Question

Magnification of a compound microscope is $30$ . The focal length of eye-piece is $5 \, cm$ and the image is formed at a distance of ‘distinct vision of $25 \, cm$ . Magnification of the objective lens is

• A. $6$
• B. $5$
• C. $7.5$
• D. $10$

Answer 1

Answer: (B)

$m=m_{1}\cdot m_{2}=30$
for eye-piece $\frac{1}{v_{1}}-\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \, \, \, \frac{- 1}{25}-\frac{1}{u}=\frac{1}{5}$
$\therefore \, \, \, u=\frac{- 25}{6}$
$\therefore \, \, \, m_{2}=\frac{- 25 \times 6}{- 25}=6$
$\therefore \, m_{1}=5$