Question

A compound microscope consists of an objective lens of focal length $1\text{cm}$ and an eye piece of focal length $5\text{cm}$ with a separation of $10\text{cm}$. The distance between an object and the objective lens, at which the strain on the eye is minimum is $\frac{n}{40}\text{cm}$. The value of $n$ is

Answer 1

Given:
Length of compound microscope, $L=10\text{cm}$
Focal length of objective $f_0=1\text{cm}$ and of eye-piece, $f_e=5\text{cm}$
$u_0=f_e=5\text{cm}$
Final image formed at infinity $\left(\infty \right)\Rightarrow v_e=\infty$
$v_0=10-5=5$

Using lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v_0}-\frac{1}{u_0}=\frac{1}{f_0}\Rightarrow \frac{1}{5}-\frac{1}{u_0}=\frac{1}{1}\Rightarrow u_0=-\frac{5}{4}\text{cm}$
$\Rightarrow \frac{5}{4}=\frac{n}{40}$

$\therefore n=\frac{200}{4}=50\text{cm}$