A compound microscope is used to enlarge an object kept at a distance $$0.03\ m$$ from it's 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, then by what distance the eye piece of the microscope must be moved to refocus the image
Correct option is D. $$9\ cm$$
If initially the objective ( focal length $$F$$) forms the image at distance $$v_o$$ then $$v_o =\dfrac{u_of_o}{u_o -f_o}=\dfrac{3\times 2}{3-2}=6\ cm$$
Now as in case of lenses in contact
$$\dfrac {1}{F_0}=\dfrac{1}{f_1}+\dfrac {1}{f_2}+\dfrac{1}{f_3}+....=\dfrac{1}{f_1}+\dfrac{1}{F_0'}$$
$$\left\{ where\ \dfrac{1}{F_0'}=\dfrac{1}{f_2}+\dfrac{1}{f_3}+....\right\}$$
So if one of the lens is removed, the focal length of the remaining lens system
$$\dfrac {1}{F_0'}=\dfrac{1}{F_0}-\dfrac{1}{f_1}=\dfrac 12-\dfrac {1}{10}\Rightarrow F_0'=2.5\ cm$$
This lens will form the image of same object at a distance $$v_0'$$ such that
$$v_0'=\dfrac{u_0F_0'}{u_0-F_0'}=\dfrac{3\times 2.5}{(3.-2.5)}=15\ cm$$
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. $$15-6=9\ cm$$.