1. Diffraction of red laser beam on the hole
Diffraction
Dr. Amit Kumar Chawla

2. Diffraction by edges

3. Diffraction Pattern, Illuminated
Penny
• This shows a diffraction pattern
pattern
created by the illumination of a
penny, positioned midway
between screen and light source.
• The bright spot at the center.
– It is a constructive interference

4. Diffraction Pattern, Object Edge
• This shows a diffraction pattern associated with light
pattern
from a single source passing by the edge of an opaque
object
• The diffraction pattern is vertical with the central
vertical
maximum at the bottom

5. Introduction to Diffraction Patterns
• Light of wavelength comparable to or
larger than the width of a slit (λ >> a)
slit
spreads out in all forward directions
upon passing through the slit.
This phenomena is called diffraction.
This indicates that light spreads beyond
the narrow path defined by the slit into
regions that would be in shadow if light
traveled in straight lines

6. Diffraction
If an opaque obstacle is placed between a source of light
and a screen (the size of obstacle must be comparable to
wavelength of light) then light bends around the corner
of the obstacle into the geometrical shadow. This
bending of light is called diffraction.
The bending or spreading of waves when they encounter
an obstacle or an opening in their path is called
diffraction.

7. Diffraction Pattern, Narrow Slit
• A single slit placed between a distant
slit
light source and a screen produces a
diffraction pattern
– It will have a broad, intense central band,
called the central maximum
– The central band will be flanked by a series
of narrower, less intense secondary bands,
called side maxima or secondary maxima
maxima
– The central band will also be flanked by a
series of dark bands, called minima

8. Diffraction vs. Interference
• Diffraction is the bending of light around an obstacle,
whereas the interference is the meeting of two waves.
• Interference pattern is obtained by the superposition of
waves coming from two different wavefronts originating
from the same source while when the waves emerging from
different parts of the same wavefont superimpose with each
other then diffraction patter is obtained.
• The widths of the diffraction fringes are not equal, but the
widths of the interference fringes may or may not be equal.
• The points of minimum intensity in interference appears
perfectly dark but these points in the case of diffraction are
not perfectly dark.

9. Classes of Diffraction
Diffraction requires a source of light, an obstacle or aperture
and the screen
Based on the distance between source, aperture and screen,
and also on the shape of wavefront, diffraction pattern is
classified into two classes
1. Fresnal Diffraction
2. Fraunhofer Diffraction

10. Fresnel Diffraction
If the source of light or the screen or both of them are
at finite distances from the diffracting aperture, then
the wavefronts falling on the aperture will not be
plane. The diffraction obtained under this type of
arrangement is called Fresnel Diffraction. This is also
called near-field diffraction. No lenses are used to
make the rays parallel or convergent

11. Fraunhofer Diffraction
If both the source of light and the screen or both of
them are effectively far enough from the aperture so
that the wavefronts reaching the aperture and the
screen can be considered plane. Then the source and
the screen are said to be at infinite distances from the
aperture. This kind of diffraction is called Fraunhofer
Diffraction. This is also called far-field diffraction.

12. Fraunhofer Diffraction is encountered in the case of
gratings that contain number of slits.
When the screen is moved, the size of the diffraction
patteren changes uniformaly while the shape of the
pattern does not change.
Fresnel Diffraction is obtained when light suffers
diffraction at a straight edge, a thin wire, a narrow slit
etc.
Both the size and shape of the pattern depends on the
distance between the diffracting aperture and the
screen.

13. Fraunhofer Diffraction at single slit
P
A θ
e θ O
θ
K θ
B
Path difference = BK = AB sinθ = e sinθ
2π 2π
Phase difference = x path difference = λ (e sin θ )
λ

14. Let the width AB of the slit be divide into n equal parts. The
amplitude of vibration at P due to the waves from each part will be
same, say a. The phase difference between the waves from any
two consecutive parts is
1  2π 
 e sin θ  = d
n λ 
Then the resultant amplitude at P is given by
 πe sin θ 
a sin  
a sin(nd 2)  λ 
R= =
sin(d 2)  πe sin θ 
sin  
 nλ 
Let us put
π 
 e sin θ  = α
λ 

15. a sin α a sin α
Then R= =
sin(α n) α n
na sin α
R=
α
When n → ∞ , a → 0 , but the product na remains finite.
Let na = A
The resultant intensity at P, being proportional to the square of
the amplitude, is
2
2  sin α 
I=R =A 
2

 α 

16. 2
 sin α 
I=R =A 
2 2

 α 
Condition for Maxima
A sin α A  α3 α5 α7 
R= = α − + − + ......
α α 3! 5! 7! 
A sin α  α2 α4 α6 
R= = A1 − + − + ......
α  3! 5! 7! 
R = A for α = 0
This is the intensity of central maximum
π 
α =  e sin θ  = 0 or sin θ = 0
λ 

17. Condition for Minima
sin α
=0
α
sin α = 0 But, α ≠0
α = ± mπ Where m has an integral value 1, 2, 3 except zero
π 
So,  e sin θ  = ± mπ ⇒ e sin θ = ± mλ
λ 
This equation gives the position of first, second, third ….
By putting m = 1, 2, 3….

18. Secondary Maxima
dI
=0
dα
d  2  sin α  2 
A   =0
dα   α  
 
 2 sin α  α cos α − sin α
A2   =0
 α  α 2
α cos α − sin α
=0
α 2
α cos α − sin α = 0
α = tan α = y ( say )

19. y =α and y = tan α
The maxima will occur when
3π 5π 7π
α= , ,
2 2 2
π n = 1,2,3.....
or α = (2n + 1)
2
These are points of secondary maxima

20. 2
 sin α 
I = I0  
 α 
3π
Put α=
2
4 4 4
I1 = 2 I 0 I2 = I
2 0
I3 = I
9π 25π 49π 2 0
The ratio of relative intensities of successive maxima are
4 4 4
1: 2 : : : .......
9π 25π 49π
2 2

21. 3π 5π (2 N − 3)π
α =± ,± , ..., ±
2N 2N 2N

22. Single-Slit Diffraction,
Intensity
• The general features of the
intensity distribution are shown
– A broad central bright fringe is
flanked by much weaker bright
fringes alternating with dark
fringes
– Each bright fringe peak lies
approximately halfway between the
dark fringes
– The central bright maximum is
twice as wide as the secondary
maxima

23. Diffraction due to grating
A P
S1 θ
e+d S2
S3 K1
O
Sn-1
Sn
B
The amplitude from each slit in the direction θ is
A sin α πe
R0 = where α = sin θ
α λ

24. The path difference between the wavelets from S1 and S2 in the
direction θ is
S 2 K1 = (e + d ) sin θ
Hence the phase difference between them
2π
(e + d ) sin θ = 2 β
λ
If N be the total number of slits in the grating, then by the method
of vector addition of amplitudes, the resultant amplitude in the
direction of θ will be
sin Nβ A sin α sin Nβ
R = R0 =
sin β α sin β

25. Thus the resultant intensity at point P is
2 2
 sin α   sin Nβ 
I=R =A 
2 2
 
 sin β 

 α   
2
 sin α 
2
The factor A   gives the intensity distribution due to single slit,
 α 
2
 sin Nβ 
while 
 sin β  gives the distribution of intensity in the diffraction

 
pattern due to the interference in the waves due to N slits.

26. Principal Maxima
2 2
 sin α   sin Nβ 
I=R =A 
2 2
 
 sin β 

 α   
The intensity will be maximum when
sin β = 0 ⇒ β = ± nπ n = 0,1,2,3.....
This results in
sin Nβ 0
= (Indeterminate)
sin β 0

27. Apply L’ Hospital rule
d
(sin Nβ )
sin Nβ dβ
Lim = Lim
β → ± nπ sin β β → ± nπ d
(sin β )
dβ
N cos Nβ
Lim ⇒ ±N
β → ± nπ cos β
2
This results in  sin α  2
I=A  2
 N
 α 
The condition for principal maxima is
sin β = 0 or β = ± nπ

28. π
(e + d ) sin θ = ± nπ
λ
(e + d ) sin θ = ± nλ
For n = 0, we get θ = 0 and this gives the direction of zero order
principal maxima. The value of n = 1, 2, 3, … gives the direction
of first, second, third …….. order principal maxima.

29. Minima
2 2
 sin α   sin Nβ 
I=R =A 
2 2
 
 sin β 

 α   
The intensity will be minimum when
sin Nβ = 0 but sin β ≠ 0
Therefore Nβ = ± mπ
Nπ
(e + d ) sin θ = ± mπ ⇒ N (e + d ) sin θ = ± mλ
λ
Here m can have all integral values except 0, N, 2N, 3N……

30. Here m can have values 1, 2, 3, …….(N-1). Thus there are (N-1)
equispaced minima between two consecutive principal maxima.
Secondary Maxima
In order to differentiate between the two consecutive minima there
should be a maximum between them. Therefore there are (N-2)
maxima between (N-1) minima. These maxima are known as
secondary maxima.
2 2
 sin α   sin Nβ 
I=A 2
 
 sin β 
 α   
dI
=0
dβ

31. A2 sin 2 α  sin Nβ   N cos Nβ sin β − sin Nβ cos β 
2
 sin β 
 =0
α2   sin β
2

N cos Nβ sin β − sin Nβ cos β = 0
tan Nβ = N tan β

32. 1 + N 2 tan 2 β tan Nβ = N tan β
N tan β
Nβ
1
N tan β
sin Nβ =
1 + N 2 tan 2 β
sin 2 Nβ N 2 tan 2 β /( 1 + N 2 tan 2 β ) 2
=
sin β
2
sin 2 β

33. N2
=
cos 2 β (1 + N 2 tan 2 β )
N2
=
cos 2 β + N 2 sin 2 β
N2
=
1 − sin 2 β + N 2 sin 2 β
sin 2 Nβ N2
=
sin β
2
1 + ( N 2 − 1) sin 2 β

34. 2 2
 sin α   sin Nβ 
I=A  2
 
 sin β 
 α   
sin 2 α N2
I=A 2
α 2 1 + ( N 2 − 1) sin 2 β
The intensity of secondary maxima is proportional to
N2
1 + ( N 2 − 1) sin 2 β
2
While the intensity of primary maxima is proportional to N

36. Intensity of secondary maxima = 1
Intensity of principal maxima 1 + ( N 2 − 1) sin 2 β
As N increases the intensity of secondary maxima decreases.
In case of diffraction grating N is very large.
Therefore the secondary maximas are not visible in the spectrum
and there is complete darkness between two successive principal
Maxima.

37. Rayleigh Criteria for Resolution
Resolving Power: The ability of an optical instrument to resolve
the images of two close point source is known as resolving power.
Limit of Resolution: The minimum separation between two objects
that can be resolved by an optical instrument is called the limit of
Resolution.
Rayleigh Criteria for Resolution: According to Rayleigh, two close
point objects are said to be just resolved if the principal maxima of
one coincides with the first minima of the other and vice-versa.

39. Rayleigh Criteria for Resolution
According to the single slit Fraunhofer diffraction
sin 2 α
I = I0
α2
First minima is formed at an angle α = π
The angle at the point of intersection will be π/2
The intensity of each curve at the dip will be
sin 2 ( π / 2) 4
I1 = I 2 = I 0 = 2 Io
( π / 2) 2
π

40. The resultant intensity at the dip is then given by
I = I1 + I 2
8
I= I = 0.81I o
2 o
π

41. Resolution, Example
• Pluto and its moon, Charon
• Left: Earth-based telescope is blurred
• Right: Hubble Space Telescope clearly resolves
the two objects

42. Resolving Power of Plane Diffraction grating
M
A
λ + dλ P2
λ P1
dθ
θ
O
B N

43. Resolving Power of Plane Diffraction grating
The resolving power of the grating is defined as the ratio of
wavelength (λ) to the difference dλ of the wavelength
The direction of nth principal maxima for wavelength λ is given by
(e + d ) sin θ = nλ
The direction of nth principal maxima for wavelength λ+dλ is given by
(e + d ) sin(θ + dθ ) = n(λ + dλ ) (1)

44. The minima for wavelength λ is given by
N (e + d ) sin θ = mλ
Here m can have all integral values except 0, N, 2N, 3N……
because for these values of m the condition of maxima is satisfied.
The first minimum adjacent to nth principal maxima in the direction
(θ+dθ) can be obtained by putting m as (nN+1)
N (e + d ) sin(θ + dθ ) = (nN + 1)λ
(nN + 1)λ (2)
(e + d ) sin(θ + dθ ) =
N

45. Comparing equation (1) and (2)
(nN + 1)λ
n ( λ + dλ ) =
N
λ
nλ + ndλ = nλ +
N
λ λ
ndλ = ⇒ = nN
N dλ
Since resolving power is directly proportional to N , it means that
larger will be the number of lines per cm of a grating greater will
be the resolving power.

46. Angular width of Principal maxima
M
A
P2
m = nN + 1
m = nN
P1
dθ n m = nN − 1
dθ n
θn
O
B N

47. θ n be the direction of nth principal maxima.
(θ n + dθ n ) and (θ n − dθ n ) be the directions of first outer and inner
sided minima adjacent to the nth maxima
The total angular width will be 2dθn
The direction of nth order principal maxima can be given by
(e + d ) sin θ n = nλ
The direction of nth minima can be given by
N (e + d ) sin θ n = mλ

48. The first order outer and inner sided minima adjacent to the nth
Maxima can be given by
N (e + d ) sin(θ n ± dθ n ) = (nN ± 1)λ
N (e + d )(sin θ n cos dθ n ± cos θ n sin dθ n ) = (nN ± 1)λ
When dθ n is small then cos dθ n = 1 and sin dθ n = dθ n . So
N (e + d )(sin θ n ± cos θ n dθ n ) = (nN ± 1)λ
N (e + d ) sin θ n ± N (e + d ) cos θ n dθ n = nNλ ± λ
Nnλ ± N (e + d ) cos θ n dθ n = nNλ ± λ
N (e + d ) cos θ n dθ n = λ

49. λ
dθ n =
N (e + d ) cos θ n
λ
2 dθ n = 2
N (e + d ) cos θ n
The angular width of the nth order principal maxima depends on
The total number of lines present on grating, grating element and
the wavelength.