Question

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)

Answer 1

let, d_i be the image distance from the lens, d_o be the object distance from the lens and f be the focan length of the lens.

The lens formula is:

which gives,

Now, linear magnification by a simple lens is given by:

Let, H be the actual diameter of Jupiter and h be the diameter of the image of Jupiter, then linear magnification can also be written as:

Now, the actual average radius of Jupiter is 69,911km = 69,911,000m

Therefore, actual diameter of Jupiter is: H = 2 X 69,911,000 = 139,822,000m

The distance between earth and Jupiter varies periodically and the distance between Jupiter and earth when it is closest to the earth is : d_o = 591,295,403km = 591,295,403,000m

Putting all the values,

One pixel is on side, therefore number of pixels in diameter the Jupiter's image is:

rounding to the upper integer, n = 699

Therefore, Jupiter's image is around 699 pixels in diameter on the CCD.

Smallest feature that we will be able to resolve is the part of Jupiter's diameter captured by a single pixel. Now, n pixels capture the entire diameter (H) of Jupiter so one pixel will capture H/n part of Jupiter's diameter.

Therefore, the smallest feature that we would expect to resolve is 200277.47m (or about 200km) in size.

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