Question

The magnifying power of an astronomical telescope is `5`. When it is set for normal adjustment, the distance between the two lenses is `24 cm`. Calculate the focal lengths of eye piece and objective lens.

Answer 1

Correct Answer - `4 cm ; 30 cm`
Here, `m = 5, L = 24 cm, f_o = ?, f_e = ?`
As `m = (f_o)/(f_e) = 5 :. f_o = 5 f_e`
Also, in normal adjustment
`L = f_o + f_e = 5 f_e + f_e = 6 f_e`
`f_e = (L)/(6) = (24)/(6) = 4 cm`
and `f_o = 5 f_e = 5 xx 4 cm = 20 cm`.