Question

In a compound microscope, the focal length of the objective lens is $0.5\text{cm}$ and that of the eyepiece is $5\text{cm}$. The intermediate real image of the object is formed at a distance of $15.5\text{cm}$ from the objective lens. If the final image is formed at a distance of $25\text{cm}$ from the eyepiece, what is the magnifying power of the compound microscope?

• A. $179$
• B. $120$
• C. $140$
• D. $160$

Answer 1

The correct option is A $179$

Given:
$f_o=0.5\text{cm}$
$f_e=5\text{cm}$
$v_o=15.5\text{cm}$
$v_e=-25\text{cm}$

For objective lens :

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

$\Rightarrow \frac{1}{0.5}=\frac{1}{15.5}-\frac{1}{u_o}$

$\Rightarrow u_o=-0.52\text{cm}$

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

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

$\Rightarrow m=\frac{15.5}{0.52}(1+\frac{25}{5})=179$

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