The primary optical element of the hubble space telescope (hst) is 3.2 m in diameter and has a focal length of 62 m. (treat it as a simple, single lens for this homework) the telescope is aimed at jupiter and the collected light is focused onto a sensitive charge coupled device (ccd) detector, similar to what is in a digital camera. each pixel in the detector is a 21 μm x 21 μm square, and the full ccd is 4096 x 4096 pixels. thus the ccd is about one square inch in size. the hst is in orbit very close to the earth (compared to other distances in the solar system).
look up the size of jupiter and the distance to jupiter when it is closest to earth. use the lens formula to determine the magnification of the image hubble takes.
how many pixels in diameter is jupiter’s image on the ccd?
given this ccd, what is the smallest feature on jupiter you would expect to be able to resolve? (another way of thinking about that question is: how large a square on the surface of jupiter does one pixel in the image represent? )
(eye piece parameters are not given for this question but it can be solved without them)