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Fresnel lens

  1. Apr 22, 2015 #1
    1. The problem statement, all variables and given/known data
    Calculate the focal length of 1D Fresnel lens, whose transmittance is given as $$T(\xi)=\frac 1 2(1+\cos(\alpha \xi ^2)).$$

    2. Relevant equations
    Anything you wish

    3. The attempt at a solution
    I have no idea. I tried to use the equation for diffraction image $$u_p=C\int _0 ^{R_p}T(\xi)e^{ik\frac{r^2}{2z}}rdr$$ which is simply a Fresnel diffraction multiplied with lens transmittance. I thought this would bring me to something useful, but I couldn't see how this would help.

    Any ideas/hints on how to solve this problem would be highly appreciated!
     
  2. jcsd
  3. Apr 23, 2015 #2

    Simon Bridge

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    ... nope, it's anything you choose ... part of the point of this section is to tell us how you are thinking about the problem. We already know how we are thinking of the problem.

    You could try looking at your notes - consider: what does the transmittance function tell you?
    How would you use the transmittance function to work out how a plane wave would be affected?
    ... so you did have some idea - did you follow through on it?
    Sometimes you won't be able to see how something will turn out until after you'd gone down the path for a bit.

    Also see:
    http://inst.eecs.berkeley.edu/~ee290f/fa04/Lenses_post.pdf
     
  4. Apr 23, 2015 #3
    The problem is here:
    $$u_f=C\int T(\xi)e^{i\frac{k}{2R}\xi ^2}d\xi=C\int \frac 1 2(1+\cos(\alpha \xi^2))e^{i\frac{k}{2R}\xi ^2}d\xi=$$ $$=C\int \frac 1 2 (1+\frac 1 2(e^{i\alpha \xi ^2}+e^{-i\alpha \xi ^2}))e^{i\frac{k}{2R}\xi ^2}d\xi=$$ $$=C\int[\frac 1 2e^{i\frac{k}{2R}\xi ^2}+\frac 1 4 e^{i(\frac{k}{2R}-\alpha)\xi ^2}+\frac 1 4 e^{i(\frac{k}{2R}+\alpha)\xi ^2}]d\xi$$
    In in the focal point the intensity (which is ##\propto |u_f|^2##) should have a huge peak. And from the equation above I should be able to already see that. Yet I don't. I am confused because of the imaginary argument in my exponents. How do I in this case find out where the peak is? :/
     
  5. Apr 24, 2015 #4
    You shouldn't be too concerned about this. These integrals are a variant of Gaussian integrals and they have the form:-
    ##\int_0^x e^{t^2}\ dt = -i\sqrt{\frac{\pi}{2}}erf(ix)##
    where ##erf(ix)## is the error function.
     
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