.50 and 1.45 respectively determine the values of
numerical aperture, acceptance angle and critical angle of
the fibre.
SECTION-C
Attempt anyone part of all questions. Each question carries 5
m a r k s . ' (5 x 5 = 25)
Deduce the Lorentz transformation equations from
Einstein's postulates. Also show that at low velocities, the
Lorentz transformations reduce' to Galilean
transformations.
Refer Q. 1.8, Page l-11A, Unit-I.
Deduce the relativistic velocity addition theorem. Show that
it is consistent with Einstein's second postulate.
Refer Q. 1.20, Page 1-20A, Unit-I.
Explain group velocity. Establish a relation between group
velocity and phase velocity and show that these velocities
are equal in non-dispersive medium.
Group Velocity: The velocity with which a wave packet moves
forward in the medium is called group velocity.
Relation between Group Velocity and Phase Velocity:
A wave packet consists of a group of waves slightly differing in their.
wavelength, velocities, phase and amplitude as shown in Fig. 1.
t wave packet
Q>
...,
J
.-E 0 loG / '. f , { , I , l ')0
v
x_
g
Fig.!.
2. Such a wave packet moves withits own velocity called group velocity
(vgl or particle velocity and velocity ofindividual waves forming the
wave packet is called phase velocity (vp )'
SP-8A (Sem.l & 2) Solved Paper (2014
3. We know that vI? = v A.
and the de-Broglie's wavelength
A.= h
mv
and E = hv = mc2
=>"v = mc2
/h
4. Eq. (1) becomes,
2 2
v = (mc (...!!-)= c
p h
)
mv v
5. Since group velocity (v ) is equal to particle velocity (v) i.e .. v,
g C
2
2
Then v = -
c
or v ' v = c
p p g
vg
C. Velocities in Non-Dispersive Medium:
1. The angular frequency ro of de-Broglie's waves associated wit h
particle of rest mass m and moving with velocity v is given by,
o
ro = 21tV and E = hv
=> v = E and E =mc2
h
21tE 2ronc2
ro = --=---=
h h
2ron c-
., [ ... In 0" - IIi
()
~ .11-.
2 V l
-1
2
21tmoc ( V
2
)2
C
ro= 1--
2
h c
2. Differentiate both sides of above equation w .r.t. v, we get
3
dro = 21t~oC2 (-=-!)(1- V
2
)-2 (_ 2V)
dv h 2 c2 c2
dro 21tmo v
...( ~!. ~
dv =
( v2)i
h 1-~
3. Also, the wave number k of de-Broglie's wave is
k = 21t = 21tm v 21tmo v
A. h g22
hx 1-
c
4. Differentiate both sides of above equation w.r.t. v, we get
:~ = (21t::)i
h 1-
2
c
5. We know that group velocity,
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Physics SP-9 A (Sem·l & 2)
...-_.__.----
dw
v = --
B ' dk
From eq, (2) and eq. (3),we get
' v2)~
2nnl" v h (1 - ~2
v = =v
g
--('-- 2 J~' 2lt1n"
h 1- .":.. ~
2
~ c
b. r.;xp"Jain Heisenberg's uncertainty principle? Describe
Heisenberg's gamma ray microscope.
ln:Ei.
A. Heisenberg's Uncertainty Principle:
1. According to this principle, "It is impossible to determine the exact
position and momentum ofa particle simultaneously".
-JV
(a) !VJ large, tix small
(b) <lx large. lip small
Fig. 2. Wave-packet: (a) Narrow, and (b) Wide.
2, If lix and lip are the uncertain position and momentum ofparticle
then according to this principle
!ixlip~!!:-
2n
or !ix lip ~ 11
The product ofuncertainty position and uncertainty momentum of
particle is greater than or equal to hI21t.
B. Heisenberg's Gamm.a-ray Microscope:
1. Let us try to measure the position and linear momentum of an
electron using an imaginary microscope with a very high resolving
power as shown in Fig. 3.
2. The electron can be observed if at least one photon is scattered by
it into the microscop~lens.
3. The limit ofresolution ofthe microscope is given by the relation.
A.
d=-
2sin9
Here d represents the distance between the two points which can
be just resolved by the microscope.
SoJv(,'d Papcr /2014·11'".
SP-IOA (Sem-l & 2)
,.. d ~I
A ( I )D
--..
Photon Electron
o
Fig. 3. Measurement of position and linear
IllOJ11l'ntUJJ1 of an eJedl'on,
4. This is the rangl-' in which the electron would IX' ,'i~ihlc when
disturbed by the phofon.
5. Therefore, the uncertainty in the measurement of till' I',,,,il ion or
the electron is
i,
' = ( = __ '.__.. ' I )
I
, 2 sin lJ ..
6. Howeyel', Ow incoming photon will intcr:ll'l ",il h :: II' "ivcl ron
through the Compton effect,
7. To see this p!c'ct:'on. the scatt<~r('(l photon :,-Iwu!d ,IIIVI" t1w
microscope within tl1C' angle 20,
8. The momentUlll imparted by the photon tu I hl' ,-,led I'"'' dutillg the
impact is of the order of hi),.
9, The cl11nponenl ofthis mO!TIpntum along 0.4. i:~ I - /iii ,-i 1 ", "JlCI (hal
,dong' OR is (hi), "in el.
la, Hence the u!lcl'I'taint)' in tht' ITwaSUI't~ll1l'n( of tilt· Ill' ',III"11UI) or
the deetron is
'h ,'I i h..' h,
'J) = I ." ,s III tl - I - - "111 (l I ,'" ',0 -II'j' I 1"I
-- " i. ! !, ' . - ,~. . . ~
11. Multiplying ('1111) by eq, (2), we obtain
A .0 h " I
LU,,j) '"' ---...---- ;'--8111" ~', !
Zsin 0 I,
A more sophisticnt0d npprnach will "hm' th:!t ' /J
12,
Describe and explain the fOI'mnt'ion ,.1' Nt'" «lit',· •"1~!S in
5. a. ref1l'cted monochromatic light. 1','0>'(' t hnt in I'PrJ. "~I ,'d light
tht~ diameter of bright dngs art' propo't ion,,1 to (Ill' "quHI'e
roots of odd natural numbers.
Refer Q, 4,12, PngC' 4-18A. Unit-4.
Ails;
b. Discuss the phenomenon of diffraction at a single slit a: ,d
shoW that the relative intensities of tiH' slIecessi VC'
maximum al'e nearly 1 : 419n
2
: 4125rr
2
...
:AUs. Refer q, 4.19, Page 4-25A, Unit-4,
"~.,,,,,:..:,
..,,,, rlr»,H1'tgx'd",t ;
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c
D
A A'
r----=
.1.3f-. ~~
SP-ll A (Sem-l & 2)
j~ Physics
,
r~
i'
i··
6. a. Describe the construction, working and application ofNicol
{ prism.
.I:~l::
:At1§;
A Construction
1. It is a calcite crystal with a principal sectionABCD whose length is
three times its breadth as shown in Fig. 4.
B
E-ray
O-ray
Fig. 4. Construction of a calcite crystal.
2. In this situation, the new blunt corners willA' and C'.
3. The crystal is then cut into two pieces from one blunt corner A' to
other blunt corner C' along a planeA'C' perpendicular to the principal
~"
sectionABCD and perpendicular to both the facesA'D' andBC'.
4. The two cut surfaces so obtained are polished optically flat and
joined by a transparent medium called Canada balsam.
5. Canada Balsam is a transparent liquid whose refractive index lies
between the refractive indices of calcite for the O-ray and E-ray.
B. Working: Refer Q. 6(b), Page SP-4A, Solved Paper 2013-14.
~j.
C. Application:
~( l. Use in microscopy and polarimetry.
~
2. Used for observing the sample placed between orthogonally oriented
'ij: polarizers.
'I!
b. Discuss the He-Ne laser with. necessary diagrams. Give its
superiority over ruby laser.
An~~ Refer Q. 5.23, Page 5-25A, Unit-5.
7. a. Explain single mode and multimode fibres. Also give the
c;haracteristics of each type of mode.
1U1Y; Refer Q. 5.4, Page 5-6A, Unit-5.
b. Explain the process of a hologram construction with
necessary diagrams. Also give some applications of
hologram.
A1fs.
A Hologram Construction
1. The monochromatic light from a laser has been passed through a
50 % beam splitter so that the amplitude division of the incident
beam into two beams takes place.
SP-12A (Sem-l & 2) Solved Paper (2014-1Ci I
2. One beam falls on mirror M 1
and the light reflect from M, falls (1
the object. This beam is known as an object beam.
3. The object scatters this beam in all directions, so that a part of Lili
scattered beam falls on the holographic plate.
4. The other beam is reflected by mirrorM 2
and falls on the holograph!
plate. This beam is known as reference beam.
5. Superposition ofthe scattered rays from the object and the referel1l'I '
beam takes place on the plane of the holographic plate, so dw'
interference pattern is formed on the plate and it is recorded.
6. The recorded interference pattern contains all the information (11
the scattered rays i.e., the phases and intensities of the scattr]'(",
rays.
7. For proper recording, the holographic plate has to be exposed to ~J;,
interference pattern for a few seconds.
8. After exposing, the holographic plate is to be developed and fixecl ;I:
like in the case of ordinary photograph.
9. The recorded holographic plate is known as hologram.
10. The hologram does not contain a distinct image of the object. it
contains information in the form ofinterference pattern.
11. Fig. 5 shows the method of recording an image on a holographi('
plate. .
Coherent laser'
light V Object beam
Reference
beam
Mirror
Beam splitter M 1
Fig. 5. Recording of hologram.
B. Applications:
1. The 3-D images produced by holograms have been used in variou:
fields, such as technical, educational also in advertising, artistic
display etc.
2. In hospitals, holography can be used to view the working of inner
organs three dimensionally.
3. Holographic interferometry is used in non-destructive testing "j
materials to find flaws in structural parts and minute distortiou:,
due to stress or vibrations, etc. iIi the objects.
4. Holography is used in information coding.
5. Many museums have made holograms of valuable articles in thell'
collections.
©©©
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Physics SP-13A (Sem-I & 2)
B.Tech.
EXAMINATION, 2015·16
ENG~ERINGPHYSICS·I
Tilne : 3 Hours Max. Marks: 100
SECTION-A
1. Attempt. all parts. All sections carry equal marks. Write answer of
each part. in short: (2 )( 10 = 20)
a.How the negative results of Micht>Json-MorJey experiment
iJlterpretl~d?
Ans. Retic')" Q. 1.20, Page SQ-3A, 2 Marks Questions, Unit-to
b. Find relativistic relation between energy and momentum.
Ans. R('f(' I' Q. 1.17. Page SQ-:3A. 2 Mnrks Questions, Unit-I.
c. If uncertainty in the position of a particle is equal to de
Broglie wavelength, what will be uncertainty in the
measurement of velocity?
Arii;l.
Given: Uncertainty in the position of a particle = de-Broglie
wavelength .
.oX=A=!!:.. ,..(1)
p
To Find: Uncertaintyinthevelocit)';.
1. According to Heisenberg's uncertainty principle,
~ h
,.':. l~P ~ -
21t
...(2)
., J....I " ]-", dw uncertainty in velocity,
~p =c. 111,v ..i :3)
,:lin~: c·q. ill andeq. (3)ineq. (2)
Ii
p
. m L v == h
:=..
2rr
f·: po=mvj
Sf'. LUlcel·t.ainLy in velocity,
0.V ==-'!....
21t
Solved Paper (2015-161
SP-l4A (Sem-l & 2)
-----------------------------_.__._-----------
d. Write the characteristics of wave function.
aum> Refer Q. 3.13, Page SQ-10A, 2 Marks Questions. LJnit-;~.
eo Why the center of Newton's ring is dark?
:IiiHO Refer Q. 4.19, Page SQ-14A, 2 Marks Questions, lJnit-4_
f. Define plane of polarization and plane of vibration.
~ Plane of polarization: The plane containing tlw direction of
propagation of the light, but containing no vibrations il" called the
plane of polarization.
Plane ofVibration : The plane containing the direction ofvibration
and direction of propagation oflight is called the pI ane of vi hrat ion
g. Define optic a#s of doubly refracting crystal.
~ A certain direction in a doubly refracting crystal along which the
speed oflightoftwo refracted light l'ays remains the sarne. is known
as optic axis of that doubly refracting crystal.
h. What is Rayleigh's criterion of resolution?
AYlii; Refer Q. 4.16, Page SQ-13A, 2 Marks Questions, U 11 i 1-1
i. Define metastable state.
~ ReferQ.5.17, PageSQ-16A,2 Marks Questi()[l";.( ,',l
j. Give few important applications of optical fih,·,'.
.A'if§;; Refer Q. 5.18, Page SQ-17A, 2 Marks ql1l'.-; inn,: [ •i~"
SECTION-B
Note: Attempt any five questions: (] 0 " 5 =50)
2. What do you mean by proper length "J),> I ;";' the expression
for relativistic length. Calculate tlw . , (. tage contraction
ofa rod moving with a velocityof(l.f .., direction inclined
at 30° to its own length.
3. Show that the relativistic inv';, .<tHce of the law of
conservation ofmomentum leads tn; 'lE' concept of variation
of mass with velocity.
~ Refer Q. 1.23, Page 1-24A, 'I. 'nit-to
4. State Heisenberg's un(~eI'tainty princi pie. Prove that
electron cannot exist inside the nucleus and proton can
exist.
~
A. Heisenberg's Uncertainty Principle: Refer Q 41 bl.
Page SP-9A, Solved Paper 2014-15.
B. Non-existence of Electrons in the Nucleus:
1. We know that the radius of nucleus is the order of H)-I,! m.
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SP-15A (Sem-l & 2)
Physics
2, Ifan electron is confined within nucleus the uncertainty position of
electron is
6x = 2 x 10-14 m
3, Now according to uncertainty principle,
h
6xt:>.p'? .
21t
34
t:>.p = _h_ = 6.63 x 10
and
21t6x 2 x 1t x 2 x 10-14
6x
Fig. 1.
=5.276 )( 10-21 kg m/s
4, lYsing relativistic formula for the energy of the electron
Eo! = p~ c2 + IIIv '2. c4
'J, r" the l'e"t "lll'J'gy Ill C2 of an electron is of the order of0.511 MeV,
v
which is much smaller than the value of first term. Hence the
~eeond term is neglected therefore,
E2 = p2 C2
E = pc = (5.276 x 10-21 ) x (3 x 108 ) J
-" 1 !l
E= 5.276x10· x3x10 eV"'97MeV
1.6 x 10-19
6, Thus. ifan electron exists inside the nucleus then its energy should
ill' of t}1(' oHlel' of 97 11eV. But the experiment shows that no
l'1l'c! mil ill 1he atom possesses kinetic energy greater than 4 MeV.
7. IlenL'l'. 110 pI.'dron can exist inside the nucleus.
C. Existence of Proton in Nucleus:
1. We know that the radius of an atom is ofthe order oflO-14
m. Thus,
ifa proton exists inside the nucleus then the maximum uncertainty
in its position is given by .
I "x)max = 2 x 10- 14
m
., ',~jllg the uncPltaiI;lty relation,
I pll.rJ = 11
I. Ill' 11!inil1111111 ullcertainty in the momentum of proton is given by
h 1.055 x 10- 3.
I.p I = .._-- = = 5.275 x 10- 21 kg-m/s
11'" lux)m", 2x10 14
" Sll'l' Hw minimum uncertainty in the momentum of a proton
,dWllld lw l'qual to its 111011lpntUlU. i.l!,
SP-16A (Sem-l & 2) Solved Paper (201 [i. Hi :
p = (t:>.p )min = 5.275 )( 10- 21 kg-m/s
4. The corresponding energy of the proton is given by
p2 (5.275xlO-21)2
E= -= J
2m 2 x 1.6 X 10-27
i.e. E = (5.275 x 10- 21 )2 eV
, 2 x 1.6 X 10-27 x 1.6 X 10.- 19
i.e., E = 52 keY
5. Thus, if a proton exists inside the nucleus then its energy ShOll!' I I
of the order of 52 keY. The experiment shows that a Pl'Otoll in 1 I"
atom possesses kinetic energy of the order of 52 kpV. Henl''
proton can exist inside the nucleus.
5. Explain the physical significance of wave function. Ded'(·
Schrodinger's time independent wave equation.
AmiO Physical Significanc~ of Wave Function: Refer Concept
Outline-2, Page 3-11A, Unit-3.
Schrodinger's Time Independent Wave Equation: kl'i'<,r
Q. 3.13, Page 3-11A, Unit-3.
6. Explain the formation of Newton's ring? If in a N«'wtG.~l
ring experiment, the air in the interspaces is replaced b~, ,
liquid of refractive index 1.33, in what proportion wouit~
the diameter of the rings changed?
A1'd': Refer Q. 4.13, Page 4-20A, Unit-4.
7. Discuss the phenomenon of diffraction at a single slit and
show that intensities of successive maxima are:
4 4' 4
1 ' - ' - - ' -
• 91t2 • 151t2 • 491t 2
~ Refer Q. 4.19, Page 4-25A, Unit-4,
8. Discuss the construction and working of a He-Ne lasi' 1'.
Compare it with ruby laser.
2&ml8 Refer Q. 5.23, Page 5-25A, Unit-5.
9. Describe the basic principle of communication of wave in
optical fibre. A step index fibre has core refractive indp),
1.468, claddingrefractive index 1.462. Compute the maxim II 11:
radius allowed for a fibre, ifit supported only one mod", ai ,;
wavelength 1300 nm.
~ Refer Q. 5.9, Page 5-12A, Unit-5.
SECTION-C
Note: Attempt any two questions from this section: (15x2=:30i
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Physics SP-17A (Sem-I & 2)
10. a. Derive the Galilean transformation equations and show
that its acceleration components are invariant.
ADs. Refer Q. 1.2, Page 1-3A, Unit-I.
b. If the kinetic energy of a body is twice its rest mass energy,
find its velocity.
AnB'; Refer Q. 1.32, Page 1-32A, Unit-I.
c. Explain de-Broglie's hypothesis. Discuss the outcome of
Davisson-Germer's experiInent in detail.
Ans.
A. de-Broglie's Hypothesis: Refer Q. 3.7, Page 3-7A, Unit-3.
B. Outcomes of Davisson-Germer's Experiment:
1. Davisson and Germer calculated the de-Broglie wavelength using
two different approaches. .
2. In the first approach, Davisson and Germer used de-Broglie's
hypothesis.
:3. They plotted the variation in the intensity ofelectron beam against
scattering angle for different accelerating voltages to study the
effect of increasing electron energy on the scattering angle <p.
4. They found tha.t a bump begins to appear in the curve for V::: 44
volts.
5. With increasing potential, the bump moves upward, and becomes
more prominunt in the curve for V =54 volts at <p =50°, thereby
indicating the maximum suffering in electron beam for V = 54 volts
as shown in Fig. 2.
lU'V
....IUSVI~54.v/.,IUov
<p •••••- <p - 5~•••••
~.. ~-_ ....-.. .. ...........
Fig. 2. Plots of intensity of electron beam against scattering
angle for different values of accelerating voltage.
6. Thus, for V::: 54 V, the de-Broglie wavelength of the electrons is
A. = 12~4::: 12~4 = 1.66 A ...(1)
" "V ,,54
7. In the second approach, Davisson and Germer calculated the
de-Broglie wavelength by treating the electron beam as a wave.
8. They used Bragg's equation, n").. = 2d sin e.
9. For nickel crystal, d = 0.91 A. Also, e '" 65°. Hence for the first order
(n = 1) reflection, we have
Solved Paper 12015-16)
SP-18A (Sem-I & 2)
t.. =2d sin e::: 2 x 0.91 x sin 65° =1.65 A · ( 21
Eq. (1) and eq. (2) show an excellent agreement between the two
10.
approaches.
Thus, the Davisson-Germer experiment provides a direct
11.
verification of wave nature of electrons and hence it also verifies
the de Broglie's hypothesis.
II. a. Explain the phenomenon of interference in thin film due to
reflected rays.
b. A diffraction grating used at normal incidence gives a
yellow line (A:' = 6000 A> in a certain spectral order
superimposed on a blue line 0.. = 4800 A) of next higher order.
If the angle of diffraction is sin-1
(3/4), calculate the grating
element.
D§.: Refer Q. 4.28, Page 4-40A, Unit-4.
c. Describe the construction and working of Nicol prism.
~ Refer Q. 6(b), Page SP-4A, Solved Paper 2013-14.
Prove that vfc x. v = c 2 , where v p = phase velocity and
12. a. g
vII = groupve OClty.
Refer Q. 4(a), Page SP-7A, Solved Paper 2014-15.
A'AO
b. Discuss the different types of optical fiber in detail.
19
c. In a Ruby Laser, total number of Cr+3
is 2.8 x 10 • If the
Laser emits radiation of wavelength 7000 A, calculate the
energy of the Laser pulse.
..... Refer Q. 5.26, Page 5-28A, Unit-5.
Physi~al Constants :
In = 9.1 X 10-81
kg
Mass of electron o
In = 1.67 X 10-27
kg
Mass of proton p
c = 3 X 108
mls
Speed of light
h = 6.63 X 10-34
J/s
Planck's constant
e = 1.67 x 10-19
C
Charge of electron
k =1.38 X 10-23 m 2 kg S-2 k- I
Boltzmann's constant
@@@
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Physics SP-19 A (Sem-l & 2)
B.Tech.
EXAMINATION, 2016-17
ENGINEERING PHYSICS-I
Time: 3 Hours Total Marks: 70
Note: A, B and C are three sections in this question paper. Attempt
all seven parts from section A, any three parts from section
B and all questions from section C.
Section-A
1. Attempt all parts ofthis section. (2 x 7 =14)
a. What is proper length of a,rod ?
Ans. Hefer Q. 1.21, Page SQ-3A, 2 Marks Questions, Unit-I.
b. Explain the concept of rest mass of photon.
Ans. TIl<' re;.;t mass is the mass ofa particle (photon) as measured by an
obs,'rvl'r who sees the particle still and with zero speed. In other
words. thl' particle is at rest as far as this observer is concerned.
'I'll lIS comes the term "rest mass". But according to special relativity,
Jight alwa.v,; t ravels with the light speed c, and is never at rest and
so particle has zero rest mass.
c. What is Wien's law?
Ans. Hefl"" Q. :3.4, Page SQ-8A, 2 Marks Q",estions, Unit-3.
d. Explain the factor responsible for changing fringe width in
wedge shaped film. .
Ans. Refer' Q. 4.3, Page SQ-11A, 2 Marks Questions, Unit-4.
e. What happens to diffraction pattern when slit width of
single slit experiment increases?
Ans. Heier Q. 4.11, Page SQ-12A, 2 Marks Questions, Unit-4.
f. What are metastable states?
Ans. Refer Q. 5.17, Page SQ-16A, 2 Marks Questions, U nit-5.
g. What precautions are needed to minimize material
dispersion?
Ans. Refer Q. 5.24, Page SQ-17A, 2 Marks Questions, Unit-5.
Section - B
2. Attempt any three parts: (7 x 3 =21)
a. Describe Michelson - Morley experiment and explain the
outcome of the experim.ent.
ADS. Refer Q. 1.4, Page 1-5A, Unit-I.
b. Derive time independent Schrodinger wave equation and
give physical interpretation of wave function. Also explain
eigen value and eigen function.
SP-20A (Sem-l & 2) Solved Paper (2016-1'1 I
.Dl1&
A. Time Independent Schrodinger Wave Equation :
B. Physical Interpretation of Wave Function: Refer Concept
Outline: Part 2, Page 3-11A, Unit-3.
C. Eigen Value and Eigen FunctioD:
1. The values of energy E for which Schrodinger's steady-statr~
equation can be solved arencalled eigenvalues and the correspondin:.;
wave functions 'l'n are called eigen functions.
2. The discrete energy levels ofthe hydrogen atom are an example ,,:
a set of eigen values.
3. Eigen value equation
G'l'=G'l' ...0 )
n n n
Where G is the operator that corresponds to G and each Gn
is Ll
real number.
4. When eq. (1) holds for the wave function of a system, it i:; :j
fundamental postulate of quantum mechanics that c1n
measurement of G can only yield one of the values of G .
5. If measurements of G are made on a number of identic~l systen,:
all in states described by the particular eigen function IF", C<Jell
measurement will yield the single value Gk
•
c.What do you understand by Newton's ring? Explain H~e'!.~'
experimental arrangement. How can you determine the
wavelength of light with this experiment?
mm;
A. Newton's Ring and their Experimental Arrangement
B. Wavelength of Light: Refer Q. 4.15, Page 4-22A, Unit-4.
d. What is the concept of four level laser system.s ? Give i.b.,;.:: .
construction and working of He-Ne laser.
.um:
A. Four Level Laser System: Refer Q. 5.20, Page 5-21A, Unit-G
B. He-Ne Laser: Refer Q. 5.23, Page 5-25A, Unit-5.
eo What do you understand by modes of an optical fibre ')'
Discuss propagation of light in single mode, multimode and
graded index fibres.
A1mO
A. Modes of an Optical Fibre: Refer Q. 5.8, Page SQ-15A, 2 Marks
Questions, Unit-5.
B. Propagation of Light in Single Mode, Multimode and Graded
Index Fibres: Refer Q. 5.4, Page 5-6A, Unit-5.
Section- C
3. Attempt any two parts: (3.5 )( 2 := 7)
a. What do you mean by length contraction ? Explain it.
b. Deduce and discuss Einstein's mass-energy relation,
E =m.c2 .
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Physics SP-21 A (Sem.l & 2)
Ans. Refer Q. 1.24, Page 1-26A, Unit-I.
c. Calculate the percentage contraction of a rod moving with
a velocity of 0.8 c in a direction at 60° to its own length.
A;ltiit. Same as Q. 1.13, Page 1-15A, Unit·I. (Ans. = 8.34%)
4. Attempt any two parts: (3.5 )( 2 =7)
a. Describe energy distribution in black body radiation.
A:ntt; Refer Q. 3.1, Page 3-2A, Unit-3.
b. Explain the modified and unmodified radiations in Compton
scattering?
Ans; Refer Q. 3.24, Page 3-23A, Unit-3.
c. Calculate the wavelength of an electron associated with
kinetic energy of6.95 )( 10-25 Joules.
Ans. Same as Q. 3.10, Page 3-9A, Unit-3. (Ans. = 5.895 x 10 - 7 m)
5. Attempt any two parts: (3.5 )( 2 = 7)
a. Explain the missing orders in the spectra of a plane
transmission grating.
AnS. Refer Q. 4.24, Page 4-37A, Unit·4.
b. Explain Rayleigh criterion of resolution.
AnI<. Refer Q. 4.25, Page 4-38A, Unit-4.
c. A plane transmission grating has 15000 lines per inch. Find
the resolving power ofgrating and the smallest wavelength
difference that can be resolved with a light of wavelength
6000 A· in the second order.
Aus;
dA = _1.._:: 6000 X 10-
10
=0.2 A
3~~ 000 30,000
6. Attempt any two parts: (3.5 )( 2 = 7)
a. Show that the plane polarized and circularly polarized light
are the special cases of elliptically polarized light.
Ans~
1. Suppose that a plane polarized light ray ofamplitude A is incident
on a uniaxial crystal at an angle of 9 as shown in Fig. 1.
-0
1>.6 -
E::::"
1::-':,'1
x
Fig. 1. Double refraction in a umaxial crystal. .. 11
x = A cos e sin (wt + 6) == a sin «M + 6) ... (2)
[For E-ray) Y = A sin e sin rot == b sin (,)t
[For O-ray) y == A cos esin rot == a sin (J)t
Where, a:: A cos e and b = A sin 8
..(:~ I
3. From eq. (2), we have
sin wt =y/b ... (4)
Hence, cos rot =[1 - sin2wt)l/2 == [1 - (y/b )2]1/2
4. Now from eq. (1), we have
x = a sin (rot + 0) = a (sin rot cos 6 + cos (1)/ sin 0) .. .( 5)
5. Using eq. (3) and eq. (4) in eq. (5), and rearranging. we have
.(x/a)2 + (y/b)2 _ [(2xy cos o)/ab) = sin
2
1) ...(6)
Eq. (6) is the general eq. of an ellipse. Now we consider the following
three cases:
Case 1: .
1. When 6 =0, then from eq. (6), we have 2
(xla)2 + (y/b)2 _ [(2xy cos O)/ab) == sin 0
(x/a)2 + (y/b)2 - [(2xy)/abJ == 0
[(x/a) - (y/b))2 == 0
Y == (b/a) x
2. This is the eq:uation ofa straight line as shown in Fig. Hal.
In this case, the emergent light is plane polarized.
y
b
x
a
Fig. 1. (a.) Plane polarized light.
Case
1.
11:When 6 '" rcl2, then from eq. (6), we have 2
(x/a)2 + (ylb)2 _ [(2xy cos rcl2)J '" sin rcl2
(x/a)2 + (y/b)2 = 1
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---------------------------
SP-23 A (Sem-l & 2)
Physics
x2 y2
_+_ = 1
2
b2
a
2, This is the equation of a symmetrical ellipse as shown in
Fig. l(b). In this case, the emergent light is elliptically
polarized. .
y
x
Fig. 1. (bl Elliptically polarized light.
Case In:
1. When 8 =0 rrJ2 and a = b, then from eq. (6), we have
(x/a)2 +- (y/a)~- [(2xy cos rrJ2)a2j = sin2 rrJ2
. (x/a)2 + (y/a)2 = 1
x2 +y2 = a2
2. This is the equation of a circle as shown in Fig. 1(c). In this
case, the emergent light is circularly polarized.
y
x
Fig. 1. (cl Circularly polarized light.
(j From the above discussion, it is clear that the plane polarized light
and thl:' circularly polarized light are the special case ofelliptically
polarized light.
b. What are Einstein's coefficients? Obtain a relation between
them.
Arts. Refer Q. 5.18, Page 5-19A, Unit-5.
c. A certain length of 5 % solution causes the optical rotation
of 20°, How much length of 10 % solution of the same
substance will cause 35° rotation?
Arts.
Given: e =20° and e2 =35° ",."" .
1 . .... ...•,. ".
To Find: Length oflO'%sal~~ij.~,} .. " '. "',';
1. Let length of 5 % solution = ll' and
Length of 10 % solution = l2
2. Since substance is same,
SP-24 A (Sem-l & 2) Solved Paper (2016-17)
. . Specific rotation S = ~ = ~
l"C1 l2C2
20° _ ~
l" x 5% - l2 X 10 %
l2 =7/8l1
7. Attempt any two parts: (3.5 x 2 ='7)
a. Describe different types of losses in optical fibre.
D'i& Refer Q. 5.10, Page 5-12A, Unit-5.
b. Explain the construction and reconstruction of image iil
holography.
iiIDiI
A. Construction: Refer Q. 7(b), Page SP-1IA, Solved Paper 2014-15.
B. Reconstruction :
1. AB shown in Fig. 2, the hologram is exposed to the laser beam from
one side and it can be viewed from the other side.
".;;:
K: •
~Firstorder
Laser beam
LaserI ... ~ Zero order
' ...
Virtual image U Real image
Hologram
Fig. 2. Image reconstruction.
2. This beam is known as reconstruction beam.
3. The reconstruction beam illuminates the hologram at the ';:l)11,
angle as the reference beam.
4. The hologram acts as a diffraction grating, so constrl1ctll('
interference takes place in some directions and destructive
interference takes place in other direction.
5. A real image is formed in front ofthe hologram and a virtual image
is formed behind the hologram.
6. It is identical to the object and hence it appears as if the object is
present. The 3-D effect in the image can be seen by moving the
head of the observer.
7. During recording, the secondary waves from every point of the
object reach complete plate.
8. So, each bit ofthe plate contains complete information ofthe object.
9. Hence, image can be constructed using a small piece of hologram
c. Calculate the acceptance angle and numerical aperture of
the optical fibre ifthe refra,ctive index ofcore and cladding
are 1.50 and 1.45 respectively.
AmI: Refer Q. 5.7, Page 5-11A, Unit-5.
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Physics SP-25A (Sem.1 & 2)
B. Tech.
EXAMINATION, 2017-18
ENGINEERING PHYSICS-I
Time: 3 Hours
Attempt all sections. T~ Max:. Marks: '10
Note: n require any missing data : then choose
suitably.
Section-A
1. Attempt all questions in brief. (2 )( 7 =14)
a. Is earth an inertial or non-inertial frame of reference ?
Justify your answer.
~ Refer Q. 1.19, Page SQ-3A, 2 Marks Questions, Unit-!.
b. What is Wien's displacement law?
AD:S. As the temperature of the body is raised the maximum energy
tends to be associated with shorter wavelength, i.e.,
A",T= Constant
c. What do you mean by group velocity?
ADS. The velocity with which the wave packet obtained by superposition
ofwave travelling in group is called group velocity
~O)
v =
g tJ{
d. Define dispersive power ofa plane transmission diffraction
grating.
AJil:§. Refer Q. 4.14, SQ-13A, 2 Marks Questions, Unit-4.
e. Differentiate between spontaneous and stimulated emission
of radiation.
An&; Refer Q. 5.25, Page SQ-17A, 2 Marks Questions, Unit-5.
f. What do you mean by specific rotation?
Aii~ The specific rotation of an optically active substance at a given
temperature for a given wavelength oflight is defined as the rotation
(in degrees) produced by a path of one decimeter length in a
substance ofunit density.
g. What do you mean by acceptance angle?
Ans' Refer Q. 5.7, Page SQ-15A, 2 Marks Questions, Unit-5.
Section-B
2. Attempt any three parts of the following: (7 )( 3 = 21)
a. Obtain Galilean transformation equations. Show that
length and acceleration are invariant under Galilean
transformations.
Solved Paper (2017-18
SP-26 A (Sem-l & 2)
Galilean Transformation Equation and Acceleration
...A. Component are Invariant: Refer Q. 1.2, Page 1-3A, Unit-!.
Length Component are Invariant: Refer Q. 1.3, Page 1-5A.
B.
Unit-I.
Derive Planck's radiation law. Show that Planck's formula
b. for the energy distribution in a thermal spectrum is
applicable for all wavelengths.
.~ Give the construction and theory of plane transmission
c. grating. Explain the formation of spectra by it.
1CJi'lO What is the advantage offour level laser systems over t.hree
do level laser systems? Describe the construction and working
of ruby laser.
~ What is holography? Explain the basic principle of holographY
e. using construction and reconstruction of image.
. Holography: It is a method ofproducing a three dimensional image ()f
~"
A. an object employing the coherence properties of a lR.ser lx'am.
Construction and Reconstruction of Image: Refer Q. 71 a).
B. Page SP-24A, Solved Paper 2016-17.
Section-C
Attempt anyone of the following: <7 x 1 = 7)
3. Deduce the relativistic velocity addition theorem. ShoW that
a. it is consistent with Einstein's second postulate.
~ What do you mean by time dilation? Establish a relation
b. f"r it. At what speed should a clock be moved so that it. may
appear to lose 1 min each hour?
mfg. Time Dilation: Refer Q. 1.16, Page 1-17A, Un1t-l
A. Numerical: Refer Q. 1.19, Page 1-19A, Unit-l
B. Attempt anyone part of the following: <7 xl = 7
4- What is the concept of de-Broglie matter waves? Describe
a. Davisson-Germer experiment and prove that electrons
possess wave nature.
De.Brogli Matter Waves: Refer Q. 3.7, Page :3-7 A. {Tnit·:1
D1h e
A. Davisson.Germer Experiment: Refer Q. lOl cl. Page S1'-17 A.
B.
Solved Paper 2015-16.
Find an expression for the energy states of a particle in a
b. one_dimensional box. Determine the probabilitY of finding
a particle trapped in a boX of length L in the region from
0.45 L to 0.55 L for the ground state.
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---------------------------
Physics SP-27A (Sem-1 & 2)
Ans.
A Expression for the Energy States of a Particle in a
One-dimensional Box: Refer Q. 3.17, Page 3-16A, Unit-3.
B. Numerical: Refer Q. 3.22, Page 2-20A, Unit-3.
5. Attempt anyone part of the following: (7)( 1 .. 7)
a. Discuss the fOrIIlatioll ofinterference fringes due toa wedge
shaped thin film seen by normally reflected monochromatic
light and obtain an expression for the fringe width.
Ans. H"f'l'r Q. -1,4, Page 4-9A, Unit-4.
Ii
b. Obtain an expression for the intensity distribution due to
Fraunhofer diffraction at a single slit. A light of wavelength
6000 A falls normally on a slit of width 0.10 mm. Calculate
the total angular width of the central maximum.
Ans.
A Fraunhofer Diffraction at Single Slit: Refer Q. 4.19,
Page 4-25A, Unit-4.
B. Numerical: Refer Q. 4.20, Page 4-29A, Unit-4.
6. A-ttempt anyone part of the following: (7)( 1 = 7)
a. Explain the phenomenon of double refraction and discuss
the various characteristics of ordinary and ex.traordinary
l'ays. ·Find the thickness of a quarter wave plate of quartz
for light of wavelength 5893 A. The refractive indices for
ordinary and extraordinary rays are 1.544 and 1.553
respectively.
Ans.
A. Phenomenon of Double Refraction : Refer Q. 6(b),
l'HgP SP-·1A, Solved Paper" 2013-14.
B. Characteristics of Ordinary Ray:
I. Ordinary ray (I'm'cls with the same velocity in all directions of the
l'J','stal.
2. '"jocity ora-my is f'ame along different directions of the crystal.
.'J. ()nlinary ray ol)('ys t J](, law of reflection.
,l, Vi brationul vectol'S of a-my are perp,endicular to the optic axis.
C. Characteristics of Extra Ordinary Ray:
1. Extraordinary ray travels with different velocities along different
dil'prtions orthe crystal.
2. Ve locity ofE-ray is different along different directions ofthe crystal.
:3. Vi brational vectors of E-ray are in the principal plane.
D. Nuinerical:
Given: 110 =1.544, 1-1, =1.553 and for sodium light
/.. =5893 A=5.893 X 10-5
cm
To Find: Thickness of quarter w.ave plate.
1. The thickness ofa quarter wave plate for positive crystal like quartz is
I.
t = .----
4(p,. - Po)
Solved Paper (20}'l.. ":
SP-28A (Sem-l & 2)
5
t ::: 5.893 x 10-
5
::: 5.893 x 10- ::: 1.624 x 10-:3 cm
4(1.553 - 1.544) 4 x 9 x 10-3
b. What do you mean by optical activity? Give Fresnel's theory
of optical activity and derive the necessary expression foll'"
the optical rotation.
~
Optical Activity:
A.
When plane-polarized light passes through certain substance>'. 1:,·,
1.
plane ofpolarization (or plane ofvibration) oflight is rotated abo i.:'
the direction of propagation oflight through a certain angle,
2. This property is known as optical activity and thesesubstances are said t<)
be optically active and this phenomenonis calledoptical rotation.
B. Fresnel's Theory of Optical Activity:
When beam of plane-polarized light incident on an optically active
1.
substance along its optic axis it splits into two oppositely directed
circular motion, one is in clockwise direction, while another is in
anticlockwise direction.
The velocity oftwo circularly polarized beams is different fol' opticalJ,
2.
active substance, and same for optically inactive substanc",
Due to different velocities the phase difference occurs between thc~n
3.
In dextrorotatory substance, the velocity of right handed compcj'lC' .
4.
greater than left handed component v R > vL •
5. In laevorotatory substance, the velocity ofvL > vR
On emergence from the substance, these two circular motiol:
6.
recombine to produce a plane polarized light.
C. Mathematical Explanation :
Let a plane polarized beam be incident normally on cI c!("i.'[';'
1.
refracting crystal like quartz plate with its faces perpl'nclicub r
the optic axis.
These beam divided into two parts, clockwise and anticlockw,"'
2.
directions, in circular motion.
ThecircuIt'll" motions travellingalongthe opticaxis have different ,'duCIc.,.'::
3.
A
B
Optic axis
:.,::'. L
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Physics SP-.29A (Sem·l &: 2)
4. Therefore a phase difference (Ii) occurs between them.
5. Now considering the case ofdextrorotatory (v > v L
).
n
AB - Optic axis.
OP and OQ - Two circular motions rotating in opposite directions.
6. At the time of emergence these vibration are represented as
XI = a cos (rot + 1i)'YI = a sin (rot + Ii) clockwise.
x 2 =- a cos (rot)'Y2 =a sin (rot) anticlockwise.
7. By superposition theorem the resultant ofx andy component are
X =Xl + x 2 =a cos (rot + Ii) - a cos rot
X = 2asin%.sin(rot + %) ...(1)
and Y = Y I +Y 2 = a sin (rot + Ii) + a sin rot
Y = 2acos%.sin(rot + %) ...(2)
8. The resultingvibrationsalongX-andY-axisare insameperiodandphase.
9. Now dividing eq. (1) by eq. (2), we get
X Ii Ii
- = tan- or X == Ytan-
Y 2 2
This equation is ofa straight line having slope tan 612 with Y-axis.
10. Iff.ln and f.lL are refractive indices ofright and left handed light and
t is thickness ofcrystal,
path difference =(f.lL - f.ln)t
and phase difference == Ii == 21t"(I-lL - I-ln)t
"
and angle ofrotation 29 = Ii
27t
29 = T(I-lL -I-ln)t
7t
() == -(f.1L - f.ln)t
"
7. Attempt anyone part of the following: (7 )( 1 = 7)
a. Explain single mode and multimode fibres. Differentiate
between characteristic properties of single mode and
multim.ode fibres.
ltIiS~
A. Single Mode and Multimode Fibres: Refer Q. 5.4, Page 5-6A,
Unit-5.
B. Difference between Single Mode and Multimode Fibres :
b. Explain dispersion and attenuation in optical fibre. The
optical powe~~after propagatingthrougb a 500 m long fibre.
reduced to 25 % of its original value. Calculate fibre loss in
dBlkm.
;tn::s: Refer Q. 5.14, Page 5-15A, Unit-5.
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