A compound microscope consists of an objective lens of focal length $2$ cm and eye-piece of focal length $6.25$ cm separated by a distance of $15$ cm.length $6.25$ cm separated by a distance Find the object distance of the image:

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Solution

Focal length of object lens is $f_{o} = 2 c m$
Focal length of eye lens $f_{e} = 6.25 c m$
Distance between eyepiece and object lens $D = 15$ cm.
From lens maker formula we have,
$\frac{1}{v_{e}} - \frac{1}{u_{e}} = \frac{1}{f_{e}}$
$\frac{1}{v_{e}} - \frac{1}{f_{e}} = \frac{1}{u_{e}}$
Considering $v_{e} = - 25$ cm (least distance of distinct vision)
$\frac{1}{- 25} - \frac{1}{- 6.25} = \frac{1}{u_{e}}$
$u_{e} = - 5$ cm
We know distance of the object from objective lens $v_{o} = 15 - | u_{e} |$
$v_{o} = 10 c m$

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