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DIFFRACTION OF LIGHT
(a) Fraunhofer diffraction Single slit; Double Slit. Multiple slits and Diffraction grating.
(b) Fresnel Diraction: Half-period zones. Zone plate

Diffraction:- The phenomena of bending of light waves around obstacle of size comparable with
the wavelength of light and resulting there by in there spreading into geometrical shadows of the object
is known as Diffraction.

There are two types of diffraction,

i) Fresnel Diffraction,
ii) Fraunhofer Diffraction.

FRAUNHOFER DIFFRACTION:
In this type of diffraction both of the source and the screen are effectively at infinite distance from the
obstacle or the aperture.

Fraunhofer diffraction at a single slit :-

Let a parallel beam of monochromatic light of wavelength λ (i.e. a planar wavefront) be incident
normally on a narrow slit AB of width 𝑎.

According to Huygens principle, each point of wave front on the slit-plane sends out secondary
wavelets in all direction. The secondary wavelets travelling normal to the slit are both to focus by a convex
lens L on the screen at C.

The wavelets travelling at an angle θ with the normal are brought to focus at Q. Let us calculate the
intensity of light at Q.

Let the complex light disturbance at any instant 𝑡 at Q due to secondary waves from the origin O, the
midpoint of the slit is represented by 𝐴𝑒 𝑖𝜔𝑡 , where A is amplitude 𝜔 is the circular frequency of the wave

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. The path difference between the waves at Q coming from O (ray 1)and from a point P (ray 2) at a distance
𝒙 from O is given by

𝑃𝑁 = 𝑥 𝑠𝑖𝑛𝜃

Hence the phase difference,

2𝜋 2𝜋
× 𝑃𝑁 = 𝑥 𝑠𝑖𝑛𝜃 = 𝑘𝑥,
𝜆 𝜆
2𝜋
Where 𝑘 = 𝑠𝑖𝑛𝜃
𝜆

The disturbance at Q due to the ray (2)

𝐴 𝑒 𝑖(𝜔𝑡−𝑘𝑥)

The disturbance at Q due to the diffracting element 𝑑𝑥 can be written as

𝑑𝑦 = 𝐶𝐴 𝑑𝑥 𝑒 𝑖(𝜔𝑡−𝑘𝑥)

Where we assume the amplitude to be proportional to the width 𝑑𝑥, C is a proportionality constant.

Resultant complex disturbance at Q due to the whole slit is,

𝑎
2

𝑦 = ∫ 𝐶𝐴 𝑑𝑥 𝑒 𝑖(𝜔𝑡−𝑘𝑥) 𝑑𝑥
𝑎
−
2
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𝑎
+
𝑒 −𝑖𝑘𝑥 2
= 𝐶𝐴𝑒 𝑖𝜔𝑡 [ ]
−𝑖𝑘 −
𝑎
2
𝐶𝐴𝑒 𝑖𝜔𝑡 𝑖𝑘𝑎 𝑖𝑘𝑎
= [𝑒 2 − 𝑒− 2 ]
𝑖𝑘
𝐶𝐴𝑒 𝑖𝜔𝑡 𝑘𝑎
= 2𝑖 𝑠𝑖𝑛 ( )
𝑖𝑘 2
𝑘𝑎
𝑠𝑖𝑛 ( 2 )
= 𝑎𝐶𝐴𝑒 𝑖𝜔𝑡
𝑘𝑎
(2)
𝑠𝑖𝑛 𝛼 𝑖𝜔𝑡
= 𝐶𝐴 𝑎 𝑒
𝛼
𝑘𝑎 𝜋
Where we substitute, α = ( 2 ) = 𝜆 a sin θ

The resultant intensity at P is given by

𝑠𝑖𝑛2 𝛼 2
𝐼 = 𝑦𝑦 ∗ = (𝐶𝐴 𝑎)
𝛼2
𝑠𝑖𝑛2 𝛼
∴ I = 𝐼0 , where 𝐼0 = (𝐶𝐴 𝑎)2
𝛼2

This is the intensity distribution function for a narrow single slit.

Central (principal) Maximum :

For the point C on the screen, θ= 0 and hence α = 0, the intensity at c

sin2 𝛼
lim 𝐼0 ( 2 ) = 𝐼0
𝛼→0 𝛼
this is the intensity of central maximum.
Type equation here.

Minima

𝐹𝑜𝑟 𝑚𝑖𝑛𝑖𝑚𝑎 , 𝑤𝑒 ℎ𝑎𝑣𝑒 ,

𝑠𝑖𝑛 𝛼 = 0 𝑏𝑢𝑡 𝛼 ≠ 0
∴ 𝛼 = 𝑚𝜋 , 𝑚 = ± 1 , ± 2 …
𝑘𝑎 𝜋
∴ 𝑎 𝑠𝑖𝑛 𝜃 = 𝑚 𝜆 ; since, α = ( 2 ) = 𝜆 a sin θ
The above equation gives the direction of the first , 2nd ….etc minima corresponding to m = ± 1 , ±
2……………..

Secondary Maxima:
𝑑𝐼
The maxima and minima are given by 𝑑𝛼
= 0
𝑑 𝑠𝑖𝑛𝛼 2
Or, [𝐼0 ( ) ]=0
𝑑𝛼 𝛼

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𝑠𝑖𝑛𝛼 𝑎 cos 𝛼−sin 𝛼
Or, 2 ( 𝛼 ) =0
𝛼2
Or, 𝑠𝑖𝑛 𝛼 (𝛼𝑐𝑜𝑠𝛼 − 𝑠𝑖𝑛𝛼) = 0

𝑠𝑖𝑛 𝛼 (𝛼𝑐𝑜𝑠𝛼 − 𝑠𝑖𝑛𝛼) = 0

𝑑2 𝐼
But it is found that 𝑑𝑥 2 is positive from 𝑠𝑖𝑛 𝛼 = 0(𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑚𝑖𝑛𝑖𝑚𝑎) and negative for 𝛼 = 𝑡𝑎𝑛 𝛼

For maxima :
𝛼 = 𝑡𝑎𝑛 𝛼
which is transcendental equation and can only be solved graphically by plotting the two curves y = α
and y = tan α and finding the points of intersection are shown below

Hence for α = 0 (central maxima) and other values of which will give maxima are less than but gradually
3𝜋 5𝜋
approaching towards ± ;± 𝑒𝑡𝑐.
2 2

The intersection of two gives α- values that satisfy y= α and y = tan α. The values are –
3𝜋 5𝜋
α=0, ± ,± … … … … . . 𝑒𝑡𝑐 ( 𝑎𝑝𝑝𝑟𝑜𝑥)
2 2

The intensity of first secondary maxima is

3𝜋
𝑠𝑖𝑛2 2 4
∴ 𝐼1 ≈ 𝐼0 3𝜋 2 = 𝐼0
( ) 9𝜋 2
2

The intensity of the second secondary maxima is

5𝜋
𝑠𝑖𝑛2 2 4
𝐼2 ≈ 𝐼0 5𝜋 = 25𝜋 2
𝐼0
( )2
2

https://www.desmos.com/calculator/mzim0qdoet
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Thus the secondary maxima are very feeble compared with the principal maximum and their intensities
diminish very rapidly with the increase in order no. The intensity distribution in the diffraction pattern

SingleSlitDiffractionPattern.cdf
due to a single slit is shown by figure below.

Width of central maximum :-

The angle of diffraction θ, for the first minima on either side of central maximum is given by
𝛼 sin 𝜃1 = 𝜆
𝜆
∴ 𝜃1 ≈ sin 𝜃1 = ; 𝜃 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙
𝑎
𝜆
Angular half width of central maximum = 𝜃1 = 𝑎
2𝜆
∴ Width of central maximum –2𝜃1 = 𝑎
∴ Width of central maximum is proportional to the slit width (𝑎).

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FRAUNHOFER DIFFRACTION IN A DOUBLE SLIT :-

Let a parallel beam of monochromatic light of wavelength λ be incident normally on a
surface containing two slit each of width a and separated by an opaque space of width b . Hence the
distance between any pair of corresponding points of two slits is 𝑑 = 𝑎 + 𝑏, which is known as a slit
constant.

According to Huygens principle, each point of the incident wave front in the plane of two
slits acts as the source of secondary wavelets.

The secondary wavelets travelling normal to the slits are brought to focus by a convex lens
L on the screen at C. The wavelets travelling at an angle θ with the normal are brought to focus at Q.

We want to calculate resultant intensity at Q.

Let the complex light disturbance at Q at any instant due to secondary waves from the
origin “ O”, the midpoint of the lower slit , be 𝐴𝑒 𝑖𝜔𝑡 where A is amplitude 𝜔 is the angular frequency of the
wave .

So the phase difference between the waves at Q coming from O and another point 𝑃 at a
distance 𝑥 from O is given by
2𝜋 2𝜋 2𝜋
𝜆
× 𝑃𝑁 = 𝜆
× 𝑥𝑠𝑖𝑛𝜃 = 𝑘𝑥, 𝑤ℎ𝑒𝑟𝑒 [ 𝑘 = 𝜆
𝑠𝑖𝑛𝜃 ]

Hence the disturbance at Q due to the ray from 𝑃 ,

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𝑖
𝑒 (𝜔𝑡 − 𝑘𝑥)

Where we assume the amplitude to be proportional to the width 𝑑𝑥, c is a proportionality

constant.

∴ Resultant complex disturbance at P due to the both slits ,

𝑎/2 𝑎
𝑑+
2
𝑦 = 𝐶𝐴 ∫ 𝑒 𝑖(𝜔𝑡−𝑘𝑥) 𝑑𝑥 + 𝐶𝐴 ∫ 𝑎
𝑒 𝑖(𝜔𝑡−𝑘𝑥) 𝑑𝑥
𝑑−
−𝑎/2 2

𝑎 𝑎
𝑑+
𝑒 −𝑖𝑘𝑥 2 𝑒 −𝑖𝑘𝑥 2
= 𝐶𝐴𝑒 𝑖𝜔𝑡 [ ] + 𝐶𝐴𝑒 𝑖𝜔𝑡 [ ]
−𝑖𝑘 −𝑎 −𝑖𝑘 𝑑−𝑎
2 2

𝑎𝑘 𝑎𝑘
2 𝑠𝑖𝑛 2 2 𝑠𝑖𝑛 2
= 𝐶𝐴 𝑒 𝑖𝜔𝑡 + 𝐶𝐴 𝑒 𝑖𝜔𝑡 𝑒 −𝑖𝑘𝑑
𝑘 𝑘
𝑎𝑘
𝑠𝑖𝑛 2
= 𝐶𝐴 𝑎 [1 + 𝑒 −𝑖𝑘𝑑 ] 𝑒 𝑖𝜔𝑡
𝑎𝑘
2
∴ Complex conjugate of 𝑦 = 𝑦 ∗

𝑎𝑘
𝑠𝑖𝑛 2
= 𝐶𝐴 𝑎 [1 + 𝑒 𝑖𝑘𝑑 ] 𝑒 𝑖𝜔𝑡
𝑎𝑘
2
The resultant intensity I at Q is obtained by multiplying y by 𝑦 ∗

𝑎𝑘
𝑆𝑖𝑛2 2
𝑥 2 −𝑖𝑘𝑑
𝑇ℎ𝑢𝑠 𝐼 = 𝑦𝑦 = (𝐶𝐴 𝑎) . 2 (1 + 𝑒 )( 1 + 𝑒 𝑖𝑘𝑑 )
𝑎𝑘
(2)
𝑎𝑘
𝑆𝑖𝑛2 2
= (𝐶𝐴 𝑎)2 . ( 2 + 𝑒 𝑖𝑘𝑑 + 𝑒 −𝑖𝑘𝑑 )
𝑎𝑘 2
(2)
𝑎𝑘
𝑆𝑖𝑛2 2
2
= (𝐶𝐴 𝑎) . 2 ( 1 + cos 𝑘𝑑 )
𝑎𝑘 2
(2)
𝑠𝑖𝑛2 𝛼
= 𝐼0 .4 . 𝑐𝑜𝑠 2 𝛽
𝛼2
𝑎𝑘 𝜋
Where α = = . 𝑎 sin 𝜃
2 𝜆

𝐼0 = (𝐶𝐴 𝑎)2

𝑘𝑑 𝜋
𝛽 = = . ( 𝑎 + 𝑏) 𝑠𝑖𝑛 𝜃
2 𝜆
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2
𝑠𝑖𝑛 𝛼
∴ 𝐼 = 4 (𝐶𝐴 𝑎)2 𝑐𝑜𝑠 2 𝛽
𝛼2
𝑠𝑖𝑛2 𝛼
= 4 𝐼𝑜 𝑐𝑜𝑠 2 𝛽 … … . . (2)
𝛼2
Thus our conclusion is resultant intensity depends on two factors,
𝑠𝑖𝑛2 𝛼
𝐼𝑜 , gives the diffraction pattern for a single slit.
𝛼2
𝑐𝑜𝑠 2 𝛽 , gives the interference pattern due the double slit.

Conditions for minima:-

The resultant intensity would be zero when either factor of equation (2) becomes zero.
𝑠𝑖𝑛2 𝛼
The factor =0 when 𝑠𝑖𝑛 𝛼 = 0 [but α ≠ o ]
𝛼2

Or 𝛼 = 𝑚𝜋 …………..(3) [ 𝑚 = ±1, ±2 … …]

i.e. 𝛼 𝑠𝑖𝑛 𝜃 = 𝑚𝜆………………..(4)

These minima are known as diffraction minima.

ii) The factors 𝑐𝑜𝑠 2 𝛽 = 0 also give minima i.e.

𝜋
𝛽 = ±(2𝑠 + 1) 2 ………….. (5) [ S = 0, ± 1, ± 2…]

𝜆
Or, (𝑎 + 𝑏) 𝑠𝑖𝑛𝜃 = ±(2𝑠 + 1) 2……………….(6)

These are called interference minima.

𝑠𝑖𝑛2 𝛼
Conditions for maxima :- (i) The diffraction term 𝛼2
gives

Central maxima for α = 0 i.e. in the direction θ = 0

ii) The maxima also occurs at,

𝜋 3𝜋 5𝜋
α= 𝑎 𝑠𝑖𝑛𝜃 = ± ,± ………….. known as secondary maxima.
𝜆 2 2

iii) The position for maxima due to interfering terms 𝑐𝑜𝑠 2 𝛽 is given by ,

𝛽 = 𝑝𝜋 𝑖. 𝑒.

(𝑎 + 𝑏) 𝑠𝑖𝑛 𝜃 = 𝑝𝜆 [𝑝 = 0 ± 1, ± 2 … ]

The intensity distribution in the resultant diffraction pattern is shown in figure . It is the plot of the product
of ordinate’s of curves (b) with those of curves (a) with constant 4𝐼0

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As the result depends on the relative values of 𝛽 and α as well as on d and a , the curve c is drawn
𝑎+𝑏
when 𝛽 = 3 α i.e. = 3. Hence the fig. shows that 3, 6, .. orders of interference maxima are absent which
𝑎
are marked by 𝑥.

Missing order in double slit pattern : -

If the condition for maxima of interference pattern and minima of diffraction pattern are satisfied
simultaneously for a given values of θ, the corresponding interference maxima will be absent or missing.
These orders are called missing order.

Let for some θ the following condition satisfied

Interference maximum (𝑎 + 𝑏) 𝑠𝑖𝑛 𝜃 = 𝑝𝜆 p, = 0, ±1 ,± 2,…….

Diffraction minima 𝑎 𝑠𝑖𝑛 𝜃 = 𝑚 𝜆 m, = ±1 ,± 2,…….

𝑎+𝑏 𝑝
∴ =
𝑎 𝑚

Hence for 𝑏 = 𝑎,

𝑝 = 2𝑚 𝑖. 𝑒. 2𝑛𝑑 , 4𝑡ℎ , 6𝑡ℎ ……………will be missing order.

similarly for b= 2a ,

p = 3m , i.e. 3rd , 6th , 9th ,……………... will be missing order.

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https://www.desmos.com/calculator/jzrqhfqrov

IntensityDistributionForMultipleSlitDiffraction.cdf MultipleSlitDiffractionPattern.cdf

PLANE DIFFRACTION GRATING :-

A plane transmission diffraction grating is an arrangement equivalent to a very large number N of
identical parallel slits of equal width , placed side by side and separated from one another by equal opaque
spaces .

If 𝒂 width of each slit and 𝒃 that of an opaque then the distance 𝒅 = 𝒂 + 𝒃 is called grating
element, which is the distance between two corresponding points of two consecutive slits.

Now let a parallel beam of monochromatic light of wavelength λ be incident normally on the
plane diffraction grating.

According to Huygens-principle each point of the incident wave front in the plane of the slit may be
regarded as the origin of secondary spherical wavelets in all direction. The secondary wavelets travelling
normal to the slits are brought to focus at “C” by a convex lens L on the screen.

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The wavelets travelling at an angle θ with the normal are focused at P . Let us calculate the intensity
of light at P. Let the complex light disturbance at P due to secondary wavelets from ‘O’ (center of 1st slit)
be represented by 𝐴𝑒 𝑖𝜔𝑡 where A is amplitude w is the circular frequency .

The phase difference between two waves at P coming from O and from a point Q at a distance 𝑥
from O is given by
2𝜋 2𝜋 2𝜋
× 𝑄𝑁 = × 𝑥 𝑠𝑖𝑛𝜃 = 𝑘𝑥, 𝑤ℎ𝑒𝑟𝑒 𝑘 = 𝑠𝑖𝑛𝜃
𝜆 𝜆 𝜆

The disturbance at P, due to the secondary waves from Q will be proportional to 𝑒 𝑖(𝜔𝑡−𝑘𝑥)

Disturbance at P due to the diffracting element,

𝑑𝑦 = 𝐶𝐴 𝑑𝑥 𝑒 𝑖(𝜔𝑡−𝑘𝑥)

Assuming amplitude to be proportional to the width 𝑑𝑥, 𝐶 being a proportionality constant.

The resultant disturbance at P due to all slits will be,

𝑎
𝑎/2 𝑑+
= 𝐶𝐴 ∫−𝑎/2 𝑒 𝑖(𝜔𝑡−𝑘𝑥) 𝑑𝑥 + 𝐶𝐴 ∫ 𝑑−
𝑎
2
𝑒 𝑖(𝜔𝑡−𝑘𝑥) 𝑑𝑥 +…………………..
2

𝑎
(𝑁−1)𝑑+
+𝐶𝐴 ∫ (𝑁−1)𝑑−
𝑎
2
𝑒 𝑖(𝜔𝑡−𝑘𝑥) 𝑑𝑥
2

𝑎 𝑎 𝑎
𝐶𝐴 𝑒 𝑖𝜔𝑡 𝑑+ (𝑁−1)𝑑+
= [∫2𝑎 𝑒 −𝑖𝑘𝑥 𝑑𝑥 + ∫𝑑−𝑎2 𝑒 −𝑖𝑘𝑥 𝑑𝑥 + ⋯ + ∫(𝑁−1)𝑑−𝑎2 𝑒 −𝑖𝑘𝑥 𝑑𝑥 ]
−𝑖𝑘 – 2 2
2

𝑎𝑘
2 𝑠𝑖𝑛
= CAa 2
[1 + 𝑒 −𝑖𝑘𝑑 + … … 𝑒 −𝑖(𝑁−1)𝑘𝑑 ] 𝑒 𝑖𝜔𝑡
𝑘

𝑎𝑘
2 𝑠𝑖𝑛 1−𝑒 −𝑖𝑁𝑘𝑑
= CAa 2
. 𝑒 𝑖𝜔𝑡
𝑘 1−𝑒 −𝑖𝑘𝑑

𝑎𝑘
2 sin 1−𝑒 𝑖𝑁𝑘𝑑
∴ 𝑦 ∗ = CAa 2
. 𝑒 𝑖𝜔𝑡
𝑘 1−𝑒 𝑖𝑘𝑑

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Resultant intensity,
𝑎𝑘
2 sin 1−𝑒 −𝑖𝑁𝑘𝑑 1−𝑒 𝑖𝑁𝑘𝑑
𝐼 = 𝑦𝑦 ∗ = (𝐶𝐴𝑎)2. 2
. .
𝑘 1−𝑒 −𝑖𝑘𝑑 1−𝑒 𝑖𝑘𝑑

𝑎𝑘
2 𝑠𝑖𝑛 2−2 cos 𝑁𝑘𝑑
= (CAa)2 . 2
. 2−2 cos 𝑁𝑘𝑑
𝑘

𝑠𝑖𝑛2 𝛼 𝑠𝑖𝑛2 𝑁𝛽
= 𝐼0 . . ………….(1)
𝛼2 𝑠𝑖𝑛2 𝛽

Taking 𝐼0 = (CAa)2
𝑎𝑘 𝜋 𝑎𝑠𝑖𝑛𝜃
α=2 = …… (2)
𝜆

𝑑𝑘 𝜋 𝑑𝑠𝑖𝑛𝜃 𝜋
𝛽= = = (𝑎 + 𝑏) sin 𝜃 … . (3)
2 𝜆 𝜆

The resultant intensity thus depends on two factors

𝑠𝑖𝑛2 𝛼
(i) 𝐼𝑜 = 𝐼1 showing diffraction pattern for single slit.
𝛼2
2
𝑠𝑖𝑛 𝑁 𝛽
(ii) = 𝐼2 showing interference pattern for diffracted light beams from N slits .
𝑠𝑖𝑛2 𝛽

[ Note by putting N=2 we can have double slit intensity pattern]

Condition for principal maxima ; minima and secondary maxima:-

If the slit width “𝑎” is very small and observation is confined near the central pattern, then the
𝑠𝑖𝑛2 𝛼 𝑠𝑖𝑛2 𝑁 𝛽
variation of is small and condition for maxima is soley controlled by 𝐼2 = ,
𝛼2 𝑠𝑖𝑛2 𝛽

𝑠𝑖𝑛2 𝑁 𝛽
For the factor 𝐼2 = the condition for maxima or minima,
𝑠𝑖𝑛2 𝛽

𝑑𝐼2
=0
𝑑𝛽

𝑑𝐼2 2𝑁 sin 𝑁𝛽 .𝑐𝑜𝑠𝑁𝛽 2𝑁 sin2 𝑁𝛽 .𝑐𝑜𝑠𝑁𝛽

Now = −
𝑑𝛽 𝑠𝑖𝑛2 𝛽 𝑠𝑖𝑛3 𝛽

𝑠𝑖𝑛2 𝑁 𝛽
= 2 ( 𝑁 𝑐𝑜𝑡𝑁 𝛽 − 𝑐𝑜𝑡𝛽)
𝑠𝑖𝑛2 𝛽

𝑠𝑖𝑛 𝑁 𝛽
∴ For maxima or minima either =0
𝑠𝑖𝑛 𝛽

Or 𝑁 𝑐𝑜𝑡𝑁 𝛽 = 𝑐𝑜𝑡𝛽 .

IntensityDistributionForMultipleSlitDiffraction.cdf
Conditions for principal maxima:-

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 Sanju 9681634157
Now for 𝛽 → 𝑚𝜋 , i.e. (𝑎 + 𝑏) 𝑠𝑖𝑛 𝜃 = 𝑚𝜆 [ 𝑚 = 0, ± 1, ± 2 ]; 𝑠𝑖𝑛 𝛽 = 𝑠𝑖𝑛 𝑁 𝛽 = 0;
𝑠𝑖𝑛 𝑁 𝛽 𝑁𝑐𝑜𝑠𝑁𝛽
𝐿𝑡 β → mπ = 𝐿𝑡 β → mπ = ±𝑁 [by L hospitals rule]
𝑠𝑖𝑛 𝛽 cos 𝛽

∴ Resultant intensity,
𝑠𝑖𝑛2 𝑎
𝐼 = 𝐼0 𝑁 2 This is known as principal maxima.
𝑎2

[Clearly intensity of principal maxima increases as the number of slits (N) increases. But due to presence
𝑠𝑖𝑛2 𝛼
of the factor , intensity of principal maxima decreases with increase with θ i.e. with the order
𝛼2
number of bands]

Conditions for Minima::-

When 𝑠𝑖𝑛 𝑁𝛽 = 0 but sin𝛽 ≠ 0 (𝑠𝑖𝑛𝛽 = 0, this corresponds to principal maxima);

𝑠𝑖𝑛2 𝑁 𝛽
𝐼2 = =0
𝑠𝑖𝑛2 𝛽

Thus for minima, N𝛽 = ± Sπ S = ± 1, ± 2…….

𝑆
Or 𝛽 = ± 𝑁 𝜋

𝑆
Or (𝑎 + 𝑏) 𝑠𝑖𝑛 𝜃 = ± 𝜆
𝑁

Where S, has integral values excepting 0, N, 2N, 3N …….etc. as for these values as S, sin 𝛽 = 0 which
corresponds to principal maxima.

[ Hence between two consecutive principal maxima there are ( N – 1) minima i.e. 𝑛 − 2 other maxima
known as a secondary maxima.]

Secondary Maxima: - The condition 𝑁 𝑐𝑜𝑡𝑁 𝛽 = 𝑐𝑜𝑡𝛽 corresponds to secondary maxima as this condition
𝑑2 𝐼
is consistent with 𝑑 𝛽22 < 0. Values of 𝛽 satisfying this condition ( except 𝛽 =mπ ) will give position of
secondary maxima.

Intensity of secondary maxima :- from the condition ;𝑁 𝑐𝑜𝑡𝑁 𝛽 = 𝑐𝑜𝑡𝛽 ;

𝑐𝑜𝑠2 𝑁𝛽 𝑠𝑖𝑛2 𝑁𝛽
𝑁2 =
𝑐𝑜𝑠2 𝛽 𝑠𝑖𝑛2 𝛽

𝑁 2 ( 1− 𝑠𝑖𝑛2 𝑁𝛽) 𝑁2 𝑠𝑖𝑛2 𝑁𝛽 𝑁2

Or , = (𝑁2) = [1+(𝑁2−1)𝑠𝑖𝑛2𝛽]
1−𝑠𝑖𝑛2 𝛽 𝑠𝑖𝑛2 𝛽

∴ Intensity of secondary maxima is given by ,

𝑁2 𝐼𝑝𝑚
𝐼𝑠𝑚 = 𝐼0 × 1+(𝑁 2 −1)𝑠𝑖𝑛2 𝛽
= 1+(𝑁2−1)𝑠𝑖𝑛2 𝛽

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 Sanju 9681634157
𝐼𝑠𝑚 1
∴ = 1+(𝑁2−1)𝑠𝑖𝑛2𝛽
𝐼𝑝𝑚

IntensityDistributionForMultipleSlitDiffraction.cdf

Absent spectra : If the condition for mth order principle maxima and sth order diffraction minima is
satisfied for the same direction (for same angle 𝜃) then,

(𝑎 + 𝑏) sin 𝜃 = 𝑚𝜆 and,

𝑎 sin 𝜃 = 𝑠𝜆
(𝑎+𝑏) 𝑚
Hence we have =
𝑎 𝑠

(𝑎+𝑏)
Hence 𝑚 = 𝑠
𝑎

GRATING ERRORS AND GHOST LINES :

While in the ideal grating the ruling should be strictly equispaced, in practice there are some
errors,

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 Sanju 9681634157
1. If the errors are random the grating gives a continuous background illumination.
2. If the error is progressive in nature the lines become sharper in plane, different from the plane of
the optical system.
3. The most common error is periodic in nature. It arises from defects in the drilling mechanism of
ruling. It results false lines with principle maxima of ideal grating. These additional lines are known
as ghost lines.

OVERLAPPING OF SPECTRAL LINES :

For a grating having grating element d(= 𝑎 + 𝑏),the angle of diffraction 𝜃 in the mth order spectrum
for a light of wavelength 𝜆 is given by,

(𝑎 + 𝑏) sin 𝜃 = 𝑚𝜆

For a given grating (𝑑 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡), 𝜃 is the same when 𝑚𝜆 is constant. If the incident light consist of
a large number of wavelengths, the spectral lines of longer wavelengths and lower order overlaps with
shorter wavelength and higher number under the condition;

𝑚1 𝜆1 = 𝑚2 𝜆2

Dispersive Power : The dispersive power of grating is defined as the rate of change of the angle of
𝑑𝜃
diffraction with wavelength i.e. 𝑑𝜆

For a grating having grating element d(= 𝑎 + 𝑏), the angle of diffraction 𝜃 in the mth order spectrum
for a light of wavelength 𝜆 is given by

(𝑎 + 𝑏) sin 𝜃 = 𝑚𝜆
𝑑𝜃 𝑚
∴ = (𝑎+𝑏) cos 𝜃 ;
𝑑𝜆

𝑑𝜃
Thus dispersive power 𝑑𝜆 ,

i) Proportional to order number,

ii) Inversely proportional to grating constant (𝑎 + 𝑏)
iii) Inversely proportional to cos 𝜃

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 Sanju 9681634157
Reighleigh’s Criterion of resolution :- According to Rayleigh’s two equally bright point source ( or two
spectral lines of equal intensities ) lying close to each other, are said to be just resolved by an instrument,
if the central maximum in the diffraction pattern of one falls on the 1st minimum due to the other.

Resolving power of a grating :- The resolving power of grating measures the ability to distinguish two very
𝜆
close spectral lines λ and 𝜆 + 𝑑𝜆 and is defined by 𝑑𝜆 ; dλ being the smallest wavelength difference for
which the spectral lines can be just resolved.

Expression for resolving power :- The 𝑚𝑡ℎ order principal maximum of a spectral line of wavelength λ of
a grating for normal incidence,

(𝑎 + 𝑏) 𝑠𝑖𝑛𝜃 = 𝑚𝜆 …………………………………(1)

Where (𝑎 + 𝑏) is the grating element, 𝜃 the diffraction angle corresponding to 𝑚𝑡ℎ order.

For wavelength (𝜆 + 𝑑𝜆) the above condition reads ,

( 𝑎 + 𝑏 ) 𝑠𝑖𝑛 (𝜃 + 𝑑𝜃 ) = 𝑚 (𝜆 + 𝑑𝜆 ) ………………………….. (2)

The corresponding diffraction angle being , ( 𝜃 + 𝑑𝜃).

Now by Rayleigh ‘s Criterion, the two wavelengths will be just resolved if the position of 𝑚𝑡ℎ order
principal maximum for λ +dλ coincides with the 1st minimum, adjacent to 𝑚𝑡ℎ order principal maximum
for 𝜆.

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 Sanju 9681634157

Condition for the minimum in a grating is,

𝑁 (𝑎 + 𝑏) sin 𝜃 = 𝑠𝜆

N = total number of slits and s can have integral values except 0, N. 2N,…………. nN; as these values satisfy
condition for different maxima .

The first minimum adjacent to 𝑚𝑡ℎ principal maximum in the direction ( θ + dθ) satisfies the
condition ,

𝑁 (𝑎 + 𝑏) sin(𝜃 + 𝑑𝜃 ) = (𝑚𝑁 + 1 )𝜆 ……………………………… (3)

Again from (2),

𝑁 (𝑎 + 𝑏) sin(𝜃 + 𝑑𝜃 ) = 𝑚𝑁(𝜆 + 𝑑𝜆)…………………………… (4)

∴ For just resolution,

(𝑚𝑁 + 1 )𝜆 = 𝑚𝑁(𝜆 + 𝑑𝜆)

Or, 𝜆 = 𝑚𝑁 𝑑𝜆
𝜆
Or, 𝑑𝜆 = 𝑚𝑁 ………………………………………….(5)

This is the required expression for resolving power of a grating .

Note :- 1. Resolving power increases with the order of spectrum .

2. increases with total number of lines.

3.Independent of grating element ( 𝑎 + 𝑏).

𝑠𝑖𝑛𝜃
=> From eq . (1) we have 𝑚 = (𝑎 + 𝑏) 𝜆

𝜆 𝑠𝑖𝑛𝜃
∴ From (5) we have = (𝑎 + 𝑏 )𝑁
𝑑𝜆 𝜆
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 Sanju 9681634157
𝑠𝑖𝑛𝜃
= W 𝜆

𝑊 = 𝑁 (𝑎 + 𝑏) , the total width of the ruled surface of the grating .

Fraunhofer diffraction due to circular aperture :-

Let AB be a narrow circular aperture of a diameter d, c is the center . A parallel beam of
monochromatic beam of light of wave length λ be incident normally on it. The secondary waves from
aperture traveling in the direction CO comes to focus at C . Hence C corresponds to the position central
maxima. All secondary waves emerging from point equidistant from O, travel the same distance and
reinforce each order at P.

Now rays undergoing diffraction by θ will meet at a point on the screen distant 𝑥 from C. The path
different ∆ between the secondary waves from two extreme ends AB on the aperture,

∆ = 𝐴𝑁 = 𝐴𝐵 𝑠𝑖𝑛 𝜃 = 𝑑 𝑠𝑖𝑛 𝜃

Hence P will be of minimum intensity if the path difference is equal to integral multiple of λ

i.e. ∆= 𝑑 𝑠𝑖𝑛 𝜃 = 𝑛𝜆;

Conversely P will be of maximum intensity if

𝜆
∆ = 𝑑 𝑠𝑖𝑛𝜃 = (2𝑛 + 1) 2

Since the circular aperture is symmetrical about axis, the resultant diffraction patterns will be same
for all the points at same distance from C and will from circular rings .Here P will trace out a circular rings
of uniform intensity , minimum or maximum , as given by above condition .

Thus diffraction pattern due to a circular aperture consists of a central bright disc , called Airy’s disc
, surrounded by alternate dark and bright concentric rings called Airey’s rings.

Radius of Airy’s disc: In case of focusing lens very closed to the aperture, or the screen at the very large
distance from the lens,
𝑥
𝑠𝑖𝑛 𝜃 ≈ 𝜃 = 𝑓 ; 𝑓 is the focal length of the lens.

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 Sanju 9681634157
For 1st secondary minima , 𝑑 𝑠𝑖𝑛 𝜃 = 𝜆
𝜆
𝑠𝑖𝑛 𝜃 ≈ 𝜃 = 𝑑

𝑥 𝜆 𝜆𝑓
∴ 𝑓 = 𝑑 thus 𝑥 = 𝑑

Although actual radius is found to be higher than the above result;

𝜆𝑓
As 𝑥 = 1.22 𝑑

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 Sanju 9681634157

Fresnel’s Diffraction: - In this type of diffraction, the source or the screen (i.e. point of
observation) or both are at finite distances from the obstacle or aperture.

Fresnel’s half period zone :- Fresnel explained the phenomena of diffraction of light on the
basis of mutual interference of secondary wavelets, originating from various parts of a wavefront. He
divided the wave front into an number of zones called Fresnel’s ½ period zone . They are so called so
𝜆
because the path of difference between the wavelets coming from the two successive zone is .
2

Theory of fresnels ½ period zone :- Let A,B,C,D be a plane wave front of

monochromatic light of wavelength λ, travelling from “O” to P, where the resultant intensity is to be found.

From P, a perpendicular Op = b is drawn to the wavefront ; The point O is known as pole w.r.t. point P .

𝜆 2𝜆
Taking P as center and radii 𝑏 + 2 , 𝑏 + … … .. a series of spheres are drawn, the sections of which by the
2
3𝜆
plane ABCD are concentric circles with centre o. and radii 𝑂 𝑀1 , 𝑀2 , 𝑀3 …… 𝑃 𝑏+ 2

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 Sanju 9681634157
The area enclosed by the inner most circle of radius OM, is called 1st
half period zone. The annular area
between 1 and 2 circle is called the second half period zone and so on.
st nd

Path difference between wavelets from successive zones.

The distance of 𝑛𝑡ℎ period zone from P,

1 𝑛𝜆 𝑛−1
𝑑𝑛= [𝑏+ + 𝑏+ 2
𝜆 ] = 𝑏 + (2𝑛 − 1)𝜆4
2 2

∴ 𝑑𝑛 − 1 = 𝑏 + { 2(𝑛 − 1) − 1} 𝜆
4

∴ Path difference ∆ = 𝑑𝑛 − 𝑑𝑛−1

= 𝜆2

Area of a zone: - The area of 𝑚𝑡ℎ zone i.e. the area between the circles of radii 𝑂𝑀𝑚 and 𝑂𝑀𝑚−1 according
to the fig.
2
S = π [ 𝑂𝑀2 -𝑂𝑀𝑚−1 ]
𝜆 𝜆
= 𝜋 [{ (𝑏 + 𝑚 2 )2 − 𝑏 2 } − { (𝑏 + (𝑚 − 1). 2 )2 − 𝑏 2 }]

𝑚2 𝜆2 (𝑚−1)2
=π[ + − ( 𝑚 − 1)𝑏𝜆 − 𝜆2]
4 4

𝜆2
= π [ 𝑏𝜆 + (2𝑚 − 1) ]
4

≈ 𝜋𝑏𝜆 ∵ b >>λ

Thus all the zone are of approximately equal area;[ though it increase slightly with ‘m’ ].

Factors governing the magnitude of amplitude of disturbance at P

Area of the zone :- Amplitude at P is directly proportional to the area of the zone .
Distance from P :- Amplitude is inversely proportional to the distance .
Oblique factor :- Amplitude is directly proportional to the oblique factor 𝑓 (𝜃𝑚 ); being an angle
which the direction of P from the 𝑚𝑡ℎ zone makes with OP . Fresnel assumed 𝑓 (𝜃𝑚 ) = 1+cos 𝜃𝑚 ;
Note: amplitude (𝐴𝑚 ) ∞ (1 + 𝑐𝑜𝑠 𝜃𝑚 );

As m increases so does (𝜃𝑚 ); and hence 𝑐𝑜𝑠 𝜃𝑚 decreases. The amplitude of the disturbance at P due to a
zone will thus gradually decrease as the order m of the zone increases.

Resultant amplitude due to entire wavefront:-

If 𝑎1 ,𝑎2 ,………….𝑎𝑚 be the amplitudes at P due to the 1st , 2nd ,…zones respectively then they are in
decreasing order in magnitude due to the oblique factor; also,have alternate – 𝑣𝑒 and +𝑣𝑒 sign as the
resultant disturbance produced by any two consecutive zones are π out of phase since the rare
𝜆
disturbance of P from them differs by 2
,

∴Resultant amplitude at P due to the whole wave front ABCD is,

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 Sanju 9681634157
𝑚−1
R= 𝑎1 − 𝑎2 + 𝑎3 − ……………..+ (−1) 𝑎m .
𝑎1 𝑎1 + 𝑎3 𝑎3 + 𝑎4 𝑎𝑚
= +( − 𝑎2 ) + ( − 𝑎4 ) + …………. + [ m = odd ]
2 2 2 2

[ or − 𝑎𝑚 ] [ m = even].

Since 𝑎2 is slightly greater than 𝑎1 and slightly smaller than 𝑎3 hence,

𝑎1+ 𝑎3
𝑎2 ≈ and so on .
2

𝑎1 𝑎𝑚
∴R= + [ m = odd ]
2 2

𝑎1 𝑎𝑚−1
Or = + − 𝑎𝑚 [𝑎𝑚 = even ]
2 2

When m is very large , greater oblique factor causes 𝑎𝑚 ≈ 𝑎𝑚−1 → 0

𝑎1
Hence R = ;
2

So , the resultant amplitude at P due to the whole wave front is equal to the half of the amplitude
due to the 1st half period zone .

Rectilinear propagation of light :- Let a small circular obstacle is a place before O ,

𝑎
the pole of a wave front which covers the 1st half period zone ; then the resultant intensity will be ( 22 )2 ;
i.e one fourth of intensity due to the 2nd period zone . If the size is gradually increased, it will successively
𝑎3 2 𝑎4 2
covers 2nd , 3rd .1/2 period zones and intensity becomes proportional to ; ……as 𝑎2 > 𝑎3 >
4 4
𝑎4 ………(due to oblique factor ) the illumination wil gradually decrease.

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 Sanju 9681634157
As the sizes of ½ period zones are very small (≈ of wavelength of light) a tiny obstacle enough to cover
large no. of zones and hence light from source is practically cut off .this is interpreted as the rectilinear
propagation of light.

Zone plate:- A zone place is an optical device based on Fresnel’s theory of ½ period
zones. It consists of a plane parallel glass plate having concentric circles of radii proportional to the square
root of natural numbers, 1,2,3,…….then even of odd order of annular spaces between the circles are made
completely dark .

A zone plate behave as a convergent lens and produces image of a light – source on a screen.

Theory of zone plate :- Let O be a point source of monochromatic light giving out
spherical wave of wave length λ; whose effect at a point I is required.

We consider an imaginary plane of a transparent medium , perpendicular to the plane of paper

and line OI.

The plane is divide into zones bounded by circles with center at P and radii 𝑃𝑀1= 𝑟; 𝑃𝑀1= 𝑟2……….
𝑃𝑀𝑚 =𝑟𝑚 such that,

𝑂𝑀1 + 𝑀1 𝐼 = OP +PI + λ/2

𝑂𝑀2 + 𝑀2 𝐼= OP + PI +2. λ/2

…………………………………………………

𝑂𝑀𝑚 + 𝑀𝑚 𝐼= OP + PI +m. λ/2………………….(1)

The annular regions thus formed are therefore half period zones for I, as the path difference between
𝜆
the corresponding points of two consecutive zone is 2.
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 Sanju 9681634157
Let OP = u and PI = v
𝑟2 𝑟2
∴ 𝑂𝑀𝑚 = ( 𝑢2 + 𝑟𝑚2 )1/2 = u (1 + 𝑢𝑚2 )1/2 = u + 2𝑢
𝑚
[ u >>𝑟𝑚 ]

2
𝑟𝑚
Similarly 𝑀𝑚 𝐼 = 𝑣 + .
2𝑣

∴ From equation . number . (1)

2
𝑟𝑚 2
𝑟𝑚 𝑚𝜆
𝑢 + + 𝑣 + = 𝑢+𝑣 +
2𝑢 2𝑣 2

1 1
Or, 𝑟𝑚2 [ 𝑣 + 𝑢 ] = 𝑚𝜆

1 1 𝑚𝜆
Or, +𝑢= 2 ……………..(2)
𝑣 𝑟𝑚

𝑢𝑣
and 𝑟𝑚 = [ 𝑢+𝑣 𝑚𝜆. ]1/2

[ thus the external radii of ½ period zones are proportional to the square root of all natural numbers
.]

The above equation is similar to the convex lens formula

1 1 1
− = 𝑓;
𝑣 𝑢

Thus the zone plate may behave like a convergent lens of focal length,

𝑓 𝑟 2 .
𝑚= 𝑚
𝑚𝜆

If odd zones are transparent and even zones are opaque , it is called a positive zone plate

If odd zones are transparent and odd zones are opaque , it is called a negative zone plate .

Focusing action :- If a zone plate the alternative zones ( say the even ones) are blocked . So
the waves from odd transparent zone differ in phase by 2x and hence one in the same phase.

∴ 𝑅𝑒𝑠𝑢𝑙𝑡𝑖𝑛𝑔 𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 𝑎𝑡 𝐴 𝑝𝑜𝑖𝑛𝑡 𝐼 ,

R = 𝑎1 + 𝑎2 + 𝑎3 + …………………………………………
𝑎1
Clearly it is many time greater them , which is according to Fresnel’s theory of ½ period zone
2
produced by entire wave from ( in absence of zone plate ). I will then present the image of “O” which is
extremely bright point.

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 Sanju 9681634157
Construction of zone plate :- To construct a zone plate , concentric circles whose
radii are proportional to the square root of natural number are drawn on a sheet of white paper the
alternate zones are pointed black and a very much reduced photograph of this drawing is obtained on a
glass plate . Thus a zone plate is formed.

If odd zones are transparent and even zones are

opaque it is said to be a positive zone plate. If even zones
are transparent and odd zones are opaque , it is said to be
negative zone plate.

Multiple Foci Of Zone Plate :- The zone plate has a number of foci depending on the number of
zones used. From the formula of zone plate,
1 1 𝑚𝜆
+𝑣= 2 where,
𝑢 𝑟𝑚

u = object distance from zone plate;

v = image distance from zone plate ;

n => no. of zones considered;

𝑟𝑛 => radius of 𝑛𝑡ℎ zone,

λ => wavelength of the ray.

It is clear that for an object at infinity (u ∞ ), the brightest focus is at a distance

𝑚 𝑟2
𝐼1 ( = v = 𝑓1 ) = 𝑚𝜆 .

𝑓1 𝑚 𝑟2
Now we consider a point 𝐼3 at a distance, = 3𝑚𝜆 ; from zone plate. For this point the “ 1st clear
3
zone “ contains 1st , 2nd , 3rd ½ period zone s, the 2nd black zone contains 4th , 5th , 6th half period zones and
so on;

∴ Resultant amplitude at 𝐼3 will be,

S = ( 𝑠1 - 𝑠2 + 𝑠3 ) + (𝑠7 - 𝑠8 + 𝑠9 ) + (𝑠13 - 𝑠14 + 𝑠15 ) + ……

𝑠1 𝑠3 𝑠1 + 𝑠3 𝑠 𝑠7 + 𝑠9
={( - 𝑠2 + ) + } + { ( 27 - 𝑠8 + 𝑠9 ) + } + ……
2 2 2 2

1
= 2 [𝑠1 + 𝑠3 + 𝑠7 + 𝑠9 + 𝑠13 + ⋯ ]

Where amplitudes 𝑠1 , 𝑠3 are roughly 1/3 of amplitudes 𝑅1 , 𝑅3 …….. regional zones. Hence 𝐼3
recieves sufficient intensity , though less than that at I , at distance f from zone plate.
2
𝑟𝑚 𝑟2 𝑟2
𝑚 𝑚
Similarly it can be shown that points 𝐼5 , 𝐼7 , 𝐼9 etc, distant , 7𝑚𝜆 , 9𝑚𝜆 from the zone plate are images
5𝑚𝜆
of “O” but of successively diminishing intensity.

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Comparison between zone plate and convex lens :-

Similarities :- (1) The relation between conjugate distance is similar for both .

(2) Focal length for both depends upon wavelength λ , hence both show chromatic aberration.

Dissimilarities :- (1) The zone plate, in contrast to convex lens , has multiple focal lengths and from
a series of point images of decreasing intensity for a point source.

(2) For a convex lens , all the rays reaching an image point have the same optical path , but for a
zone plate , the path difference between the rays from two successive transparent zone is λ .

(3) For a zone plate focal length decreases with λ hence focal length for red light is shorter than
that for violet .But reverse is the case for convex lens.

Phase reversal zone plate :- To make the image produced by a zone plate more intense this
modified form of zone plate was devised. If instead of blocking the light alternating zones an additional
𝜆
path of is introduce between the light waves from consecutive zones, the amplitudes from each
2
successive zone will be rain forced . Such a zone is known as a phase reversal zone.

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Single slit:
1. Derive the expression for the intensity distribution of Fraunhofer diffraction pattern formed by a
single slit. Cu 2005,07
2. Explain the case when the slit width is changed.
3. Explain the pattern with the white light.
4. Suppose that the slit with is 0.03cm and wavelength of light used is 6 × 10−5 cm. Find the
diffraction angle for the first dark band. Cu 2005
5. A convex lens of focal length 20𝑐𝑚 is placed after a slit of slit width 0.6 𝑚𝑚. If a plane wave of 𝜆 =
6000𝐴̇ falls on the slit normally, calculate the angular and linear width and the angular and linear
separation between second minima on the either side of the central maxima.
6. D
7. What is Airys disc ? Find its radius. For 𝜆 = 6000𝐴̇, calculate the radius due to a circular hole of
radius 10𝜇𝑚 using focal length 1m.

Double slit:
8. Obtain the intensity expression for the Fraunhofer diffraction pattern of a double slit. Cu 2001
,03,06,13
9. Explain the condition for the maxima and minima. What is the missing order?
10. Draw the intensity distribution plot takin 𝑎 = 0.1𝑚𝑚, 𝑏 = 0.2𝑚𝑚.
11. If there are 9 fringes at the diffraction central fringes obtain the ratio of the opaque and slit width.
12. What is the angular width of central maxima and interreference maxima in term of 𝑎, 𝑏?
13. Explain your observations if 𝑎, is increased keeping b fixed and visa versa.
14. Explain the case when b is finite and b>>a.
15. What is the basic difference that makes the difference between Young’s double slit interference
and double slit diffraction pattern?-explain

Grating:
16. A parallel beam of monochromatic light is incident normally on a plane transmission grating.
Find an expression for the intensity pattern due to diffraction by grating at an angle 𝜃 . Cu
2000,16
17.A parallel beam of monochromatic wavelength𝜆 s incident normally on a plane diffraction grating
of grating element (a+b). Intensity of Fraunhofer diffraction pattern in the 𝜃 direction is given
by,
𝑠𝑖𝑛2 𝛼 𝑠𝑖𝑛2 𝑁𝛽
𝐼0
𝛼 2 𝑠𝑖𝑛2 𝛽
The terms have their usual meaning

i) Find the angular position of maxima and minima in double slit diffraction pattern indicating
position of diffraction minima and interference maxima.

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 Sanju 9681634157
ii) Find the intensity of a secondary Maxima of a grating diffraction pattern, as a function of intensity
of the principal maxima and hence explain why secondary Maxima are not generally observed.
Cu 2008

18.In the diffraction pattern produced by in N slits, Find the expression of width of the principal
maximum. Cu 09, 11
19.What are missing spectra in the diffraction pattern of a plane transmission grating. Cu 2004,09,15
20. Draw the pattern taking 𝑁 = 6; 𝑏 = 3𝑎 including at least 2 principal maxima on either side if the
0-order principal maxima.
21. Explain the errors in grating spectrum Cu 20011
22. Write down the differences between grating and the prism spectra.
23.A plane diffraction grating at normal incidence gives a green line 5.4 × 10−5 𝑐𝑚. In a certain
spectral order superpose on a violet line4.05 × 10−5 𝑐𝑚 Of the next higher order. If the angle of
deflection be 300, Find the grating constant. Cu 13

24.A transmission type diffraction grating having 250 lines /mm is illuminated by visible light(4000-
7000)𝐴̇ at normal incidence. What wavelengths appear at diffraction angle of 30 degree and
what colours are they? Cu 2002 ,15
25.How is the resolving power of a diffraction grating define? 04,15
26.What is meant by Rayleigh's criteria of resolution . Cu 2000, 04,12 ,10
27.Obtain an expression for the resolving power of a plane transmission grating and hence give an
estimate of the number of lines required to resolve D1 D2 line of sodium in the second order. Cu
2000,02

28. Find the least width that a diffraction grating (820/cm)must have to resolve Na D lines.
29. Sodium light is incident normally on a plane transmission grating having 3000 line/cm. Find the
direction of the first order or 30 lines and the weight of the grating necessary to resolve Na 𝐷1 , 𝐷2

30.A transmission grating is 4 cm long and having 4000 line/cm. Compute the resolving power of
the greeting for 𝜆 = 5900𝐴.̇ 𝑖𝑛 the first order spectrum. Will this greeting separate the Na D-
lines? Cu 07
31.Discuss the statement a grating having higher dispersive power than other does not necessarily
have a higher resolving power.

32.You are given two plane transmission grating of 3 cm having 3000 line and 2 cm having 2000 line
. Compare the resolving power of these two grating. Cu 2004

33.Consider the case of double slit diffraction where slit width is 88 × 10−3 separation between the
stre Wavelength of light 𝑑 = 7 × 10−2. Find the number of interference minimum according in
the central difference in maxima. Cu 2003

Zone plate:
34. Distinguish between Fresnel’s and Fraunhofer diffraction. Cu 10,15

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 Sanju 9681634157
35.What are Fresnel half period zones? Find the area of a nth half period zones with reference to
plane wavefront. Cu 2008
36. What are the factors on which the amplitude of light waves from a half period zone at
observation point depends?
37.Find the expression of the resultant amplitude at any point due to the half period zone. Pot the
intensity with number of zones open.
38.Explain rectilinear propagation of light fron the concept of half period zone.

39.Outline the theory of zone plate . Cu 08

40.Explain multiple foci of zone plate.
41.A zone plate is constructed by drawing concentric circles of radii equal to that of dark Newton's
ring formed by a convex lens of radius of curvature R =2m. Find the first focal length of the zone
plate. Cu 10

42.Prove that the radius of a particular zone is proportional to the square root of the number of the
zone.
43. Write down the differences between convex lens and zoneplate.
44.Let us consider plane light wave of wavelength 𝜆 incident perpendicular on an opaque screen
with a circular aperture, the radius 𝑟,of which can be varied. Calculate radii of the maximum and
minimum intensities. Cu 2006
45.If zone plate is constructed in such a way that the radii of the zones are √𝑛𝑘 and even zones are
𝑘2
blackened. Prove that the points at the distance 𝑚𝜆 will be maximum when m =1 and minimum
m=2 Cu 2017
46.Calculate the 1st zone of a zone plate having total 500 zones, having focal length 0.2m using 𝜆 =
500𝑛𝑚 ̇ .
47.An object is illuminated by 5000𝐴̇ wavelength of light placed at 60 cm from a zone plate and its
image is found at a 30 cm from the zone plate. Calculate radius of the 1st and 100th zone. Also
estimate total number zones in a radius of 5cm.
48.A zone plate is drawn and the copied on a reduced scale such that the diameter of the central
zone is 2𝑚𝑚. If the source of the monochromatic light 𝜆 = 500𝑛𝑚 is placed 2m from the plate
find the position of the primary and the weaker next secondary images .
49.A plane wave front is advancing towards a point is divided in number of half period zones
contributing amplitudes as 1.0.98,0.96,0.94…….. Compares the intensities at the point when only
first 31 and 36 zones are exposed.
50.What is the phase reversal zone plate?

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