A compound microscope is used to enlarge an object kept at a distance 0.03m from its objective which consists of several convex lenses in contact and has focal length 0.02m. If a lens of focal length 0.1m is removed from the objective, find out the distance by which the eyepiece of the microscope must be moved to refocus the image.

If initially the objective form the image at distance   $v_1 \frac{1}{v_1}-\frac{1}{-3}=\frac{1}{2} i.e.,v_1=6cm$
Now as in case of lenses in contact $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}+\mathrm{....}or\frac{1}{F}=\frac{1}{f_1}+\frac{1}{F\text{'}}$
 With  $\frac{1}{F\text{'}}=\frac{1}{f_2}+\frac{1}{f_3}+\mathrm{....}$
So if one of the lenses is removed, the focal length of the remaining lens system 
$\frac{1}{F\text{'}}=\frac{1}{F}-\frac{1}{f_1}=\frac{1}{2}-\frac{1}{10} i.e.,F\text{'}=2.5cm$

This lens will form the image of same object at a distance $v_2$ such that 
 $\frac{1}{v_2}-\frac{1}{-3}=\frac{1}{2.5} i.e.,v_2=15cm$
So to refocus the image, eyepiece must be moved by the same distance through which the image formed by the objective has shifted, i.e., 15-6=9 cm away from the objective.