Optics Question 341

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

[Pb. PMT 2002]

Options:

A) 1.8 cm

B) 1.5 cm

C) 2.1 cm

D) 2.4 cm

Show Answer

Answer:

Correct Answer: A

Solution:

$ {L _{\infty }}=v _{o}+f _{e} $

$ \Rightarrow 14=v _{o}+5 $

$ \Rightarrow v _{o}=9\ cm $ Magnifying power of microscope for relaxed eye $ m=\frac{v _{o}}{u _{o}}.\frac{D}{f _{e}} $ or $ 25=\frac{9}{u _{o}}.\frac{25}{5} $ or $ u _{o}=\frac{9}{5}=1.8\ cm $



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