Question:
A compound microscope consists of an objective lens of focal length $1 \mathrm{~cm}$ and an eye piece of focal length $5 \mathrm{~cm}$ with a separation of $10 \mathrm{~cm}$.
The distance between an object and the objective lens, at which the strain on the eye is minimum is $\frac{\mathrm{n}}{40} \mathrm{~cm}$. The value of $\mathrm{n}$ is
Solution:
Final image at $\infty$
$\Rightarrow$ obj. for eye piece at $5 \mathrm{~cm}$
$\Rightarrow$ image for objective at $5 \mathrm{~cm}$
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{5}+\frac{1}{x}=1$
$\frac{1}{x}=1-\frac{1}{5}=\frac{4}{5} \Rightarrow x=\frac{5}{4}$