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Q.

The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the objective and eyepiece is 20 cm. The focal length of lenses are :

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a

18 cm, 2 cm

b

15 cm, 5 cm

c

11 cm, 9 cm

d

10 cm, 10 cm

answer is C.

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Detailed Solution

M.P. $= 9 = \frac{f_{0}}{f_{e}}$

$\begin{aligned} \Rightarrow f_{0} = 9 f_{e} \dots (1) \\ f_{0} + f_{e} = 20 \dots (2) \end{aligned}$

on solving

$f_{0} = 18 cm =$ focal length of the objective

$f_{e} = 2 cm =$ focal length of the eyepiece

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