A compound microscope consists of an objective of focal length 1 cm and an eyepiece of focal length 5 cm. An object is placed at a distance of 0.5 cm from the objective. What should be the separation between the lenses so that the microscope projects an inverted real image of the object on a screen 30 cm behind the eyepiece?

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Solution

Here, $f_{0}$ = 1cm, $f_{e}$ = 5 cm,

$μ_{0}$ = 0.5 cm, $v_{e}$ = 30 cm

$∴ \frac{1}{v_{0}} - \frac{1}{u_{0}} = \frac{1}{f_{0}}$

$\Rightarrow \frac{1}{v_{0}} + \frac{1}{0.5} = \frac{1}{1}$

$\Rightarrow \frac{1}{v_{0}} = 1 - \frac{10}{0.5} = - 1$

$\Rightarrow v_{0} = - 1 c m$

Again, $\frac{1}{v_{e}} - \frac{1}{u_{e}} = \frac{1}{f_{e}}$

$\frac{1}{30} - \frac{1}{u_{e}} = \frac{1}{5}$

$\Rightarrow - \frac{1}{u_{e}} = \frac{1}{5} - \frac{1}{30} = \frac{6 - 1}{30} = \frac{1}{6}$

$\Rightarrow u_{e} = - 6 c m$

Here, separation between objective and eye piece = 6 - 1 = 5 cm

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Here, f0 = 1cm, fe = 5 cm, μ0 = 0.5 cm, ve = 30 cm ∴ 1v0−1u0=1f0 ⇒ 1v0+10.5=11 ⇒ 1v0=1−100.5=−1 ⇒ v0=−1 cm Again, 1ve−1ue=1fe 130−1ue=15 ⇒ −1ue=15−130=6−130=16 ⇒ ue=−6 cm Here, separation between objective and eye piece = 6 - 1 = 5 cm