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Q.

An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eyepiece is 36cm and the final image is formed at infinity. The focal length $f_{0}$ of the objective and the focal length $f_{e}$ of the eyepiece are

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a

$f_{0} = 45 c m and f_{e} = - 9 c m$

b

$f_{0} = 50 c m and f_{e} = 10 c m$

c

$f_{0} = 7.2 c m and f_{e} = 5 c m$

d

$f_{0} = 30 c m and f_{e} = 6 c m$

answer is D.

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Detailed Solution

In an astronomical telescope when the final image is formed at infinity:
Angular magnification
$∴ M = \frac{f_{0}}{f_{e}} = 5 \Rightarrow f_{0} = 5 f_{e}$
And the separation between the objectives and eye-piece.
$L = f_{0} + f_{e} = 36 \Rightarrow 5 f_{e} + f_{e} = 36 or f_{e} = 6 c m ∴ f_{0} = 5 f_{e} = 5 \times 6 = 30 c m$

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