Question

In a compound microscope, the objective lens and eyepiece have a focal length of $0.95\,cm$ and $5\,cm$ respectively and are kept at a distance of $20\,cm$ . If the last image is formed at a distance of $25\,cm$ from the eyepiece, calculate magnifying power.

• A. $94$
• B. $84$
• C. $75$
• D. $88$

Answer 1

Answer: (A)

$f_{0}=0.95\,\,m=\frac{V_{0}}{u_{0}}\left(1 + \frac{D}{f_{e}}\right)$
$f_{e}=5\,cm \,u_{e}=\frac{25}{6}$
For eyepiece
$m_{e}=1+\frac{D}{f_{e}}=\frac{v_{e}}{u_{e}},u_{e}=\frac{v_{e}}{6}$
$v_{0}=20-\frac{25}{6}=\frac{95}{6}$
$m_{0}=\frac{v_{o}}{u_{o}}=\frac{v_{o}}{f_{o}}=\frac{\frac{95}{6}}{0.95}\approx\frac{94}{6}$
$m_{n e t}=\frac{94}{6}\times 6=94$