Question

1. 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 23.0 and consists of two converging lenses, the objective and the eyepiece, fixed at either end of a tube 81.0cm 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?
Note that to view the crater with a completely relaxed eye, the eyepiece must form its image at infinity and that images formed by telescopes of this nature are always inverted.
fe = ?cm
2. What is the focal length of the objective lens? (fo = ?cm)

Answer 1

Answer:

a)  $f_{e}$ = 3,375 cm , b)  f₀ = 77.625 cm

Explanation:

The magnification of a telescope is, to see at the far point of vision (infinity image)

        m = - f₀ /  $f_{e}$

The length of the tube is

        L = f₀ + $f_{e}$

a) The focal length of the eyepiece

       L = - m  $f_{e}$ +  $f_{e}$

       L = $f_{e}$ (1-m)

        $f_{e}$ = L / (1-m)

Let's calculate

        $f_{e}$ = 81.0 / (1 - (-23.0)

        $f_{e}$ = 3,375 cm

b) the focal length of the target

       f₀ = -m  $f_{e}$

       f₀ = 23 (3.68)

       f₀ = 77.625 cm