Question

The magnifying power of an astronomical telescope in the normal adjustment position is `100`. The distance between the objective and eye piece is `101 cm` . Calculate the focal lengths of objective and eye piece.

Answer 1

Correct Answer - `100 cm` and `1 cm`
`m = - 100, f_(0) + f_(e) = 101 cm, f_(0) = ?, f_(e) = ?`
`m = -(f_(0))/(f_(e)) = - 100 :. F_(0) = 100 f_(e)`
Now `f_(0) + f_(e) = 101`
`100 f_(e) + f_(e) = 101`,
`f_(e) = 1 cm`
`f_(0) = 100 f_(e) = 100 cm`