Aperture function of Diffraction Grating

[makotech222]

Homework Statement

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.

Homework Equations

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

 

[fzero]

If you write

$$\cos (kx) = \frac{e^{ikx}+e^{-ikx}}{2},$$

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 $2\pi$ correct). Mathematica probably doesn't know this trick for some reason.

 

[makotech222]

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.

 

[fzero]

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.

Since the grating is large, you should integrate from $-\infty$ to $\infty$.

As for the shift, if the Fourier transform is

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

then the Fourier transform of $e^{ikx} f(x)$ is

$$\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-i\omega x} e^{ikx}f(x)dx =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-i(\omega-k) x} f(x)dx = \hat{f}(\omega-k).$$

 

[makotech222]

Sorry, I'm still not getting it, I don't understand what I'm taking the Fourier transform of.

 

[sunjin09]

simply take the limit a-> infinity, the Gaussian part becomes constant.