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# Fourier Optics

Module by: Paul Padley. E-mail the author

Summary: Introduces the fact that the diffraction pattern through an aperture is the Fourier transform of the aperture.

## Fourier Transforms

The Fourier Transform can be used to represent any well behaved function f ( x ) . f ( x ) .

f ( x ) = 1 π 0 [ A ( k ) cos ( k x ) + B ( k ) s i n ( k x ) ] d k f ( x ) = 1 π 0 [ A ( k ) cos ( k x ) + B ( k ) s i n ( k x ) ] d k where A ( k ) = f ( x ) cos ( k x ) x A ( k ) = f ( x ) cos ( k x ) x B ( k ) = f ( x ) sin ( k x ) x B ( k ) = f ( x ) sin ( k x ) x I can now substitute for A A and B B in the original expression and write:

f ( x ) = 1 π 0 cos k x f ( x ) cos k x x k +    + 1 π 0 sin k x f ( x ) sin k x x k f ( x ) = 1 π 0 cos k x f ( x ) cos k x x k +    + 1 π 0 sin k x f ( x ) sin k x x k and then use cos ( x x ) = c o s k x cos k x + sin k x sin k x cos ( x x ) = c o s k x cos k x + sin k x sin k x

f ( x ) = 1 π 0 f ( x ) cos ( k [ x x ] ) x k f ( x ) = 1 π 0 f ( x ) cos ( k [ x x ] ) x k Since the inner integral is an even function we can write f ( x ) = 1 2 π f ( x ) cos ( k [ x x ] ) x k f ( x ) = 1 2 π f ( x ) cos ( k [ x x ] ) x k Now consider the fact that i 2 π f ( x ) sin ( k [ x x ] ) x k = 0 i 2 π f ( x ) sin ( k [ x x ] ) x k = 0 because sin sin is an odd function, ie. sin ( k [ x x ] ) k = 0 sin ( k [ x x ] ) k = 0 So we could have written f ( x ) = 1 2 π f ( x ) cos ( k [ x x ] ) x k +    + i 2 π f ( x ) sin ( k [ x x ] ) x k f ( x ) = 1 2 π f ( x ) cos ( k [ x x ] ) x k +    + i 2 π f ( x ) sin ( k [ x x ] ) x k or f ( x ) = 1 2 π f ( x ) e i k ( x x ) x k f ( x ) = 1 2 π f ( x ) e i k ( x x ) x k or f ( x ) = 1 2 π g ( k ) e i k x k f ( x ) = 1 2 π g ( k ) e i k x k where g ( k ) = f ( x ) e i k x x g ( k ) = f ( x ) e i k x x is the Fourier transform of f ( x ) f ( x ) .

Symbolically we write g ( k ) = Ϝ { f ( x ) } g ( k ) = Ϝ { f ( x ) } f ( x ) = Ϝ 1 { g ( k ) } = Ϝ 1 { Ϝ { f ( x ) } } f ( x ) = Ϝ 1 { g ( k ) } = Ϝ 1 { Ϝ { f ( x ) } }

Now these concepts are easily extended to two dimensions:

f ( x , y ) = 1 ( 2 π ) 2 g ( k x , k y ) e i ( x k x + y k y ) k x k y f ( x , y ) = 1 ( 2 π ) 2 g ( k x , k y ) e i ( x k x + y k y ) k x k y where g ( k x , k y ) = f ( x , y ) e i ( x k x + y k y ) x y . g ( k x , k y ) = f ( x , y ) e i ( x k x + y k y ) x y .

This tells us is that any nonperiodic function of two variables f ( x , y ) f ( x , y ) can be synthesized from a distribution of plane waves each with amplitude g ( k x , k y ) g ( k x , k y ) .

Lets consider Fraunhofer diffraction through an aperture again. For example consider a rectangular aperture as show in the figure. If ɛ A ɛ A is the source strength per unit area (assumed to be constant over the entire area in this example) and d S = x z d S = x z is an infinitesmal area at a point in the aperture then we have:

We can define a source strength per unit area d E = ɛ A R e i ( ω t k r ) d E = ɛ A R e i ( ω t k r )

Notice that I flipped the sign in the exponential from what I used in the earlier lectures on diffraction. This does not change the physics content of what we are doing in any way, however it allows our notation to follow standard convention. E = A p e r t u r e ɛ A R e i ( ω t k r ) z y = ɛ A e i ( ω t k R ) R A p e r t u r e e i k y Y / R y e i k z Z / R z E = A p e r t u r e ɛ A R e i ( ω t k r ) z y = ɛ A e i ( ω t k R ) R A p e r t u r e e i k y Y / R y e i k z Z / R z

If we define k y = k Y / R k y = k Y / R and k z = Z / R k z = Z / R and we see that

E = ɛ A e i ( ω t k R ) R A p e r t u r e e i y k y e i z k z y z E = ɛ A e i ( ω t k R ) R A p e r t u r e e i ( y k y + z k z ) y z E = ɛ A e i ( ω t k R ) R A p e r t u r e e i y k y e i z k z y z E = ɛ A e i ( ω t k R ) R A p e r t u r e e i ( y k y + z k z ) y z

That is, it is equal to the Fourier transform. In fact one can define an "Aperture Function" A ( y , z ) . A ( y , z ) . Such that

E = A ( y , z ) e i ( y k y + z k z ) y z E = A ( y , z ) e i ( y k y + z k z ) y z

For a rectangular aperture A ( y , z ) = ɛ A e i ( ω t k R ) R A ( y , z ) = ɛ A e i ( ω t k R ) R inside the aperture and zero outside it. The aperture function can be much more complex (literally) allowing for changes in source strength and phase through the aperture. The resulting E field is the Fourier transform of the aperture function.

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