Question

A compound microscope consists of an objective lens of focal length $2 \,cm$ and eyepiece of focal length $6.25\, cm$ separated by a distance of $15\, cm$. How far (in $cm )$ from the objective lens should an objeçt be placed in order to obtain the final image at the least distance of distinct vision?

Answer 1

For eyepiece:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f} $
$\Rightarrow-\frac{1}{25}-\frac{1}{u}=\frac{1}{6.25}$
$ \Rightarrow u=-5\, cm$
For Objective:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f} $
$\Rightarrow \frac{1}{10}-\frac{1}{u}=\frac{1}{2}$
$ \Rightarrow u=-2.5\, cm$