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Q.

Focal length of objective lens and eye lens in compound microscope is 1cm and 5cm respectively. They are separated by distance 10 cm. Then magnifying power of normal adjustment is (object very close to focus of objective) D=25cm

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a

60

b

50

c

25

d

30

answer is A.

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Detailed Solution

$M = \frac{- v}{u} (\frac{D}{f_{e}}) \Rightarrow M = \frac{- L}{f_{0}} (\frac{D}{f_{e}}) = - \frac{10}{1} \times \frac{25}{5} = - 50$

$(v ≅ L; u ≅ f_{0})$

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