On one of the shelves in your physics lab is displayed an antique telescope. a sign underneath the instrument says that the telescope has a magnification of 20 and consists of two converging lenses, the objective and the eyepiece, fixed at either end of a tube 60.0 cm long. assuming that this telescope would allow an observer to view a lunar crater in focus with a completely relaxed eye, what is the focal length fe of the eyepiece?

The focal length fe of the eyepiece is 2.86 cm

Explanation:

Since we are given the telescope's magnification and the length of the tube, we can use the expressions

M = f_o/fe (1) and

l = f_o + fe   (2)

where

M is the telescope's magnificationl is the length of the tubefe is the focal length of the eye-piece

Rearranging equation (2) to make f_o the subject of the formula, we get

f_o = l - fe

Substituting the above equation into equation (1) we get

M = (l - fe)/fe ⇒ fe = l/(M +1)

                     ⇒ fe = 60/(20 + 1)

                     ⇒ fe  = 2.86 cm

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