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The magnifying power of a telescope with tube length 60 cm is 5. What is the focal length of its objective lens?

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$30 c m$

$40 c m$

$50 c m$

$20 c m$

answer is C.

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Detailed Solution

$L = f_{0} + f_{e} = 60 c m$

$M = \frac{f_{0}}{f_{e}} = 5; f_{0} = 5 f_{e} \Rightarrow 5 f_{e} + f_{e} = 60 c m \Rightarrow f_{e} = 10 c m$

$f_{0} = 50 c m$

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