Question

A compound microscope has a magnification of $ 30 $ . The focal length of the eye-piece is $ 5\,cm $ If the final image is formed at the least distance of distinct vision $ (25\,\,cm) $ the magnification produced by the objective is

• A. $ 10 $
• B. $ 7.5 $
• C. $ 5 $
• D. $ 15 $

Answer 1

Answer: (C)

$m=30 f_{e}=5 \,cm$
$m=m_{o} \times m_{e}$
$m=m_{o} \times\left(1+\frac{D}{f_{e}}\right)$
$30=m_{0}\left(1+\frac{25}{5}\right)$
$ \Rightarrow m_{o}=5$