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 $
BETA
