Diffraction of light

CHAPTER 3. DIFFRACTION OF LIGHT
By
Dr. Vishal Jain
Assistant Professor

Diffraction of Light
Diffraction refers to various phenomena that occur when a wave
encounters an obstacle or a slit. It is defined as the bending of light
around the corners of an obstacle or aperture into the region of
geometrical shadow of the obstacle.
Diffraction pattern of red
laser beam made on a
plate after passing
through a small circular
aperture in another plate
Thomas Young's sketch of two-slit diffraction,
which he presented to the Royal Society in 1803.

Fresnel's Diffraction Fraunhofer diffraction
Cylindrical wave fronts Planar wave fronts
Source of screen at finite distance from the
obstacle
Observation distance is infinite. In practice,
often at focal point of lens.
Move in a way that directly corresponds with
any shift in the object.
Fixed in position
Fresnel diffraction patterns on flat surfaces Fraunhofer diffraction patterns on spherical
surfaces.
Change as we propagate them further
‘downstream’ of the source of scattering
Shape and intensity of a Fraunhofer diffraction
pattern stay constant.
Classification of Diffraction
Diffraction phenomena of light can be divided into two different classes

Fraunhofer Diffraction at a Single Slit
Let us consider a slit be rectangular aperture
whose length is large as compared to its
breath. Let a parallel beam of wavelength be
incident normally upon a narrow slit of AB.
And each point of AB send out secondary
wavelets is all direction. The rays proceeding
in the same direction as the incident rays are
focused at point O and which are diffracted at
angle θ are focused at point B. The width of
slit AB is small a.
The path difference between AP and BP is
calculated by draw a perpendicular BK.
According to figure the path difference
sinaBK 
a
BK
AB
BK
sin
the phase difference





 sin
22
a
……………..eq 1
……………..eq 2

According to Huygens wave theory each point of slit AB spread out secondary wavelets which
interfere and gives diffraction phenomena. Let n be the secondary wavelets of the wave front
incident on slit AB . The resultant amplitude due to all equal parts of slit AB at the point P can
be determine by the method of vector addition of amplitude. This method is known as
polygon method
For this construct a polygon of vector that magnitude Ao represent the amplitude of each
wavelets and direction of vector represented the phase of each wavelets
nɸ=δ
δ/2
δ/2 N
A
B
C
δ
2δ
nδ
r
nɸ=δ
A
B
R
R/2
R/2

r
nA
radius
arc o


onA
r 
Now a perpendicular CN is draw from the center C of an arc on the line A, which will
divide the amplitude R in two parts
from triangle ACN and BCN
By assuming polygon as a arc of a circle of radius r we can calculate the angle
,
2/
2
sin
AC
R


BC
R 2/
2
sin 

,
2
sin
2

AC
R

AC=BC = r so
2
sin
2

BC
R

2
sin2

rR 
2
sin
2/2
sin2




oo nAnA
R 
By putting the values of r
By assuming δ/2 =α=π/λ a sinθ

sin
onAR 
and the intensity I is given by
2
RI 
2
2
sin


oII 
22
oo AnI
where


Intensity distribution by single slit diffraction
Central Maxima
For the central point P on the screen θ = 0 and hence α = 0
Hence intensity at point P will be
oo III  2
2
sin


1
sin
0  


Lim
Hence intensity at point P will be maximum
Principal Minima
For the principal minima intensity should be zero
,0
sin
2
2



oII ,0sin   n
Where n = 1, 2, 3,4……
n = 0

Secondary Maxima
To find out direction of secondary maxima we
differentiate intensity equation with respect to
α and equivalent to zero
 
  0sincos
0sincos
sin2
0
sin1
cossin2
0
sin
0
3
3
2
2
2
2

































o
o
o
I
I
I
d
d
d
dI
 tan
This is the condition for secondary
maximas and can be solved by plotting a
graph between y= tanα and y=α as shown
The point of intersection of two curves gives
the position of secondary maxima. The
positions are α1 = 0, α2 = 1.43π, α3 = 2.46π,
α4 = 3.47π,..
2
)12(sin




  na
22
22
2
)12(
4
,
4/)12(
2/)12(sin








n
I
I
so
n
n
II
o
o
.......
61
,
22
, 21
oo
o
I
I
I
III 

Intensity distribution by single slit diffraction
Central maxima
Secondary maxima’s
Principal Minima’s

Width of the central maximum
The width of he central maximum can be
derived as the separation between the first
minimum on either side of the central
maximum.
If he first maximum is at distance x from the
central maximum then
x
x
D
f
xw 2
We know that
a
n
for
na







1
1
sin
,1
sin
From the diagram
θ1
f
x
D
x
1tan
If θ is very small sinθ1 = tanθ1 = θ1
a
f
x
f
x
a




a
f
w
2

……………..eq 1
…..eq 2
…..eq 3
…..eq 4

Diffraction Grating
A diffraction grating is an arrangement equivalent to a large number of parallel slits of
equal widths and separated from one another by equal opaque spaces.
Construction
Diffraction grating can be made by
drawing a large number of equidistant
and parallel lines on an optically plane
glass plate with the help of a sharp
diamond point. The rulings scatter the
light and are effectively opaque, while
the unruled parts transmit light and act
as slits. The experimental arrangement
of diffraction grating is shown
They are two type refection and
transmission gratings

A very large reflecting
diffraction grating
An incandescent light
bulb viewed through a
transmissive diffraction
grating.

Theory for transmission grating
(resultant intensity and amplitude)
Let AB be the section of a grating having
width of each slit as a, and b the width of
each opaque space between the slits. The
quantity (a + b) is called grating element,
and two consecutive slits separated by the
distance (a + b) are called corresponding
points.
The schematic ray diagram of grating has
been shown in figure

sin
onAR 
Let a parallel beam of light of wavelength λ incident normally on the grating
using the theory of single slit & Huygens principle, the amplitude of the wave
diffracted at angle θ by each slit is given by
……………………eq 1

Diffraction by n parallel slit at an angle θ is equivalent to N parallel waves of
amplitude R

sin
onAR 
That emitted from each slit s1, s2, s3 …..sN Where α = π/λ (a sinθ), These N parallel
waves interfere and gives diffraction pattern consisting of maxima and minima on
the screen.
The path difference between the waves emitted from two consecutive slits given
by.
sin)(2
21
2
baKS
KSSfrom
KS



The corresponding Phase Difference





 sin)(
22
ba 
Thus there are N equal waves of equal amplitude and with a increasing phase difference of δ
……………………eq 2
……………………eq 3
……………………eq 4

To find the resultant amplitude of these N parallel waves we use the vector
polygon method. Waves from each slit is represented by vectors where its
magnitude represented by amplitude and direction represents the phase.
Thus joining the N vectors tail to tip we get a polygon of N equal sides and the
angle between two consecutive sides is δ
The phase difference between waves from first to last slit is Nδ obtained by
drawing the tangents at A and B
Nδ
Nδ/2
Nδ/2
A
B
C
δ
2δ
Nδ
r
Nδ
A
B
RN
RN
N

Diffraction by n parallel slit at an angle θ is equivalent to N parallel waves of
amplitude RN.
Consider a triangle ACN and DCN C
A N D
δ/2 δ/2
R/2 R/2
AC
R 2/
2
sin 

CD
R 2/
2
sin 

Since AC=CD we can rewrite
2/sin2 
R
AC  ……………………eq 5
In triangle ABC








sin
sinsin
2/sin
2/sinsin
2/sin2
2/sin2
2/sin2
2/sin2/
NnA
R
NnA
R
NR
R
NACR
NACR
o
N
o
N
N
N
N





……………………eq 6
Here

sin
onAR 



 sin)(2/ ba 

So the resultant intensity




2
2
2
2
sin
sinsin NI
I o
 ……………………eq 7
The above equation gives the resultant intensity of N parallel waves diffracted at an
angle θ. The resultant intensity is the product of two terms
2
2
sin
.1

oI


2
2
sin
sin
...2
N
Due to diffraction from each slit
Due to Interference of N slits
Intensity distribution by diffraction Grating
Central Maxima
oo III  2
2
sin


1
sin
0  


Lim
Hence intensity at point P will be maximum
Principal Minima
,0
sin
2
2



oII ,0sin 
 m
Where m = 1, 2, 3,4……
1. Due to Diffraction from Each Slit 2
2
sin
.1

oI

Secondary Maxima
 tan 2
)12(sin




  ma22
)12(
4


m
I
I o
2. Due to Interference of N slits
Principal Maxima’s
Position of Principal maxima’s obtained when


2
2
sin
sin
...2
N


n
 0sin
Where n= 0, 1, 2, 3………..
Then sinNβ is also equal to zero and becomes indeterminate so by using L’ Hosptal
Rule 

2
2
sin
sin N
N
n
nNNNN
Lim
d
d
N
d
d
Lim
n
n





)cos(
)(cos
cos
cos
sin
sin










Hence the intensity of principal maxima is given by
.......2,1,0
sin 2
2
2


n
where
N
I
I o
p



Manima’s
The intensity will be minimum when I = 0 i.e. sinNβ = 0 but sinβ ≠ 0
Nβ = ±pπ here p = 1, 2, 3………..(N-1)(N+1)…….(2N-1)(2N+1)…….. i.e. p ≠ N, 2N……
hence
N
p
ba
pN





sin)(
Secondary Maxima’s
To find out direction of secondary maxima we differentiate intensity equation with respect to
α and equivalent to zero























NN
NNN
NNNN
NNNN
NNNN
I
N
I
d
d
d
dI
o
o
tantan
sin
cossin
1
cos
sin
cos*sin2
sin
cos**sin2
0
sin
cos*sin2
sin
cos**sin2
0
sin
cos*sin2
sin
cos**sin2sin
0
sin
sinsin
0
3
2
2
3
2
2
3
2
22
2
2
2
2
2
























 NN tantan 
The solution of the above equation
except p=±nπ gives the position of the
secondary maxima’s

Intensity of Secondary Maxima’s
The position of secondary maxima is given by using this equation a
right triangle can be formed as shown 1
tan
tan


N
N 
Nβ
A
B
C
Ntanβ
From figure























22
2
2
2
2
22
2
2
2
22
2
2
2
22
2
222
2
2
2
222
2
2
2
222
2
2
2
222
22
2
2
22
22
2
22
sin)1(1sin
sin)1(1
sin
sin)1(1
sin
sin)1(1)sinsin1(sin
sin
)sin(cossin
sin
cos)tan1(sin
sin
sin)tan1(
tan
sin
sin
tan1
tan
sin
tan1
tan
sin




















N
N
NI
N
N
I
I
I
N
N
II
N
N
N
NN
N
NN
N
NN
N
NN
N
N
N
N
N
N
o
o
p
s
os
22
2
sin)1(1 

N
N
I
I
p
s
22
2
sin)1(1 

N
N
I
I
p
s
As increase in number of slit the
number of secondary maxima also
increases

Intensity distribution by Diffraction Gratings

Formation of Spectra with Diffraction Grating
With White Light
With Monochromatic Light

Characteristics of Grating Spectra
If the angle of diffraction is such that, the minima due to diffraction component in the
intensity distribussion falls at the same position of principal maxima due to interference
component, then the order of principal maxima then absent. If mth order minima fall on nth
order principal maxima then






m
n
a
ba
nba
ma




sin
sin)(
sin)(
sin
m
n
a
ba

 )(
Now we consider some cases
A. If b=a, i.e. width of opaque space in equal to width of slit then from equation 3. n = 2m
since m=1, 2, 3 …. Then n = 2nd, 4th , 6th ….spectra will be absent
B. If b=2a, i.e. width of opaque space in equal to width of slit then from equation 3. n = 3m
since m=1, 2, 3 …. Then n = 3rd, 6th , 9th ….spectra will be absent
1. ABSENT SPECTRA
……………eq 1
……………eq 2
……………eq 3

2. Maximum Number of Order Observed by Grating
Principal maximum in grating spectrum is given by
 nba  sin)(

sin)( ba
n


Maximum possible angle of diffraction is 90 degree therefore

o
ba
n
90sin)(
max

 
)(
max
ba
n

So
Q.1. A plane transmission grating has 6000 lines/cm. Calculate the higher order of spectrum
which can be seen with white light of wavelength 4000 angstrom
Sol. Given a+b=1/6000, Wavelength 4000X 10-8cm
As we know that gratings equation written as
For maximum order
Maximum order will be 4th

sin)( ba
n


16.4
24
100
1040006000
1)(
8max 



 
cm
cmba
n

……………..eq 1
……………..eq 2

3. Width of principal maxima
The angular width of principal maxima of nth order is defined as the angular separation between
the first two minima lying adjacent to principal maxima on either side
θn
2δθn
θn-δθn
θn+δθn
A O
If θn is the position of nth order principal maxima
θn+δθn, θn-δθn are positions of first minima
adjacent to principal minima then the width of
nth principal maxima = 2δθn
From the Grating Equation nth order maxima
And the position of minima is given by
Hence equation rewritten as
 nba n  sin)( ……………eq 1
)1(,,sin)(  nNpwhere
N
p
ba n 
 




 

N
nN
dba nn
1
)sin()( …eq 2
On dividing eq 2 by eq 1















nN
d
d
nN
dd
n
nn
n
n
nnnn
1
1
)sin(
sincos
cos
1
1
)sin(
)sincoscos(sin










 


nN
nNd
n
nn 1
)sin(
)sin(


If dθn is very small than cosdθn = 1, sindθn = dθn
nN
d
nN
d
nN
d
n
nnn
n
nn




tan
,
1
cot
1
1
)sin(
cos
1








nN
d n
n


tan2
2 

4. Dispersive Power of Diffraction Gratings
For a definite order of spectrum, the rate of change of angle of diffraction θ with respect to the
wavelength of light ray is called dispersive power of Grating.
Dispersive Power = dθ/dλ
We know that gratings equation
Also written as
 nba n  sin)(





cos)(
cos)(
ba
n
d
d
n
d
d
ba



222
2222
)(
sin)()(sin1)(




nba
n
d
d
baba
n
ba
n
d
d






……………..eq 1
……………..eq 2
……………..eq 3

5. Experimental demonstration of diffraction grating to determine
wavelength

Resolving Power
Resolving Power
The ability of an optical instrument to produce two distinct separate images of two objects located very close to
each other is called the resolution power
Limit of resolution
The smallest distance between two object, when images
are seen just as separate is known as limit of resolution
For eye limit of resolution is 1 minutes
Resolution
When two objects or their images are very close to each other they appeared as a one and it not be possible for the
eye to seen them separate. Thus to see two close objects just as separate is called resolution

Rayleigh Criterion for Resolution
Lord Rayleigh (1842-1919) a British Physicist proposed a criterion
which can manifest when two object are seen just separate this criterion
is called Rayleigh’s Criterion for Resolution
Well Resolved
Just resolved
Not resolved

Resolving power of a telescope
Resolving power of telescope is defined as the reciprocal of the smallest angle sustained
at the object by two distinct closely spaced object points which can be just seen as
separate ones through telescope.
Let a is the diameter of objective telescope as shown in fig and P1 and P2 are the
positions of the central maximum of two images. According to Rayleigh criterion these
two images are said to be separated if the position of central maximum of the second
images coincides with the first minimum or vice versa.
P1
P2
A
B
dθ
dθ
dθ
a
The path difference between AP2 and
BP2 is zero and the path difference
between AP1 and BP1 is given by
If dθ is very small sin dθ = dθ
C


daBC
dABBC
sin
sin






a
dRP
a
dad


/1
,
For rectangular aperture 22.1
a
RP  For circular aperture
……………..eq 1
………eq 2

Resolving power of a Diffraction Grating
The resolving power of a grating is the ability to separate the spectral lines which are
very close to each other.
When two spectral lines in spectrum produced by diffraction grating are just resolved,
then in this position the ratio of the wavelength difference and the mean wave length of
spectral lines are called resolution limit of diffraction Grating
Q
dθ
Let parallel beams of light of wavelength λ and λ+dλ
be incident normally on the diffraction grating. If nth
principal maxima of λ and λ+dλ are formed in the
direction of θn , θn+dθn respectively
For the principal maximum by wavelength λ the
gratings equation written as
for wavelength d λ
We know that for minima
By eq 2 and 3


d
RP 
θn
λ+dλ
δθn
A
λ P
 nba n  sin)(
)()sin()(  dndba nn 
N
nN
N
p
dba nn


)1(
)sin()(


N
n
NN
nN
N
nN
dn



 


)1(
)(
Nnd
N
nndn
/




nN
d
RP 


……………..eq 1
..eq 2
..eq 3 ..eq 4

Diffraction of light

More Related Content

What's hot (20)

Similar to Diffraction of light (20)

More from Dr. Vishal Jain (17)

Recently uploaded (20)

Diffraction of light