Question

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

• A. 2.5 cm
• B. 6 cm
• C. 15 cm
• D. 9 cm

Answer 1

The correct option is D 9 cm

If initially the objective (focal length $F_0$) forms the image at distance $v_0$ then $v_0=\frac{u_0{f}_0}{u_0-f_0}=\frac{3\times 2}{3-2}=6cm$
Now as in case of lenses in contact $\frac{1}{F_0}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}+.....=\frac{1}{f_1}+\frac{1}{F_0^{\prime }}\{where\frac{1}{F_0^{\prime }}=\frac{1}{f_2}+\frac{1}{f_3}+...\}$
So if one of the lens is removed, the focal length of the remaining lens system
$\frac{1}{F_0^{\prime }}=\frac{1}{F_0}-\frac{1}{f_1}=\frac{1}{2}-\frac{1}{10}\Rightarrow F_0^{\prime }=2.5cm$
This lens will form the image of same object at a distance $v_0^{\prime }$ such that $v_0^{\prime }=\frac{u_0{F}_0^{\prime }}{u_0-F_0^{\prime }}=\frac{3\times 2.5}{(3-2.5)}=15cm$
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 = 9 cm.