Question

In a compound microscope, the focal length of the objective lens is $2\text{cm}$ and that of the eyepiece is $5\text{cm}$. When the object is kept at $3\text{cm}$ in front of the objective lens, the final image is formed at a distance of $50\text{cm}$ from the eyepiece. Magnifying power of the compound microscope is -

• A. $14$
• B. $11$
• C. $16$
• D. $18$

Answer 1

The correct option is B $11$

Given :
$f_o=2\text{cm}$
$f_e=5\text{cm}$
$u_o=-3\text{cm}$
$v_e=-50\text{cm}$

Applying lens formula for objective lens,

$\frac{1}{f_o}=\frac{1}{v_o}-\frac{1}{u_o}$

$\Rightarrow \frac{1}{2}=\frac{1}{v_o}-\frac{1}{(-3)}$

$\Rightarrow v_o=6\text{cm}$

The general formula for magnifying power of compound microscope is,

$m=\frac{v_o}{|u_o|}(\frac{D}{|v_e|}+\frac{D}{f_e})$

$\Rightarrow m=\frac{6}{3}\times (\frac{25}{50}+\frac{25}{5})=11$

Hence, option $\left(B\right)$ is the correct answer.