The linear magnification produced by a telescope having an objective of 50cm and eyepiece of 5cm focal length when used to read a scale 3m away is

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1.2

0.6

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{1}{f_{0}} = \frac{1}{v_{0}} - \frac{1}{u_{0}} \\ \frac{1}{50} = \frac{1}{v_{0}} - \frac{1}{(- 300)} \Rightarrow v_{0} = 60 cm; m_{0} = \frac{v_{0}}{u_{0}} = 0 .2 \\ m = \frac{v_{0}}{u_{0}} [1 + \frac{D}{fe}] = 0 .2 [1 + \frac{25}{5}] = 0 .2 \times 6 = 1 .2 \end{array}$

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