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16 April, 12:02

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?

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  1. 16 April, 12:15
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    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
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