Question

This is another question regarding Fourier Optics.
May a bit long but worth to solve.
Thanks in advance to have a look and try to solve.

Answer 1

So the Fraunhofer diffraction pattern is the square of the absolute value of the Fourier transform of the transmission function.

The transmission function here is
\[ f(x) = \delta(x) + \delta(x-a) + \delta(x+a) \]
where the coordinate system is centered around the center slit. Now, what's the Fourier transform of each of those delta functions?

Answer 2

Ok.Here we go .Fourier transform.
\[F(w)=1+e ^{-iaw}+e ^{iaw}\]

Answer 3

right. Now, you'll want to write that as some sort of sum of cos and sin

Answer 4

O.K.So it will be then :
\[1+2Cos(aw)\]

Answer 5

Now we should make its square value ?

Answer 6

For intensity, yes.

Answer 7

It's a nice pattern:
http://www.wolframalpha.com/input/?i=%281+%2B+2cos%28x%29%29%5E2

Answer 8

Wow nice.

Answer 9

Now, what happens if you phase shift the middle wave by pi?

Answer 10

I think if we put that plate there then we should give a pi shift to central delta function and it will become \[\delta(x-\pi)\] and again we should start the same procedure.

Answer 11

No, that's like moving the slit over by pi. Phase shifting is something altogether different.

Answer 12

If you had a propagating wave, say
\[ U(x,t) = Ae^{i(kx - \omega t)} \] how would you phase shift it?

Answer 13

OOOOOOOOOH yeah you are right I made a bad mistake.As we saw this kind of difference is showing itself in a over or under the central slit .Sorry for mistake

Answer 14

If I wanted to phase shift U(x,t) by \( \phi \), I multiply it by \( e^{i\phi} \). Hence
\[ \hat{U}(x,t) = Ae^{i(kx - \omega t + \phi)} \]

Answer 15

Ok so here it will show itself in the argument and will be \[U(x,t)=Ae ^{i(kx-wt+\pi)}\] here it seems propagation direction is Z so it is better to use \[U(z,t)=Ae^{i(kz−wt+π)}\]

Answer 16

What will that do to the Fourier transform?

Answer 17

It should multiply a \[e^{(-\pi)} \] expression to Fourier transform.

Answer 18

e^(-pi.i) yes

Answer 19

and that simplifies to ...

Answer 20

-1

Answer 21

right. Hence the intensity now is proportional to ...

Answer 22

intensity which is square value of Fourier transform will be the same but why we didn't consider the other slits effect??those of which doesn't face any phase plate.

Answer 23

No, it's NOT the same.

Answer 24

You only multiply one term of the Fourier transform by \( e^{i\pi} \)

Answer 25

my problem is exactly I don't know how to go through diffraction pattern with the general wave equation.

Answer 26

The Fourier transform is the waveform as it reaches the observation plate. Hence it is
\[ \hat{U}(x) = -1 + 2\cos(ax) \]
and therefore the intensity is proportional to
\[ | \hat{U}(x) |^2 \ \ \alpha \ \ (2\cos(ax)-1)^2 \]

Answer 27

I can't distinguish final part.
We told that plate makes a pi phase shift in the wave equation.
So how again we calculated diffraction pattern by the use of slits function?!

Answer 28

The Fourier transform is now not just
\[ F(\omega) = 1 + e^{ia\omega} + e^{-ia\omega} \]
The wave associated with the center slit, 1, is phase shifted by \( e^{-i\pi} \). Hence the new response function, the new Fraunhofer diffraction is
\[ \hat{F}(\omega) = 1e^{-i\pi} + e^{ia\omega} + e^{-ia\omega} \]
\[ = -1 + e^{ia\omega} + e^{-ia\omega} \]
\[ = 2 \cos(a\omega) - 1 \]

That is why the intensity for the diffraction patter changes. Now you need to ask yourself what that means: what will we now see on the screen vs. what we saw before that center wave was phase shifted.