A compound microscope consists of an objective lens of focal length 1 cm
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 $\frac{n}{40} \mathrm{~cm}$.

The value of $n$ is___________

Solution:

(50)

Given : Length of compound microscope, $L=10 \mathrm{~cm}$

Focal length of objective $f_{0}=1 \mathrm{~cm}$ and of eye-piece, $f_{e}=5$

$\mathrm{cm}$

$u_{0}=f_{e}=5 \mathrm{~cm}$

Final image formed at infinity $(\infty), v_{e}=\infty$

$v_{0}=10-5=5$

Using lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}} \Rightarrow \frac{1}{5}-\frac{1}{u_{0}}=\frac{1}{1} \Rightarrow u_{0}=-\frac{5}{4} \mathrm{~cm}$

or, $\frac{5}{4}=\frac{N}{40}$

$\therefore N=\frac{200}{4}=50 \mathrm{~cm}$