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Q.

An astronomical telescope has objective and eyepiece of focal lengths $40 cm$ and $4 cm$ respectively. To view an object $200 cm$ away from the objective, the lenses must be separated by a distance

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a

50 cm

b

54 cm

c

46 cm

d

37.3 cm

answer is B.

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Detailed Solution

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Given, $f_{0} = 40 cm and f_{e} = 4 cm,$ $u_{0} = - 200 c m$ .
For objective lens,
$\begin{array}{l} \frac{1}{f_{0}} = \frac{1}{v_{0}} - \frac{1}{u_{0}} \\ \Rightarrow \frac{1}{40} = \frac{1}{v_{0}} - \frac{1}{(- 200)} \\ \Rightarrow v_{0} = 50 c m \end{array}$
The length of telescope,
$L = v_{0} + f_{e} = 50 + 4 = 54 cm$

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