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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.
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.

This Question is Closed

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1So the Fraunhofer diffraction pattern is the square of the absolute value of the Fourier transform of the transmission function. The transmission function here is \[ f(x) = \delta(x) + \delta(xa) + \delta(x+a) \] where the coordinate system is centered around the center slit. Now, what's the Fourier transform of each of those delta functions?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok.Here we go .Fourier transform. \[F(w)=1+e ^{iaw}+e ^{iaw}\]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1right. Now, you'll want to write that as some sort of sum of cos and sin

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0O.K.So it will be then : \[1+2Cos(aw)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now we should make its square value ?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1It's a nice pattern: http://www.wolframalpha.com/input/?i=%281+%2B+2cos%28x%29%29%5E2

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Now, what happens if you phase shift the middle wave by pi?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1No, that's like moving the slit over by pi. Phase shifting is something altogether different.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1If you had a propagating wave, say \[ U(x,t) = Ae^{i(kx  \omega t)} \] how would you phase shift it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0OOOOOOOOOH 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

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1If I wanted to phase shift U(x,t) by \( \phi \), I multiply it by \( e^{i\phi} \). Hence \[ \hat{U}(x,t) = Ae^{i(kx  \omega t + \phi)} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok so here it will show itself in the argument and will be \[U(x,t)=Ae ^{i(kxwt+\pi)}\] here it seems propagation direction is Z so it is better to use \[U(z,t)=Ae^{i(kz−wt+π)}\]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1What will that do to the Fourier transform?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It should multiply a \[e^{(\pi)} \] expression to Fourier transform.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1and that simplifies to ...

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1right. Hence the intensity now is proportional to ...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0intensity 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.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1You only multiply one term of the Fourier transform by \( e^{i\pi} \)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0my problem is exactly I don't know how to go through diffraction pattern with the general wave equation.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1The 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 \[  \hat{U}(x) ^2 \ \ \alpha \ \ (2\cos(ax)1)^2 \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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?!

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1The 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.
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