In compound microscope, the focal length of objective lens is 1.2 cm and focal length of eye piece is 3.0 cm. When object is kept at 1.25 cm in front of objective, final image is formed at infinity. Magnifying power of the compound microscope should be

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150

100

400

200

answer is A.

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Detailed Solution

For normal adjustment (final image at $\infty$ ) the magnifying power of microscope is given as
$M = | \frac{V_{0}}{u_{0}} . \frac{D}{f_{e}} | .... (i)$
By using lens formula for objective lens, we have
$\frac{1}{f_{0}} = \frac{1}{v_{0}} - \frac{1}{u_{0}}$
$\Rightarrow \frac{1}{1.2} = \frac{1}{v_{0}} + \frac{1}{1.25}$

$\Rightarrow \frac{1}{v_{0}} = \frac{1}{1.2} - \frac{1}{1.25} = \frac{1}{30}$

$\Rightarrow v_{0} = 30 c m$
From (i) , we have
$M = \frac{30}{1.25} \times \frac{25}{3} = 200$

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