The primary optical element of the Hubble Space Telescope (HST) is 3.2 m in diameter and...

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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 asimple, single lens for this homework) The telescope is aimed atJupiter and the collected light is focused onto a sensitive ChargeCoupled Device (CCD) detector, similar to what is in a digitalcamera. Each pixel in the detector is a 21 Î¼m x 21 Î¼m square, andthe full CCD is 4096 x 4096 pixels. Thus the CCD is about onesquare 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 itis closest to Earth. Use the lens formula to determine themagnification of the image Hubble takes.
How many pixels in diameter is Jupiterâ€™s image on the CCD?
Given this CCD, what is the smallest feature on Jupiter youwould expect to be able to resolve? (Another way of thinking aboutthat question is: How large a square on the surface of Jupiter doesone pixel in the image represent?)
(eye piece parameters are not given for this question but it canbe solved without them)

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4.0 Ratings (509 Votes)

let di be the image distance from the lensdo be the object distance from the lens and f be thefocan length of the lensThe lens formula iswhich givesNow linear magnification by a simple lens is given byLet H be the actual diameter of Jupiter See Answer

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