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___________
(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, $\frac{5}{4}=\frac{N}{40}$
$\therefore N=\frac{200}{4}=50 \mathrm{~cm}$
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