Question

Calculate the diffraction pattern within the framework of Fraunhofer diffraction by Fourier transformation of the transmission function of the following arrangement of slits:

| | | |
<....><.........><..>
a 3a a
------------------->x
0
The width of the single slits can be neglected , so that the transmission function can be expressed as sum of delta-functions . Sketch the resulting diffraction pattern (hint : express your solution as product of cosine functions)

Answer 1

*bookmark. I'm going to try and figure this out.

Answer 2

Nice .I hope this can be a good challenge for Fourier optics readers. :)

Answer 3

The aperture function is
\[ A(x) = \delta(x-5a/2) + \delta(x-3a/2) + \delta(x+3a/2) + \delta(x+5a/2) \]
The formula for the amplitude of the diffraction pattern of the Fourier transform of that function, which is pretty straight-forward. In fact, this whole exercise is just a straight-forward extension of the case for two slits.

So, here we go ...

Answer 4

Write F(f(x)) for the Fourier transform of f(x).

Then as
\[ F(\delta(x)) = 1 \] using the 'exponential-shift formula',
\[ F(\delta(x-a)) = e^{-2\pi i a \xi} \]
and linearity of of the Fourier transform, the Fourier transform of A(x) is
\[ F(A(x))(\xi) = e^{-5\pi i a\xi} + e^{-3\pi i a\xi} + e^{3\pi i a\xi} + e^{5\pi i a\xi} \]
\[ = 2 ( \cos(3\pi a \xi) + \cos(5 \pi a \xi) ) \]
Now let z be the distance to the screen and \( \lambda \) the wavelength, then \( \xi = x/(\lambda z) \). Rewriting now the Fourier transform as function \(U(\theta) \) of the angle of deflection \( \theta \) where \( \sin\theta = x/z \), we have

\[ U(\theta) = F(A(x))(\theta) = 2\left( \cos(3\pi a \frac{\sin \theta}{\lambda}) + \cos(5\pi a \frac{\sin \theta}{\lambda}) \right) \]

Answer 5

Finally, to sketch this, the question is suggesting you use a basic trig identity that I just wrote over in Mathematics for a high-school student:

2 cos x . cos y = cos(x+y) + cos(x-y).

Re arrange the expression above for U(θ) into the product of two cos functions.

(And if that student asks me why we should learn these trig identities, I'll tell them I have a great example why. ;-) )

Answer 6

*correction (three posts up): "The formula for the amplitude of the diffraction pattern IS [not of] the Fourier transform of that function [the aperture function], which is pretty straight-forward."

Answer 7

I'm sure in lectures or your text book, you'll find an example for two slits. Check that against this example. If nothing else, you'll probably want to harmonize the notation of your solution with the notation you are using in your course.

Answer 8

Just let me see what have you done here sounds like it's perfect :) but I will put my perfect answer on Tuesday

Answer 9

Because now I'm just studying and perusing through some literature and I hope I'll solve it in 2 days with my own way.

Answer 10

BTW, I should also squared that answer to finally arrive at:
\[ |U(\theta)|^2 = 16 \cos^2(4\pi \sin \theta / \lambda)\cos^2(\pi \sin \theta / \lambda) \]

Answer 11

yeah of course for intensity it should be done in this way.

Answer 12

but the last answer should be written in this way :\[\left| U(\theta) \right|^2=16\cos ^2(4\pi a \sin \theta/\lambda)\cos^2(\pi asin \theta/\lambda)\]

Answer 13

yes, I dropped the a from the expression.