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Magnification in Thin Lens

A compound mi...

Question

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

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Solution

Given:
Length of compound microscope, $L = 10 cm$
Focal length of objective $f_{0} = 1 cm$ and of eye-piece, $f_{e} = 5 cm$
$u_{0} = f_{e} = 5 cm$
Final image formed at infinity $(\infty) \Rightarrow v_{e} = \infty$
$v_{0} = 10 - 5 = 5$

Using lens formula,
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v_{0}} - \frac{1}{u_{0}} = \frac{1}{f_{0}} \Rightarrow \frac{1}{5} - \frac{1}{u_{0}} = \frac{1}{1} \Rightarrow u_{0} = - \frac{5}{4} cm$
$\Rightarrow \frac{5}{4} = \frac{n}{40}$

$∴ n = \frac{200}{4} = 50 cm$

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