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Moving Object, Stationary Lens

An astronomic...

Question

An astronomical telescope set for normal adjustment has a magnifying power $10$ . If the focal length of the objective is $1.20 m$ , what is the focal length of the eye-piece?

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Solution

magnifying power $m = \frac{f_{0}}{f_{e}}$
this gives , $f_{e} = \frac{f_{0}}{m} = \frac{1.20}{10} = 0.12 m = 12 c m$

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