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Q. The focal length of the objective and the eyepiece of a microscope are $4 \,mm$ and $25 \,mm$ respectively. If the final image is formed at infinity and the length of the tube is $16 \,cm$, then the magnifying power of microscope will be

Ray Optics and Optical Instruments

Solution:

When final image is formed at $\infty$, then
$M=\frac{v_{o}}{u_{o}}\left(\frac{d}{f_{e}}\right)=\frac{v_{o}}{f_{o}}\left(\frac{d}{f_{e}}\right)$
Now, $v_{o}=16-f_{ e }=16-2.5=13.5 \,cm$
$\therefore M=\frac{13.5}{-0.4} \times \frac{25}{2.5}=-337.5$