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 eyepiece of the microscope must be moved to refocus the image

A compound microscope is used to enlarge an object kept at a dist…

MCQ

A compound microscope is used to enlarge an object kept at a distance $0.03 \mathrm{~m}$ from it's objective which consists of several convex lenses in contact and has focal length $0.02 \mathrm{~m}$. If a lens of focal length $0.1 \mathrm{~m}$ is removed from the objective, then by what distance the eyepiece of the microscope must be moved to refocus the image

A
$2.5 \mathrm{~cm}$
B
$6 \mathrm{~cm}$
C
$15 \mathrm{~cm}$
✓
$9 \mathrm{~cm}$

✓

Answer

Correct option: D.

$9 \mathrm{~cm}$

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