## anonymous 4 years ago 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.

1. anonymous

2. JamesJ

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?

3. anonymous

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

4. JamesJ

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

5. anonymous

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

6. anonymous

Now we should make its square value ?

7. JamesJ

For intensity, yes.

8. JamesJ

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

9. anonymous

Wow nice.

10. JamesJ

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

11. anonymous

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.

12. JamesJ

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

13. JamesJ

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

14. anonymous

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

15. JamesJ

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

16. anonymous

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 $U(z,t)=Ae^{i(kz−wt+π)}$

17. JamesJ

What will that do to the Fourier transform?

18. anonymous

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

19. JamesJ

e^(-pi.i) yes

20. JamesJ

and that simplifies to ...

21. anonymous

-1

22. JamesJ

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

23. anonymous

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.

24. JamesJ

No, it's NOT the same.

25. JamesJ

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

26. anonymous

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

27. JamesJ

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 $| \hat{U}(x) |^2 \ \ \alpha \ \ (2\cos(ax)-1)^2$

28. anonymous

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?!

29. JamesJ

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.