01 ray-optics-mm

genius PHYSICS
A collection by Pradeep Kshetrapal for Physics students at genius, Maxwell and Gupta classes Only
Reflection of Light 1
When a ray of light after incidenting on a boundary separating two media comes back into the
same media, then this phenomenon, is called reflection of light.
Note : After reflection velocity, wavelength and frequency of light remains same but
intensity decreases.
If light ray incident normally on a surface, after reflection it retraces the path.
Real and virtual images
If light rays, after reflection or refraction, actually meets at a point then real image is formed
and if they appears to meet virtual image is formed.
Plane Mirror.
The image formed by a plane mirror is virtual, erect, laterally inverted, equal in size that of
the object and at a distance equal to the distance of the object in front of the mirror.
⇒ ∠i = ∠r
⇒ After reflection, velocity, wave length and
frequency of light remains same but intensity
decreases
⇒ There is a phase change of π if reflection takes
place from denser medium
Boundary
Reflected ray
Normal
i r
Incident
ray
x x
IO
(Real object) (Virtual image)
O
I
(Real image) (Virtual object)
Real image
(Real image)
(Virtual object)
I
O
(Virtual image)
(Real object)
(Virtual image)
O I

genius PHYSICS
2 Reflection of Light
(1) Deviation : Deviation produced by a plane mirror and by two inclined plane mirrors.
Note : If two plane mirrors are inclined to each other at 90o, the emergent ray is anti-
parallel to incident ray, if it suffers one reflection from each. Whatever be the angle
to incidence.
(2) Rotation : If a plane mirror is rotated in the plane of incidence through angle θ, by
keeping the incident ray fixed, the reflected ray turned through an angle 2θ.
(3) Images by two inclined plane mirrors : When two plane mirrors are inclined to each
other at an angle θ, then number of images (n) formed of an object which is kept between them.
(i) 





−= 1
360
θ
n ; If =
θ
360
even integer
(ii) If =
θ
360
odd integer then there are two possibilities
(a) Object is placed symmetrically (b) Object is placed asymmetrically






−= 1
360
θ
n
θ
360
=n
Note : If θ = 0o i.e. mirrors are parallel to each other so ∞=n i.e. infinite images will be
formed.
If θ = 90o, 31
90
360
=−=n
If θ = 72o, 41
72
360
=−=n (If nothing is said object is supposed to be symmetrically
placed).
(4) Other important informations
(i) When the object moves with speed u towards (or away) from the plane mirror then image
also moves toward (or away) with speed u. But relative speed of image w.r.t. object is 2u.
i r
δ
δ = (180 – 2i) δ = (360 – 2θ)
θ
δ
δ
Final path
Original
path
2θ
θ
θ
θ
IR
RR
IR RR
θ/2
θ/2
Object
α
β
Object

genius PHYSICS
(ii) When mirror moves towards the stationary object with speed u, the image will move with
speed 2u.
(iii) A man of height h requires a mirror of length at least equal to h/2, to see his own
complete image.
(iv) To see complete wall behind himself a person requires a plane mirror of at least one third
the height of wall. It should be noted that person is standing in the middle of the room.
Example: 1 A plane mirror makes an angle of 30o with horizontal. If a vertical ray strikes the mirror,
find the angle between mirror and reflected ray
(a) 30o (b) 45o (c) 60o (d) 90o
Solution : (c) Since angle between mirror and normal is 90o and reflected ray (RR) makes
an angle of 30o with the normal so required angle will be o
60=θ .
Concepts
The reflection from a denser medium causes an additional phase change of π or path change of λ/2 while reflection from
rarer medium doesn't cause any phase change.
We observe number of images in a thick plane mirror, out of them only second is brightest.
To find the location of an object from an inclined plane mirror, you have to see the perpendicular distance of the object from the
mirror.
Example
s
30o
30o
30o
θ = 60o
IR
RR
H
E
L
M '
M'
h
2
h
H
E
B
M '
M'
h
3
h
d d
E
2u
O I
Rest
u
Mirror is
moving
O I
Mirror at rest
u u
Incident light
(100%)
10%
80%
9%
0.9%
Brightest image
M
O
I
Correct
OM = MI
M
O I
Wrong
OM = MI

genius PHYSICS
Example: 2 Two vertical plane mirrors are inclined at an angle of 60o with each other. A ray of light
travelling horizontally is reflected first from one mirror and then from the other. The
resultant deviation is
(a) 60o (b) 120o (c) 180o (d) 240o
Solution : (d) By using )2360( θδ −= ⇒ o
240602360 =×−=δ
Example: 3 A person is in a room whose ceiling and two adjacent walls are mirrors. How many images
are formed
[AFMC 2002]
(a) 5 (b) 6 (c) 7 (d) 8
Solution : (c) The walls will act as two mirrors inclined to each other at 90o and so sill form 31
90
360
=−
images of the person. Now these images with object (Person) will act as objects for the
ceiling mirror and so ceiling will form 4 images as shown. Therefore total number of images
formed = 3 + 4 = 7
Note : The person will see only six images of himself ),,,,,( '
3
'
2
'
1321 IIIIII
Example: 4 A ray of light makes an angle of 10o with the horizontal above it and strikes a plane mirror
which is inclined at an angle θ to the horizontal. The angle θ for which the reflected ray
becomes vertical is
(a) 40o (b) 50o (c) 80o (d) 100o
Solution : (a) From figure
9010 =++ θθ
o
40=⇒ θ
Example: 5 A ray of light incident on the first mirror parallel to the second and is reflected from the
second mirror parallel to first mirror. The angle between two mirrors is
(a) 30o (b) 60o (c) 75o (d) 90o
Solution : (b) From geometry of figure
o
180=++ θθθ
o
60=⇒ θ
Example: 6 A point object is placed mid-way between two plane mirrors distance 'a' apart. The plane
mirror forms an infinite number of images due to multiple reflection. The distance between
the nth order image formed in the two mirrors is
(a) na (b) 2na (c) na/2 (d) n2 a
Solution : (b)
I1
I2 I3
O
Three images by walls
I′1
I′2 I′3
O′
Four images by
ceiling
10o
θ
θ
θ
IR
Vertical RR
Horizontal line
Plane mirror
θ θ
θ
θ
θ

genius PHYSICS
From above figure it can be proved that seperation between nth order image formed in the
two mirrors = 2na
Example: 7 Two plane mirrors P and Q are aligned parallel to each other, as shown in the figure. A light
ray is incident at an angle of θ at a point just inside one end of A. The plane of incidence
coincides with the plane of the figure. The maximum number of times the ray undergoes
reflections (including the first one) before it emerges out is
(a)
θtand
l
(b)
θtanl
d
(c) θtanld
(d) None of these
Solution : (a) Suppose n = Total number of reflection light ray undergoes before exist out.
x = Horizontal distance travelled by light ray in one reflection.
So nx = l also
d
x
=θtan
θtand
l
n =⇒
Example: 8 A plane mirror and a person are moving towards each other with same velocity v. Then the
velocity of the image is
(a) v (b) 2v (c) 3v (d) 4v
Solution : (c) If mirror would be at rest, then velocity of image should be 2v. but due to the motion of
mirror, velocity of image will be 2v + v = 3v.
Example: 9 A ray reflected successively from two plane mirrors inclined at a certain angle undergoes a
deviation of 300o. The number of images observable are
(a) 10 (b) 11 (c) 12 (d) 13
Solution : (b) By using θθδ 2360300)2360( −=⇒−=
o
30=⇒ θ . Hence number of images 111
30
360
=−=
A small plane mirror placed at the centre of a spherical screen of radius R. A beam of light is falling
on the mirror. If the mirror makes n revolution. per second, the speed of light on the screen after
reflection from the mirror will be
(a) 4πnR (b) 2πnR (c)
π2
nR
(d)
π4
nR
Solution : (a) When plane mirror rotates through an angle θ, the reflected ray rotates through an angle 2θ. So
spot on the screen will make 2n revolution per second
∴∴∴∴ Speed of light on screen nRRnRv ππω 4)2(2 ===
d
l
θ
I1' a/2 a/2
a
M' M
I order
image
II order
image
III
order
image
III
order
image
II order
image
I order
image
I1 I2 I3I2'I3'
a/2
3a/2
a/2
3a/2
5a/25a/2
O
d
l
θθ
x
Tricky example: 1Tricky example: 1
Tricky example: 2

genius PHYSICS
A watch shows time as 3 : 25 when seen through a mirror, time appeared will be
[RPMT 1997; JIPMER 2001, 2002]
(a) 8 : 35 (b) 9 : 35 (c) 7 : 35 (d) 8 : 25
Solution : (a) For solving this type of problems remember
Actual time = 11 : 60 – given time
So here Actual time = 11 : 60 – 3 : 25 = 8 : 35
When a plane mirror is placed horizontally on a level ground at a distance of 60 m from the foot of
a tower, the top of the tower and its image in the mirror subtend an angle of 90o at the eye. The
height of the tower will be [CPMT 1984]
(a) 30 m (b) 60 m (c) 90 m (d) 120 m
Solution : (b) Form the figure it is clear that oh
45tan
60
=
⇒ h = 60 m
Curved Mirror.
It is a part of a transparent hollow sphere whose one surface is polished.
(1) Some definitions :
(i) Pole (P) : Mid point of the mirror
(ii) Centre of curvature (C) : Centre of the sphere of which the mirror is a part.
(iii) Radius of curvature (R) : Distance between pole and centre of curvature.
(Rconcave = –ve , Rconvex = +ve , Rplane = ∞)
(iv) Principle axis : A line passing through P and C.
(v) Focus (F) : An image point on principle axis for which object is at ∞
(vi) Focal length (f) : Distance between P and F.
Principle axis
C
F
P
Converges the light rays Diverges the light rays
C
F
P
Tricky example: 3
45o
45o
Image
Tower
h
60 m

genius PHYSICS
(vii) Relation between f and R :
2
R
f = (fconcare = –ve , fconvex = + ve , fplane = ∞ )
(viii) Power : The converging or diverging ability of mirror
(ix) Aperture : Effective diameter of light reflecting area.
Intensityofimage∝Area∝(Aperture)2
(x) Focal plane : A plane passing from focus and perpendicular to principle
axis.
(2) Rules of image formation and sign convention :
Rule (i) Rule (ii) Rule (iii)
(3) Sign conventions :
(i) All distances are measured from the pole.
(ii) Distances measured in the direction of incident rays are taken
as positive while in the direction opposite of incident rays are taken
negative.
(iii) Distances above the principle axis are taken positive and
below the principle axis are taken negative.
Note : Same sign convention are also valid for lenses.
Use following sign while solving the problem :
Concave mirror
Convex mirror
Real image (u ≥≥≥≥ f) Virtual image (u< f)
Distance of object u → –
Distance of image v → –
Focal length f → –
Height of object O → +
Height of image I → –
Radius of curvature R → –
Magnification m → –
u → –
v → +
f → –
O → +
I → +
R → –
m → +
u → –
v → +
f → +
O → +
I → +
R → +
m → +
F F F F C C
+–
+
–Mirror or Lens
Incident ray
Principle
axis

genius PHYSICS
(4) Position, size and nature of image formed by the spherical mirror
Mirror Location of the
object
Location of the
image
Magnification,
Size of the
image
Nature
Real
virtual
Erect
inverted
(a) Concave At infinity
i.e. u = ∞
At focus i.e. v = f m << 1,
diminished
Real inverted
Away from centre
of curvature (u >
2f)
Between f and 2f
i.e.
f < v < 2f
m < 1, diminished Real inverted
At centre of
curvature u = 2f
At centre of
curvature i.e. v =
2f
m = 1, same size as
that of the object
Real inverted
Between centre of
curvature and
focus :
F < u < 2f
Away from the
centre of curvature
v > 2f
m > 1, magnified Real inverted
At focus i.e. u = f At infinity i.e. v =
∞
m = ∞, magnified Real inverted
Between pole and
focus u < f
v > u m > 1 magnified Virtual erect
(b) Convex At infinity i.e. u =
∞
At focus i.e., v = f m < 1, diminished Virtual erect
Anywhere between
infinity and pole
Between pole and
focus
m < 1, diminished Virtual erect
Note : In case of convex mirrors, as the object moves away from the mirror, the image
becomes smaller and moves closer to the focus.
Images formed by mirrors do not show chromatic aberration.
For convex mirror maximum image distance is it’s focal length.
In concave mirror, minimum distance between a real object and it's real image is
zero.
(i.e. when u = v = 2f)
Mirror formula and magnification.
For a spherical mirror if u = Distance of object from pole, v = distance of image from pole, f =
Focal length, R = Radius of curvature, O = Size of object, I = size of image, m = magnification (or
linear magnification ), ms = Areal magnification, oA = Area of object, iA = Area of image
(1) Mirror formula :
uvf
111
+= ; (use sign convention while solving the problems).
Note : Newton’s formula : If object distance (x1) and image distance (x2) are measured
from focus instead of pole then 21
2
xxf =
PFC
∞
CFP
∞

genius PHYSICS
(2) Magnification : m =
imageofSize
objectofSize
Linear magnification
Areal magnification
Transverse Longitudinal
When a object is placed
perpendicular to the
principle axis, then linear
magnification is called lateral
or transverse magnification.
It is given by
f
vf
uf
f
u
v
O
I
m
−
=
−
=−==
(* Always use sign
convention while solving the
problems)
When object lies along the
principle axis then its longitudinal
magnification
)(
)(
12
12
uu
vv
O
I
m
−
−−
==
If object is small;
2






=−=
u
v
du
dv
m
Also Length of
image =
)(objectofLength 0
2
L
u
v
×





oi L
fu
f
L .)(
2






−
=
If a 2D-object is placed with it's plane
perpendicular to principle axis
It's Areal magnification
)(objectofArea
)(imageofArea
o
i
s
A
A
M =
2
m
ab
mbma
=
×
=
o
i
s
A
A
mm ==⇒ 2
Note : Don't put the sign of quantity which is to be determined.
If a spherical mirror produces an image ‘m’ times the size of the object (m =
magnification) then u, v and f are given by the followings
u
m
m
ffmvf
m
m
u 





−
=−−=




 −
=
)1
and)1(,
1
(use sign convention)
(3) Uses of mirrors
(i) Concave mirror : Used as a shaving mirror, In search light, in cinema projector, in
telescope, by E.N.T. specialists etc.
(ii) Convex mirror : In road lamps, side mirror in vehicles etc.
Note : Field of view of convex mirror is more than that of concave mirror.
b
mb
a
ma

genius PHYSICS
Different graphs
Graph between
v
1
and
u
1
(a) Real image formed by
concave mirror
(b) Virtual image formed by
concave mirror
(c) Virtual image formed by
convex mirror
Graph between u and v
for real image of concave
mirror
Graph between u and m
for virtual image by
concave mirror
Graph between u and m
for virtual image by
convex mirror.
Concepts
Focal length of a mirror is independent of material of mirror, medium in which it is placed, wavelength of
incident light
Divergence or Convergence power of a mirror does not change with the change in medium.
If an object is moving at a speed vo towards a spherical mirror along it’s axis then speed of image away from
mirror is oi v
fu
f
v .
2






−
−= (use sign convention)
When object is moved from focus to infinity at constant speed, the image will move
faster in the beginning and slower later on, towards the mirror.
As every part of mirror forms a complete image, if a part of the mirror is obstructed,
full image will be formed but intensity will be reduced.
Can a convex mirror form real images?
yes if (distance of virtual object) u < f (focal length)
Example: 10 A convex mirror of focal length f forms an image which is 1/n times the object. The distance
of the object from the mirror is
v
1
u
1
P
F
C
∞
O
I
O
CF
I
Real
image
Virtual object
m
1
f u
m
1
u
2f
f
f 2f
Hyperbola
Example
s
v
1
u
1
v
1
u
1

genius PHYSICS
(a) (n – 1) f (b) f
n
n





 − 1
(c) f
n
n





 + 1
(d) (n + 1) f
Solution : (a) By using
uf
f
m
−
=
Here
n
m
1
+= , ff +→ So,
uf
f
n −+
+
=+
1
⇒ fnu )1( −−=
Example: 11 An object 5 cm tall is placed 1 m from a concave spherical mirror which has a radius of
curvature of 20 cm. The size of the image is
(a) 0.11 cm (b) 0.50 cm (c) 0.55 cm (d) 0.60 cm
Solution : (c) By using
uf
f
O
I
−
=
Here ,5 cmO + cm
R
f 10
2
−=−= , cmmu 1001 −=−=
So,
)100(10
10
5 −−−
−
=
+
I
⇒ I = – 0.55 cm.
Example: 12 An object of length 2.5 cm is placed at a distance of 1.5 f from a concave mirror where f is the
magnitude of the focal length of the mirror. The length of the object is perpendicular to the
principle axis. The length of the image is
(a) 5 cm, erect (b) 10 cm, erect (c) 15 cm, erect (d) 5 cm,
inverted
Solution : (d) By using
uf
f
O
I
−
= ; where I = ? , O = + 2.5 cm. ff −→ , u = – 1.5 f
∴
)5.1(5.2 ff
fI
−−−
−
=
+
⇒ .5 cmI −= (Negative sign indicates that image is
inverted.)
Example: 13 A convex mirror has a focal length f. A real object is placed at a distance f in front of it from
the pole produces an image at
(a) Infinity (b) f (c) f / 2 (d) 2f
Solution : (c) By using
uvf
111
+= ⇒
( ) 2
111 f
v
fvf
=⇒
−
+=
+
Example: 14 Two objects A and B when placed one after another infront of a concave mirror of focal
length 10 cm from images of same size. Size of object A is four times that of B. If object A is
placed at a distance of 50 cm from the mirror, what should be the distance of B from the
mirror
(a) 10 cm (b) 20 cm (c) 30 cm (d) 40 cm
Solution : (b) By using
A
B
A
B
B
A
uf
uf
O
O
I
I
uf
f
O
I
−
−
=×⇒
−
=
( )5010
10
4
1
1
1
−−−
−−
=×⇒ Bu
⇒ cmuB 20−= .
Example: 15 A square of side 3 cm is placed at a distance of 25 cm from a concave mirror of focal length
10 cm. The centre of the square is at the axis of the mirror and the plane is normal to the
axis. The area enclosed by the image of the wire is
(a) 4 cm2 (b) 6 cm2 (c) 16 cm2 (d) 36 cm2

genius PHYSICS
o
i
A
A
m =2
; where
uf
f
m
−
=
Hence from given values
( ) 3
2
2510
10 −
=
−−−
−
=m and 2
9 cmAo = ∴
2
2
49
3
2
cmAi =×




 −
=
Example: 16 A convex mirror of focal length 10 cm is placed in water. The refractive index of water is 4/3.
What will be the focal length of the mirror in water
(a) 10 cm (b) 40/3 cm (c) 30/4 cm (d) None of
these
Solution : (a) No change in focal length, because f depends only upon radius of curvature R.
Example: 17 A candle flame 3 cm is placed at distance of 3 m from a wall. How far from wall must a
concave mirror be placed in order that it may form an image of flame 9 cm high on the wall
(a) 225 cm (b) 300 cm (c) 450 cm (d) 650 cm
Solution : (c) Let the mirror be placed at a distance x from wall
By using
( )
( )33
9
−−
−−
=
+
−
⇒
−
=
x
x
u
v
O
I
.4505.4 cmmx −=−=⇒
Example: 18 A concave mirror of focal length 100 cm is used to obtain the image of the sun which
subtends an angle of 30'. The diameter of the image of the sun will be
(a) 1.74 cm (b) 0.87 cm (c) 0.435 cm (d) 100 cm
Solution : (b) Diameter of image of sun θfd =
18060
30
100
π
×





×=⇒ d
⇒ cmd 87.0= .
Example: 19 A thin rod of length f / 3 lies along the axis of a concave mirror of focal length f. One end of
its magnified image touches an end of the rod. The length of the image is [MP PET 1995]
(a) f (b) f
2
1
(c) 2 f (d) f
4
1
Solution : (b) If end A of rod acts an object for mirror then it's image will be A' and if
3
5
3
2
ff
fu =−=
So by using
uvf
111
+=
3
5
111
fvf −
+=
−
⇒ fv
2
5
−=⇒
∴ Length of image
2
2
2
5 f
ff =−=
Example: 20 A concave mirror is placed on a horizontal table with its axis directed vertically upwards. Let
O be the pole of the mirror and C its centre of curvature. A point object is placed at C. It has
a real image, also located at C. If the mirror is now filled with water, the image will be
(a) Real, and will remain at C (b) Real, and located at a point between
C and ∞
x
(x–
3)m
3cm
3m
Image of
sun
θ
d
F
o






==
60
30
'30θ
CA'
F
A
2f
f / 3
v
u = 2f – (f/3)

genius PHYSICS
(c) Virtual and located at a point between C and O (d) Real, and located at a point between
C and O
Solution : (d)
An object is placed infront of a convex mirror at a distance of 50 cm. A plane mirror is introduced
covering the lower half of the convex mirror. If the distance between the object and plane mirror is
30 cm, it is found that there is no parallel between the images formed by two mirrors. Radius of
curvature of mirror will be
(a) 12.5 cm (b) 25 cm (c) cm
3
50
(d) cm18
Solution : (b) Since there is no parallel, it means that both images (By plane mirror
and convex mirror) coinciding each other.
According to property of plane mirror it will form image at a distance of 30 cm behind it. Hence for
convex mirror u = – 50 cm, v = + 10 cm
By using
uvf
111
−=
50
4
50
1
10
11
=
−
+
+
=⇒
f
⇒ cmf
2
25
= .252 cmfR ==⇒
A convergent beam of light is incident on a convex mirror so as to converge to a
distance 12 cm from the pole of the mirror. An inverted image of the same size
is formed coincident with the virtual object. What is the focal length of the mirror
(a) 24 cm (b) 12 cm (c) 6 cm (d) 3 cm
Solution : (c) Here object and image are at the same position so this position must
be centre of curvature
∴ R = 12 cm
2
R
f =⇒
PPrraaccttiiccee QQuueessttiioonnss BBaassiicc LLeevveell
1. A light bulb is placed between two mirrors (plane) inclined at an angle of 60o. Number of images formed are
[NCERT 1980; CPMT 1996, 97; SCRA 1994; AIIMS 1997; RPMT 1999; AIEEE 2002; Orissa JEE 2003; MP PET 2004]
(a) 2 (b) 4 (c) 5 (d) 6
2. Two plane mirrors are inclined at an angle of o
72 . The number of images of a point object placed between them
will be
[KCET (Engg. & Med.)1999; BCECE 2003]
Tricky example: 4
A
Object
50 cm
30 cm 20 cm
10 cm
Tricky example: 5
C
Object
image
O
Initiall
y
C Object
O
Finally
Image
C

genius PHYSICS
(a) 2 (b) 3 (c) 4 (d) 5
3. To get three images of a single object, one should have two plane mirrors at an angle of [AIEEE 2003]
(a) o
30 (b) o
60 (c) o
90 (d) o
120
4. A man of length h requires a mirror of length at least equal to, to see his own complete image [MP PET 2003]
(a)
4
h
(b)
3
h
(c)
2
h
(d) h
5. Two plane mirrors are at 45o to each other. If an object is placed between them then the number of images will be [MP PMT 2003
(a) 5 (b) 9 (c) 7 (d) 8
6. An object is at a distance of 0.5 m in front of a plane mirror. Distance between the object and image is [CPMT 2002]
(a) 0.5 m (b) 1 m (c) 0.25 m (d) 1.5 m
7. A man runs towards a mirror at a speed 15 m/s. The speed of the image relative to the man is [RPMT 1999;
Kerala PET 2002]
(a) 15 1−
ms (b) 30 1−
ms (c) 35 1−
ms (d) 20 1−
ms
8. The light reflected by a plane mirror may form a real image [KCET (Engg. & Med.) 2002]
(a) If the rays incident on the mirror are diverging (b) If the rays incident on the mirror are
converging
(c) If the object is placed very close to the mirror (d) Under no circumstances
9. A man is 180 cm tall and his eyes are 10 cm below the top of his head. In order to see his entire height right from
toe to head, he uses a plane mirror kept at a distance of 1 m from him. The minimum length of the plane mirror
required is [MP PMT 1993; DPMT 2001]
(a) 180 cm (b) 90 cm (c) 85 cm (d) 170 cm
10. A small object is placed 10 cm infront of a plane mirror. If you stand behind the object 30 cm from the object and
look at its image, the distance focused for your eye will be
(a) 60 cm (b) 20 cm (c) 40 cm (d) 80 cm
11. Two plane mirrors are at right angles to each other. A man stands between them and combs his hair with his right
hand. In how many of the images will he be seen using his right hand
(a) None (b) 1 (c) 2 (d) 3
12. A man runs towards mirror at a speed of 15 m/s. What is the speed of his image [CBSE PMT 2000]
(a) 7.5 m/s (b) 15 m/s (c) 30 m/s (d) 45 m/s
13. A ray of light is incidenting normally on a plane mirror. The angle of reflection will be [MP PET 2000]
(a) 0o (b) 90o (c) Will not be reflected (d) None of
these
14. A plane mirror produces a magnification of [MP PMT/PET 1997]
(a) – 1 (b) + 1 (c) Zero (d) Between 0
and + ∞
15. When a plane mirror is rotated through an angle θ, then the reflected ray turns through the angle 2θ, then the size
of the image [MP PAT 1996]
(a) Is doubled (b) Is halved (c) Remains the same (d) Becomes
infinite
16. What should be the angle between two plane mirrors so that whatever be the angle of incidence, the incident ray
and the reflected ray from the two mirrors be parallel to each other
(a) 60o (b) 90o (c) 120o (d) 175o
17. Ray optics is valid, when characteristic dimensions are [CBSE PMT 1994]
(a) Of the same order as the wavelength of light (b) Much smaller than the wavelength of light
(c) Of the order of one millimeter (d) Much larger than the wavelength of light
18. It is desired to photograph the image of an object placed at a distance of 3 m from the plane mirror. The camera
which is at a distance of 4.5 m from the mirror should be focussed for a distance of
(a) 3 m (b) 4.5 m (c) 6 m (d) 7.5 m
19. Two plane mirrors are parallel to each other an spaced 20 cm apart. An object is kept in between them at 15 cm
from A. Out of the following at which point an image is not formed in mirror A (distance measured from mirror A)
(a) 15 cm (b) 25 cm (c) 45 cm (d) 55 cm

genius PHYSICS
20. Two plane mirrors A and B are aligned parallel to each other, as shown in the figure. A light ray is incident at an
angle of 30o at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The
maximum number of times the ray undergoes reflections (including the first one) before it emerges out is
(a) 28
(b) 30
(c) 32
(d) 34
21. A point source of light B is placed at a distance L in front of the centre of a mirror of width d hung vertically on a
wall. A man walks in front of the mirror along a line parallel to the mirror at a distance 2L from it as shown. The
greatest distance over which he can see the image of the light source in the mirror is
(a) d/2
(b) d
(c) 2d
(d) 3d
22. The figure shows two rays A and B being reflected by a mirror and going as A' and B'. The mirror is
(a) Plane (b) Concave
(c) Convex (d) May be any spherical mirror
23. An object is initially at a distance of 100 cm from a plane mirror. If the mirror approaches the object at a speed of
5 cm/s, then after 6 s the distance between the object and its image will be
(a) 60 cm (b) 140 cm (c) 170 cm (d) 150 cm
24. An object placed in front of a plane mirror is displaced by 0.4 m along a straight line at an angle of 30o to mirror
plane. The change in the distance between the object and its image is
(a) 0.20 m (b) 0.40 m (c) 0.25 m (d) 0.80 m
25. A ray of light travels from A to B with uniform speed. On its way it is reflected by the surface XX'. The path
followed by the ray to take least time is
(a) 1 (b) 2
(c) 3 (d) 4
26. A point object O is placed between two plan mirrors as shown is fig. The distance of the first three images formed
by mirror 2M from it are
(a) 2 mm, 8 mm, 18 mm
(b) 2 mm, 18 mm, 28 mm
(c) 2 mm, 18 mm, 22 mm
(d) 2 mm, 18 mm, 58 mm
27. A plane mirror is placed at the bottom of the tank containing a liquid of refractive index µ . P is a small object at a
height h above the mirror. An observer O-vertically above P outside the liquid see P and its image in the mirror.
The apparent distance between these two will be
m32
30o
0.2m
d
L
2L
A
B
B′
A′
XX′
BA
1 2 3 4
O
P
h
10mm
M1 M2
O
AAddvvaannccee LLeevveell

genius PHYSICS
(a) hµ2 (b)
µ
h2
(c)
1
2
−µ
h
(d) 





+
µ
1
1h
28. One side of a glass slab is silvered as shown. A ray of light is incident on the other side at angle of incidence
o
i 45= . Refractive index of glass is given as 1.5. The deviation of the ray of light from its initial path when it
comes out of the slab is
(a) 90o
(b) 180o
(c) 120o
(d) 45o
29. If an object moves towards a plane mirror with a speed v at an angle θ to the perpendicular to the plane of the
mirror, find the relative velocity between the object and the image
(a) v
(b) 2v
(c) θcos2v
(d) θsin2v
30. Figure shows a cubical room ABCD will the wall CD as a plane mirror. Each side of the room is 3m. We place a
camera at the midpoint of the wall AB. At what distance should the camera be focussed to photograph an object
placed at A
(a) 1.5 m (b) 3 m
(c) 6 m (d) More than 6 m
31. A man having height 6 m, want to see full height in mirror. They observe image of 2m height erect, then used mirror
is [J & K CET 2004]
(a) Concave (b) Convex (c) Plane (d) None of
these
32. An object of length 6cm is placed on the principal axis of a concave mirror of focal length f at a distance of 4 f. The
length of the image will be [MP PET 2003]
(a) 2 cm (b) 12 cm (c) 4 cm (d) 1.2 cm
33. Convergence of concave mirror can be decreased by dipping in [AFMC 2003]
(a) Water (b) Oil (c) Both (d) None of these
34. In an experiment of find the focal length of a concave mirror a graph is drawn between the magnitudes of u and v.
The graph looks like
(a) (b) (c) (d)
35. An object 2.5 cm high is placed at a distance of 10 cm from a concave mirror of radius of curvature 30 cm The size
of the image is
[BVP 2003]
(a) 9.2 cm (b) 10.5 cm (c) 5.6 cm (d) 7.5 cm
36. A diminished virtual image can be formed only in [MP PMT 2002]
BBaassiicc LLeevveell
Reflection of light at spherical surface
v
u
v
u
v
u
v
u
A B
CD
3m
45o
µ = 1.5
O I
y
vO
→
vI
→
θ θ
x

genius PHYSICS
(a) Plane mirror (b) A concave mirror (c) A convex mirror (d) Concave-
parabolic mirror
37. A point object is placed at a distance of 30 cm from a convex mirror of focal length 30cm. The image will form at [JIPMER 2002
(a) Infinity (b) Focus (c) Pole (d) 15 cm
behind the mirror
38. The focal length of a convex mirror is 20 cm its radius of curvature will be [MP PMT 2001]
(a) 10 cm (b) 20 cm (c) 30 cm (d) 40 cm
39. A concave mirror of focal length 15 cm forms an image having twice the linear dimensions of the object. The
position of the object when the image is virtual will be
(a) 22.5 cm (b) 7.5 cm (c) 30 cm (d) 45 cm
40. Under which of the following conditions will a convex mirror of focal length f produce an image that is erect,
diminished and virtual
[AMU (Engg.) 2001]
(a) Onlywhen2f >u>f (b) Only when u = f (c) Only when u < f (d) Always
41. A concave mirror gives an image three times as large as the object placed at a distance of 20 cm from it. For the
image to be real, the focal length should be [SCRA 1998; JIPMER 2
(a) 10 cm (b) 15 cm (c) 20 cm (d) 30 cm
42. A point object is placed at a distance of 10 cm and its real image is formed at a distance of 20cm from a concave
mirror. If the object is moved by 0.1cm towards the mirror, the image will shift by about
(a) 0.4 cm away from the mirror (b) 0.4 cm towards the mirror
(c) 0.8 cm away from the mirror (d) 0.8 cm towards the mirror
43. The minimum distance between the object and its real image for concave mirror is [RPMT 1999]
(a) f (b) 2f (c) 4f (d) Zero
44. An object is placed at 20 cm from a convex mirror of focal length 10 cm. The image formed by the mirror is [JIPMER 1999]
(a) Real and at 20 cm from the mirror (b) Virtual and at 20 cm from the mirror
(c) Virtual and at 20/3 cm from the mirror (d) Real and at 20/3 cm from the mirror
45. An object is placed 40 cm from a concave mirror of focal length 20 cm. The image formed is [MP PET 1986; MP PMT/PET 199
(a) Real, inverted and same in size (b) Real, inverted and smaller
(c) Virtual, erect and larger (d) Virtual, erect and smaller
46. Match List I with List II and select the correct answer using the codes given below the lists [SCRA 1998]
List I List II
(Position of the object) (Magnification)
(I) An object is placed at focus before a convex mirror (A) Magnification is – ∞
(II) An object is placed at centre of curvature before a concave mirror (B) Magnification is 0.5
(III) An object is placed at focus before a concave mirror (C) Magnification is + 1
(IV) An object is placed at centre of curvature before a convex mirror (D) Magnification is – 1
(E) Magnification is 0.33
Codes :
(a) I-B, II-D, III-A, IV-E (b) I-A, II-D, III-C, IV-B (c) I-C, II-B, III-A, IV-E (d) I-B, II-E,
III-D, IV-C
47. In a concave mirror experiment, an object is placed at a distance 1x from the focus and the image is formed at a
distance 2x from the focus. The focal length of the mirror would be
(a) 21xx (b) 21xx (c)
2
21 xx +
(d)
2
1
x
x
48. Which of the following forms a virtual and erect image for all positions of the object [IIT-JEE 1996]
(a) Convex lens (b) Concave lens (c) Convex mirror (d) Concave
mirror
49. A convex mirror has a focal length f. A real object is placed at a distance f in front of it from the pole produces an
image at
[MP PAT 1996]
(a) Infinity (b) f (c) f / 2 (d) 2f
50. Radius of curvature of concave mirror is 40 cm and the size of image is twice as that of object, then the object
distance is [AFMC 1995]
(a) 60 cm (b) 20 cm (c) 40 cm (d) 30 cm
51. All of the following statements are correct except [Manipal MEE 1995]

genius PHYSICS
(a) The magnification produced by a convex mirror is always less than one
(b) A virtual, erect, same-sized image can be obtained using a plane mirror
(c) A virtual, erect, magnified image can be formed using a concave mirror
(d) A real, inverted, same-sized image can be formed using a convex mirror
52. Radius of curvature of convex mirror is 40 cm and the size of object is twice as that of image, then the image
distance is [AFMC 1995]
(a) 10 cm (b) 20 cm (c) 40 cm (d) 30 cm
53. If an object is placed 10 cm in front of a concave mirror of focal length 20 cm, the image will be [MP PMT 1995]
(a) Diminished, upright, virtual (b) Enlarged, upright, virtual (c)
Diminished, inverted, real (d) Enlarged, upright, real
54. An object 1 cm tall is placed 4 cm in front of a mirror. In order to produce an upright image of 3 cm height one needs
a [SCRA 1994]
(a) Convex mirror of radius of curvature 12 cm (b) Concave mirror of radius of curvature 12
cm
(c) Concave mirror of radius of curvature 4 cm (d) Plane mirror of height 12 cm
55. The image formed by a convex mirror of a real object is larger than the object [CPMT 1994]
(a) When u < 2f (b) When u > 2f (c) For all values of u (d) For no
value of u
56. An object 5 cm tall is placed 1 m from a concave spherical mirror which has a radius of curvature of 20 cm. The
size of the image is
[MP PET 1993]
(a) 0.11 cm (b) 0.50 cm (c) 0.55 cm (d) 0.60 cm
57. A virtual image three times the size of the object is obtained with a concave mirror of radius of curvature 36 cm.
The distance of the object from the mirror is
(a) 5 cm (b) 12 cm (c) 10 cm (d) 20 cm
58. Given a point source of light, which of the following can produce a parallel beam of light [CPMT 1974]
(a) Convex mirror (b) Concave mirror
(c) Concave lens (d) Two plane mirrors inclined at an angle of
90o
59. A convex mirror is used to form the image of an object. Then which of the following statements is wrong
(a) The images lies between the pole and the focus (b) The image is diminished in size
(c) The images is erect (d) The image is real
60. A boy stands straight infront of a mirror at a distance of 30 cm away from it. He sees his erect image whose height
is
5
1 th of his real height. The mirror he is using is
(a) Plane mirror (b) Convex mirror (c) Concave mirror (d) Plano-
convex mirror
61. For the largest distance of the image from a concave mirror of focal length 10cm, the object should be kept at
(a) 10 cm (b) Infinite (c) 40 cm (d) 60 cm
62. A dentist uses a small mirror that gives a magnification of 4 when it is held 0.60 cm from a tooth. The radius of
curvature of the mirror is
(a) 1.60 cm (convex) (b) 0.8 cm (concave) (c) 1.60 cm (concave) (d) 0.8 cm
(convex)
63. A dice is placed with its one edge parallel to the principal axis between the principal focus and the centre of the
curvature of a concave mirror. Then the image has the shape of
(a) Cube (b) Cuboid (c) Barrel shaped (d) Spherical
64. A short linear object of length l lies along the axis of a concave mirror of focal length f at a distance u form the pole
of the mirror. The size of the image is approximately equal to [IIT 1988; BHU 2003]
(a)
2/1





 −
f
fu
l (b)
2





 −
f
fu
l (c)
2/1






− fu
f
l (d)
2






− fu
f
l
65. A point object is moving on the principal axis of a concave mirror of focal length 24 cm towards the mirror. When
it is at a distance of 60 cm from the mirror, its velocity is 9 cm/sec. What is the velocity of the image at that instant
AAddvvaannccee LLeevveell

genius PHYSICS
(a) 5 cm/sec towards the mirror (b) 4 cm/sec
towards the mirror
(c) 4 cm/sec away from the mirror (d) 9 cm/sec away from the mirror
66. A convex mirror of focal length 10 cm forms an image which is half of the size of the object. The distance of the
object from the mirror is
(a) 10 cm (b) 20 cm (c) 5 cm (d) 15 cm
67. A concave mirror is used to focus the image of a flower on a nearby well 120 cm from the flower. If a lateral
magnification of 16 is desired, the distance of the flower from the mirror should be
(a) 8 cm (b) 12 cm (c) 80 cm (d) 120 cm
68. A thin rod of 5 cm length is kept along the axis of a concave mirror of 10 cm focal length such that its image is real
and magnified and one end touches the rod. Its magnification will be
(a) 1 (b) 2 (c) 3 (d) 4
69. A luminous object is placed 20 cm from surface of a convex mirror and a plane mirror is set so that virtual images
formed in two mirrors coincide. If plane mirror is at a distance of 12 cm from object, then focal length of convex
mirror, is
(a) 5 cm (b) 10 cm (c) 20 cm (d) 40 cm
70. A rear mirror of a vehicle is cylindrical having radius of curvature 10 cm. The length of arc of curved surface is also
10 cm. If the eye of driver is assumed to be at large distance, from the mirror, then the field of view in radian is
(a) 0.5 (b) 1 (c) 2 (d) 4
71. A vehicle has a driving mirror of focal length 30 cm. Another vehicle of dimension 3
75.142 m×× is 9 m away
from the mirror of first vehicle. Position of the second vehicle as seen in the mirror of first vehicle is
(a) 30 cm
(b) 60 cm
(c) 90 cm
(d) 9 cm
72. A cube of side 2 m is placed in front of a concave mirror focal length 1m with its face P at a distance of 3 m and
face Q at a distance of 5 m from the mirror. The distance between the images of face P and Q and height of images
of P and Q are
(a) 1 m, 0.5 m, 0.25 m
(b) 0.5 m, 1 m, 0.25 m
(c) 0.5 m, 0.25 m, 1m
(d) 0.25 m, 1m, 0.5 m
73. A concave mirror of radius of curvature 60 cm is placed at the bottom of tank containing water upto a height of 20
cm. The mirror faces upwards with its axis vertical. Solar light falls normally on the surface of water and the image
of the sun is formed. If
3
4
=wa µ then with the observer in air, the distance of the image from the surface of water
is
(a) 30 cm (b) 10 cm (c) 7.5 cm above (d) 7.5 cm
below
74. A concave mirror forms an image of the sun at a distance of 12 cm from it
(a) The radius of curvature of this mirror is 6 cm
(b) To use it as a shaving mirror, it must be held at a distance of 8-10 cm from the face
(c) If an object is kept at a distance of 12 cm from it, the image formed will be of the same size as the object
(d) All the above a alternatives are correct
75. A small piece of wire bent into an L shape with upright and horizontal portions of equal lengths, is placed with the
horizontal portion along the axis of the concave mirror whose radius of curvature is 10 cm. If the bend is 20 cm
from the pole of the mirror, then the ratio of the lengths of the images of the upright and horizontal portions of
the wire is
(a) 1 : 2 (b) 3 : 1 (c) 1 : 3 (d) 2 : 1
76. As the position of an object (u) reflected from a concave mirror is varied, the position of the image (v) also varies.
By letting the u changes from 0 to +∞ the graph between v versus u will be
77.
9m
2m
2m
PQ
3m

genius PHYSICS
(a) b c d
78. A concave mirror has a focal length 20 cm. The distance between the two positions of the object for which the
image size is double of the object size is
(a) 20 cm (b) 40 cm (c) 30 cm (d) 60 cm
79. A concave mirror of focal length 10 cm and a convex mirror of focal length 15 cm are placed facing each other 40
cm apart. A point object is placed between the mirrors, on their common axis and 15 cm from the concave mirror.
Find the position and nature of the image produced by the successive reflections, first at concave mirror and then
at convex mirror
(a) 2 cm (b) 4 cm (c) 6 cm (d) 8 cm
AAAAnswer Sheet
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
c c c c c b b b b c b b a b c b d d c b
21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
d a b b c c b a c d b a d c d c d d b d
41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
b a d c a a b b, c c d d a b b d c b b d b
61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78
a c b d c a a a a b a d c b b a a c
v
u
v
u
v
u
v
u
Assignments

genius PHYSICS
F′ F
The bending of the ray of light passing from one medium to the other medium is called refraction.
Snell’s law
The ratio of sine of the angle of incidence to the angle of refraction (r) is a constant called
refractive index
i.e. µ=
r
i
sin
sin
(a constant). For two media, Snell's law can be written as
r
i
sin
sin
1
2
21 ==
µ
µ
µ
⇒ ri sinsin 21 ×=× µµ i.e. =θµ sin constant
Also in vector form : )( nrµni ˆˆˆˆ ×=×
Refractive Index.
Refractive index of a medium is that characteristic which decides speed of light in it. It is a scalar, unit
less and dimensionless quantity
.(1) Types : It is of following two types
Absolute refractive index Relative refractive index
(i) When light travels from air to any transparent
medium then R.I. of medium w.r.t. air is called it’s
absolute R.I. i.e.
v
c
=mediumair µ
(i) When light travels from medium (1) to medium (2)
then R.I. of medium (2) w.r.t. medium (1) is called it’s
relative R.I. i.e.
2
1
1
2
21
v
v
==
µ
µ
µ (where v1 and v2 are the
speed of light in medium 1 and 2 respectively).
(ii) Some absolute R.I.
5.1
2
3
glass ==µa , 33.1
3
4
==watera µ
,4.2diamond =µa 62.12
=Csa µ
,52.1crown =µa 1vacuum =µ , 10003.1air ≈=µ
(ii) Some relative R.I.
(a) When light enters from water to glass :
8
9
3/4
2/3
===
w
g
gw
µ
µ
µ
(b) When light enters from glass to diamond :
5
8
5.1
4.2
===
g
D
Dg
µ
µ
µ
Note : Cauchy’s equation : ......42
+++=
λλ
µ
CB
A violetRed( λλ > so )violetRed µµ <
i
r δ
Incident ray
Rarer medium
Denser medium Refracted ray
Deviation δ = (i – r)
i
r
δ
Denser medium
Rarer medium
Deviation δ = (r – i )
λ
µ
1
∝
v
1
∝µ
λ∝v

genius PHYSICS
If a light ray travels from medium (1) to medium (2), then
2
1
2
1
1
2
21
v
v
===
λ
λ
µ
µ
µ
(2) Dependence of Refractive index
(i) Nature of the media of incidence and refraction.
(ii) Colour of light or wavelength of light.
(iii) Temperature of the media : Refractive index decreases with the increase in temperature.
(3) Principle of reversibility of light and refraction through several media :
Principle of reversibility
12
21
1
µ
µ =
Refraction through several media
1133221 =×× µµµ
Refraction Through a Glass Slab and Optical Path
(1) Lateral shift
The refracting surfaces of a glass slab are parallel to each other. When a light ray passes
through a glass slab it is refracted twice at the two parallel faces
and finally emerges out parallel to it's incident direction i.e. the
ray undergoes no deviation δ = 0. The angle of emergence (e) is
equal to the angle of incidence (i)
The Lateral shift of the ray is the perpendicular distance
between the incident and the emergent ray, and it is given by
MN = t sec r sin (i – r)
Normal shift
Normal shift txOO 





−==
µ
1
1'
Or the object appears to be shifted towards the slab by the distance x
(2) Optical path :
It is defined as distance travelled by light in vacuum in the same time in which it travels a
given path length in a medium.
Time taken by light ray to pass through the medium
c
xµ
= ;
where x = geometrical path and µx = optical path
µµµµ
Light
x
Incident ray 1
2
3
1
i
r
µ
r δ
N
M
t
1
2
r
i
O O'
µµµµ
t
Glass
slabx

genius PHYSICS
For two medium in contact optical path = 2211 xx µµ +
Note : Since for all media ,1>µ so optical path length )( xµ is always greater than the
geometrical path length (x).
Real and Apparent Depth.
If object and observer are situated in different medium then due to refraction, object appears
to be displaced from it’s real position. There are two possible conditions.
(1) When object is in denser medium and observer is
in rarer medium
(1) Object is in rarer medium and observer is in denser
medium.
(2) 'depthApparent
depthReal
h
h
==µ
Real depth >Apparent depth that's why a coin at the
bottom of bucket (full of water) appears to be
raised)
(2)
h
h'
=µ
Real depth < Apparent depth that's why high flying
aeroplane appears to be higher than it's actual height.
(3) Shift hhhd 





−=−=
µ
1
1'
(3) hd )1( −= µ
(4) For water
43
4 h
d =⇒=µ
For glass
32
3 h
d =⇒=µ
(4) Shift for water
3
h
dw =
Shift for glass
2
h
dg =
Note : If a beaker contains various immisible liquids as shown then
Apparent depth of bottom ....
3
3
2
2
1
1
+++=
µµµ
ddd
µ combination =
....
.....
2
2
1
1
21
. ++
++
=
µµ
dd
dd
d
d
App
AC
(In case of two liquids if 21 dd = than
21
212
µµ
µµ
µ
+
= )
µµµµ2
Light
x1
µµµµ1
x2
d1
d2
d3
µ1
µ2
µ3
µ
h
h′
d
O
O′
O
O′
h′
d
h

genius PHYSICS
Total Internal Reflection.
When a ray of light goes from denser to rarer medium it bends away from the normal and as
the angle of incidence in denser medium increases, the angle of refraction in rarer medium also
increases and at a certain angle, angle of refraction becomes 90o, this angle of incidence is called
critical angle (C).
When Angle of incidence exceeds the critical angle than light ray comes back in to the same medium
after reflection from interface. This phenomenon is called Total internal reflection (TIR).
Important formula C
C
µ cosec
sin
1
== ; where DenserRerer µµ →
Note : When a light ray travels from denser to rarer medium, then deviation of the ray is
max.2 →⇒−= δθπδ when C=→ min.θ
i.e. ;)2(max C−= πδ C → critical angle
(1) Dependence of critical angle
(i) Colour of light (or wavelength of light) : Critical angle depends upon wavelength as
Csin
1
∝∝
µ
λ
(a) VRVR CC >⇒> λλ
(b) Sin C
R
D
R
D
D
R
DR v
v
====
λ
λ
µ
µ
µ
1
(for two media) (c) For TIR from boundary of two
media
D
R
i
µ
µ1
sin−
>
(ii) Nature of the pair of media : Greater the refractive index lesser will be the critical angle.
(a) For (glass- air) pair o
C 42glass =→ (b) For (water-air) pair o
C 49water =→
(c) For (diamond-air) pair o
C 24amonddi =→
(iii) Temperature : With temperature rise refractive index of the material decreases therefore
critical angle increases.
(2) Examples of total internal reflection (TIR)
(i)
θ θ
δ
i = C
90o
θ θ
θ > C TIR
i
rRarer
Denser
Looming : An optical illusion in cold
countries
Earth
Desner
Rarer
Sky
i>θC
O
I
Mirage : An optical illusion in deserts
Denser
Rarer
Earth
I
O
i>θC

genius PHYSICS
(ii) Brilliance of diamond : Due to repeated internal reflections diamond sparkles.
(iii) Optical fibre : Optical fibres consist of many long high quality composite glass/quartz
fibres. Each fibre consists of a core and cladding. The refractive index of the material of the core
(µ1) is higher than that of the cladding (µ2).
When the light is incident on one end of the fibre at a small angle, the light passes inside,
undergoes repeated total internal reflections along the fibre and finally comes out. The angle of
incidence is always larger than the critical angle of the core material with respect to its cladding.
Even if the fibre is bent, the light can easily travel through along the fibre
A bundle of optical fibres can be used as a
'light pipe' in medical and optical examination. It
can also be used for optical signal transmission.
Optical fibres have also been used for transmitting
and receiving electrical signals which are
converted to light by suitable transducers.
(iv) Field of vision of fish (or swimmer) : A fish (diver) inside the water can see the
whole world through a cone with.
(a) Apex angle o
C 982 ==
(b) Radius of base
1
tan
2
−
==
µ
h
Chr
(c) Area of base A
)1( 2
2
−
=
µ
πh
Note : For water
3
4
=µ so
7
3h
r = and
7
9 2
h
A
π
= .
(v) Porro prism : A right angled isosceles prism, which is used in periscopes or binoculars.
It is used to deviate light rays through o
90 and o
180 and also to erect the image.
Example: 1 A beam of monochromatic blue light of wavelength 4200 Å in air travels in water )3/4( =µ .
Its wavelength in water will be
(a) 2800 Å (b) 5600 Å (c) 3150 Å (d) 4000 Å
Solution: (c)
λ
µ
1
∝ ⇒
1
2
2
1
λ
λ
µ
µ
= ⇒
4200
3
4
1 2λ
= ⇒ Å31502 =λ
Example: 2 On a glass plate a light wave is incident at an angle of 60o. If the reflected and the refracted
waves are mutually perpendicular, the refractive index of material is [MP
PMT 1994; Haryana CEE 1996]
(a)
2
3
(b) 3 (c)
2
3
(d)
3
1
45o
45o
45o90o
45o
45o
90o
A′
B′
A
B
45o
45o
45o
45o
90o
Cladding
Core
µ2
µ1
Example
s
C
C
h
r
C
θθθθ >C

genius PHYSICS
Solution: (b) From figure o
r 30=
∴ 3
30sin
60sin
sin
sin
=== o
o
r
i
µ
Example: 3 Velocity of light in glass whose refractive index with respect to air is 1.5 is 2 × sm /10 8
and in
certain liquid the velocity of light found to be sm /1050.2 8
× . The refractive index of the
liquid with respect to air is
[CPMT 1978; MP
PET/PMT 1988]
(a) 0.64 (b) 0.80 (c) 1.20 (d) 1.44
Solution: (c)
v
1
∝µ ⇒ 8
8
105.2
102
5.1 ×
×
=⇒= l
l
g
g
li
v
v µ
µ
µ
⇒ 2.1=lµ
Example: 4 A ray of light passes through four transparent media with refractive indices
4321 and,,. µµµµ as shown in the figure. The surfaces of all media are parallel. If the
emergent ray CD is parallel to the incident ray AB, we must have
(a) 21 µµ =
(b) 32 µµ =
(c) 43 µµ =
(d) 14 µµ =
Solution: (d) For successive refraction through difference media =θµ sin constant.
Here as θ is same in the two extreme media. Hence 41 µµ =
Example: 5 A ray of light is incident at the glass–water interface at an angle i, it emerges finally parallel
to the surface of water, then the value of gµ would be
(a) (4/3) sin i
(b) 1/ sin i
(c) 4/ 3
(d) 1
Solution: (b) For glass water interface
r
i
g
sin
sin
=ωµ ......(i) and For water-air interface
90sin
sinr
a =µω
.....(ii)
∴ iag sin=× µµ ωω ⇒
i
g
sin
1
=µ
Example: 6 The ratio of thickness of plates of two transparent mediums A and B is 6 : 4. If light takes
equal time in passing through them, then refractive index of B with respect to A will be
(a) 1.4 (b) 1.5 (c) 1.75 (d) 1.33
Solution: (b) By using
c
x
t
µ
=
⇒
4
6
==
B
A
A
B
x
x
µ
µ
5.1
2
3
==⇒ BA µ
Example: 7 A ray of light passes from vacuum into a medium of refractive index µ, the angle of incidence
is found to be twice the angle of refraction. Then the angle of incidence is
A
B C
µ1 µ µ µ
D
60 60
90
r
Wate
r µw=4/3
r
i
Glass

genius PHYSICS
(a) ( )2/cos 1
µ−
(b) ( )2/cos2 1
µ−
(c) ( )µ1
sin2 −
(d)
( )2/sin2 1
µ−
r
i
sin
sin
=µ ⇒
r
rr
r
r
sin
cossin2
sin
2sin
==µ ( θθθ cossin22sin = )
⇒ 





= −
2
cos 1 µ
r . So, 





== −
2
cos22 1 µ
ri .
Example: 8 A ray of light falls on the surface of a spherical glass paper weight making an angle α with
the normal and is refracted in the medium at an angle β . The angle of deviation of the
emergent ray from the direction of the incident ray is
(a) ( )βα − (b) ( )βα −2 (c) ( ) 2/βα − (d) ( )βα +
Solution: (b) From figure it is clear that OBC∆ is an isosceles triangle,
Hence β=∠OCB and emergent angle is α
Also sum of two in terior angles = exterior angle
∴ ( ) ( )βαβαβαδ −=−+−= 2)(
Example: 9 A rectangular slab of refractive index µ is placed over another slab of refractive index 3, both
slabs being identical in dimensions. If a coin is placed below the lower slab, for what value of
µ will the coin appear to be placed at the interface between the slabs when viewed from the
top
(a) 1.8 (b) 2 (c) 1.5 (d) 2.5
Solution: (c) Apparent depth of coin as seen from top x
xx
=+=
21 µµ
1
11
21
=+⇒
µµ
1
1
3
1
=+⇒
µ
5.1=⇒ µ
Example: 10 A coin is kept at bottom of an empty beaker. A travelling microscope is focussed on the coin
from top, now water is poured in beaker up to a height of 10 cm. By what distance and in
which direction should the microscope be moved to bring the coin again in focus
(a) 10 cm up ward (b) 10 cm down ward (c) 2.5 cm up wards (d) 2.5 cm
down wards
Solution: (c) When water is poured in the beaker. Coin appears to shift by a distance cm
h
d 5.2
4
10
4
===
Hence to bring the coil again in focus, the microscope should be moved by 2.5 cm in upward
direction.
Example: 11 Consider the situation shown in figure. Water 





=
3
4
wµ is filled in a breaker upto a height
of 10 cm. A plane mirror fixed at a height of 5 cm from the surface of water. Distance of
image from the mirror after reflection from it of an object O at the bottom of the beaker is
(a) 15 cm (b) 12.5 cm (c) 7.5 cm (d) 10 cm
Solution: (b) From figure it is clear that object appears to be raised by ( )cmcm 5.2
4
10
Hence distance between mirror and cmO 5.125.75' =+=
So final image will be formed at 12.5 cm behind the plane mirror.
5 cm
10 cm
O
4
10
c
O'
µ2 = µ
µ1 = 3
x
x
α α –β
β β
δ
B
O
C
αβ –α
A D

genius PHYSICS
Example: 12 The wavelength of light in two liquids 'x' and 'y' is 3500 Å and 7000 Å, then the critical angle of x
relative to y will be
(a) 60o (b) 45o (c) 30o (d) 15o
Solution: (c)
o
CC 30
2
1
7000
3500
sin
2
1
1
2
=⇒====
λ
λ
µ
µ
Example: 13 A light ray from air is incident (as shown in figure) at one end of a glass fiber (refractive
index µ = 1.5) making an incidence angle of 60o on the lateral surface, so that it undergoes a
total internal reflection. How much time would it take to traverse the straight fiber of length
1 km [Orissa JEE 2002]
(a) 3.33 µ sec
(b) 6.67 µ sec
(c) 5.77 µ sec
(d) 3.85 µ sec
Solution: (d) When total internal reflection just takes place from lateral surface then i = C i.e. C = 60o
From
Csin
1
=µ
3
2
60sin
1
==⇒ µ
Hence time taken by light traverse some distance in medium
C
x
t
µ
=
( )
.85.3
103
101
3
2
8
3
sect µ=
×
××
=⇒
Example: 14 A glass prism of refractive index 1.5 is immersed in water )3/4( =µ . A light beam incident
normally on the face AB is totally reflected to reach the face BC if
(a) 9/8sin >θ
(b) 9/8sin3/2 << θ
(c) 3/2sin ≤θ
(d) 9/8cos ≥θ
Solution: (a) From figure it is clear that
Total internal reflection takes place at AC, only if θ > C
Csinsin >⇒ θ
gµ
θ
ω
1
sin >⇒
8/9
1
sin >⇒ θ
9
8
sin >⇒ θ
Example: 15 When light is incident on a medium at angle i and refracted into a second medium at an
angle r, the graph of sin i vs sin r is as shown in the graph. From this, one can conclude that
(a) Velocity of light in the second medium is 1.73 times the
velocity of light in the I medium
(b) Velocity of light in the I medium is 1.73 times the velocity in
the II medium
(c) The critical angle for the two media is given by sin
3
1
=ci
(d) sin
2
1
=ci
B A
θ
C
B A
θ
C
θ
30o
sin
i
sin
r
60Air
Air
Glas
s

genius PHYSICS
Solution: (b, c) From graph
21
1
sin
sin
30tan
µ
==
i
ro
⇒ 321 =µ ⇒ 73.1
2
1
1
2
==
v
v
µ
µ
⇒ 21 75.1 vv =
Also from
Csin
1
=µ ⇒
DenserRarer
C
µ
1
sin = ⇒
3
11
sin
21
==
µ
C .
Example: 16 A beam of light consisting of red, green and blue colours is incident on a right angled prism.
The refractive indices of the material of the prism for the above red, green and blue
wavelength are 1.39, 1.44 and 1.47 respectively. The prism will
(a) Separate part of red colour from the green and the blue
colours
(b) Separate part of the blue colour from the red and green
colours
(c) Separate all the colours from one another
(d) Not separate even partially any colour from the other two colours
Solution: (a) At face AB, i = 0 so r = 0, i.e., no refraction will take place. So light will be incident on face
AC at an angle of incidence of 45o. The face AC will not transmit the light for which ,Ci θ>
i.e., Ci θsinsin >
or ( )µ/145sin >o
i.e., )41.1(2 =>µ
Now as µµ <R while Gµ and ,µµ >B so red will be transmitted
through the face AC while green and blue will be reflected. So the
prism will separate red colour from green and blue.
Example: 17 An air bubble in a glass slab ( )5.1=µ is 6 cm deep when viewed from
one face and 4 cm deep when viewed from the opposite face. The thickness of the glass plate
is
(a) 10 cm (b) 6.67 cm (c) 15 cm (d) None of
these
Solution: (c) Let thickness of slab be t and distance of air bubble from one side is x
When viewed from side (1) : cmx
x
9
6
5.1 =⇒=
When viewed from side (2) :
( ) cmt
txt
15
4
9
5.1
4
)(
5.1 =⇒
−
=⇒
−
=
One face of a rectangular glass plate 6 cm thick is silvered. An object held 8 cm in front of the first
face, forms an image 12 cm behind the silvered face. The refractive index of the glass is
(a) 0.4 (b) 0.8 (c) 1.2 (d) 1.6
Solution : (c) From figure thickness of glass plate t = 6 cm.
Let x be the apparent position of the silvered surface.
According to property of plane mirror
x + 8 = 12 + 6 – x ⇒ x = 5 cm
Also
x
t
=µ ⇒ 2.1
5
6
==µ
45
Object Image
x
8 cm 12 cm
12 +(6–x)
t
Tricky example: 1
45°
45°
A
B C
6 cm 4 cm
Side 1 Side 2
Air
bubble
t
x

genius PHYSICS
A ray of light is incident on a glass sphere of refractive index 3/2. What should be the angle of incidence so that the
ray which enters the sphere doesn't come out of the sphere
(a) 




−
3
2
tan 1
(b) 




−
3
2
sin 1
(c) 90o (d) 




−
3
1
cos 1
Solution : (c) Ray doesn't come out from the sphere means TIR takes place.
Hence from figure COABABO =∠=∠
∴
Csin
1
=µ ⇒
3
21
sin ==
µ
C
Applying Snell's Law at A
2
3
sin
sin
=
C
i
⇒ 1
3
2
2
3
sin
2
3
sin =×== Ci ⇒ o
i 90=
The image of point P when viewed from top of the slabs will be
(a) 2.0 cm above P (b) 1.5 cm above P (c) 2.0 cm below P (d) 1 cm above P
Solution: (d) The two slabs will shift the image a distance
( ) cmtd 15.1
5.1
1
12
1
12 =





−=





−=
µ
Therefore, final image will be 1 cm above point P.
Refraction From Curved Surface.
=1µ Refractive index of the medium from which light rays are coming (from object).
=2µ Refractive index of the medium in which light rays are entering.
u = Distance of object, v = Distance of image, R = Radius of curvature
Refraction formula :
uvR
1212 µµµµ
−=
−
(use sign convention while solving the problem)
Note : Real image forms on the side of a refracting surface that is opposite to the object,
and virtual image forms on the same side as the object.
Lateral (Transverse) magnification
u
v
O
I
m
2
1
µ
µ
== .
Specific Example
In a thin spherical fish bowl of radius 10 cm filled with water of refractive index 4/3 there is a small
fish at a distance of 4 cm from the centre C as shown in figure. Where will the image of fish appears, if
seen from E
O P I
1 2 21
O P I
Tricky example: 2
C C
O
BA
i
Tricky example: 3
1.5 cm
1.5 cm
1.5 cm
2.0 cm
P

genius PHYSICS
(a) 5.2 cm (b) 7.2 cm (c) 4.2 cm (d) 3.2 cm
Ruv
1212 µµµµ −
=−
where ,
3
4
1 =µ ,12 =µ ,6 cmu −= v = ?
On putting values cmv 2.5−=
Lens.
Lens is a transparent medium bounded by two refracting surfaces, such that at least one
surface is spherical.
(1) Type of lenses
Convex lens (Converges the light rays) Concave lens (Diverges the light rays)
Double convex Plano convex Concavo convex Double concave Plane concave Convexo concave
Thick at middle Thin at middle
It forms real and virtual images both It forms only virtual images
(2) Some definitions
(i) Optical centre (O) : A point for a given lens through which light ray passes undeviated
(Light ray passes undeviated through optical centre).
(ii) Principle focus
First principle focus Second principle focus
Note : Second principle focus is the principle focus of the lens.
When medium on two sides of lens is same then |||| 21 FF = .
C1, C2 – Centre of curvature,
R1, R2 – Radii of curvature
F1
F1
F2 F2
C E
4 cm
Optical axis
Principle
axis
C2
+R1– R2
C1
O
C1
+R2– R1
C2
O

genius PHYSICS
If medium on two sides of lens are not same then the ratio of two focal lengths
2
1
2
1
µ
µ
=
f
f
(iii) Focal length (f) : Distance of second principle focus from optical centre is called focal
length
→convexf positive, →concavef negative, ∞→planef
(iv) Aperture : Effective diameter of light transmitting area is called aperture.
2
(Aperture)imageofIntensity ∝
(v) Power of lens (P) : Means the ability of a lens to converge the light rays. Unit of power
is Diopter (D).
)(
100
)(
1
cmfmf
P == ; positive,convex →P negative,concave →P zeroplane →P .
Note : Thick lens Thin lens
P ↑ f ↓ R ↓ P ↓ f ↑ R ↑
(2) Image formation by lens
Lens Location of
the object
Location of the
image
Nature of image
Magnificatio
n
Real
virtual
Erect
inverted
Convex At infinity
i.e. ∞=u
At focus i.e. fv = 1<m
diminished
Real Inverted
Away from 2f
i.e. )2( fu >
Between f and 2f
i.e. fvf 2<<
1<m
diminished
Real Inverted
At 2f or )2( fu = At 2f i.e. )2( fv = 1=m
same size
Real Inverted
Between f and 2f
i.e. fuf 2<<
Away from 2f i.e.
)2( fv >
1>m
magnified
Real Inverted
At focus
i.e. fu =
At infinity i.e. ∞=v ∞=m
magnified
Real Inverted
Between optical
centre and
focus, fu <
At a distance
greater than that of
object uv >
1>m
magnified
Virtual Erect
Concave At infinity
i.e. ∞=u
At focus i.e. v = f 1<m
diminished
Virtual Erect
2f f f 2f
µµµµ1 µµµµ2

genius PHYSICS
Anywhere
between infinity
and optical
centre
Between optical
centre and focus
1<m
diminished
Virtual Erect
Note : Minimum distance between an object and it’s real image formed by a convex lens is
4f. Maximum image distance for concave lens is it’s focal length.
(4) Lens maker’s formula
The relation between f, µ, R1 and R2 is known as lens maker’s formula and it is






−−=
21
11
)1(
1
RRf
µ
Equiconvex lens Plano convex lens Equi concave lens Plano concave lens
RR =1 and RR −=2
)1(2 −
=
µ
R
f
for 5.1=µ , Rf =
,1 ∞=R RR −=2
)1( −
=
µ
R
f
for 5.1=µ , Rf 2=
RR −=1 , RR +=2
)1(2 −
−=
µ
R
f
for 5.1=µ Rf −=
∞=1R , RR =2
)1(2 −
=
µ
R
f
for ,5.1=µ Rf 2−=
(5) Lens in a liquid
Focal length of a lens in a liquid )( lf can be determined by the following formula
1)(
1)(
−
−
=
gl
ga
a
l
µ
µ
f
f
(Lens is supposed to be made of glass).
Note : Focal length of a glass lens )5.1( =µ is f in air then inside the water it’s focal length
is 4f.
In liquids focal length of lens increases (↑) and it’s power decreases (↓).
(6) Opposite behaviour of a lens
In general refractive index of lens >)( Lµ refractive index of medium surrounding it )( Mµ .
µµµµL > µµµµM µµµµL < µµµµM µµµµL = µµµµM

genius PHYSICS
(7) Lens formula and magnification of lens
(i) Lens formula :
uvf
111
−= ; (use sign convention)
(ii) Magnification : The ratio of the size of the image to the size of object is called
magnification.
(a) Transverse magnification :
f
vf
uf
f
u
v
O
I
m
−
=
+
=== (use sign convention while solving
the problem)
(b) Longitudinal magnification :
12
12
uu
vv
O
I
m
−
−
== . For very small object
222





 −
=





+
=





==
f
vf
uf
f
u
v
du
dv
m
(c) Areal magnification :
2
2






+
===
uf
f
m
A
A
m
o
i
s , (Ai = Area of image, Ao = Area of object)
(8) Relation between object and image speed
If an object move with constant speed )( oV towards a convex lens from infinity to focus, the
image will move slower in the beginning and then faster. Also oi V
uf
f
V .
2






+
=
(9) Focal length of convex lens by displacement method
(i) For two different positions of lens two images )and( 21 II of an object is formed at the same
location.
(ii) Focal length of the lens
21
22
4 mm
x
D
xD
f
−
=
−
=
where
O
I
m 1
1 = and
O
I
m 2
2 =
(iii) Size of object 21 . IIO =
(10) Cutting of lens
(i) A symmetric lens is cut along optical axis in two equal parts. Intensity of image formed by
each part will be same as that of complete lens.
(ii) A symmetric lens is cut along principle axis in two equal parts. Intensity of image formed
by each part will be less compared as that of complete lens.(aperture of each part is
2
1
times that
of complete lens)
Object
O
x
D > 4f
I2
I1
Screen
f, P
⇒
2f 2f
P/2 P/2
f, P
⇒
f, P

genius PHYSICS
(11) Combination of lens
(i) For a system of lenses, the net power, net focal length and magnification given as follows :
..........321 PPPP ++= , ...........
1111
321
+++=
fffF
,
............321 ×××= mmmm
(ii) In case when two thin lens are in contact : Combination will behave as a lens, which have
more power or lesser focal length.
21
111
ffF
+= ⇒
21
21
ff
ff
F
+
= and 21 PPP +=
(iii) If two lens of equal focal length but of opposite nature are in contact then combination
will behave as a plane glass plate and ∞=ncombinatioF
(iv) When two lenses are placed co-axially at a distance d from each other then equivalent
focal length (F).
2121
111
ff
d
ffF
−+= and 2121 PdPPPP −+=
(v) Combination of parts of a lens :
(12) Silvering of lens
On silvering the surface of the lens it behaves as a mirror. The focal length of the silvered lens
is
ml ffF
121
+= where =lf focal length of lens from which refraction takes place (twice)
=mf focal length of mirror from which reflection takes place.
(i) Plano convex is silvered
,
2
R
fm =
)1( −
=
µ
R
fl so
µ2
R
F = ,∞=mf
)1( −
=
µ
R
fl so
)1(2 −
=
µ
R
F
(ii) Double convex lens is silvered
Since
2
,
)1(2
R
f
R
f ml =
−
=
µ
f
⇒
F = fF = f
and
f
and
F = f/2
F = ∞
⇒
F fl fm
+
fl
+⇒
F fm
fl fm
+
F
⇒
d
f1 f2

genius PHYSICS
So
)12(2 −
=
µ
R
F
Note : Similar results can be obtained for concave lenses.
(13) Defects in lens
(i) Chromatic aberration : Image of a white object is coloured and blurred because µ
(hence f) of lens is different for different colours. This defect is called chromatic aberration.
RV µµ > so VR ff >
Mathematically chromatic aberration
= yVR ωfff =−
ω = Dispersion power of lens.
fy = Focal length for mean colour VR ff=
Removal : To remove this defect i.e. for Achromatism we use two or more lenses in contact in
place of single lens.
Mathematically condition of Achromatism is : 0
2
2
1
1
=+
ff
ωω
or 1221 ff ωω −=
Note : Component lenses of an achromatic doublet cemented by canada blasam because it
is transparent and has a refractive index almost equal to the refractive of the glass.
(ii) Spherical aberration : Inability of a lens to form the point image of a point object on
the axis is called Spherical aberration.
In this defect all the rays passing through a lens are not focussed at a single point and the
image of a point object on the axis is blurred.
Removal : A simple method to reduce spherical aberration is to use a stop before and infront
of the lens. (but this method reduces the intensity of the image as most of the light is cut off). Also
by using plano-convex lens, using two lenses separated by distance d = F – F', using crossed lens.
Note : Marginal rays : The rays farthest from the principal axis.
Paraxial rays : The rays close to the principal axis.
Spherical aberration can be reduced by either stopping paraxial rays or marginal
rays, which can be done by using a circular annular mask over the lens.
Parabolic mirrors are free from spherical aberration.
(iii) Coma : When the point object is placed away from the principle axis and the image is
received on a screen perpendicular to the axis, the shape of the image is like a comet. This defect is
called Coma.
It refers to spreading of a point object in a plane ⊥ to principle axis.
Real
FV FR
White
light
fV
fR
Violet
F′ F
Marginal rays
Paraxial ray F′ F

genius PHYSICS
Removal : It can be reduced by properly designing radii of curvature of the lens surfaces. It
can also be reduced by appropriate stops placed at appropriate distances from the lens.
(iv) Curvature : For a point object placed off the axis, the image is spread both along and
perpendicular to the principal axis. The best image is, in general, obtained not on a plane but on a
curved surface. This defect is known as Curvature.
Removal : Astigmatism or the curvature may be reduced by using proper stops placed at
proper locations along the axis.
(v) Distortion : When extended objects are imaged, different portions of the object are in
general at different distances from the axis. The magnification is not the same for all portions of
the extended object. As a result a line object is not imaged into a line but into a curve.
(vi) Astigmatism : The spreading of image (of a point object placed away from the principal
axis) along the principal axis is called Astigmatism.
Concepts
If a sphere of radius R made of material of refractive index 2µ is placed in a medium of refractive index 1µ , Then if the
object is placed at a distance R







− 12
1
µµ
µ
from the pole, the real image formed is equidistant from the sphere.
The lens doublets used in telescope are achromatic for blue and red colours, while these used in camera are achromatic for
violet and green colours. The reason for this is that our eye is most sensitive between blue and red colours, while the
photographic plates are most sensitive between violet and green colours.
Position of optical centre
Equiconvex and equiconcave Exactly at centre of lens
Convexo-concave and concavo-convex Outside the glass position
Plano convex and plano concave On the pole of curved surface
Composite lens : If a lens is made of several materials then
Number of images formed = Number of materials used
Here no. of images = 5
Distorted imagesObject
P
P ′
Image of P
Axis
µ1
µ2
µ3
µ4
µ5
Example
s
µ1 µ2
O IP1 P2
µ2
x 2x x

genius PHYSICS
Example: 18 A thin lens focal length 1f and its aperture has diameter d. It forms an image of intensity I.
Now the central part of the aperture upto diameter d/2 is blocked by an opaque paper. The
focal length and image intensity will change to
(a)
2
f
and
2
I
(b) f and
4
I
(c)
4
3 f
and
2
I
(d) f and
4
3I
Solution: (d) Centre part of the aperture up to diameter
2
d
is blocked i.e. th
4
1
area is blocked








=
4
2
d
A
π
. Hence remaining area AA
4
3
=′ . Also, we know that intensity ∝ Area ⇒
4
3
=
′
=
′
A
A
I
I
⇒ II
4
3
=′ .
Focal length doesn't depend upon aperture.
Example: 19 The power of a thin convex lens )5.1( =ga µ is + 5.0 D. When it is placed in a liquid of
refractive index ,la µ then it behaves as a concave lens of local length 100 cm. The refractive
index of the liquid la µ will be
(a) 5 / 3 (b) 4 / 3 (c) 3 (d) 5 / 4
Solution: (a) By using
1
1
−
−
=
gl
ga
a
l
f
f
µ
µ
; where
ll
g
gl
µµ
µ
µ
5.1
== and cmm
P
fa 20
5
11
===
⇒
1
5.1
15.1
20
100
−
−
=
−
lµ
⇒ 3/5=lµ
Example: 20 A double convex lens made of a material of refractive index 1.5 and having a focal length of
10 cm is immersed in liquid of refractive index 3.0. The lens will behave as [NCERT 1973]
(a) Diverging lens of focal length 10 cm (b) Diverging lens of focal length 10 / 3
cm
(c) Converging lens of focal length 10 / 3 cm (d) Converging lens of focal length 30
cm
1
1
−
−
=
gl
ga
a
l
f
f
µ
µ
⇒
1
3
5.1
15.1
10
−
−
=
+
lf
⇒ cmfl 10−= (i.e. diverging lens)
Example: 21 Figure given below shows a beam of light converging at point P. When a concave lens of
focal length 16 cm is introduced in the path of the beam at a place O shown by dotted line,
the beam converges at a distance x from the lens. The value x will be equal to
(a) 12 cm
(b) 24 cm
(c) 36 cm
(d) 48 cm
Solution: (d) From the figure shown it is clear that
For lens : u = 12 cm and v = x = ?
By using
uvf
111
−= ⇒
12
11
16
1
+
−=
+ x
⇒ x = 48 cm.
Example: 22 A convex lens of focal length 40 cm is an contact with a concave lens of focal length 25 cm.
The power of combination is
(a) – 1.5 D (b) – 6.5 D (c) + 6.5 D (d) + 6.67 D
21
111
ffF
+= ⇒
25
1
40
11
−
+
+
=
F
O 12 cm
P
PP'
x
12c
m

genius PHYSICS
⇒ cmF
3
200
−= , hence D
cmf
P 5.1
3/200
100
)(
100
−=
−
==
Example: 23 A combination of two thin lenses with focal lengths 1f and 2f respectively forms an image
of distant object at distance 60 cm when lenses are in contact. The position of this image
shifts by 30 cm towards the combination when two lenses are separated by 10 cm. The
corresponding values of 1f and 2f are [AIIMS 1995]
(a) 30 cm, – 60 cm (b) 20 cm, – 30 cm (c) 15 cm, – 20 cm (d) 12 cm, –
15 cm
Solution: (b) Initially cmF 60= (Focal length of combination)
Hence by using
21
111
ffF
+= ⇒
60
111
21
=+
ff
⇒
21
21
ff
ff
+
......(i)
Finally by using
2121
111
ff
d
ffF
−+=
′
where cmF 30=′ and d = 10 cm ⇒
2121
1011
30
1
ffff
−+= ......(ii)
From equations (i) and (ii) .60021 −=ff
From equation (i) 1021 −=+ ff …..(iii)
Also,difference of focal lengths can written as 21
2
2121 4)( ffffff −+=− ⇒ 5021 =− ff
.(iv)
From (iii) × (iv) 201 =f and 302 −=f
Example: 24 A thin double convex lens has radii of curvature each of magnitude 40 cm and is made of
glass with refractive index 1.65. Its focal length is nearly
(a) 20 cm (b) 31 cm (c) 35 cm (d) 50 cm
( )12 −
=
µ
R
f ⇒
( )
.317.30
165.12
40
cmcmf ≈=
−
=
Example: 25 A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass
(refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is
found to have a real image Q in the glass. The line PQ cuts the surface at a point O and
OQPO = . The distance PO is equal to
[MP PMT 1994; Haryana CEE 1996]
(a) 5 R (b) 3 R (c) 2 R (d) 1.5 R
Ruv
1212 µµµµ −
=−
Where ,11 =µ ,5.12 =µ u = – OP, v = OQ
Hence
( )ROPOQ +
−
=
−
−
15.115.1
⇒
ROPOP
5.015.1
=+
⇒ OP = 5 R
Example: 26 The distance between an object and the screen is 100 cm. A lens produces an image on the
screen when placed at either of the positions 40 cm apart. The power of the lens is
(a) 3 D (b) 5 D (c) 7 D (d) 9 D
QP O

genius PHYSICS
D
xD
f
4
22
−
= ⇒ cmf 21
1004
40100 22
=
×
−
=
Hence power
( )
D
cmF
P 5
21
100100
+≈==
Example: 27 Shown in figure here is a convergent lens placed inside a cell filled with a liquid. The lens has
focal length +20 cm when in air and its material has refractive index 1.50. If the liquid has
refractive index 1.60, the focal length of the system is
(a) + 80 cm (b) – 80 cm
(c) – 24 cm (d) – 100 cm
Solution: (d) Here ( )
100
3
20
11
16.1
1
1
−
=





−
∞
−=
f
.......(i)
( )
20
1
20
1
20
1
15.1
1
2
=





−
−−=
f
.......(ii)
( )
100
31
20
1
16.1
1
3
−
=





∞
−
−
−=
f
.......(iii)
By using
321
1111
fffF
++= ⇒
100
3
20
1
100
31
−+
−
=
F
⇒ cmF 100−=
Example: 28 A concave lens of focal length 20 cm placed in contact with a plane mirror acts as a RA 1998]
(a) Convex mirror of focal length 10 cm (b) Concave mirror of focal length 40 cm
(c) Concave mirror of focal length 60 cm (d) Concave mirror of focal length 10 cm
ml ffF
121
+=
Since ∞=mf ⇒ cm
f
F l
10
2
20
2
===
(After silvering concave lens behave as convex mirror)
Example: 29 A candle placed 25 cm from a lens, forms an image on a screen placed 75 cm on the other
end of the lens. The focal length and type of the lens should be
(a) + 18.75 cm and convex lens (b) – 18.75 cm and concave lens
(c) + 20.25 cm and convex lens (d) – 20.25 cm and concave lens
Solution: (a) In concave lens, image is always formed on the same side of the object. Hence the given lens
is a convex lens for which u = – 25 cm, v = 75 cm.
By using
uvf
111
−= ⇒
( ) ( )25
1
75
11
−
−
+
=
f
⇒ f = + 18.75 cm.
Example: 30 A convex lens forms a real image of an object for its two different positions on a screen. If
height of the image in both the cases be 8 cm and 2 cm, then height of the object is [KCET (Engg./Med.) 20
(a) 16 cm (b) 8 cm (c) 4 cm (d) 2 cm
Solution: (c) By using 21 IIO = ⇒ cmO 428 =×=
Example: 31 A convex lens produces a real image m times the size of the object. What will be the distance
of the object from the lens [JIPMER 2002]
(a) f
m
m





 + 1
(b) fm )1( − (c) f
m
m





 − 1
(d)
f
m 1+
uf
f
m
+
= here
( )
( ) uf
f
m
++
+
=− ⇒
f
u
f
uf
m
+=
+
=− 1
1
⇒ f
m
m
u .
1





 +
−=
Lens
Liquid
⇒ +
F Fe
Fm
⇒ + +
f1 f2 f3
F

genius PHYSICS
Example: 32 An air bubble in a glass sphere having 4 cm diameter appears 1 cm from surface nearest to
eye when looked along diameter. If 5.1=ga µ , the distance of bubble from refracting surface
is [CPMT 2002]
(a) 1.2 cm (b) 3.2 cm (c) 2.8 cm (d) 1.6 cm
Ruv
1212 µµµµ −
=−
where u = ? , v = – 1 cm, 5.11 =µ , 12 =µ , R = – 2 cm.
( )2
5.115.1
1
1
−
−
=−
− u
⇒ .2.1
5
6
cmu −=−=
Example: 33 The sun's diameter is m9
104.1 × and its distance from the earth is m11
10 . The diameter of its
image, formed by a convex lens of focal length 2m will be
(a) 0.7 cm (b) 1.4 cm (c) 2.8 cm (d) Zero (i.e.
point image)
Solution: (c) From figure
2
1011
=
d
D
⇒ .8.2
10
104.12
11
9
cmd =
××
=
Example: 34 Two point light sources are 24 cm apart. Where should a convex lens of focal length 9 cm be
put in between them from one source so that the images of both the sources are formed at
the same place
(a) 6 cm (b) 9 cm (c) 12 cm (d) 15 cm
Solution: (a) The given condition will be satisfied only if one source (S1) placed on one side such that u < f
(i.e. it lies under the focus). The other source (S2) is placed on the other side of the lens such
that u > f (i.e. it lies beyond the focus).
If 1S is the object for lens then
xyf −
−
−
=
111
⇒
fxy
111
−= ........(i)
If 2S is the object for lens then
)24(
111
xyf −−
−
+
= ⇒
)24(
111
xfy −
−= ........(ii)
From equation (i) and (ii)
)24(
1111
xffx −
−=− ⇒
9
22
)24(
11
==
−
+
fxx
⇒ 0108242
=+− xx
On solving the equation cmx 18= , 6 cm
Example: 35 There is an equiconvex glass lens with radius of each face as R and 2/3=ga µ and
3/4=wa µ . If there is water in object space and air in image space, then the focal length is
(a) 2R (b) R (c) 3 R/2 (d) 2
R
Solution: (c) Consider the refraction of the first surface i.e. refraction from rarer medium to denser
medium
I1
I2
S1 S2
x (24 –
4)24 cm
y
µ1=1.5
µ2=1
C
u
R = 2cm
v =1cm
Sun
(D)
(d)
Image
α
α
f
1011 m

genius PHYSICS
1
2112
vuR
µµµµ
+
−
=
−
⇒ Rv
vR
92
3
3
4
3
4
2
3
1
1
=⇒+
∞
=






−





Now consider the refraction at the second surface of the lens i.e.
refraction from denser medium to rarer medium
Rv
vRR






=⇒+−=
−
−
2
31
9
2
3
2
3
1
2
2
The image will be formed at a distance do R
2
3
. This is equal to the focal length of the lens.
A luminous object is placed at a distance of 30 cm from the convex lens of focal length 20 cm. On
the other side of the lens. At what distance from the lens a convex mirror of radius of curvature 10
cm be placed in order to have an upright image of the object coincident with it
[CBSE PMT 1998; JIPMER 2001, 2002]
(a) 12 cm (b) 30 cm (c) 50 cm (d) 60 cm
Solution : (c) For lens cmfcmu 20,30 == , hence by using
uvf
111
−= ⇒ cmv
v
60
30
11
20
1
=⇒
−
−=
+
The final image will coincide the object, if light ray falls normally
on convex mirror as shown. From figure it is seen clear that
reparation between lens and mirror is 60 – 10 = 50 cm.
A convex lens of local length 30 cm and a concave lens of 10 cm focal length are placed so as to have
the same axis. If a parallel beam of light falling on convex lens leaves concave lens as a parallel
beam, then the distance between two lenses will be
(a) 40 cm (b) 30 cm (c) 20 cm (d) 10 cm
Solution : (c) According to figure the combination behaves as plane glass plate (i.e., F= ∞)
Hence by using
2121
111
ff
d
ffF
−+=
⇒
( )( )103010
1
30
11
−
−
−
+
+
=
∞
d
⇒ d = 20 cm
I
O
30 cm
60 cm
10 cm
d
Tricky example: 4
Tricky example: 5
I1I
Water Air

genius PHYSICS
Prism
Prism is a transparent medium bounded by refracting surfaces, such that the incident surface (on which
light ray is incidenting) and emergent surface (from which light rays emerges) are plane and non
parallel.
Commonly used prism :
(1) Refraction through a prism
21 rrA += and δ+=+ Aei
For surface
1sin
sin
r
i
AC =µ ; For surface AB
e
r
sin
sin 2
=µ
(2) Deviation through a prism
For thin prism A)1( −= µδ . Also deviation is different for different colour light e.g. VR µµ <
so VR δδ < . And CrownFlint µµ > so CF δδ >
Maximum deviation Minimum deviation
In this condition of maximum deviation
,90 o
i =∠ ,1 Cr = CAr −=2 and from
Snell’s law on emergent surface





 −
= −
C
CA
e
sin
)sin(
sin 1
It is observed if
ei ∠=∠ and rrr =∠=∠ 21
then :
(i) Refracted ray inside the prism is parallel to the base
of the prism
(ii)
2
A
r = and
2
mA
i
δ+
=
(iii)
2/sin
sin
A
i
=µ or
2/sin
2
sin
A
A mδ
µ
+
=
Note : If Am =δ then 2/cos2 A=µ
Equilateral prism Right angle prism Right angled isosceles
prism
A
A
B
δ
i r1 r2
C
e
µ
i – Angle of incidence, e – Angle of emergence,
A – Angle of prism or refracting angle of prism,
r1 and r2 – Angle of refraction,δ – Angle of
deviation
ei
r r
δ
δm
i
e
r1 = C
r2
i = 90o
δma
x

genius PHYSICS
(3) Normal incidence on a prism
If light ray incident normally on any surface of prism as shown
In any of the above case use
A
i
sin
sin
=µ and Ai −=δ
(4) Grazing emergence and TIR through a prism
When a light ray falls on one surface of prism, it is not necessary that it will exit out from the
prism. It may or may not be exit out as shown below
Normal incidence
A = angle of prism and C = Critical angle for the material of
the prism
Grazing incidence
Note : For the condition of grazing emergence. Minimum angle of incidence



 −−= −
cosAsinAµsinimin 121
.
(5) Dispersion through a prism
The splitting of white light into it’s constituent colours is called dispersion of light.
(i) Angular dispersion (θ ) : Angular separation between extreme colours i.e.
Aµµδδθ RVRV )( −=−= . It depends upon µ and A.
(ii) Dispersive power (ω) :





 +
=
−
−
==
2
ewher
1
RV
y
y
RV
y
µµ
µ
µ
µµ
δ
θ
ω
⇒ It depends only upon the material of the prism i.e. µ and it doesn't depends upon angle of
prism A
Note : Remember CrownFlint ωω > .
er2
i = 0o
r1 = 0o
i r1
e = 0o
r2 = 0o
orand
Ray – 3: TIR
A > C and
µ > cosec A
Ray – 2: Grazing emergence
A = C and
µ = cosec A
Ray –1 : General emergence
A < C and
µ < cosec A
Ray – 3: TIR
A > 2C and
µ > cosec (A/2)
Ray – 2: Grazing emergence
A = 2C and
µ = cosec (A/2)
Ray –1 : General emergence
A < 2C and
µ < cosec (A/2)
Ray - 1
Ray - 2
Ray - 3
A
Ray - 1
Ray - 2
Ray - 3
A
V
Y
R
Screen
Incident
white light

genius PHYSICS
(6) Combination of prisms
Two prisms (made of crown and flint material) are combined to get either dispersion only or
deviation only.
Dispersion without deviation (chromatic combination) Deviation without dispersion (Achromatic
combination)
(i)
)1'(
)1('
−
−
−=
y
y
A
A
µ
µ
(i)
)''(
)('
RV
RV
A
A
µµ
µµ
−
−
−=
(ii) )''(
'
1net δωωδ
ω
ω
θθ −=





−= (ii) 





−=
'
1net
ω
ω
δδ
Scattering of Light
Molecules of a medium after absorbing incoming light radiations, emits them in all direction.
This phenomenon is called Scattering.
(1) According to scientist Rayleigh : Intensity of scattered light 4
1
λ
∝
(2) Some phenomenon based on scattering : (i) Sky looks blue due to scattering.
(ii) At the time of sunrise or sunset it looks reddish. (iii) Danger signals are made from red.
(3) Elastic scattering : When the wavelength of radiation remains unchanged, the
scattering is called elastic.
(4) Inelastic scattering (Raman’s effect) : Under specific condition, light can also suffer
inelastic scattering from molecules in which it’s wavelength changes.
Rainbow
Rainbow is formed due to the dispersion of light suffering refraction and TIR in the droplets present in
the atmosphere.
(1) Primary rainbow : (i) Two refraction and one
TIR. (ii) Innermost arc is violet and outermost is red. (iii)
Subtends an angle of o
42 at the eye of the observer. (iv)
More bright
(2) Secondary rainbow : (i) Two refraction and two
TIR. (ii) Innermost arc is red and outermost is violet.
(iii) It subtends an angle of o
5.52 at the eye. (iv)
Comparatively less bright.
A
A′
Flint
Crown
R
V
V
R A
A′
Flint
Crown
R
V
Re
dViolet
42o40o

genius PHYSICS
Colours
Colour is defined as the sensation received by the eye (cone cells of the eye) due to light coming from
object.
(1) Types of colours
Spectral colours Colours of pigment and dyes
(i) Complementary colours :
Green and magenta
Blue and yellow
Red and cyan
(i) Complementary colours :
yellow and mauve
Red and green
Blue and orange
(ii) Combination :
Green + red + blue = White
Blue + yellow = White
Red + cyan = White
Green + magenta = White
(ii) Combination :
Yellow + red + blue = Black
Blue + orange = Black
Red + green = Black
Yellow + mauve = Black
(2) Colours of object : The perception of a colour by eye depends on the nature of object
and the light incident on it.
Colours of opaque object Colours of transparent object
(i) Due to selective reflection. (i) Due to selective transmission.
(ii) A rose appears red in white light because it reflects
red colour and absorbs all remaining colours.
(ii) A red glass appears red because it absorbs all
colours, except red which it transmits.
(iii) When yellow light falls on a bunch of flowers, then
yellow and white flowers looks yellow. Other flowers
looks black.
(iii) When we look on objects through a green glass or
green filter then green and white objects will appear
green while other black.
Note : A hot object will emit light of that colour only which it has observed when it was
heated.
Spectrum.
The ordered arrangements of radiations according to wavelengths or frequencies is called
Spectrum. Spectrum can be divided in two parts (I) Emission spectrum and (II) Absorption
spectrum.
(1) Emission spectrum : When light emitted by a self luminous object is dispersed by
a prism to get the spectrum, the spectrum is called emission spectra.
white
Yellow (S)Cyan (S)
Magenta (S)
Blue (P) Red (P)
Green (P)
Black
Orange (S)Green (S)
Radish violet (S)
(Mauve)
Blue (P) Red (P)
Yellow (P)

genius PHYSICS
Continuous emission
spectrum
Line emission spectrum Band emission
spectrum
(i) It consists of continuously
varying wavelengths in a definite
wavelength range.
(i) It consist of distinct bright lines. (iii) It consist of district bright
bands.
(ii) It is produced by solids, liquids
and highly compressed gases heated
to high temperature.
(ii) It is produced by an excited
source in atomic state.
(ii) It is produced by an
excited source in molecular
state.
(iii) e.g. Light from the sun, filament
of incandescent bulb, candle flame
etc.
(iii) e.g. Spectrum of excited
helium, mercury vapours, sodium
vapours or atomic hydrogen.
(iii) e.g. Spectra of molecular
,2H CO, 3NH etc.
(2) Absorption spectrum : When white light passes through a semi-transparent solid, or
liquid or gas, it’s spectrum contains certain dark lines or bands, such spectrum is called absorption
spectrum (of the substance through which light is passed).
(i) Substances in atomic state produces line absorption spectra. Polyatomic substances such
as ,2H 2CO and 4KMnO produces band absorption spectrum.
(ii) Absorption spectra of sodium vapour have two (yellow lines) wavelengths )5890(1 ÅD and
)5896(2 ÅD
Note : If a substance emits spectral lines at high temperature then it absorbs the same lines
at low temperature. This is Kirchoff’s law.
(3) Fraunhoffer’s lines : The central part (photosphere) of the sun is very hot and emits all
possible wavelengths of the visible light. However, the outer part (chromosphere) consists of
vapours of different elements. When the light emitted from the photosphere passes through the
chromosphere, certain wavelengths are absorbed. Hence, in the spectrum of sunlight a large
number of dark lines are seen called Fraunhoffer lines.
(i) The prominent lines in the yellow part of the visible spectrum were labelled as D-lines,
those in blue part as F-lines and in red part as C-line.
(ii) From the study of Fraunhoffer’s lines the presence of various elements in the sun’s
atmosphere can be identified e.g. abundance of hydrogen and helium.
(4) Spectrometer : A spectrometer is used for obtaining pure spectrum of a source in
laboratory and calculation of µ of material of prism and µ of a transparent liquid.
It consists of three parts : Collimator which provides a parallel beam of light; Prism Table for
holding the prism and Telescope for observing the spectrum and making measurements on it.
The telescope is first set for parallel rays and then collimator is set for parallel rays. When
prism is set in minimum deviation position, the spectrum seen is pure spectrum. Angle of prism
(A) and angle of minimum deviation )( mδ are measured and µ of material of prism is calculated
using prism formula. For µ of a transparent liquid, we take a hollow prism with thin glass sides.
Fill it with the liquid and measure )( mδ and A of liquid prism. µ of liquid is calculated using prism
formula.
(5) Direct vision spectroscope : It is an instrument used to observe pure spectrum. It
produces dispersion without deviation with the help of n crown prisms and )1( −n flint prisms
alternately arranged in a tabular structure.

genius PHYSICS
For no deviation ')1'()1()1( AnAn −−=− µµ .
Concepts
When a ray of white light passes through a glass prism red light is deviated less than blue light.
For a hollow prism 0≠A but 0=δ
If an opaque coloured object or crystal is crushed to fine powder it will appear white (in sun light) as it will lose it's
property of selective reflection.
Our eye is most sensitive to that part at the spectrum which lies between the F line (sky green) one the C-line (red)
of hydrogen equal to the refractive index for the D line (yellow) of sodium. Hence for the dispersive power, the
following formula is internationally accepted
1−
−
=
D
CF
µ
µµ
ω
Sometimes a part of prism is given and we keep on thinking whether how should we proceed ? To solve such
problems first complete the prism then solve as the problems of prism are solved
Some other types of prism
Example: 36 When light rays are incident on a prism at an angle of 45o, the minimum deviation is
obtained. If refractive index of the material of prism is 2 , then the angle of prism will be
(a) 30o (b) 40o (c) 50o (d) 60o
Solution: (d)
2
sin
sin
A
i
=µ ⇒
2
sin
45sin
2
A
= ⇒
2
1
2
2
1
2
sin ==
A oo
A
A
6030
2
=⇒=⇒
Example: 37 Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism.
The angle of prism is )75.041(cos =o
(a) 62o (b) 41o (c) 82o (d) 31o
Solution: (c) Given Am =δ , then by using
2
sin
2
sin
A
A mδ
µ
+
=
2
sin
2
sin
A
AA +
=⇒ µ
2
cos2
2
sin
sin A
A
A
==






=
2
cos
2
sin2sin
AA
A
⇒ oo
A
AAA
82
2
41
2
cos75.0
2
cos25.1 =⇒=⇒=⇒= .
Example: 38 Angle of glass prism is 60o and refractive index of the material of the prism is 1.414,then
what will be the angle of incidence, so that ray should pass symmetrically through prism
(a) 38o 61' (b) 35o 35' (c) 45o (d) 53o 8'
60o 70o
⇒
60o 70o
50o
A
CB

genius PHYSICS
Solution: (c) incident ray and emergent ray are symmetrical in the cure, when prism is in minimum
deviation position.
Hence in this condition
2
sin
sin
A
i
=µ ⇒ 





=
2
sinsin
A
i µ ⇒ oo
ii 45
2
1
30sin414.1sin =⇒=×=
Example: 39 A prism )5.1( =µ has the refracting angle of 30o. The deviation of a monochromatic ray
incident normally on its one surface will be )75.0'3648(sin =o
(a) 18o 36' (b) 20o 30' (c) 18o (d) 22o 1'
Solution: (a) By using '364875.0sin
30sin
sin
5.1
sin
sin o
ii
i
A
i
=⇒=⇒=⇒=µ
Also from '361830'3648 ooo
Ai =−=⇒−= δδ
Example: 40 Angle of a prism is 30o and its refractive index is 2 and one of the surface is silvered. At
what angle of incidence, a ray should be incident on one surface so that after reflection from
the silvered surface, it retraces its path
(a) 30o (b) 60o (c) 45o (d) 5.1sin 1−
Solution: (c) This is the case when light ray is falling normally an second surface.
Hence by using
A
i
sin
sin
=µ ⇒ o
o
ii
i
45
2
1
2sin
30sin
sin
2 =⇒×=⇒=
Example: 41 The refracting angle of prism is A and refractive index of material of prism is
2
cot
A
. The
angle of minimum deviation is
(a) Ao
3180 − (b) Ao
2180 + (c) Ao
−90 (d) Ao
2180 −
Solution: (d) By using
2
sin
2
sin
A
A mδ
µ
+
= ⇒
2
sin
2
sin
2
sin
2
cos
2
sin
2
sin
2
cot
A
A
A
A
A
A
A
mm δδ +
=⇒
+
=
⇒ 




 +
=





−
2
sin
2
90sin mAA δ
22
90 mAA δ+
=−⇒ Am 2180 −=⇒ δ
Example: 42 A ray of light passes through an equilateral glass prism in such a manner that the angle of
incidence is equal to the angle of emergence and each of these angles is equal to 3/4 of the
angle of the prism. The angle of deviation is
(a) 45o (b) 39o (c) 20o (d) 30o
Solution: (d) Given that o
A 60= and o
Aei 4560
4
3
4
3
=×===
By using δ+=+ Aei ⇒ o
30604545 =⇒+=+ δδ
Example: 43 PQR is a right angled prism with other angles as 60o and 30o. Refractive index of prism is
1.5. PQ has a thin layer of liquid. Light falls normally on the face PR. For total internal
reflection, maximum refractive index of liquid is
(a) 1.4
(b) 1.3
(c) 1.2
(d) 1.6
Solution: (c) For TIR at PQ C<θ
R
P Q
60
°
30°
θ

genius PHYSICS
From geometry of figure 60=θ i.e. CC sin60sin60 >⇒>
⇒ ismLiquid
ism
Liquid
Pr
Pr 2
3
2
3
µµ
µ
µ
×<⇒> 3.15.1
2
3
<⇒×<⇒ LiquidLiquid µµ .
Example: 44 Two identical prisms 1 and 2, each will angles of 30o, 60o and 90o are placed in contact as
shown in figure. A ray of light passed through the combination in the position of minimum
deviation and suffers a deviation of 30o. If the prism 2 is removed, then the angle of
deviation of the same ray is [PMT (Andhra) 1995]
(a) Equal to 15o
(b) Smaller than 30o
(c) More than 15o
(d) Equal to 30o
Solution: (a) A)1( −= µδ as A is halved, so δ is also halves
Example: 45 A prism having an apex angle 4o and refraction index 1.5 is located in front of a vertical plane
mirror as shown in figure. Through what total angle is the ray deviated after reflection from
the mirror
(a) 176o (b) 4o
(c) 178o (d) 2o
Solution: (c) oo
ism A 24)15.1()1(Pr =−=−= µδ
∴ MirrorismTotal δδδ += Pr
oo
iA 178)22180(2)2180()1( =×−+=−+−= µ
Example: 46 A ray of light is incident to the hypotenuse of a right-angled prism
after travelling parallel to the base inside the prism. If µ is the refractive index of the
material of the prism, the maximum value of the base angle for which light is totally
reflected from the hypotenuse is [EAMCET
2003]
(a) 




−
µ
1
sin 1
(b) 




−
µ
1
tan 1
(c) 




 −−
µ
µ 1
sin 1
(d) 




−
µ
1
cos 1
Solution: (d) If =α maximum value of vase angle for which light is totally reflected from hypotenuse.
==− C)90( α minimum value of angle of incidence an hypotenuse for TIR
µ
α
1
sin)90sin( ==− C ⇒ 





= −
µ
α
1
cos 1
Example: 47 If the refractive indices of crown glass for red, yellow and violet colours are 1.5140, 1.5170
and 1.5318 respectively and for flint glass these are 1.6434, 1.6499 and 1.6852 respectively,
then the dispersive powers for crown and flint glass are respectively
(a) 0.034 and 0.064 (b) 0.064 and 0.034 (c) 1.00 and 0.064 (d) 0.034 and 1.0
Solution: (a) 034.0
)15170.1(
5140.15318.1
1
Crown =
−
−
=
−
−
=
y
rv
µ
µµ
ω and
1'
''
Flint
−
−
=
y
rv
µ
µµ
ω 064.0
16499.1
6434.16852.1
=
−
−
=
Example: 48 Flint glass prism is joined by a crown glass prism to produce dispersion without deviation.
The refractive indices of these for mean rays are 1.602 and 1.500 respectively. Angle of
prism of flint prism is 10o, then the angle of prism for crown prism will be
(a) '4.212 o
(b) '412 o
(c) o
24.1 (d) 12o
2°
2°
2°
4°
(90-α)
(90-α)
90° α
30o
30o
60o
60o90o 90o

genius PHYSICS
Solution: (a) For dispersion without deviation
)1(
)1(
−
−
=
C
F
F
C
A
A
µ
µ
⇒
)1500.1(
)1602.1(
10 −
−
=
A
⇒ '4.21204.12 oo
A ==
An achromatic prism is made by crown glass prism )19( o
CA = and flint glass prism
)6( o
FA = . If 5.1=v
Cµ and 66.1=v
Fµ , then resultant deviation for red coloured ray will be
(a) 1.04o (b) 5o (c) 0.96o (d) 13.5o
Solution : (d) For achromatic combination FC ww −= ⇒ FrvCrv AA ])[(])[( µµµµ −−=−
⇒ FvCvFrCr AAAA ][][][][ µµµµ +=+ 5.3866.16195.1 =×+×=
Resultant deviation FrCr AA ])1[(])1[( −+−= µµδ
)(][][ FCFrCr AAAA +−+= µµ o
5.13)619(5.38 =+−=
The light is incident at an angle of 60o on a prism of which the refracting angle of prism is
30o. The refractive index of material of prism will be
(a) 2 (b) 32 (c) 2 (d) 3
Solution : (d) By using δ+=+ Aei ⇒ 0303060 =⇒+=+ ee .
Hence ray will emerge out normally so by using the formula
3
30sin
60sin
sin
sin
===
A
i
µ
60°
Tricky example: 6
Tricky example: 7

genius PHYSICS
Human Eye.
(1) Eye lens : Over all behaves as a convex lens of 437.1=µ
(2) Retina : Real and inverted image of an object, obtained at retina, brain sense it erect.
(3) Yellow spot : It is the most sensitive part, the image formed at yellow spot is brightest.
(4) Blind spot : Optic nerves goes to brain through blind spot. It is not sensitive for light.
(5) Ciliary muscles – Eye lens is fixed between these muscles. It’s both radius of curvature
can be changed by applying pressure on it through ciliary muscles.
(6) Power of accomodation : The ability of eye to see near objects as well as far objects is
called power of accomodation.
Note : When we look distant objects, the eye is relaxed and it's focal length is largest.
(7) Range of vision : For healthy eye it is 25 cm (near point) to ∞ (far point).
A normal eye can see the objects clearly, only if they are at a distance greater than 25 cm. This
distance is called Least distance of distinct vision and is represented by D.
(8) Persistence of vision : Is 1/10 sec. i.e. if time interval between two consecutive light
pulses is lesser than 0.1 sec., eye cannot distinguish them separately.
(9) Binocular vision : The seeing with two eyes is called binocular vision.
(10) Resolving limit : The minimum angular displacement between two objects, so that
they are just resolved is called resolving limit. For eye it is
o






=
60
1
1'
.
Specific Example
A person wishes to distinguish between two pillars located at a distances of 11 Km. What should be the minimum
distance between the pillars.
Solution : As the limit of resolution of eye is
o






60
1
So
o






>
60
1
θ md
d
2.3
18060
1
1011 3
>⇒×





>
×
⇒
π
(11) Defects in eye
Myopia (short sightness) Hypermetropia (long sightness)
(i) Distant objects are not seen clearly but nearer
objects are clearly visible.
(i) Distant objects are seen clearly but nearer object
are not clearly visible.
(ii) Image formed before the retina. (ii) Image formed behind the retina.
d
11 km
θ

genius PHYSICS
(iii) Far point comes closer. (iii) Near point moves away
(iv) Reasons :
(a) Focal length or radii of curvature of lens reduced
or power of lens increases.
(b) Distance between eye lens and retina increases.
(iv) Reasons :
(a) Focal length or radii of curvature of lens increases
or power of lens decreases.
(b) Distance between eye lens and retina decreases.
(v) Removal : By using a concave lens of suitable focal
length.
(v) Removal : By using a convex lens.
(vi) Focal length :
(a) A person can see upto distance → x
wants to see → ∞, so
focal length of used lens xf −= = – (defected far point)
(b) A person can see upto distance → x
wants to see distance → y (y > x)
so
yx
xy
f
−
=
(vi) Focal length :
(a) A person cannot see before distance → d
wants to see the object place at distance → D
so
Dd
dD
f
−
=
Presbyopia : In this defect both near and far objects are not clearly visible. It is an old age
disease and it is due to the loosing power of accommodation. It can be removed by using bifocal
lens.
Astigmatism : In this defect eye cannot see horizontal and vertical lines clearly,
simultaneously. It is due to imperfect spherical nature of eye lens. This defect can be removed by
using cylindrical lens (Torric lenses).
Microscope.
It is an optical instrument used to see very small objects. It’s magnifying power is given by
)(visiondistinctofdistanceleastatplacedisobjectwhenangleVisual
)(instrumentwithangleVisual
α
β
=m
(1) Simple miscroscope
(i) It is a single convex lens of lesser focal length.
(ii) Also called magnifying glass or reading lens.
(iii) Magnification’s, when final image is formed at D
and ∞ (i.e. Dm and ∞m )
max
1 





+=
f
D
mD and
min






=∞
f
D
m
Note : 1.min.max =− mm
Concave
Convex

genius PHYSICS
If lens is kept at a distance a from the eye then
f
aD
m D
−
+= 1 and
f
aD
m
−
=∞
(2) Compound microscope
(i) Consist of two converging lenses called
objective and eye lens.
(ii) objectivelenseye ff > and
objectivelenseye )diameter(diameter)( >
(iii) Final image is magnified, virtual and
inverted.
(iv) =0u Distance of object from objective (o),
=0v Distance of image )( BA ′′ formed by objective from objective, =eu Distance of BA ′′ from eye
lens, ve = Distance of final image from eye lens, f0 = Focal length of objective, fe = Focal length of eye
lens.
Magnification : 





+
−
−=





+
−
−=





+−=
eee
D
f
D
f
fv
f
D
fu
f
f
D
u
v
m 1
)(
1
)(
1
0
00
00
0
0
0
eee F
D
f
fv
f
D
fu
f
F
D
u
v
m .
)(
)(
.
0
00
00
0
0
0 −
−=





−
−
=−=∞
Length of the tube (i.e. distance between two lenses)
When final image is formed at D ;
Df
Df
fu
fu
uvL
e
e
eD
+
+
−
=+=
00
00
0
When final images is formed at ∞ ; ee f
fu
fu
fvL +
−
=+=∞
00
00
0
(Do not use sign convention while solving the problems)
Note :
e
e
ff
DffL
m
0
0 )( −−
= ∞
∞
For maximum magnification both 0f and ef must be less.
lenseyeobjective mmm ×=
If objective and eye lens are interchanged, practically there is no change in
magnification.
(3) Resolving limit and resolving power : In reference to a microscope, the minimum
distance between two lines at which they are just distinct is called Resolving limit (RL) and it’s
reciprocal is called Resolving power (RP)
θµ
λ
sin2
.. =LR and ⇒=
λ
θµ sin2
..PR
λ
1
.. ∝PR
λ = Wavelength of light used to illuminate the object,
µ = Refractive index of the medium between object and objective,
θ = Half angle of the cone of light from the point object, θµ sin = Numerical aperture.

genius PHYSICS
Note : Electron microscope : electron beam )1( Å≈λ is used in it so it’s R.P. is approx 5000
times more than that of ordinary microscope )5000( Å≈λ
Telescope.
By telescope distant objects are seen.
(1) Astronomical telescope
(i) Used to see heavenly bodies.
(ii) eyelensobjective ff > and lenseyeobjective dd > .
(iii) Intermediate image is real, inverted and small.
(iv) Final image is virtual, inverted and small.
(v) Magnification : 





+−=
D
f
f
f
m e
e
D 10
and
e
o
f
f
m −=∞
(vi) Length :
Df
Df
fufL
e
e
eD
+
+=+= 00 and effL +=∞ 0
(2) Terrestrial telescope
(i) Used to see far off object on the earth.
(ii) It consists of three converging lens : objective,
eye lens and erecting lens.
(iii) It’s final image is virtual erect and smaller.
(iv) Magnification : 





+=
D
f
f
f
m e
e
D 10
and
ef
f
m 0
=∞
(v) Length :
Df
Df
ffuffL
e
e
eD
+
++=++= 44 00 and efffL ++=∞ 40
(3) Galilean telescope
(i) It is also a terrestrial telescope but of much smaller field of view.
(ii) Objective is a converging lens while eye lens is diverging lens.
(iii) Magnification : 





−=
D
f
f
f
m e
e
D 10
and
ef
f
m 0
=∞
(iv) Length : eD ufL −= 0 and effL −=∞ 0
(4) Resolving limit and resolving power
Smallest angular separations (dθ) between two distant objects, whose images are separated in
the telescope is called resolving limit. So resolving limit
a
d
λ
θ
22.1
=
and resolving power ⇒==
λθ 22.1
1
)(
a
d
RP
λ
1
.. ∝PR where a = aperture of objective.
B Q B'
A'
O
A
B''
A''
Erecting
lens
ue = D to ∞
P
fo 2f 2f uo

genius PHYSICS
Note :Minimum separation (d) between objects, so they can just resolved by a telescope is –
..PR
r
d = where r = distance of objects from telescope.
(5) Binocular
If two telescopes are mounted parallel to each other so that an object can be seen by both the eyes
simultaneously, the arrangement is called 'binocular'. In a binocular, the
length of each tube is reduced by using a set of totally reflecting prisms which
provided intense, erect image free from lateral inversion. Through a
binocular we get two images of the same object from different angles at same
time. Their superposition gives the perception of depth also along with length
and breadth, i.e., binocular vision gives proper three-dimensional (3D)
image.
Concepts
As magnifying power is negative, the image seen in astronomical telescope is truly inverted, i.e., left is turned right
with upside down simultaneously. However, as most of the astronomical objects are symmetrical this inversion does
not affect the observations.
Objective and eye lens of a telescope are interchanged, it will not behave as a microscope but object appears very small.
In a telescope, if field and eye lenses are interchanged magnification will change from (fo / fe) to (fe / fo), i.e., it will
change from m to (1/m), i.e., will become (1/m2) times of its initial value.
As magnification for normal setting as (fo / fe), so to have large magnification, fo must be as large as practically
possible and fe small. This is why in a telescope, objective is of large focal length while eye piece of small.
In a telescope, aperture of the field lens is made as large as practically possible to increase its resolving power as
resolving power of a telescope ∝ (D/λ)*. Large aperture of objective also helps in improving the brightness of
image by gathering more light from distant object. However, it increases aberrations particularly spherical.
For a telescope with increase in length of the tube, magnification decreases.
In case of a telescope if object and final image are at infinity then :
d
D
f
f
m
e
o
==
If we are given four convex lenses having focal lengths 4321 ffff >>> . For making a good telescope and
microscope. We choose the following lenses respectively. Telescope )(),( 41 efof Microscope )(),( 34 efof
If a parrot is sitting on the objective of a large telescope and we look towards (or take a photograph)of distant
astronomical object (say moon) through it, the parrot will not be seen but the intensity of the image will be slightly
reduced as the parrot will act as obstruction to light and will reduce the aperture of the objective.
Example: 1 A man can see the objects upto a distance of one metre from his eyes. For correcting his eye
sight so that he can see an object at infinity, he requires a lens whose power is
or
A man can see upto 100 cm of the distant object. The power of the lens required to see far
objects will be
[MP PMT 1993, 2003]
fe
fo
Example
s

genius PHYSICS
(a) +0.5 D (b) +1.0 D (c) +2.0 D (d) –1.0 D
Solution: (d) f = –(defected far point) = – 100 cm. So power of the lens D
f
P 1
100
100100
−=
−
==
Example: 2 A man can see clearly up to 3 metres. Prescribe a lens for his spectacles so that he can see clearly
up to 12 metres
[DPMT 2002]
(a) – 3/4 D (b) 3 D (c) – 1/4 D (d) – 4 D
Solution: (c) By using
yx
xy
f
−
= ⇒ mf 4
123
123
−=
−
×
= . Hence power D
f
P
4
11
−==
Example: 3 The diameter of the eye-ball of a normal eye is about 2.5 cm. The power of the eye lens varies
from
(a) 2 D to 10 D (b) 40 D to 32 D (c) 9 D to 8 D (d) 44 D to 40
D
Solution: (d) An eye sees distant objects with full relaxation so
f
11
105.2
1
2
=
∞−
−
× −
or
D
f
P 40
1025
11
2
=
×
== −
An eye sees an object at 25 cm with strain so
f
1
1025
1
105.2
1
22
=
×−
−
× −−
or
D
f
P 44440
1
=+==
Example: 4 The resolution limit of eye is 1 minute. At a distance of r from the eye, two persons stand
with a lateral separation of 3 metre. For the two persons to be just resolved by the naked eye,
r should be
(a) 10 km (b) 15 km (c) 20 km (d) 30 km
Solution: (a) From figure
r
d
=θ ; where rad
o
18060
1
60
1
'1
π
θ ×





=





==
⇒
r
3
18060
1
1 =××
π
⇒ r = 10 km
Example: 5 Two points separated by a distance of 0.1 mm can just be resolved in a microscope when a
light of wavelength 6000 Å is used. If the light of wavelength 4800 Å is used this limit of
resolution becomes
[UPSEAT 2002]
(a) 0.08 mm (b) 0.10 mm (c) 0.12 mm (d) 0.06 mm
Solution: (a) By using resolving limit (R.L.) ∝ λ ⇒
2
1
2
1
.).(
.).(
λ
λ
=
LR
LR
⇒
4800
6000
.).(
1.0
2
=
LR
⇒ mmLR 08.0.).( 2 = .
Example: 6 In a compound microscope, the focal lengths of two lenses are 1.5 cm and 6.25 cm an object
is placed at 2 cm form objective and the final image is formed at 25 cm from eye lens. The
distance between the two lenses is
[EAMCET (Med.) 2000]
(a) 6.00 cm (b) 7.75 cm (c) 9.25 cm (d) 11.00 cm
Solution: (d) It is given that fo = 1.5 cm, fe = 6.25 cm, uo = 2 cm
When final image is formed at least distance of distinct vision, length of the tube
Df
Df
fu
fu
L
e
e
oo
oo
D
+
+
−
=
⇒ cmLD 11
)2525.6(
2525.6
)5.12(
5.12
=
+
×
+
−
×
= .
d = 3m
r
θ

genius PHYSICS
Example: 7 The focal lengths of the objective and the eye-piece of a compound microscope are 2.0 cm
and 3.0 cm respectively. The distance between the objective and the eye-piece is 15.0 cm.
The final image formed by the eye-piece is at infinity. The two lenses are thin. The distances
in cm of the object and the image produced by the objective measured from the objective
lens are respectively [IIT-JEE 1995]
(a) 2.4 and 12.0 (b) 2.4 and 15.0 (c) 2.3 and 12.0 (d) 2.3 and
3.0
Solution: (a) Given that cmfo 2= , cmfe 3= , cmL 15=∞
By using eo fvL +=∞ ⇒ 315 += ov ⇒ cmvo 12= . Also
o
oo
o
o
f
fv
u
v −
= ⇒
2
21212 −
=
ou
⇒
cmuo 4.2= .
Example: 8 The focal lengths of the objective and eye-lens of a microscope are 1 cm and 5 cm
respectively. If the magnifying power for the relaxed eye is 45, then the length of the tube is
(a) 30 cm (b) 25 cm (c) 15 cm (d) 12 cm
Solution: (c) Given that cmfo 1= , cmfe 5= , 45=∞m
By using
eo
eo
ff
ffL
m
)( −−
= ∞
∞ ⇒
51
25)51(
45
×
×−−
= ∞L
⇒ cmL 15=∞
Example: 9 If the focal lengths of objective and eye lens of a microscope are 1.2 cm and 3 cm respectively
and the object is put 1.25 cm away from the objective lens and the final image is formed at
infinity, then magnifying power of the microscope is
(a) 150 (b) 200 (c) 250 (d) 400
Solution: (b) Given that cmfo 2.1= , cmfe 3= , cmuo 25.1=
By using
eoo
o
f
D
fu
f
m ⋅
−
−=∞
)(
⇒ 200
3
25
)2.125.1(
2.1
−=×
−
−=∞m .
Example: 10 The magnifying power of an astronomical telescope is 8 and the distance between the two
lenses is 54cm. The focal length of eye lens and objective lens will be respectively [MP PMT 1991; CPMT 1991
(a) 6 cm and 48 cm (b) 48 cm and 6 cm (c) 8 cm and 64 cm (d) 64 cm and
8 cm
Solution: (a) Given that 8=∞m and 54=∞L
By using
e
o
f
f
m =∞ || and eo ffL +=∞ we get cmfo 6= and cmfe 48= .
Example: 11 If an object subtend angle of 2° at eye when seen through telescope having objective and
eyepiece of focal length cmfo 60= and cmfe 5= respectively than angle subtend by image at
eye piece will be [UPSEAT 2001]
(a) 16° (b) 50° (c) 24° (d) 10°
Solution: (c) By using
e
o
f
f
=
α
β
⇒
5
60
20
=
β
⇒ °= 24β
Example: 12 The focal lengths of the lenses of an astronomical telescope are 50 cm and 5 cm. The length
of the telescope when the image is formed at the least distance of distinct vision is
(a) 45 cm (b) 55 cm (c)
6
275
cm (d)
6
325
cm
Solution: (d) By using cm
Df
Df
fufL
e
e
oeoD
6
325
)255(
255
50 =
+
×
+=
+
+=+=
Example: 13 The diameter of moon is km3
105.3 × and its distance from the earth is km5
108.3 × . If it is
seen through a telescope whose focal length for objective and eye lens are 4 m and 10 cm
respectively, then the angle subtended by the moon on the eye will be approximately
(a) 15° (b) 20° (c) 30° (d) 35°

genius PHYSICS
Solution: (b) The angle subtended by the moon on the objective of telescope
2
5
3
10
8.3
5.3
108.3
105.3 −
×=
×
×
=α rad
Also
α
β
==
e
o
f
f
m ⇒
α
β
=
10
400
⇒ αβ 40= ⇒ °=×
×
×
×= 20
180
108.3
105.3
40 5
3
π
β
Example: 14 A telescope has an objective lens of 10 cm diameter and is situated at a distance one
kilometre from two objects. The minimum distance between these two objects, which can be
resolved by the telescope, when the mean wavelength of light is 5000 Å, is of the order of
(a) 0.5 m (b) 5 m (c) 5 mm (d) 5cm
Solution: (b) Suppose minimum distance between objects is x and their distance from telescope is r
So Resolving limit
r
x
a
d ==
λ
θ
22.1
⇒ mmm
a
r
x 1.6101.6
)10(
)101()105000(22.122.1 3
310
=×=
−
××××
=
×
= −
−
λ
Hence, It’s order is ≈ 5 mm.
Example: 15 A compound microscope has a magnifying power 30. The focal length of its eye-piece is 5
cm. Assuming the final image to be at the least distance of distinct vision. The magnification
produced by the objective will be
(a) +5 (b) – 5 (c) +6 (d) – 6
Solution (b) Magnification produced by compound microscope eo mmm ×=
where mo = ? and 6
5
25
11 =+=







+=
e
e
f
D
m ⇒ 630 ×−= om ⇒ 5−=om .
Tricky Example 1 : A man is looking at a small object placed at his least distance of distinct vision. Without
changing his position and that of the object he puts a simple microscope of magnifying power 10 X and just sees the
clear image again. The angular magnification obtained is
(a) 2.5 (b) 10.0 (c) 5.0 (d) 1.0
Solution : (d) Angular magnification
O
I
DO
DI
====
/
/
tan
tan
α
β
α
β
Since image and object are at the same position, 1==
u
v
O
I
⇒ Angular magnification = 1
Tricky Example 2: A compound microscope is used to enlarge an object kept at a distance 0.03m from it’s
objective which consists of several convex lenses in contact and has focal length 0.02m. If a lens of focal length 0.1m
is removed from the objective, then by what distance the eye-piece of the microscope must be moved to refocus the
image
(a) 2.5 cm (b) 6 cm (c) 15 cm (d) 9 cm
Solution : (d) If initially the objective (focal length Fo) forms the image at distance vo then
cm
fu
fu
v
oo
oo
o 6
23
23
=
−
×
=
−
=
Now as in case of lenses in contact
oo FffffF ′
+=+++=
11
.....
1111
1321 





++=
′
.....
111
where
32o ffF
So if one of the lens is removed, the focal length of the remaining lens system
10
1
2
1111
10
−=−=
′ fFFo
⇒ cmFo 5.2=′
This lens will form the image of same object at a distance ov′ such that cm
Fu
Fu
v
oo
oo
o 15
)5.23(
5.23
=
−
×
=
′−
′
=′
So to refocus the image, eye-piece must be moved by the same distance through which the image
formed by the objective has shifted i.e. 15 – 6 = 9 cm.

genius PHYSICS
AAAAssignment
80. Near and far points of human eye are [EAMCET (Med.) 1995; MP PET 2001; Bihar CECE 2004]
(a) 25 cm and infinite (b) 50 cm and 100 cm (c) 25 cm and 50 cm (d) 0 cm and
25 cm
81. A defective eye cannot see close objects clearly because their image is formed [MP PET 2003]
(a) On the eye lens (b) Between eye lens and retina
(c) On the retina (d) Beyond retina
82. Retina of eye acts like ….. of camera [AFMC 2003]
(a) Shutter (b) Film (c) Lens (d) None of
these
83. A person who can see things most clearly at a distance of 10 cm. Requires spectacles to enable him to see clearly
things at a distance of 30 cm. What should be the focal length of the spectacles [BHU 2003]
(a) 15 cm (concave) (b) 15 cm (convex) (c) 10 cm (d) 0
84. An astronaut is looking down on earth's surface from a space shuttle at an altitude of .400km Assuming
that the astronaut's pupil diameter is mm5 and the wavelength of visible light is .500nm The astronaut
will be able to resolve linear object of the size of about [AIIMS 2003]
(a) 0.5 m (b) m5 (c) 50 m (d) 500 m
85. A person uses a lens of power + 3D to normalise vision. Near point of hypermetropic eye is [CPMT 2002]
(a) 1 m (b) 1.66 m (c) 2 m (d) 0.66 m
86. The separation between two microscopic particles is measured PA and PB by two different lights of wavelength
2000 Å and 3000 Å respectively, then [AIEEE 2002]
(a) BA PP > (b) BA PP < (c) BA PP 2/3< (d) BA PP =
87. To remove myopia (short sightedness) a lens of power 0.66 D is required. The distant point of the eye is
approximately
[MP PMT 2001]
(a) 100 cm (b) 150 cm (c) 50 cm (d) 25 cm
88. A person suffering from ‘presbyopia’ should use [MP PET 2001]
(a) A concave lens (b) A convex lens
(c) A bifocal lens whose lower portion is convex (d) A bifocal lens whose upper portion is
convex
89. The resolving limit of healthy eye is about [MP PET 1999; RPMT 1999; AIIMS 2001]
(a) 1′ (b) 1′′ (c) 1° (d)
60
1
′′
90. A person uses spectacles of power + 2D. He is suffering from [MP PET 2000]
(a) Short sightedness or myopia (b) Long sightedness or hypermetropia
(c) Presbyopia (d) Astigmatism
91. The hyper metropia is a [CBSE PMT 2000]
(a) Short-side defect (b) Long-side defect
(c) Bad vision due to old age (d) None of these
92. A man cannot see clearly the objects beyond a distance of 20 cm from his eyes. To see distant objects clearly he
must use which kind of lenses and of what focal length [MP PMT 2000]
(a) 100 cm convex (b) 100 cm concave (c) 20 cm convex (d) 20 cm
concave
Human eye

genius PHYSICS
93. An eye specialist prescribes spectacles having a combination of convex lens of focal length 40 cm in contact with a
concave lens of focal length 25 cm. The power of this lens combination in diopters is [IIT 1997 Cancelled; DPMT 2000]
(a) + 1.5 (b) –1.5 (c) +6.67 (d) –6.67
94. Two parallel pillars are 11 km away from an observer. The minimum distance between the pillars so that they can
be seen separately will be [RPET 1997; RPMT 2000]
(a) 3.2 m (b) 20.8 m (c) 91.5 m (d) 183 m
95. A person cannot see objects clearly beyond 2.0 m. The power of lens required to correct his vision will be
[MP PMT/PET 1998; JIPMER 2000; KCET (Engg./Med.) 2000]
(a) + 2.0 D (b) – 1.0 D (c) + 1.0 D (d) – 0.5 D
96. When objects at different distances are seen by the eye, which of the following remains constant [MP PMT 1999]
(a) The focal length of the eye lens (b) The object distance from the eye lens
(c) The radii of curvature of the eye lens (d) The image distance from the eye lens
97. A person wears glasses of power –2.0 D. The defect of the eye and the far point of the person without the glasses
will be
[MP PMT 1999]
(a) Nearsighted, 50 cm (b) Farsighted, 50 cm (c) Nearsighted, 250 cm (d)
Astigmatism, 50 cm
98. A person is suffering from the defect astigmatism. Its main reason is [MP PMT 1997]
(a) Distance of the eye lens from retina is increased (b) Distance of the eye lens from retina is
decreased
(c) The cornea is not spherical (d) Power of accommodation of the eye is
decreased
99. Myopia is due to [AFMC 1996]
(a) Elongation of eye ball (b) Irregular change in focal length
(c) Shortening of eye ball (d) Older age
100. Human eye is most sensitive to visible light of the wavelength [CPMT 1996]
(a) 6050 Å (b) 5500 Å (c) 4500 Å (d) 7500 Å
101. Match the List I with the List II from the combinations shown [ISM Dhanbad 1994]
(I) Presbiopia (A) Sphero-cylindrical lens
(II) Hypermetropia (B) Convex lens of proper power may be used close to the eye
(III) Astigmatism (C) Concave lens of suitable focal length
(IV) Myopia (D) Convex spectacle lens of suitable focal length
(a) I-A; II-C; III-B; IV-D (b) I-B; II-D; III-C; IV-A (c) I-D; II-B; III-A; IV-C (d) I-D; II-A;
III-C; IV-B
102. The human eye has a lens which has a [MP PET 1994]
(a) Soft portion at its centre (b) Hard surface
(c) Varying refractive index (d) Constant refractive index
103. A man with defective eyes cannot see distinctly object at the distance more than 60 cm from his eyes. The power
of the lens to be used will be [MP PMT 1994]
(a) + 60D (b) – 60D (c) – 1.66D (d) D
66.1
1
104. A person’s near point is 50 cm and his far point is 3 m. Power of the lenses he requires for
(i) Reading and (ii) For seeing distant stars are [MP PMT 1994]
(a) – 2D and 0.33D (b) 2D and – 0.33D (c) – 2D and 3D (d) 2D and –
3D
105. The focal length of a simple convex lens used as a magnifier is 10 cm. For the image to be formed at a
distance of distinct vision (D = 25 cm), the object must be placed away from the lens at a distance of [CPMT 1991]
(a) 5 cm (b) 7.14 cm (c) 7.20 cm (d) 16.16 cm

genius PHYSICS
106. A person is suffering from myopic defect. He is able to see clear objects placed at 15 cm. What type and of what
focal length of lens he should use to see clearly the object placed 60 cm away [MP PMT 1991]
(a) Concave lens of 20 cm focal length (b) Convex lens of 20 cm focal length
(c) Concave lens of 12 cm focal length (d) Convex lens of 12 cm focal length
107. A person can see a thing clearly when it is at a distance of 1 metre only. If he wishes to see a distance star, he needs
a lens of focal length [MP PET 1990]
(a) +100 cm (b) – 100 cm (c) +50 cm (d) –50 cm
108. A man suffering from myopia can read a book placed at 10 cm distance. For reading the book at a distance of 60
cm with relaxed vision, focal length of the lens required will be [MP PMT 1989]
(a) 45 cm (b) – 20 cm (c) –12 cm (d) 30 cm
109. A person can see clearly objects at 100 cm distance. If he wants to see objects at 40 cm distance, then the power of
the lens he shall require is [MP PET 1989]
(a) +1.5 D (b) – 1.5 D (c) +3.0 D (d) –3.0 D
110. If the distance of the far point for a myopia patient is doubled, the focal length of the lens required to cure it will
become [MP PET 1989]
(a) Half (b) Double
(c) The same but a convex lens (d) The same but a
concave lens
111. Image is formed for the short sighted person at [AFMC 1988]
(a) Retina (b) Before retina (c) Behind the retina (d) Image is
not formed at all
112. A man who cannot see clearly beyond 5 m wants to see stars clearly. He should use a lens of focal length
[MP PET/PMT 1988]
(a) – 100 metre (b) + 5 metre (c) – 5 metre (d) Very large
113. Far point of myopic eye is 250 cm, then the focal length of the lens to be used will be [CPMT 1986; DPMT 2002]
(a) + 250 cm (b) – 250 cm (c) + 250/9 cm (d) – 250/9 cm
114. One can take pictures of objects which are completely invisible to the eye using camera film which are invisible to [MNR 1985]
(a) Ultra-violet rays (b) Sodium light (c) Visible light (d) Infra-red
rays
115. In human eye the focussing is done by [CPMT 1983]
(a) To and fro movement of eye lens (b) To and fro movement of the retina
(c) Change in the convexity of the lens surface (d) Change in the refractive index of the eye
fluids
116. The minimum light intensity that can be perceived by the eye is about 210
/10 metrewatt−
. The number of photons
of wavelength 7
106.5 −
× metre that must enter per second the pupil of area 24
10 metre−
for vision, is
approximately equal to sec)106.6( 34
−×= −
jouleh [NCERT 1982]
(a) 2
103 × photons (b) 6
103 × photons (c) 4
103 × photons (d) 5
103 ×
photons
117. A far sighted man who has lost his spectacles, reads a book by looking through a small hole (3-4 mm) in a sheet of
paper. The reason will be [CPMT 1977]
(a) Because the hole produces an image of the letters at a longer distance
(b) Because in doing so, the focal length of the eye lens is effectively increased
(c) Because in doing so, the focal length of the eye lens is effectively decreased
(d) None of these
118. The maximum focal length of the eye-lens of a person is greater than its distance from the retina. The eye is
(a) Always strained in looking at an object (b) Strained for objects at large distances only
(c) Strained for objects at short distances only (d) Unstrained for all distances
119. The focal length of a normal eye-lens is about
(a) 1 mm (b) 2 cm (c) 25 cm (d) 1
120. The distance of the eye-lens from the retina is x. For normal eye, the maximum focal length of the eye-lens is

genius PHYSICS
(a) = x (b) < x (c) > x (d) = 2x
121. A man wearing glasses of focal length +1m can clearly see beyond 1m
(a) If he is farsighted (b) If he is nearsighted (c) If his vision is normal (d) In each of
these cases
122. The near point of a person is 50 cm and the far point is 1.5 m. The spectacles required for reading purpose and for
seeing distance are respectively
(a) DD 





−+
3
2
,2 (b) DD 2
3
2
−





+ (c) DD 





+−
3
2
,2 (d)
DD 2
3
2
+





−
123. A man, wearing glasses of power +2D can read clearly a book placed at a distance of 40 cm from the eye. The
power of the lens required so that he can read at 25 cm from the eye is
(a) +4.5 D (b) +4.0 D (c) +3.5 D (d) +3.0 D
124. A person can see clearly between 1 m and 2m. His corrective lenses should be
(a) Bifocals with power –0.5D and additional +3.5D (b) Bifocals with power –1.0D and additional
+3.0 D
(c) Concave with power 1.0 D (d) Convex with power 0.5 D
125. While reading the book a man keeps the page at a distance of 2.5 cm from his eye. He wants to read the book by
holding the page at 25 cm. What is the nature of spectacles one should advice him to use to completely cure his
eye sight
(a) Convex lens of focal length 25 cm (b) Concave lens of focal length 25 cm
(c) Convex lens of focal length 2.5 cm (d) Concave lens of focal length 2.5 cm
126. The blades of a rotating fan can not be distinguished from each other due to
(a) Parallex (b) Power of accommodation (c) Persistence of vision (d) Binocular
vision
127. Aperture of the human eye is 2 mm. Assuming the mean wavelength of light to be 5000 Å, the angular resolution
limit of the eye is nearly
(a) 2 minutes (b) 1 minute (c) 0.5 minute (d) 1.5 minutes
128. If there had been one eye of the man, then
(a) Image of the object would have been inverted (b) Visible region would have decreased
(c) Image would have not been seen three dimensional (d) (b) and (c) both
129. A man can see the object between 15cm and 30cm. He uses the lens to see the far objects. Then due to the lens
used, the near point will be at
(a)
3
10
cm (b) 30 cm (c) 15 cm (d)
3
100
cm
130. A presbyopic patient has near point as 30 cm and far point as 40 cm. The dioptric power for the corrective lens for
seeing distant objects is
(a) 40 D (b) 4 D (c) 2.5 D (d) 0.25 D
131. A man swimming under clear water is unable to see clearly because
(a) The size of the aperture decreases (b) The size of the aperture increases
(c) The focal length of eye lens increases (d) The focal length of eye lens decreases
132. The distance between retina and eye-lens in a normal eye is 2.0 cm. The accommodated power of eye lens range
from
(a) 45 D to 50 D (b) 50 D to 54 D (c) 10 D to 16 D (d) 5 D to 8 D
133. If the eye is taken as a spherical ball of radius 1 cm, the range of accommodated focal length of eye-lens is
(a) 1.85 cm to 2.0 cm (b) 1.0 cm to 2.8 cm (c) 1.56 cm to 2.5 cm (d) 1.6 cm to
2.0 cm

genius PHYSICS
134. A person cannot read printed matter within 100 cm from his eye. The power of the correcting lens required to
read at 20 cm from his eye if the distance between the eye lens and the correcting lens is 2 cm is
(a) 4.8 D (b) 1.25 D (c) 4.25 D (d) 4.55 D
135. A student having –1.5 D spectacles uses a lens of focal length 5 cm as a simple microscope to read minute scale
divisions in the laboratory. The least distance of distinct vision without glasses is 20 cm for the student. The
maximum magnifying power he gets with spectacles on is
(a) 6 (b) 9 (c) 5 (d) 4
136. In a compound microscope the object of fo and eyepiece of fe are placed at distance L such that L equals [Kerala PMT 2004]
(a) eo ff + (b) eo ff −
(c) Much greater than fo or fe (d) Need not depend either value of focal
lengths
137. In a simple microscope, if the final image is located at infinity then its magnifying power is [CPMT 1985; MP PMT 2004]
(a)
f
25
(b)
25
D
(c)
25
f
(d) 1+
D
f
138. In a simple microscope, if the final image is located at 25 cm from the eye placed close to the lens, then the
magnifying power is
[BVP 2003]
(a)
f
25
(b)
f
25
1 + (c)
25
f
(d) 1
25
+
f
139. The maximum magnification that can be obtained with a convex lens of focal length 2.5 cm is (the least
distance of distinct vision is 25 cm) [MP PET 2003]
(a) 10 (b) 0.1 (c) 62.5 (d) 11
140. In a compound microscope, the intermediate image is [IIT-JEE (Screening) 2000; AIEEE 2003]
(a) Virtual, erect and magnified (b) Real, erect and
magnified
(c) Real, inverted and magnified (d) Virtual, erect and reduced
141. A compound microscope has two lenses. The magnifying power of one is 5 and the combined magnifying power is
100. The magnifying power of the other lens is [Kerala PMT 2002]
(a) 10 (b) 20 (c) 50 (d) 25
142. Wavelength of light used in an optical instrument are Å40001 =λ and Å50002 =λ , then ratio of their respective
resolving power (corresponding to 1λ and 2λ ) is [AIEEE 2002]
(a) 16 : 25 (b) 9 : 1 (c) 4 : 5 (d) 5 : 4
143. The angular magnification of a simple microscope can be increased by increasing [Orissa JEE 2002]
(a) Focal length of lens (b) Size of object (c) Aperture of lens (d) Power of
lens
144. The magnification produced by the objective lens and the eye lens of a compound microscope are 25 and 6
respectively. The magnifying power of this microscope is [Manipal MEE 1995; DPMT 2002]
(a) 19 (b) 31 (c) 150 (d) 150
145. The length of the compound microscope is 14 cm. The magnifying power for relaxed eye is 25. If the focal length of
eye lens is 5 cm, then the object distance for objective lens will be [Pb. PMT 2002]
(a) 1.8 cm (b) 1.5 cm (c) 2.1 cm (d) 2.4 cm
146. The magnifying power of a simple microscope is 6. The focal length of its lens in metres will be, if least distance of
distinct vision is 25 cm [MP PMT 2001]
(a) 0.05 (b) 0.06 (c) 0.25 (d) 0.12
Microscope

genius PHYSICS
147. Relative difference of focal lengths of objective and eye lens in the microscope and telescope is given as
[MH CET (Med.) 2001]
(a) It is equal in both (b) It is more in telescope (c) It is more in microscope (d) It may be
more in any one
148. Three objective focal lengths (fo) and two eye piece focal lengths (fe) are available for a compound microscope. By
combining these two, the magnification of microscope will be maximum when [RPMT 2001]
(a) eo ff = (b) eo ff >> (c) fo and fe both are small (d) eo ff >>
149. If the red light is replaced by blue light illuminating the object in a microscope the resolving power of the
microscope
[DCE 2001]
(a) Decreases (b) Increases (c) Gets halved (d) Remains
unchanged
150. In case of a simple microscope, the object is placed at [UPSEAT 2000]
(a) Focus f of the convex lens (b) A position between f and 2f (c) Beyond 2f (d) Between
the lens and f
151. In a compound microscope cross-wires are fixed at the point [EAMCET (Engg.) 2000]
(a) Where the image is formed by the objective (b) Where the image is formed by the eye-piece
(c) Where the focal point of the objective lies (d) Where the focal point of the eye-piece lies
152. The length of the tube of a microscope is 10 cm. The focal lengths of the objective and eye lenses are 0.5 cm and
1.0 cm. The magnifying power of the microscope is about [MP PMT 2000]
(a) 5 (b) 23 (c) 166 (d) 500
153. Least distance of distinct vision is 25 cm. Magnifying power of simple microscope of focal length 5 cm is
[EAMCET (Engg.) 1995; Pb. PMT 1999]
(a) 1/5 (b) 5 (c) 1/6 (d) 6
154. The objective of a compound microscope is essentially [SCRA 1998]
(a) A concave lens of small focal length and small aperture (b) Convex lens of small focal length and large
aperture
(c) Convex lens of large focal length and large aperture (d) Convex lens of small focal length and small
aperture
155. For relaxed eye, the magnifying power of a microscope is [CBSE PMT 1998]
(a)
eo
o
f
D
u
v
×− (b)
D
f
u
v e
o
o
×− (c)
eo
o
f
D
v
u
× (d) 







−×
eo
o
f
D
v
u
156. A person using a lens as a simple microscope sees an [AIIMS 1998]
(a) Inverted virtual image (b) Inverted real magnified image
(c) Upright virtual image (d) Upright real magnified image
157. The focal length of the objective lens of a compound microscope is [CPMT 1985; MNR 1986; MP PET 1997]
(a) Equal to the focal length of its eye piece (b) Less than the focal length of eye piece
(c) Greater than the focal length of eye piece (d) Any of the above three
158. To produce magnified erect image of a far object, we will be required along with a convex lens, is
[MNR 1983; MP PAT 1996]
(a) Another convex lens (b) Concave lens (c) A plane mirror (d) A concave
mirror
159. An object placed 10 cm in front of a lens has an image 20 cm behind the lens. What is the power of the lens (in
dioptres)
[MP PMT 1995]
(a) 1.5 (b) 3.0 (c) – 15.0 (d) +15.0

genius PHYSICS
160. Resolving power of a microscope depends upon [MP PET 1995]
(a) The focal length and aperture of the eye lens (b) The focal lengths of the objective and the
eye lens
(c) The apertures of the objective and the eye lens (d) The wavelength of light illuminating the
object
161. If the focal length of the objective lens is increased then [MP PMT 1994]
(a) Magnifying power of microscope will increase but that of telescope will decrease
(b) Magnifying power of microscope and telescope both will increase
(c) Magnifying power of microscope and telescope both will decrease
(d) Magnifying power of microscope will decrease but that of telescope will increase
162. If in compound microscope 1m and 2m be the linear magnification of the objective lens and eye lens respectively,
then magnifying power of the compound microscope will be [CPMT 1985; KCET 1994]
(a) 21 mm − (b) 21 mm + (c) 2/)( 21 mm + (d) 21 mm ×
163. The magnifying power of a microscope with an objective of 5 mm focal length is 400. The length of its tube is 20
cm. Then the focal length of the eye-piece is [MP PMT 1991]
(a) 200 cm (b) 160 cm (c) 2.5 cm (d) 0.1 cm
164. In a compound microscope, if the objective produces an image Io and the eye piece produces an image Ie, then
[MP PET 1990]
(a) Io is virtual but Ie is real (b) Io is real but Ie is virtual (c) Io and Ie are both real (d) Io and Ie are
both virtual
165. In an electron microscope if the potential is increased from 20 kV to 80 kV, the resolving power of the microscope
will change from R to [CPMT 1988, 89]
(a) R/4 (b) 4R (c) 2R (d) R/2
166. When the length of a microscope tube increases, its magnifying power [MNR 1986]
(a) Decreases (b) Increases (c) Does not change (d) May
decrease or increase
167. An electron microscope is superior to an optical microscope in [CPMT 1984]
(a) Having better resolving power (b) Being easy to handle
(c) Low cost (d) Quickness of observation
168. In a compound microscope magnification will be large, if the focal length of the eye piece is [CPMT 1984]
(a) Large (b) Smaller (c) Equal to that of objective (d) Less than
that of objective
169. An electron microscope gives better resolution than optical microscope because [CPMT 1982]
(a) Electrons are abundant (b) Electrons can be focused nicely
(c) Effective wavelength of electron is small (d) None of these
170. A man is looking at a small object placed at his near point. Without altering the position of his eye or the object, he
puts a simple microscope of magnifying power 5X before his eyes. The angular magnification achieved is
(a) 5 (b) 2.5 (c) 1 (d) 0.2
171. The focal length of the objective of a compound microscope is 0f and its distance from the eyepiece is L. The
object is placed at a distance u from the objective. For proper working of the instrument
(a) L < u (b) L > u (c) 00 2 fLf << (d) 02 fL >
172. Find the maximum magnifying power of a compound microscope having a 25 diopter lens as the objective, a 5
diopter lens as the eyepiece and the separation 30 cm between the two lenses. The least distance for clear vision is
25 cm
(a) 8.4 (b) 7.4 (c) 9.4 (d) 10.4

genius PHYSICS
173. The focal length of the objective and the eye-piece of a microscope are 2 cm and 5 cm respectively and the distance
between them is 30 cm. If the image seen by the eye is 25 cm from the eye-piece, the distance of the object from
the objective is
(a) 0.8 cm (b) 2.3 cm (c) 0.4 cm (d) 1.2 cm
174. The focal length of objective and eye-piece of a microscope are 1 cm and 5 cm respectively. If the magnifying
power for relaxed eye is 45, then length of the tube is
(a) 6 cm (b) 9 cm (c) 12 cm (d) 15 cm
175. A microscope has an objective of focal length 1.5 cm and an eye-piece of focal length 2.5 cm. If the distance
between objective and eye-piece is 25 cm. What is the approximate value of magnification produced for relaxed
eye is
(a) 75 (b) 110 (c) 140 (d) 25
176. The magnifying power of a microscope is generally marked as 10X, 100 X, etc. These markings are for a normal
relaxed eye. A microscope marked 10X is used by an old man having his near point at 40 cm. The magnifying
power of the microscope for the old man with his eyes completely relaxed is
(a) 10 (b) 18 (c) 12 (d) 16
177. If the focal length of objective and eye lens are 1.2 cm and 3 cm respectively and the object is put 1.25 cm away
from the objective lens and the final image is formed at infinity. The magnifying power of the microscope is
(a) 150 (b) 200 (c) 250 (d) 400
178. A compound microscope is adjusted for viewing the distant image of an object, the distance of the object from the
object glass is now slightly increased, what re-adjustment of the instrument would be necessary for obtaining a
distant image again
(a) Objective should be moved away from the eye-piece (b) Eye-piece should be moved towards the
objective
(c) Both should be moved towards each other (d) Both should be moved away from each
other
179. When the object is self-luminous, the resolving power of a microscope is given by the expression
(a)
λ
θµ sin2
(b)
λ
θµ sin
(c)
λ
θµ cos2
(d)
λ
µ2
180. In a compound microscope, maximum magnification is obtained when the final image
(a) Is formed at infinity (b) Is formed at the least of distinct vision
(c) Coincides with the object (d) Coincides with the objective lens
181. How should people wearing spectacles work with a microscope
(a) They should keep on wearing their spectacles
(b) They should take off their spectacles
(c) They may keep on wearing or take off their spectacles, It makes no difference
(d) They cannot use a microscope at all
182. The focal length of the objective and eyepiece of an astronomical telescope for normal adjustments are 50 cm and
5 cm. The length of the telescope should be [MP PMT 2004]
(a) 50 cm (b) 55 cm (c) 60 cm (d) 45 cm
183. The resolving power of an astronomical telescope is 0.2 seconds. If the central half portion of the objective lens is
covered, the resolving power will be [MP PMT 2004]
(a) 0.1 sec (b) 0.2 sec (c) 1.0 sec (d) 0.6 sec
Telescope

genius PHYSICS
184. If oF and eF are the focal length of the objective and eye-piece respectively of a telescope, then its magnifying
power will be
[CPMT 1977, 82, 97, 99, 2003; SCRA 1994; KCET (Engg./Med.) 1999; Pb. PMT 2000; BHU 2001; BCECE 2003, 2004]
(a) eo FF + (b) eo FF × (c) eo FF / (d) )(
2
1
eo FF +
185. The length of an astronomical telescope for normal vision (relaxed eye) (fo = focal length of objective lens and fe =
focal length of eye lens) is [EAMCET (Med.) 1995; MP PAT 1996; CPMT 1999; BVP 2003]
(a) eo ff × (b)
e
o
f
f
(c) eo ff + (d) eo ff −
186. A telescope of diameter 2m uses light of wavelength 5000 Å for viewing stars. The minimum angular separation
between two stars whose image is just resolved by this telescope is [MP PET 2003]
(a) 4
104 −
× rad (b) 6
1025.0 −
× rad (c) 6
1031.0 −
× rad (d) 3
100.5 −
×
rad
187. The aperture of the objective lens of a telescope is made large so as to [AIEEE 2003; KCET 2003]
(a) Increase the magnifying power of the telescope (b) Increase the resolving power of the
telescope
(c) Make image aberration less (d) Focus on distant
objects
188. The distance of the moon from earth is 5
108.3 × km. The eye is most sensitive to light of wavelength 5500 Å. The
separation of two points on the moon that can be resolved by a 500 cm telescope will be [AMU (Med.) 2002]
(a) 51 m (b) 60 m (c) 70 m (d) All of the
above
189. To increase both the resolving power and magnifying power of a telescope [Kerala PET 2002; KCET (Engg.) 2002]
(a) Both the focal length and aperture of the objective has to be increased
(b) The focal length of the objective has to be increased
(c) The aperture of the objective has to be increased
(d) The wavelength of light has to be decreased
190. The focal lengths of the objective and eye lenses of a telescope are respectively 200cm and 5cm. The maximum
magnifying power of the telescope will be [MP PMT/PET 1998; JIPMER 2001, 2002]
(a) – 40 (b) – 48 (c) – 60 (d) – 100
191. A telescope has an objective of focal length 50 cm and an eye piece of focal length 5 cm. The least distance of
distinct vision is 25 cm. The telescope is focussed for distinct vision on a scale 200 cm away. The separation
between the objective and the eye-piece is
[Kerala PET 2002]
(a) 75 cm (b) 60 cm (c) 71 cm (d) 74 cm
192. In a laboratory four convex lenses 321 ,, LLL and 4L of focal lengths 2, 4, 6 and 8cm respectively are available.
Two of these lenses form a telescope of length 10cm and magnifying power 4. The objective and eye lenses are [MP PMT 2001]
(a) 32, LL (b) 41, LL (c) 23, LL (d) 14 , LL
193. Four lenses of focal length + 15 cm, + 20 cm, + 150 cm and + 250 cm are available for making an
astronomical telescope. To produce the largest magnification, the focal length of the eye-piece should
be [CPMT 2001; AIIMS 2001]
(a) + 15 cm (b) + 20 cm (c) + 150 cm (d) + 250 cm
194. In a terrestrial telescope, the focal length of objective is 90 cm, of inverting lens is 5 cm and of eye lens is 6 cm. If
the final image is at 30 cm, then the magnification will be [DPMT 2001]
(a) 21 (b) 12 (c) 18 (d) 15
195. The focal lengths of the objective and the eyepiece of an astronomical telescope are 20 cm and 5 cm respectively. If
the final image is formed at a distance of 30 cm from the eye piece, find the separation between the lenses for
distinct vision [BHU (Med.) 2000]
(a) 32.4 cm (b) 42.3 cm (c) 24.3 cm (d) 30.24 cm
196. Resolving power of reflecting type telescope increases with [DPMT 2000]

genius PHYSICS
(a) Decrease in wavelength of incident light (b) Increase in wavelength of incident light
(c) Increase in diameter of objective lens (d) None of these
197. A planet is observed by an astronomical refracting telescope having an objective of focal length 16 m and an eye-
piece of focal length 2cm [IIT-JEE 1992; Roorkee 2000]
(a) The distance between the objective and the eye-piece is 16.02 m
(b) The angular magnification of the planet is 800
(c) The image of the planet is inverted
(d) All of the above
198. The astronomical telescope consists of objective and eye-piece. The focal length of the objective is [AIIMS 1998; BHU 2000]
(a) Equal to that of the eye-piece (b) Greater than that of the eye-piece
(c) Shorter than that of the eye-piece (d) Five times shorter than that of the eye-
piece
199. The diameter of the objective of a telescope is a, the magnifying power is m and wavelength of light is λ. The
resolving power of the telescope is [MP PMT 2000]
(a) (1.22λ)/a (b) (1.22a)/λ (c) λm/(1.22a) (d) a/(1.22λm)
200. An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation
between the objective and the eyepiece is 36 cm and final image is formed at infinity. The focal lengths of the
objective and eyepiece are respectively
[IIT-JEE 1989; MP PET 1995; JIPMER 2000]
(a) 20 cm, 16 cm (b) 50 cm, 10 cm (c) 30 cm, 6 cm (d) 45 cm, –9
cm
201. A photograph of the moon was taken with telescope. Later on, it was found that a housefly was sitting on the
objective lens of the telescope. In photograph [NCERT 1970; MP PET 1999]
(a) The image of housefly will be reduced (b) There is a reduction in the intensity of the
image
(c) There is an increase in the intensity of the image (d) The image of the housefly will be enlarged
202. The magnifying power of a telescope is M. If the focal length of eye piece is doubled, then the magnifying power
will become
[Haryana CEET 1998]
(a) 2 M (b) M/2 (c) M2 (d) 3 M
203. The minimum magnifying power of a telescope is M. If the focal length of its eyelens is halved, the magnifying
power will become
[MP PMT/PET 1998]
(a) M/2 (b) 2 M (c) 3 M (d) 4 M
204. The final image in an astronomical telescope is [EAMCET (Engg.) 1998]
(a) Real and errect (b) Virtual and inverted (c) Real and inverted (d) Virtual and
errect
205. The astronomical telescope has two lenses of focal powers 0.5 D and 20 D. Its magnifying power will be [CPMT 1997]
(a) 40 (b) 10 (c) 100 (d) 35
206. An astronomical telescope of ten-fold angular magnification has a length of 44 cm. The focal length of the objective is[CBSE PMT 19
(a) 4 cm (b) 40 cm (c) 44 cm (d) 440 cm
207. A telescope consisting of an objective of focal length 100 cm and a single eyes lens of focal length 10 cm is focussed
on a distant object in such a way that parallel rays emerge from the eye lens. If the object subtends an angle of 2°
at the objective, the angular width of the image is [JIPMER 1997]
(a) 20° (b) 1/6° (c) 10° (d) 24°
208. When diameter of the aperture of the objective of an astronomical telescope is increased, its [MP PMT 1997]
(a) Magnifying power is increased and resolving power is decreased
(b) Magnifying power and resolving power both are increased
(c) Magnifying power remains the same but resolving power is increased
(d) Magnifying power and resolving power both are decreased

genius PHYSICS
209. The focal length of objective and eye-piece of a telescope are 100 cm and 5 cm respectively. Final image is formed
at least distance of distinct vision. The magnification of telescope is [RPET 1997]
(a) 20 (b) 24 (c) 30 (d) 36
210. A simple telescope, consisting of an objective of focal length 60 cm and single eye lens of focal length 5 cm is
focussed on a distant object in such a way that parallel rays comes out from the eye lens. If the object subtends an
angle 2° at the objective, the angular width of the image [CPMT 1979; NCERT 1980; MP PET 1992; JIPMER 1997]
(a) 10° (b) 24° (c) 50° (d) 1/6°
211. The diameter of the objective of the telescope is 0.1 metre and wavelength of light is 6000 Å. Its resolving power
would be approximately [MP PET 1997]
(a) 6
1032.7 −
× radian (b) 6
1036.1 × radian (c) 5
1032.7 −
× radian (d)
5
1036.1 × radian
212. A Gallilean telescope has objective and eye-piece of focal lengths 200 cm and 2 cm respectively. The magnifying
power of the telescope for normal vision is [MP PMT 1996]
(a) 90 (b) 100 (c) 108 (d) 198
213. All of the following statements are correct except [Manipal MEE 1995]
(a) The total focal length of an astronomical telescope is the sum of the focal lengths of its two lenses
(b) The image formed by the astronomical telescope is always erect because the effect of the combination of the
two lenses its divergent
(c) The magnification of an astronomical telescope can be increased by decreasing the focal length of the eye-
piece
(d) The magnifying power of the refracting type of astronomical telescope is the ratio of the focal length of the
objective to that of the eye-piece
214. The length of a telescope is 36 cm. The focal length of its lenses can be [Bihar MEE 1995]
(a) 30 cm, 6 cm (b) – 30 cm, – 6 cm (c) – 30 cm, – 6 cm (d) – 30 cm, 6
cm
215. The diameter of the objective lens of telescope is 5.0 m and wavelength of light is 6000 Å. The limit of resolution
of this telescope will be [MP PMT 1994]
(a) 0.03 sec (b) 3.03 sec (c) 0.06 sec (d) 0.15 sec
216. If tube length of astronomical telescope is 105 cm and magnifying power is 20 for normal setting, calculate the
focal length of objective [AFMC 1994]
(a) 100 cm (b) 10 cm (c) 20 cm (d) 25 cm
217. Radio telescope is used to see [AFMC 1994]
(a) Distant start and planets (b) Sun and to
measure its temperature
(c) Stars and to measures diameters (d) None of these
218. Four lenses with focal lens ± 15 cm and ± 150 cm are being placed for used as a telescopic objective. The focal
length of the lens which produces the largest magnification with a given eye-piece is [CBSE PMT 1994]
(a) –15 cm (b) +150 cm (c) –150 cm (d) +15 cm
219. The image of a star (effectively a point source) is made by convergent lens of focal length 50 cm and diameter of
aperture 5.0 cm. If the lens is ideal, and the effective wavelength in image formation is taken as cm5
105 −
× , the
diameter of the image formed will be nearest to [NSEP 1994]
(a) Zero (b) cm6
10−
(c) cm5
10−
(d) cm3
10−
220. To increase the magnifying power of telescope (fo = focal length of the objective and fe = focal length of the eye
lens)
[MP PET/PMT 1988; MP PMT 1992, 94]
(a) fo should be large and fe should be small (b) fo should be small and fe should be large
(c) fo and fe both should be large (d) fo and fe both should be small

genius PHYSICS
221. The limit of resolution of a 100 cm telescope )105.5( 7
m−
×=λ is [BHU 1993]
(a) "14.0 (b) "3.0 (c) 1′ (d) "1
222. In a reflecting astronomical telescope, if the objective (a spherical mirror) is replaced by a parabolic mirror of the
same focal length and aperture, then [IIT-JEE 1993]
(a) The final image will be erect (b) The larger
image will be obtained
(c) The telescope will gather more light (d) Spherical aberration will be absent
223. A planet is observed by an astronomical refracting telescope having an objective of focal length 16 m and an
eyepiece of focal length 2 cm [IIT-JEE 1993]
(a) The distance between the objective and the eyepiece is 16.02 m
(b) The angular magnification of the planet is 800
(c) The image of the planet is inverted
(d) The objective is larger than the eyepiece
224. The average distance between the earth and moon is 4
106.38 × km. The minimum separation between the two
points on the surface of the moon that can be resolved by a telescope whose objective lens has a diameter of 5 m
with λ = 6000 Å is [MP PMT 1993]
(a) 5.65 m (b) 28.25 m (c) 11.30 m (d) 56.51 m
225. The focal length of the objective and eye piece of a telescope are respectively 60 cm and 10 cm. The magnitude of
the magnifying power when the image is formed at infinity is [MP PET 1991]
(a) 50 (b) 6 (c) 70 (d) 5
226. The focal length of an objective of a telescope is 3 metre and diameter 15 cm. Assuming for a normal eye, the
diameter of the pupil is 3 mm for its complete use, the focal length of eye piece must be [MP PET 1989]
(a) 6 cm (b) 6.3 cm (c) 20 cm (d) 60 cm
227. An opera glass (Gallilean telescope) measures 9 cm from the objective to the eyepiece. The focal length of the
objective is 15 cm. Its magnifying power is [DPMT 1988]
(a) 2.5 (b) 2/5 (c) 5/3 (d) 0.4
228. The focal length of objective and eye lens of a astronomical telescope are respectively 2 m and 5 cm. Final image is
formed at (i) least distance of distinct vision (ii) infinity. The magnifying power in both cases will be [MP PMT/PET 1988]
(a) – 48, – 40 (b) – 40, – 48 (c) – 40, 48 (d) – 48, 40
229. An optical device that enables an observer to see over or around opaque objects, is called [CPMT 1986]
(a) Microscope (b) Telescope (c) Periscope (d)
Hydrometer
230. The magnifying power of a telescope can be increased by [CPMT 1979]
(a) Increasing focal length of the system (b) Fitting eye piece of high power
(c) Fitting eye piece of low power (d) Increasing the distance of objects
231. An achromatic telescope objective is to be made by combining the lenses of flint and crown glasses. This proper
choice is [CPMT 1977]
(a) Convergent of crown and divergent of flint (b) Divergent of crown and convergent of flint
(c) Both divergent (d) Both convergent
232. An observer looks at a tree of height 15 m with a telescope of magnifying power 10. To him, the tree appears [CPMT 1975]
(a) 10 times taller (b) 15 times taller (c) 10 times nearer (d) 15 times
nearer
233. The magnification produced by an astronomical telescope for normal adjustment is 10 and the length of the
telescope is 1.1 m. The magnification when the image is formed at least distance of distinct vision (D = 25 cm) is
(a) 14 (b) 6 (c) 16 (d) 18
234. The objective of a telescope has a focal length of 1.2 m. it is used to view a 10.0 m tall tower 2 km away. What is
the height of the image of the tower formed by the objective
(a) 2 mm (b) 4 mm (c) 6 mm (d) 8 mm
235. A giant telescope in an observatory has an objective of focal length 19 m and an eye-piece of focal length 1.0 cm. In
normal adjustment, the telescope is used to view the moon. What is the diameter of the image of the moon formed
by the objective? The diameter of the moon is 6
105.3 × m and the radius of the lunar orbit round the earth is
8
108.3 × m
(a) 10 cm (b) 12.5 cm (c) 15 cm (d) 17.5 cm

genius PHYSICS
236. The aperture of the largest telescope in the world is ≈ 5 metre. If the separation between the moon and the earth is
5
104 ×≈ km and the wavelength of the visible light is Å5000~= , then the minimum separation between the
objects on the surface of the moon which can be just resolved is
(a) 1 metre approximately (b) 10 metre approximately (c) 50 metre approximately (d) 200 metre
approximately
237. In Galileo’s telescope, magnifying power for normal vision is 20 and power of eye-piece is –20 D. Distance
between the objective and eye-piece should be
(a) 90 cm (b) 95 cm (c) 100 cm (d) 105 cm
238. The least resolve angle by a telescope using objective of aperture 5 m and light of wavelength = 4000 A.U. is
nearly
(a) °
50
1
(b)
50
1
sec (c)
50
1
minute (d)
500
1
sec
239. The limit of resolution of a 10 cm telescope for visible light of wavelength 6000 Å is approximately
(a) 0.1 s or arc (b) 30° (c)
o






6
1
(d) None of
these
240. An eye-piece of a telescope with a magnification of 100 has a power of 20 diopters. The object of this telescope has
a power of
(a) 2 diopters (b) 0.2 diopters (c) 2000 diopters (d) 20 diopters
241. The Yerkes Observatory telescope has a large telescope with objective of diameter of about 1 m. Assuming
wavelength of light to be m7
106 −
× , the angular distance θ between two stars which can just be resolved is
(a) °× −
)103.7( 7
(b) 7
103.7 −
× rad (c)
40
1
of a second (d) None of
these
242. A Galilean telescope measures 9 cm from the objective to the eye-piece. The focal length of the objective is 15 cm.
Its magnifying power is
(a) 2.5 (b) 2/5 (c) 5/3 (d) 0.4
243. For seeing a cricket match, we prefer binoculars to the terrestrial telescope, because
(a) Binoculars give three-dimensional view (b) Terrestrial telescope gives inverted image
(c) To avoid chromatic aberration (d) To have larger magnification
244. A simple two lens telescope has an objective of focal length 50 cm and an eye-piece of 2.5 cm. The telescope is
pointed at an object at a very large distance which subtends at an angle of 1 milliradian on the naked eye. The eye
piece is adjusted so that the final virtual image is formed at infinity. The size of the real image formed by the
objective is
(a) 5 mm (b) 1 mm (c) 0.5 mm (d) 0.1 mm
245. The objective of a telescope, after focussing for infinity is taken out and a slit of length L is placed in its position. A
sharp image of the slit is formed by the eye-piece at a certain distance from it on the other side. The length of this
image is l, then magnification of telescope is
(a)
L
l
2
(b)
l
L2
(c)
L
l
(d)
l
L
246. An astronomical telescope in normal adjustment receives light from a distant source S. The tube length is now
decreased slightly
(a) A virtual image of S will be formed at a finite distance
(b) No image will be formed
(c) A small, real image of S will be formed behind the eye-piece, close to it
(d) A large, real image of S will be formed behind the eye-piece, far away from it
247. A telescope consisting of object glass of power + 2 D and eye-glass of power + 20 D is focussed on an object 1m
from the object glass. The final image is seen with completely relaxed eye. The magnifying power of the telescope
is
(a) 20 (b) 41 (c) 24 (d) 49.2
248. An astronomical telescope and a Galilean telescope use identical objective lenses. They have the same
magnification, when both are in normal adjustment. The eye-piece of the astronomical telescope has a focal length
f

genius PHYSICS
(a) The tube lengths of the two telescopes differ by f (b) The tube lengths of the two telescopes
differ by 2f
(c) The Galilean telescope has a shorter tube length (d) The Galilean telescope has a longer tube
length
79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98
a d b a c a b b c a b b d b a d d a c a
99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118
b c c c b b c b c a b b c b d c c a a b
119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138
a d a c a d c b d b c c b a d a c a b d
139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158
c b d d c a a b c b d a d d d a c b b d
159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178
d d d c b c a a b c c b,d a b d c d b b a
179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198
b b b c c c c b a a b c d a c c a,
c
d b d
199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218
c b b b b a b a c b b d b b a a a a b d
219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238
a a d a d b a a a c b a c a c d c b b a
239 240 241 242 243 244 245 246 247
b b a a c d a b b,
c

01 ray-optics-mm

More Related Content

Viewers also liked (20)

Similar to 01 ray-optics-mm (20)

Recently uploaded (20)

01 ray-optics-mm