# Diffraction limit on resolution

1. Mar 25, 2004

### lovelylm1980

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

2. Mar 25, 2004

### Chi Meson

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?

3. Mar 25, 2004

### lovelylm1980

got it thanks

4. Mar 25, 2004

### Integral

Staff Emeritus
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

5. Mar 26, 2004

### Chi Meson

So the math involve in producing this factor of 1.22 is absurdly difficult?

6. Mar 26, 2004

### Integral

Staff Emeritus
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.

7. Mar 29, 2004

### Chi Meson

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.

8. Mar 29, 2004

### Integral

Staff Emeritus
My thoughts exactly. We have some common ground!