Huygens’ wave construction
• wavefronts propagate from initial disturbance
in all directions
• each point on the wavefront acts as a
secondary source
• further wavefronts propagate from the
secondary sources in the same fashion
• where wavefronts coincide (strong constructive
interference), a new wavefront is formed
Christiaan Huygens (1629-1695)
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Diffraction and Fourier Transform
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Interference from Two Equal
Sources of Separation f
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Young’s Double Slit Experiment
The distance on the
screen between two bright fringes
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Fraunhofer diffraction
• Two (smaller) apertures
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Interference from Linear Array
of N Equal Sources
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Interference from Linear Array
of N Equal Sources
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Diffraction from a Slit
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Single slit diffraction
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Intensity Distribution for Interference
with Diffraction from N Identical Slits
Example: N=2
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The Missing Orders
The missing orders will be n = 2; 4; 6; 8,
etc. for m = 1; 2; 3; 4, etc.
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Device example: diffraction gratings
Rayleigh’s criterion
newport.com
Resolving power for nth order
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More device examples
Phased-array loudspeakers
Phased-array radar
• 2560 transducers, 25m diameter, 420-450 MHz
Spatial light modulators
• 1920 x 1200 pixels, LCD technology
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Fraunhofer vs. Fresnel Diffraction
from a Single Slit
Far from the slit:
Close to the slit:
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Diffraction and Fourier Transform
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Fresnel and Fraunhofer Diffraction
We wish to find the light electric field after a screen with a hole in it.
This is a very general problem with far-reaching applications.
Incident
wave
This region is assumed to be
much smaller than this one.
What is E(x0,y0) at a distance z from the plane of the aperture?
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Diffraction Solution
The field in the observation plane, E(x0,y0), at a distance z from the
aperture plane is given by a convolution:
E ( x0 , y0 ) 

h( x0  x1 , y0  y1 ) E ( x1 , y1 )dx1dy1
Aperature(x1 , y1 )
where:
1 exp(ikr01 )
h( x0  x1 , y0  y1 ) 
i
r01
and:
r01  z   x0  x1    y0  y1 
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2
2
2
Fresnel Diffraction: Approximations
But we can approximate r01 in the denominator by z.
And, we can write:
z 2   x0  x1    y0  y1 
r01 
2
2
 x  x1   y0  y1 
 z 1  0
 

 z   z 
2
2
But 1    1   / 2 if  <<1
 1  x0  x1  2 1  y0  y1  2 
 z 1  
  
 
2
z
2
z



 

 x  x1 
 z 0
2z
2
 y  y1 
 0
2
2z
This yields:

2
2

x0  x1 
y0  y1   


1
 

E ( x0 , y0 ) 
exp ik  z 

  E ( x1 , y1 )dx1dy1
i z
2z
2z

 



Aperture ( x , y )
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1
1
Fresnel Diffraction: Approximations
Multiplying out the squares:
E  x0 , y0  

  ( x02  2 x0 x1  x12 ) ( y02  2 y0 y1  y12 )  
1
exp ik  z 

 E ( x1 , y1 )dx1dy1

i z
2z
2z
 
 
Aperture ( x1 , y1 )
Factoring out the quantities independent of x1 and y1:

x02  y02 
1
E  x0 , y0  
exp  ikz 

i z
z 


  (2 x0 x1  2 y0 y1 ) ( x12  y12 )  
exp ik 

 E  x1 , y1  dx1dy1
2
z
2
z
 
 
Aperture(x1 , y1 )
This is the Fresnel integral. It's complicated!
Note that it looks a bit like a Fourier Transform, except for the exp of
the quadratics in x1 and y1.
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Fresnel Diffraction: Approximations
We’ll typically assume that a plane wave is incident on the aperture.
E ( x1 , y1 )  const
And we’ll explicitly write the aperture function in the integral:

x02  y02 
1
E  x0 , y0  
exp  ikz 

i z
z 


  (2 x0 x1  2 y0 y1 ) ( x12  y12 )  
exp ik 

 Aperture(x1 , y1 ) dx1dy1

2z
2 z  
 
And we’ll usually neglect the phase factors in front:
E  x0 , y0  
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
  (2 x0 x1  2 y0 y1 ) ( x12  y12 )  
exp ik 

 Aperture(x1 , y1 ) dx1dy1

2z
2 z  
 
Fraunhofer Diffraction: The Far Field
Recall the Fresnel diffraction result:

x02  y02 
1
E  x0 , y0  
exp  ikz 

i z
z 


  (2 x0 x1  2 y0 y1 ) ( x12  y12 )  
exp ik 

 E  x1 , y1  dx1dy1
2
z
2
z
 
 
Aperture(x1 , y1 )
Let D be the size of the aperture: D = x12 + y12.
When kD2/2z << 1, the quadratic terms <<1, so we can neglect them:

x02  y02 
1
E  x0 , y0  
exp  ikz 

i z
z



Aperture ( x , y1 )
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 ik

exp   x0 x1  y0 y1   E  x1 , y1  dx1dy1
 z

Fraunhofer Diffraction Conventions
As in Fresnel diffraction, we’ll typically assume a plane wave incident
field, we’ll neglect the phase factors, and we’ll explicitly write the
aperture function in the integral:
E  x0 , y0  

 ik

exp   x0 x1  y0 y1   Aperture( x1 , y1 ) dx1dy1
 z

This is just the Fourier Transform!
So, the far-field light field is the Fourier Transform of the aperture field!
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Fraunhofer Diffraction
from a Square Aperture
Diffracted field
is a sinc function in
both x0 and y0
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Diffraction from a
Circular Aperture
A circular aperture
yields a diffracted
"Airy Pattern,"
which involves a
Bessel function.
Diffracted Irradiance
Diffracted field
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