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Aperture function of Diffraction Grating

  1. Mar 24, 2012 #1
    1. The problem statement, all variables and given/known data
    A very large diffraction grating has an amplitude transmission aperture function A(x) = 1/2[1+cos(kx)]e^(-x^2/a^2)

    a) What is the far-field irradiance diffraction pattern when a >>1/k
    B) Plot the pattern
    c) Using the appropriate resolving criteria, find the grating's chromatic resolving power in terms of k and a.


    2. Relevant equations



    3. The attempt at a solution

    Well i know the irradiance pattern is proportional to the fourier transform of the aperture function, but trying to take the fourier of that monster function is pretty hard. Putting it in mathematica gives off a much too complicated function to plot. I assume the function simplifies when a>>1/k, but I'm not sure how. I have no idea how to get the resolving power, so some help there would be nice. Thanks
     
  2. jcsd
  3. Mar 24, 2012 #2

    fzero

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    If you write

    [tex]\cos (kx) = \frac{e^{ikx}+e^{-ikx}}{2},[/tex]

    then you can express the Fourier transform in terms of shifted Gaussian functions (you might want to do a coordinate rescaling to get the factors of [itex]2\pi[/itex] correct). Mathematica probably doesn't know this trick for some reason.
     
  4. Mar 24, 2012 #3
    Can you clarify on how to change it to shifted Gaussian function?

    Another question. Since it's not specified, what limits of integration do i take? I only assume it is -a/2 to a/2, from other examples i've seen.
     
  5. Mar 24, 2012 #4

    fzero

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    Since the grating is large, you should integrate from [itex]-\infty[/itex] to [itex]\infty[/itex].

    As for the shift, if the Fourier transform is

    [tex] \hat{f}(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-i\omega x} f(x)dx,[/tex]

    then the Fourier transform of [itex]e^{ikx} f(x)[/itex] is

    [tex] \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-i\omega x} e^{ikx}f(x)dx =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-i(\omega-k) x} f(x)dx = \hat{f}(\omega-k).[/tex]
     
  6. Mar 25, 2012 #5
    Sorry, I'm still not getting it, I don't understand what i'm taking the fourier transform of.
     
  7. Mar 26, 2012 #6
    simply take the limit a-> infinity, the Gaussian part becomes constant.
     
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