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

Sol: If initially the objective forms the image at distance v_{1} ,

$\large \frac{1}{v_1} – \frac{1}{-3} = \frac{1}{2} $

v_{1} = 6 cm

$\large \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} + ….$

$\large \frac{1}{F} = \frac{1}{f_1} + \frac{1}{F’} $

with , $\large \frac{1}{F’} = \frac{1}{f_2} + \frac{1}{f_3}+ …. $

So if one of the lenses is removed, the focal length of the remaining lens system

$\large \frac{1}{F’} = \frac{1}{F} – \frac{1}{f’} $

$\large \frac{1}{F’} = \frac{1}{2} – \frac{1}{10} $

F’ = 25 cm

This lens will form the image of same object at a distance v_{2} such that

$\large \frac{1}{v_2} – \frac{1}{-3} = \frac{1}{2.5} $

i.e. v_{2} = 15 cm ;

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.