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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» 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` |
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