My professor gave us the following solution to a practice exam question: I\'m ha
ID: 3896328 • Letter: M
Question
My professor gave us the following solution to a practice exam question:
I'm having trouble understanding his explanation, and I would like to see a more thorough one. Thank you!
Consider a uniform plane wave of wavelength A illuminating a Fresnel zone plate with small openings located at radial positions pm = 2m f.m = 0,1,2, - such that light emerging from each opening interferes constructively at positions {z = f,f/2,f/3,f/ 4," Assuming that the Fresnel zone plate openings remain at their initial locations, but the wavelength of illumination is increased by a factor of 2 (i.e. Right arrow 2 ), at which of the above foci will the new wavelength also come to a focus? Please circle YES or NO for each focus. Will light with a wavelength 2A focus at = f ? Will light with a wavelength 2A focus at = f/2 ? Will light with a wavelength 2A focus at = f/3 ? Will light with a wavelength 2A focus at = f/4 ?Explanation / Answer
Well basically what happens here is that after the openings all rays illuminate al the points given, f/4.f/3. f/2, f, 2f. etc. in all cases.
However, what modifies with wavelength is that the condition of maxima of diffraction. To be in focus in a certain position one needs a maxima of interference-diffraction in that point.
If initially the focus is at distance f, the difference of path delta(L) for the n-th ray at f is
delta(L) = ro_n*sin(theta_n)
sin(theta_n) = tan(theta_n) = ro_n / f for small angles
delta(L) = ro_n^2 /f = 2*n*lambda*f0 /f = 2n (if f0=f , focus at f)
If lambda ---> 2*lambda
to keep the same difference of path delta(L) =(2n)
one needs the focus to become f0 ----> f/2
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