The focal lengths of objective lens and eye lens of a Galilean telescope are respectively 30cm and 3.0 cm. Telescope produces virtual, erect image of an object situated far away from it at least distance of distinct vision from the eye lens. In this condition, the magnifying power of the Galilean telescope should be:

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-11.2

-8.8

+11.2

+8.8

answer is D.

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Detailed Solution

Given, Focal length of objective, $f_{0} = 30 c m$
Focal length of eye lens, $f_{e} = 3.0 c m$
Magnifying power, M =?
Magnifying power of the Galilean telescope,
$M_{D} = \frac{f_{0}}{f_{e}} (1 - \frac{f_{e}}{D}) = \frac{30}{3} (1 - \frac{3}{25}) [∵ D = 25 c m]$

$10 \times \frac{22}{25} = 8.8$

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