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Magnifying power of an astronomical telescope in near point adjustment is 110% of its magnifying power in normal adjustment. Then find the focal length of objective when the length of the telescope in normal adjustment is 10 cm. (Take least distance of clear vision = 25 cm)

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7.5cm

2.5cm

2cm

8cm

answer is C.

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Detailed Solution

$\frac{f_{0}}{f_{e}} [1 + \frac{f_{e}}{D}] = \frac{110}{100} (- \frac{f_{0}}{f_{e}})$

$\Rightarrow f_{e} = 2.5 cm$
$L = F_{e} + F_{0} (in normal adjustment)$

$10 cm = 2.5 cm + f_{0} \Rightarrow f_{0} = 7.5 cm$

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