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)

$5$

Answer 2

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