Question

The focal length of the objective lens of a compound microscope is $0.02\text{m}$ and that of the eyepiece is $0.04\text{m}$. The distance between the objective lens and the eyepiece is $0.20\text{m}$. If the final image is formed at the least distance of distinct vision, then calculate the magnifying power of the compound microscope.

• A. $68$
• B. $50$
• C. $53$
• D. $83$

Answer 1

The correct option is C $53$

For the eyepiece :

$\frac{1}{f_e}=\frac{1}{v_e}-\frac{1}{u_e}$

$\Rightarrow \frac{1}{4}=\frac{1}{(-25)}-\frac{1}{u_e}$

$\Rightarrow u_e=-3.45\text{cm}$

Given, the distance between the objective lens and the eyepiece is $0.20\text{m}$

So, $|v_o|+|u_e|=20\text{cm}$

$\Rightarrow |v_o|=20-3.45=16.55\text{cm}$

For the objective lens :

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

$\Rightarrow \frac{1}{2}=\frac{1}{16.55}-\frac{1}{u_o}$

$\Rightarrow u_o=-2.28\text{cm}$

Now, magnification of the microscope when the final image is formed at the least distance of distinct vision is given by,

$m=\frac{v_o}{|u_o|}(1+\frac{D}{f_e})$

$\Rightarrow m=\frac{16.55}{2.28}(1+\frac{25}{4})\approx 53$

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