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How does the Fraunhofer irradiance distribution look for a double slit aperture with $d$ = integer multiples of $b$?

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The more mathematical wikipedia page has all the tools, but the technique is exactly the same as the base case: you draw all the various geometric paths that contribute, figure the phase difference for each path and sum or integrate the contributions as appropriate. –  dmckee May 21 '12 at 19:14
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It may be worth a few words to define your notation. Is "d" the spacing and "b" the width? or is "b" something else? (finite slit height possibly). As stated the question is a bit unclear. –  tmac May 21 '12 at 20:41

1 Answer 1

The intensity is just, up to geometric factors, the Fourier transform of the function that tells you how much light gets through at each point. I will assume you are asking about long slits of length 2b and separation 2d.

The 1d two slit problem is very simple--- the Fourier transform of two delta-functions with coefficient 1/2 is the cosine function,

$$ f(k_x) = {1\over 2} (e^{idk_x} + e^{-idk_x}) = \cos(dk_x)$$

Now you are asking about extending a 1d system over a finite long length b in the y-direction. When b is infinite, you reproduce a delta function in the y-direction. When b is finite, you multiply by the Fourier transform of the function which is 1 from -b to b.

$$ g(k_y) \propto {\sin(b k_y) \over k_y} $$

The full Fourier transform is the product of the two, since the transmission function is a product.

$$ A(k_x,k_y) \propto {\cos(d k_x) \sin(b k_y)\over k_y} $$

The proportionality constant includes Fourier transform factors and the widths of the slits together, and it only determines the overall intensity, so I won't include it. When you are looking for diffraction in the outoing direction (k_x,k_y) (the light is coming in along z), you evaluate A(k_x,k_y) where

$$|k|={2\pi\over\lambda}$$

The formula should remind you of the Born approximation in quantum mechanics, because it's the same thing.

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