The magnifying power of an astronomical telescope for normal adjustment is 10 and the length of the telescope is 110 cm. Find the magnifying power of the telescope when the image is formed at the least distance of distinct vision for normal eye.

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Detailed Solution

$m = \frac{f_{0}}{f_{e}} = 10; L = f_{0} + f_{e} = 110 c m$
$f_{0} = 100 c m; f_{e} = 10 c m$
when image is at near point; $V_{e} = - 25 c m$
$\frac{1}{f_{e}} = \frac{1}{V_{e}} - \frac{1}{U_{e}}; \frac{1}{10} = \frac{1}{- 25} - \frac{1}{U_{e}} \frac{1}{10} + \frac{1}{25} = - \frac{1}{U_{e}}; \frac{- 1}{U_{e}} = \frac{5 + 2}{50} = \frac{7}{50}$
-Ve sig indicates object is infront of lens
$m = \frac{f_{0}}{U_{e}} = f_{0} \times \frac{1}{U_{e}}; = 100 \times \frac{7}{50} = 14$

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