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  4. 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

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Q. 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

Ray Optics and Optical Instruments

Solution:

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