The focal length of the objective and the eyepiece of a telescope are 50cm and 5cm respectively. If the telescope is focussed for distinct vision on a scale distant 2m from its objective, then its magnifying power will be :

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-4

-2

-8

answer is D.

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Detailed Solution

$u_{0} = - 200 cm; \frac{1}{f} = \frac{1}{v_{0}} - \frac{1}{u_{0}} \frac{1}{50} = \frac{1}{v_{0}} - \frac{1}{(- 200)} v_{0} = \frac{200}{3} cm, m_{0} = \frac{- (200 / 3)}{200} = \frac{1}{3}$

$\frac{1}{f_{e}} = \frac{1}{v_{e}} - \frac{1}{u_{e}} \Rightarrow v_{e} = - 25 cm u_{e} = \frac{- 25}{6} cm, f_{e} = 5 cm, m_{e} = \frac{- 25}{(- 25 / 6)} = 6 m = m_{0} m_{e}; m = \frac{- 1}{3} x 6 = - 2$

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