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Q.

The length of the compound microscope is 14cm and the magnifying power is 25 for relaxed eye. If the focal length of eye lens is 5 cm, then the object distance for objective lens will be

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a

1.8 cm

b

1.5 cm

c

2.1 cm

d

2.4 cm

answer is A.

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Detailed Solution

$L_{\infty} = v_{0} + f_{e} \Rightarrow 14 = v_{0} + 5 \Rightarrow v_{0} = 9 cm$
Magnifying power of microscope for relaxed eye
$m = \frac{v_{0}}{u_{0}} \cdot \frac{D}{f_{e}} or 25 = \frac{9}{u_{0}} \cdot \frac{25}{5} or u_{0} = \frac{9}{5} = 1.8 cm$

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