Diffraction
A Phenomenon of bending of light or spreading of light by an
obstacle is known as diffraction.
Italian scientist Grimaldi
A’
A
S
B
Slit
B’
Screen
Difference between interference and diffraction
Interference
Diffraction
1) All maxima have same intensity.
2) Interference fringes have almost
same width.
3) The region of minimum intensity is
usually almost dark.
4) This phenomenon is the result of
interaction taking place between
two separate wave fronts obtaining
from two coherent sources.
1) All maxima have varying intensity.
2) The diffraction fringes don’t have
same width.
3) The region of minimum intensity is
not perfectly dark.
4) This phenomenon is the result of
interaction of light between the
secondary wavelets originating from
different pints of the same wave
front.
Type of diffraction:
Fraunhofer Diffraction
Fresnel Diffraction
Fraunhofer Diffraction
Fresnel Diffraction
1) The source and the screen are at
infinite distance from the diffracting
aperture.
2) Diffracted light is collected by a lens
as in a telescope.
3) The centre of the diffraction pattern
is always bright for all paths parallel
to the axis of the lens.
4) The wave fronts are plane which is
realized by using convex lens.
5) For this, single, double slit or plane
diffraction gratings are used.
1) The source and the screen are at
finite distance from the diffracting
aperture.
2) No mirror and lenses are used for
observation.
3) The centre of the diffraction pattern
may be bright or dark depending
upon the number of Fresnel Zone.
4) The wave fronts are divergent either
spherical or cylindrical.
5) For obtaining Fresnel diffraction
zone plates are used.
Fresnel’s Zone: A wave front can be divided into a large number of elements or
zones called Fresnel’s zone.
Fraunhofer’s diffraction at a single slit:
X
O
θ
e
A
θθ
e
e
S
Monochromatic Source
B
P
θθNN
e e
Convex lens
Slit
Y
Screen
The path difference (Δ) = BN.
BN  AB sin 
So that the path difference = e sin θ
Therefore, phase difference =
2

 e sin
Let the width AB of the slit be divided into n equal parts. So that n is the no.
of vibrating particles and let d is the constant phase difference between
those particles.
d
1
 phase difference 
n
d
2 e sin 
n
This is the constant phase difference of each vibrating particle
To calculate the resultant amplitude (like that: the resultant amplitude
for a simple harmonic vibration having the common phase difference).
Draw a polygon, to determine the resultant amplitude R and resultant
phase δ.
R
3d
δ
2d
d
a
Resolved part of a system of vibration in the horizontal direction:
R cos  a cos0  a cosd  a cos2d  a cos3d        a cosn  1d
and for vertical direction:
R sin   a sin 0  a sin d  a sin 2d  a sin 3d        a sinn  1d
By solving these two equations
a sin nd 2
R
sin d 2
  e sin  
a sin 




R
  e sin  
sin 

 n 
or
Let us put
 e sin 


then
R
a sin 
sin

n
n is the no. of vibrating particles, when n is very large then
should be very small, so that: sin
R
a sin 


n
R
n
Let
na  A , then
R


n
na sin 

A sin 


n
Therefore, the resultant intensity at point P is being proportional to the
square of the amplitude.
I  R2
I
A 2 sin 2 
(1)
2
Positions of Minima:
It is clear from above equation that the intensity should be minimum (zero)
when, sin 
sin   0 , (but   0)
0

e sin   m
,
Finally, we have
  tan 
This equation is solved graphically by plotting the curves
y  tan 
y = tan α
y = tan α
and
y = tan α
y 
y
min
45◦
π/2
min
1.43π
(max)
3π/2
2.46π
(max)
5π/2
α
Form the graph the point of intersection gives the values of α , these values
are approximately given
  0,
3 5 7
,
,
,       
2 2 2
or more exactly as
  0, 1.43 , 2.46 , 3.47 ,       
Put the approximately values of α in equation (1), we get the intensities of
various maxima. Thus the intensities of the central maxima is
 sin  
IA 

  
2
(i)
For principal maxima   0 then
I 0  A2
2
(ii)
Intensity for the first secondary maxima   3 

I
4
9 2
I0 
2 
I0
22
Now intensity for the second secondary maxima   5 

I
I
4
I  0
2 0
61
25
2 
So that the ratios of successive maxims are:
4
4
:
:
2
2
2
9 25 49
1 1 1
1:
: :
:
22 61 121
1:
or
4
:
where m = 1, 2, 3,……………..
  m
The minima lie at
   ,  2 ,  3 ,  4 ,        
It is clear that most of the incident light is concentrated in the principal maxima,
which occur when   0
I
-4π
-3π
-2π
-π
0
+π
+2π
+3π
+4π
α
Fraunhofer Diffraction at N-slit (Diffraction grating):
X
e
A
P
d
θθ
K θ
O
θ
θθ
B
Y
By the theory of Fraunhofer diffraction at a single slit, the secondary
waves from all points of a slit in the direction θ are equivalent to a single
wave of amplitude  Ao 
Ao 
Ao
A sin 

is the resultant amplitude due to a single slit, in the normal direction.
A = na, where A is the amplitude of n number of vibrating particles in a slit,
α is the phase difference between the wavelets.


e sin 

If N is the total number of slits in grating, the diffracted rays from all the slits
act as N number of parallel rays.
Draw an equal lengths (like vector) MP1, P1P2, P2P3 -------------, the closing
side MPN of the polygon represents the resultant amplitude (R). If O is the
centre of polygon, then from geometry:
PN
2Nβ
O
2β
β
6β
MPN 2OM sin N

MP1
2OM sin 
MPN  MP1
Resultant amplitude (R)
4β
2β
P2
M N P1
sin N
sin 
Ao or amplitude of each vibration
Resultant amplitude (R)
R
A sin  sin N


sin 
Then the resultant intensity (I) is:
A2 sin 2  sin 2 N
I

2

sin 2 
The first factor
A 2 sin 2 
2
while the second factor
gives the diffraction pattern due to single slit;
sin 2 N
sin 2 
gives the intensity distribution pattern due
to interference pattern for N slits.
Let us consider the intensity distribution due to the second factor.
Principal Maxima:
When sin   0
i. e.
   n
where n = 0, 1, 2, 3, -----------
Then sin N  0
and thus sin N  0
sin 
0
Let us find the value of usual method of differentiating the numerator and the
denominator. By using ‘L’ Hospital rule:
sin N
 N
   n sin 
lim
So that the intensity from equation (1), we have,
I
A 2 sin 2 
2
N2
This is the maximum intensity.
   n
e  d sin   n
where n = 0, 1, 2, 3, -----------
For n = 0, we get the zero order maxima. For n =1 this is
e  d sin   
For n = ±1, ±2, ±3, -------------, we obtain the first, second, third order principal
maxima respectively.
For Minima:
When
sin N  0
but sin   0
sin N
0
sin 
Hence, from equation (1) we get
I=0
This is minimum intensity. These minima are obtained in the direction
given by
or
N  m
N e  d sin   m
Where m takes all integral values except 0, N, 2N, 3N, -------- nN, because these
value of m make sin   0 which gives the principal maxima.
It is clear from above that m = 0 gives a principal maxima and m = 1, 2, 3, -------- (N-1)
give minima.Then m = N gives again a principal maxima. Thus, there are (N-1) minima
between two consecutive principal maxima.
Secondary Maxima:
As there are (N-1) minima between to consecutive principal maxima, there must be
(N-2) secondary maxima between two principal maxima.
Principal Maxima
Secondary Maxima
Minima
The position of the secondary maxima are obtained by differentiate equation (1) with
respect to β and equating it to zero.
dI
0
d
N tan   tan N
1  N 2 tan2  
N tan 
Nβ
1
sin 2 N
sin 2 


N2

1  N 2  1 sin 2 
This equation shows that the intensity of the secondary maxima is proportional to

N2

1  N 2  1 sin 2 
therefore
where as the intensity of the principal maxima is proportional to N 2
Intensity of the Secondarymaxima
1

Intensity of the principalmaxima 1  N 2  1 sin 2 


Hence, greater value of N (Number of slits), the weaker are secondary maxima,
In actual grating, N is very large. Hence these secondary maxima are not visible
in the grating spectrum.
Formation of Multiple Spectra by the grating:
When a beam of wavelength  fall normally on the grating,
the principal maxima are formed in the direction given by
e  d sin   n
R2
V2
n = 0, 1, 2, 3, -------------------
IInd order
R1
Ist order
V1
Zero order
R1
V1
Ist order
R2
V2
IInd order
Condition for absent spectra:
ed
n

e
m
This is the condition for the spectrum of the order n to be absent.
Maximum number of orders:
nmax 
e  d 

Dispersive power of a grating:
The dispersive power of a grating is defined as “the rate of change of
d
the angle of diffraction with the wavelength of light”. It is expressed as d 
The angle of diffraction  , for the principal maxima is related to the
corresponding wavelength by grating equation
e  d sin   n
Differentiating the above equation with respect to  we get
d
n

d e  d  cos
The above equation shows that the dispersive power is1.
2.
3.
directly proportional to the order n.
inversely proportional to the grating element
inversely proportional to cos  It means larger value of 
smaller value of cos  then the value of dispersive power is higher.
Example- The angle of diffraction for red is greater than the
violet in a given order spectrum, the dispersion in the red region
is greater then the violet region.
Rayleigh’s Criterion for resolution:
Resultant intensity
A
B
λ'
λ
Widely separated
C
A
B
λ
λ'
Very close
In this figure, the resultant intensity
shows a maxima C and two
spectral lines appear to be one.
Dip
A
λ
B
λ'
θ
θ'
In this case, the resultant intensity
shows a dip and two wavelengths
are said to be just resolved.
Resolving Power of a Diffraction grating:
It defined as the ratio of the wavelength of light λ to the difference
between two wavelengths to be resolved. If λ and λ+dλ are the two

wavelengths to be resolved then the resolving power is given as
d
Grating
Ist minima of
the wavelength
λ overlapping
the maxima of
wavelength
λ+dλ.
dθn
θn
Resolving power of a diffraction grating using Rayleigh’s criterion of
resolution
Let the angle of diffraction for the nth order principal maxima of wavelength

be θn and that of wavelength   d be θn+dθn.
Using Rayleigh Criterion

 nN
d
A grating with larger numbers of slits will have larger resolving power
Relation between Resolving Power and Dispersive Power:
The resolving power of a grating is equal to the product of the order
of the spectrum and total number of lines on the grating.

 nN
d
d
n
We know that the dispersive power of the grating is d  e  d cos
so that from the above equation, we have

 Re solving Power  Total Apperature  Dispersive Power
d
N e  d cos is total aperture width.