Optimal Aperture Width for Solar Eclipse Pinhole Camera

[Pacopag]

Homework Statement

Consider a pinhole camera with depth R = 0.1 m used to observe a solar eclipse. Suppose the sunlight has wavelength \lambda = 5.8 \times 10^{-7} m. Estimate the optimal value of the aperture width w.

Homework Equations

Hint: balance the geometetrical distortion of the image, which goes to zero as w goes to zero, against the spread of an image point due to single-slit diffraction.

The Attempt at a Solution

I determined that the approximate spread due to diffraction is {2R\lambda \over w}.
But I can't seem to find any mathematical formalism for the geometrical distortion. All I can find are statements about how the image becomes sharper as the aperture becomes smaller (i.e. the statement in 'relevant equation'). But this really doesn't give the precise dependence on w. For example, w and w^2, w^3, w^4... are all possible since they all go to zero as w goes to zero.

 

[mgb_phys]

The answer is approx, hole radius <= sqrt(f lambda) where f is focal length (R in this case)

Some pretty smart cookies have spent a lot of time calculating the optimal size, see http://photo.net/pinhole/pinhole.htm

 

[Pacopag]

Awesome. Thank you very much.

 

[Marty]

I think MGB gave you the answer without explaining how to get it. You already did the hard part of the job by finding the spread due to diffraction. The spread due to hole diameter is actually easy if you think about it: it's just the diameter of the hole. Think of the image you would get from a point at infinity passing through a hole of diameter w.

Then just add together the two spreads and use calculus to minimize the total, which is the direction you seemed to be heading in.

 

[Pacopag]

Thank you Marty. That's exactly what I need.

 

[Pacopag]

But when I do as you said, I get an extra factor of \sqrt{2}. This comes from my diffraction calculation.

 

[Marty]

You can't worry TOO much about small constant factors. The idea of spread caused by diffraction is an inexact concept, since the boundaries of the spread are indistinct. So I would say that your basic calculation is sound.

 

[Pacopag]

Cool. Thank you very much for your replies.