The magnifying power of a telescope is `9.` When it is adjusted for parallel rays the distance between the objective and eyepiece is `20cm`. The focal lengths of lenses are

Options

A. 10 cm, 10 cm
B. 15 cm, 5 cm
C. 18 cm, 2 cm
D. 11 cm, 9 cm

Answer

C

Explanation

Here, `M = (f_0)/(f_e) = 9 :. f_0 = 9 f_e` …..(i)
When adjusted for parallel rays, the distance between the objective anf eye piece is
`f_0 + f_e = 20 cm`
Using (i), `9 f_e + f_e = 20` or `f_e = 2 cm`
From (i), `f_0 = 9 xx 2 cm = 18 cm`.