Diffraction limit on resolution on camera lens

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Discussion Overview

The discussion revolves around the diffraction limit on resolution in camera lenses, specifically in the context of spy planes operating at high altitudes. Participants explore the mathematical relationships involved in determining the minimum aperture required for a given resolution, referencing the Rayleigh Criterion and the complexities of diffraction patterns.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a calculation for the minimum aperture needed for a camera lens to resolve features at a specific distance and resolution, expressing uncertainty about the correctness of their approach.
  • Another participant suggests consulting the "Rayleigh Criterion" and notes that for circular apertures, a factor of 1.22 is involved, questioning its origin.
  • A later reply clarifies that the factor of 1.22 relates to the radius of the first dark ring in the Fraunhofer diffraction pattern for circular apertures, indicating that the computation is complex.
  • One participant expresses skepticism about the difficulty of the mathematics involved in deriving the factor of 1.22, while another argues that it requires evaluating an involved integral.
  • Another participant mentions finding the derivation in a textbook, describing it as complicated and expressing a sense of nostalgia for their ability to navigate such mathematics in the past.

Areas of Agreement / Disagreement

Participants express varying levels of confidence regarding the complexity of the mathematics involved in deriving the factor of 1.22, with some suggesting it is difficult while others believe it is manageable. There is no consensus on the ease or difficulty of the calculations.

Contextual Notes

Participants reference different optics textbooks and the specific mathematical processes involved in diffraction theory, indicating a reliance on various sources and interpretations of the material.

lovelylm1980
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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
 
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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?
 
got it thanks:smile:
 
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
 
Integral said:
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?
 
Chi Meson said:
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.
 
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.
 
Chi Meson said:
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.

My thoughts exactly. We have some common ground!
 

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