anonymous
  • anonymous
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.
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
JamesJ
  • 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?
anonymous
  • anonymous
Ok.Here we go .Fourier transform. \[F(w)=1+e ^{-iaw}+e ^{iaw}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

JamesJ
  • JamesJ
right. Now, you'll want to write that as some sort of sum of cos and sin
anonymous
  • anonymous
O.K.So it will be then : \[1+2Cos(aw)\]
anonymous
  • anonymous
Now we should make its square value ?
JamesJ
  • JamesJ
For intensity, yes.
JamesJ
  • JamesJ
It's a nice pattern: http://www.wolframalpha.com/input/?i=%281+%2B+2cos%28x%29%29%5E2
anonymous
  • anonymous
Wow nice.
JamesJ
  • JamesJ
Now, what happens if you phase shift the middle wave by pi?
anonymous
  • 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.
JamesJ
  • JamesJ
No, that's like moving the slit over by pi. Phase shifting is something altogether different.
JamesJ
  • JamesJ
If you had a propagating wave, say \[ U(x,t) = Ae^{i(kx - \omega t)} \] how would you phase shift it?
anonymous
  • 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
JamesJ
  • JamesJ
If I wanted to phase shift U(x,t) by \( \phi \), I multiply it by \( e^{i\phi} \). Hence \[ \hat{U}(x,t) = Ae^{i(kx - \omega t + \phi)} \]
anonymous
  • 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 \[U(z,t)=Ae^{i(kz−wt+π)}\]
JamesJ
  • JamesJ
What will that do to the Fourier transform?
anonymous
  • anonymous
It should multiply a \[e^{(-\pi)} \] expression to Fourier transform.
JamesJ
  • JamesJ
e^(-pi.i) yes
JamesJ
  • JamesJ
and that simplifies to ...
anonymous
  • anonymous
-1
JamesJ
  • JamesJ
right. Hence the intensity now is proportional to ...
anonymous
  • 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.
JamesJ
  • JamesJ
No, it's NOT the same.
JamesJ
  • JamesJ
You only multiply one term of the Fourier transform by \( e^{i\pi} \)
anonymous
  • anonymous
my problem is exactly I don't know how to go through diffraction pattern with the general wave equation.
JamesJ
  • 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 \[ | \hat{U}(x) |^2 \ \ \alpha \ \ (2\cos(ax)-1)^2 \]
anonymous
  • 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?!
JamesJ
  • 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.