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 magnification l is the length of the tube fe 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