Optical power lost at a pinhole

[lalligagger]

Hi,
I am a physics undergrad designing a solar spectrometer as part of a senior design type course. If we have light incident from the sun onto some sort of telescope objective (say a single lens for simplicity, focal length "f" and aperture radius "R") and then a pinhole at the focal point (radius "r"), how do we model the power loss at the pinhole? I was thinking it'd be
I_sun*(pi*R²)*[(.01*f/2)²/(r)²]
In other words, intensity times the area of the entrance aperture gives you power in, and taking the ratio of your image area over the pinhole area gives you the fraction of power that gets through. The .01 comes from the sun being approximately .01 radians in the sky. This made sense to me, but it seems like we would barely get any power through with a 1 meter telescope into a 20-50 micron pinhole. Any help would be greatly appreciated.

 

[Andy Resnick]

For an ideal lens used the way you are thinking (object at infinity), the distribution of light at the focal plane is given by an Airy function, which can be written as J_1(ax)/ax, just like sinc(ax) = sin(ax)/ax.

The width of the Airy disk is given by the wavelength and f-number (or numerical aperture) of the lens:

http://en.wikipedia.org/wiki/Airy_disk

For your application, simply compare the size of the Airy disk with the size of the pinhole.