Diffraction

RajeshKumar
Roll − 161148
Indian Institute Of Technology, Kanpur
July 2017

Abstract
The image pattern formed on screen from the light emanating from aperture of various shape
and size can be predicted using Huygens-Fresnel principle.It turns out that electric field at any
point is the Fourier transform of the aperture function.This principle is applied to predict the im-
age form due to light diffraction from various aperture shape,viz.rectangular,circular,triangular.3-
D interference using finite plane wave and Gaussian wave is also discussed.The image pattern
is plotted using python programming.

1 Huygens-Fresnel Principle
The Huygens-Fresnel principle states that every point on a wavefront is a source of wavelets. These
wavelets spread out in all directions with the speed of wave.These secondary wavelets mutually in-
terfere and gives the shape of the new wavefront.

Figure 1: Image taken from [http://imagebank.osa.org]

We consider a plane wave whose amplitude variation on the aperture plane is given by A(ξ, η).Using
Huygens-Fresnel principle,the field at point Po is given by
R R A(ξ,η)eikr
u(Po ) = C r
dξdη.
Where, C is proportionality constant and integration is over the entire aperture.From more general
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theory C is given by iλ .

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RR A(ξ,η)eikr
u(Po ) = iλ r
dξdη. ......(1)

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1.1 The Fraunhofer Approximation
Thepquantity r is the distance between
√ P1 andPo .It is expressed as
r= (x − ξ)2 + (y − η)2 + z 2 = z 1 + α
For α << 1 2 2
r= z + (x−ξ)
2z
+ (y−η)
2z
ikz R R ik 2 2
With this approximation equn.(1) becomes u(x, y, z) = eiλz A(ξ, η)e 2z [(x−ξ) +(y−η) ] dξdη
The above equation can be written in the form
ik
eikz 2z 2 2 ik 2 +η 2 )
e−i(uξ+vη) dξdη
RR
u(x, y, z) = iλz
e (x +y ) A(ξ, η)e 2z (ξ

Where, u = 2πx λz
andv = 2πy
λz
2 2
For z >> [ξ +ηλ ]max
ikz ik 2 2 R R
u(x, y, z) = eiλz e 2z (x +y ) A(ξ, η)e−i(uξ+vη) dξdη . ..............................................(a)
2 2
Thus for Fraunhofer Approximation to be valid α << 1 and z >> [ξ +ηλ ]max .
Thus we get the very important result:

Fraunhofer diffraction pattern is the Fourier transform of the aperture function.

ikz ik 2 2 R R
u(x, y, z) = eiλz e 2z (x +y ) A(ξ, η)e−i(uξ+vη) dξdη

1.2 Fraunhofer Diffraction Examples

First we consider diffraction from long narrow slit.Light is incident on it normally and we consider
phase and aperture function of the wave to be constant on the slit plane.The amplitude at any point
is calculated using equn(a) of section 1.1.The calculated amplitude is given by
ikz ik 2 2
u(x, y, z) = Abeiλz
e 2z (x +y ) sin(β)
β
2πδ(ν)
Where b is slid width and δ(ν) is the dirac delta function.The Intensity pattern is given by
I = Io ( sin(β)
β
)2
The normalized intensity is plotted in python as shown below.

Single-Slit Diffraction Pattern

Parameters: λ = 600nm; dslitwidth = .1mm

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Plotting N-Slit Diffraction Pattern in Python

1-Slit

2-Slit

3-Slit

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4-Slit

10-Slit

20-Slit

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1.3 Circular aperture
Taking Fourier transform of circular aperture function A(=constant) given by section 1.1 we obtain
amplitude as
ikz ik 2 2
u(x, y, z) = Aeiλz e 2z (x +y ) πa2 [ 2J1ν(ν) ]
Where ν = kasin(θ); a = apertureradius; J1 (ν) = besselsf unction
Intensity distribution is given by Io [ 2J1ν(ν) ]2
Example
Parameters
Aperture radius=.5mm
Distance z between screen and aperture=57cm
Wavelength of light=600nm

1.4 Triangular Diffraction

Fourier transform of the aperture function A(=constant) is given by
ikz ik 2 +y 2 ) R a R (a−ξ)
u(x, y, z) = A eiλz e 2z (x 0
2
−(a−ξ) e−i(uξ+vη) dξdη .......(2)
2
kx ky
where, a=aperture side(equilateral triangle ); u = z
;v =u= z

On evaluating equn.(2) we get

ikz ik 2 2 v a v a
u(x, y, z) = −A eiλz e 2z (x +y ) av [e−i( 2 +u) 2 sinc(( v2 − u) a2 )] − ei( 2 −u) 2 sinc(( v2 + u) a2 )]

Intensity I(x, y, z) = |u(x, y, z)|2

Example
Parameters:
λ = 600nm; a = .5mm; z = 30cm

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 Since Fraunhofer approximation is used the center of image pattern is at the center.With no
approximation the pattern would have shifted to the right such that perpendicular line drawn from
center of image passes through the center of triangle(whose plane is parallel to screen)

2 Volume Interference
2.1 Finite Plane Wave

Let’s assume the two beam are arriving from two identical circular aperture of equal radii ’a’.
The electric field due to beam ’B1’ at any point is given by Fourier transform of aperture function
as obtained in section 1.3 . Let’s consider reference frame F1,F2 for beam B1,B2 respectively where
’z’ axis of respective frame align along beam direction and ’y’ axis of two are parallel to each other.
A third reference frame ’F’ lies midway between two circular slits,with its z axis bisecting the angle
of intersection and y axis parallel to the other two frame,as shown in fig. below.

The following relation exists between them

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X1 = (X − d2 )cos(α) + Zsin(α)
Y1=Y
Z1 = −(X − d2 )sin(α) + Zcos(α)

X2 = (X + d2 )cos(α) − Zsin(α)
Y2=Y
Z2 = (X + d2 )sin(α) + Zcos(α)

Now the Electric field amplitude due to B1 and B2 is given by

ik
AeikZ1 2z 2 2
u1(X1, Y 1, Z1) = iλZ1
e (X1 +Y 1 ) πa2 [ 2J1ν1(ν1) ]
ikZ2 ik 2 2 2 2
u2(X2, Y 2, Z2) = Ae iλZ2
e 2z (X2 +Y 2 ) πa2 [ 2J1ν2(ν2) ]. respectively. where ν = ka x +y
z
The net electric field at any point in the interference region is given by their sum.
U(X,Y,Z)=u1+u2

Intensity I(X, Y, Z) = |U (X, Y, Z)|2

Example
Two identical circular slits separated by distance d=2cm and their plane are oriented such that
their normal makes an angle of θ = 2◦ with one another.As shown in fig. above with F1 and F2 as
the two apertures frame.Now the interference at any distance z in F frame of reference is plotted in
python as shown below.
The plot is made serially from left to right for increasing distance Z.

Parameters: d=2cm ; a(aperture radius)=.5mm ;α = θ/2 = 1◦ ; λ = 600nm

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 Fourth figure shows the pattern where z-axis of apertures frame(F1 and F2) intersect and plane
of screen is perpendicular to z-axis of F frame.

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2.2 Gaussian Beam
In Gaussian beam the aperture function is not constant.It’s amplitude falls exponentially as square
of distance from center.Taking the same arrangement as in section 2.2 and only replacing finite plane
wave with Gaussian wave. 2 2
ξ1 +η1
−
In F1 frame at Z1=0,let the amplitude distribution is given by A(ξ, η) = ae wo2
Its Fourier transform as given in section 1.1 gives electric field u1(X1,Y1,Z1) at any point in F1
frame.Similarly we obtain the electric field due to beam from F2 frame as u2(X2,Y2,Z2).The net
electric field at any point as seen from any point is given by
U(X,Y,Z)=u1+u2 , the coordinate transformation of aperture frame to F frame is same as given in
section 2.1

Example Two identical Gaussian beam ,of spot size wo = .38mm,making an angle of 2◦ inter-
fere.The plot of pattern obtained on screen oriented perpendicular to z-axis in F frame at various
position in the interference region is shown below.

The first figures below are zoomed on going from left towards right.
2.2.1 Screen is at the center of intersection of two beams

2.2.2 Screen is placed +6 cm ahead than pervious position

The full image view of pattern of section 2.2.1 and 2.2.2 are:

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References
[1] Ajoy Ghatak, ”OPTICS”

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