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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.

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  1. anonymous
    • 4 years ago
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  2. JamesJ
    • 4 years ago
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    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
    • 4 years ago
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    Ok.Here we go .Fourier transform. \[F(w)=1+e ^{-iaw}+e ^{iaw}\]

  4. JamesJ
    • 4 years ago
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    right. Now, you'll want to write that as some sort of sum of cos and sin

  5. anonymous
    • 4 years ago
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    O.K.So it will be then : \[1+2Cos(aw)\]

  6. anonymous
    • 4 years ago
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    Now we should make its square value ?

  7. JamesJ
    • 4 years ago
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    For intensity, yes.

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

  9. anonymous
    • 4 years ago
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    Wow nice.

  10. JamesJ
    • 4 years ago
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    Now, what happens if you phase shift the middle wave by pi?

  11. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    No, that's like moving the slit over by pi. Phase shifting is something altogether different.

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

  14. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    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
    • 4 years ago
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    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
    • 4 years ago
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    What will that do to the Fourier transform?

  18. anonymous
    • 4 years ago
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    It should multiply a \[e^{(-\pi)} \] expression to Fourier transform.

  19. JamesJ
    • 4 years ago
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    e^(-pi.i) yes

  20. JamesJ
    • 4 years ago
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    and that simplifies to ...

  21. anonymous
    • 4 years ago
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    -1

  22. JamesJ
    • 4 years ago
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    right. Hence the intensity now is proportional to ...

  23. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    No, it's NOT the same.

  25. JamesJ
    • 4 years ago
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    You only multiply one term of the Fourier transform by \( e^{i\pi} \)

  26. anonymous
    • 4 years ago
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    my problem is exactly I don't know how to go through diffraction pattern with the general wave equation.

  27. JamesJ
    • 4 years ago
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    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
    • 4 years ago
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    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
    • 4 years ago
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    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.

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