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I am currently attempting a physics question on optics.

A Fresnel Zone Plate has overall radius a and contains N half-period zones. The odd-numbered zones are open but the even -numbered zones are covered by glass so as to invert the phase of light passing through them. The zone plate is illuminated with monochromatic, collimated light of wavelength lambda and intensity I0

There are a number of prior steps where the focal length is found to be

                         f=a^2/(N*lambda)

the intensity of the focal spot was found to be 

                         I=4I0a^4/(f^2*lambda^2)

and that the radius of the focal spot was approximately equal to the width of the outermost half-period zone at 

                            a/2N

However, in the final part the question now asks:

Show that along the optic axis the focal spot has a length of 4f/N . Hint the intensity on axis will be zero if the phase change across the first open zone is 

                            pi(1+2/N)

However, after several attempts I haven't got anywhere and can't quite picture what is going on. I couldn't find any help online so I was wondering could anyone explain how the length of the focal point occurs and point me in the right direction to solve the problem .

Edit//

I believe I solved my problem. I found that the two minimums around the focal spot where at f(1+2/N) and f(1-2/N) thus giving my a length of 4f/N.

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  • $\begingroup$ The length of the focal spot is probably given by the length of constructive interference along the optical axis around the focal point. Did you try to calculate the intensity as a function of distance along the optical axis? $\endgroup$
    – CuriousOne
    Dec 27 '14 at 21:32
  • $\begingroup$ I was trying to find how the phase across the first open zoned changed with position along the optical axis in the hope i could find where destructive interference occurred. I can't see how to calculate intensity as a function of distance. How could that be done? $\endgroup$
    – SiberianSloth
    Dec 27 '14 at 21:35
  • $\begingroup$ How did you calculate the other results? You should be able to calculate the phase shift for the N-th zone, which should be the largest and most sensitive to a displacement along the optical axis. When does that start to interfere destructively? Will that give the right answer? If not, then you will have to find a formula to add up all the phase shifts. Is f assumed to be large compared to the diameter of the plate? $\endgroup$
    – CuriousOne
    Dec 27 '14 at 21:40
  • $\begingroup$ The focal length was found by equating the area of the circular plate to the area of N fresnal half period zones and using the fact R is approximately f is we assume that the source is far away. The intensity was found by usinga phasor diagram and saying the amplitude at P was equal to 2N omega0 and subbing in for N using the first result. No comment was made relating a and f but that could be reasonably assumed. $\endgroup$
    – SiberianSloth
    Dec 27 '14 at 21:44
  • $\begingroup$ So what happens when you try to do what you did for the intensity as a function along the axis using the approximation? $\endgroup$
    – CuriousOne
    Dec 27 '14 at 21:47
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I believe I solved my problem. I found that the two minimums around the focal spot where when the phase across the first open zone was pi(1+2/N) and pi(1-2/N).

This corresponds to lengths along the optic axis as f(1+2/N) and f(1-2/N) thus giving my a length of 4f/N.

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