Photographing Palomar 4

Adapted from IOAA 2024 (Rio de Janeiro), theory problem 6. We have restated it in our own words and kept the original data. The solution was last checked on 20 October 2026.

An astronomer photographs the globular cluster Palomar 4 in the V band from a site with no light pollution. The cluster appears as a disc 72.0″ across with an even surface brightness of 20.6 magnitudes per square arcsecond, which means that each square arcsecond of it is as bright as a star of magnitude 20.6. The telescope is a reflector 305 mm across at f/5, with a camera at its prime focus whose quantum efficiency is 80% and whose square pixels are 3.80 µm wide. The V band is centred on 550 nm and is 88.0 nm wide, and a star of magnitude 0 delivers 10 000 photons per nanometre of the band, per square centimetre, per second.

(a) What is the plate scale, in arcminutes per millimetre? (3 points)

(b) About how many pixels does the image of the cluster cover? (4 points)

(c) A 15 s exposure gives the cluster a signal-to-noise ratio of 225. The camera has a read noise of 5 electrons per pixel and a dark current of 6 electrons per pixel per minute. Find the surface brightness of the sky, in magnitudes per square arcsecond. (13 points)

AnswerThe plate scale is 2.25′ per millimetre, the cluster covers about 15 400 pixels, and the sky's surface brightness is about 21.9 magnitudes per square arcsecond, as expected for a dark site. The official solution, which rounds the photon flux in step 4 more coarsely, finds a sky count of 55 250 electrons and the same surface brightness to one decimal place.