Calculating Focal Length of 1D Fresnel Lens

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skrat
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Homework Statement


Calculate the focal length of 1D Fresnel lens, whose transmittance is given as $$T(\xi)=\frac 1 2(1+\cos(\alpha \xi ^2)).$$

Homework Equations


Anything you wish

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!
 
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Anything you wish
... 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.

I have no idea.
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?
I thought this would bring me to something useful, but I couldn't see how this would help.
... 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
 
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? :/
 
skrat said:
I am confused because of the imaginary argument in my exponents.
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.