Diffraction limit on resolution

by lovelylm1980
Tags: diffraction, limit, resolution
 P: 18 Spy planes fly at extremely high altitudes (24.0 km) to avoid interception. Their cameras are reportedly able to discern features as small as 5.48 cm. What must be the minimum aperture of the camera lens to afford this resolution? (Use lambda = 550 nm.) This problem is unfamiliar to me, I'm not sure if the eqtion I used is correct a= lambda*length/d= (550e-9m)*(24e+3m)/5.48e-2m= 0.241m
 Sci Advisor HW Helper P: 1,772 Go to your text and look up "Rayleigh Criterion." YOu have the general idea already, but for circular aperatures, there is a factor of 1.22 that is multiplied to the wavelength. By the way, anyone, where does this 1.22 come from?
 P: 18 got it thanks
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PF Gold
P: 7,334
Diffraction limit on resolution

The given expression is the radius of the first dark ring in the Fraunhofer diffraction pattern for circular aperture.

The computation involves summing intensity contributions from each point of the aperture accounting for path length differences.

In Optics by Rossi 6 pages are committed to doing this computation for a rectangular aperture. For a circular aperture he simply presents the results and comments
 The theory of Fraunhofer diffraction by a circular aperture requires more elaborate mathematical computations then that of the Fraunhofer diffraction by a rectangular aperture
HW Helper
P: 1,772
 Quote by Integral In Optics by Rossi 6 pages are committed to doing this computation for a rectangular aperture. For a circular aperture he simply presents the results and comments
So the math involve in producing this factor of 1.22 is absurdly difficult?
Emeritus
PF Gold
P: 7,334
 Quote by Chi Meson So the math involve in producing this factor of 1.22 is absurdly difficult?
I do not think it is absurdly difficult, it requires setting up and evaluating an involved integral. Rossi must have felt that going through the process for the simpler case, of a rectangular aperture, was more productive then doing the same math only more difficult integrals.

I am sure that if you searched through enough optics texts you will find it worked out some where.

I have the tools to figure it out in Rossi, but frankly it would take more time then I have to commit to it now.
 Sci Advisor HW Helper P: 1,772 I found it in my Blaker & Rosenblum textbook "Optics." It comes from the Fraunhoffer diffraction integral after it is crammed through a 2-dimensional Fourier transform. TO me, it looks nasty. There was a day I could blunder my way through it, but ... the sun went down.
Emeritus