Rules of Thumb.

PSF's

One arcsecond is 4.86 microradians

Typical field of view for amateur astronomy telescopes is a few arcminutes. LSST has an unusually wide field, which is why it requires a three-mirror configuration to correct for aberrations. 

Diffraction limit (angular size in radians) for an image made through an aperture is lambda/D where lambda is wavelength and D is diameter.

Achieving seeing-limited images with 1 arcsecond FWHM requires an aperture of D> Lambda/seeing, or D>0.5E-6/4.86E-6 > 0.1 m, or D>4 inches.

The f-number or f/# of a diverging or converging optical beam is the ratio of the beam diameter to distance from focus. An f/18 beam 1 meter away from focus is 1/18 meter in diameter. 
Low f-numbers are "fast" beams and high f-numbers are "slow" beams, taken from the language of photography and how long it takes to accumulate a decent image. 

The spatial FWHM of a diffraction limited beam on a detector is the wavelength times the f-number of the beam that is converging to form the image. 
An example is an f/4 telescope operating at 1 micron- this makes a spot with FWHM ~ 4 microns. 

Surface brightness

If you point an imager at a region of surface brightness, and move away, the light per pixel stays the same (ignoring attenuation along the line of sight). A camera pointed at the moon from the ground gets just as many photons per pixel as one orbiting the moon on a satellite. The difference is spatial resolution (better if you're closer). Light from a point source falls off as 1/r^2 of course. This is exactly compensated by each pixel seeing more emitting area as you move away. 

Photon fluxes

A single photon at 500 nm carries 4 e-19 joules. So producing 10^19 photons/sec at that wavelength requires 4 Watts of optical power. Typical Thor Labs LED-to-fiber sources have a few mW of optical power, so that corresponds to (lambda/500 nm)*2.5^15 photons per sec per mW of optical power. 

CBP fluxes into a telescope generate around 1E6 photoelectrons per second, which is an optical power of 4 e-13 W. 

Solar cell wants to see 10 nA or more of photocurrent. That corresponds to around 1E-8/1.6e-19 = 6.25e10 photons per second, or 25 nW of optical power. So we need to attenuate beam into CBP by a factor of about 10,000 or more. Two reflections off glass would attenuate by (4e-2)^2 = 16e-4, so pretty close. Three reflections might be even better. 

The star Vega is a 0th magnitude star, and provides us with ~1000 photons/sec/cm^2/Angstrom. 

Almost-collimated projector

Imagine we are trying to illuminate a telescope with diameter D from a projector optic with diameter d, a distance L away, using a source of size s. The collimator has a focal length CFL, and the telescope has a focal length TFL. To achieve full-pupil illumination the beam must hit the edges of the input pupil of the telescope. For d<< D this means that we need to have a beam divergence of at least D/L. If the telescope is focused at infinity, that's also the angular extent of the image of the source on the focal plane. If we install a Hartmann screen with subapertures of size DH separated a distance LH, we get multiple images on the focal plane separated by an angle LH/L (the angular separation in direction of arrival) with an angular size that is the larger of DH/L or lambda/DH, i.e. the geometrical and diffraction limits respectively. Amusingly, setting those two equal to each other gives DH as the geometrical mean of lambda and L. For L=1km, that implies DH=sqrt(0.5E-6*1e3) = 2.24 cm or about an inch. The resulting PSF would be sqrt(2)*lambda/D or 6.5 arcsec. The beam spread for L=1 km and D = 1.2 m is 4 arcmin so it fits on the Aux Tel detector, barely. We'd want no more than a 10 x 10 array of holes in the screen, to sustain good spacing between the rather large PSF's. 

SO- for a point source at the Gemini site 1 km away we'd want a Hartmann screen with 1 inch holes. 

Rainbow CPB collimator:

If we want diffraction-limited 1 arc sec PSF, then the minimum aperture is set by theta=lambda/D or  5E-6=0.5e-6/D, so D_min = 0.1m = 10 cm = 4 inches. To get a decent amount of light we need a pinhole of 10 microns or larger. That subtends an angle of theta2=1e-5/FL which requires FL > 1E-5/5e-6 = 2m.