Question

A compound microscope is used to enlarge an object kept at a distance of 0.03 $m$ from it's objective which consists of several convex lenses in contact and has focal length $0.02m$. If a lens of focal length 0.1 $m$ is removed from the objective, then to refocus the image the eye-piece of the microscope must be moved by (in cm)

Answer 1

If initially the objective (focal length $F_0$) forms the image at distance $v_0$ then,
$v_0=\frac{u_0{f}_0}{u_0-f_0}=\frac{3\times 2}{3-2}=6cm$
So if one of the lens is removed, the focal length of the remaining lens system
$\frac{1}{F{\text{'}}_0}=\frac{1}{F_0}-\frac{1}{f_1}=\frac{1}{2}-\frac{1}{10}\Rightarrow F{\text{'}}_0=2.5cm$
This lens will form the image of same object at a distance $v{\text{'}}_0$such that
$v{\text{'}}_0=\frac{u_{0}F{\text{'}}_0}{u_0-F{\text{'}}_0}=\frac{3\times 2.5}{\left(}=15cm$
So to refocus the image, eye piece must be moved by the same distance through which the image formed by the objective has shifted.
i.e. $15-6=9cm$