Optics Question 129
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
Options:
A) 2.5 cm
B) 6 cm
C) 15 cm
D) 9 cm
Show Answer
Answer:
Correct Answer: D
Solution:
If initially the objective (focal length Fo) forms the image at distance vo then $ v _{o}=\frac{u _{o}f _{o}}{u _{o}-f _{o}}=\frac{3\times 2}{3-2}=6cm $
Now as in case of lenses in contact $ \frac{1}{F _{o}}=\frac{1}{f _{1}}+\frac{1}{f _{2}}+\frac{1}{f _{3}}+…..=\frac{1}{f _{1}}+\frac{1}{F _{o}’} $
$ \frac{1}{v}-\frac{2}{(-15)}=\frac{(1-2)}{10} $
So if one of the lens is removed, the focal length of the remaining lens system $ \frac{1}{{{{{F}’}} _{o}}}=\frac{1}{F _{0}}-\frac{1}{f _{1}}=\frac{1}{2}-\frac{1}{10} $
Therefore $ F_{o}’ = 2.5,\text{cm} $
This lens will form the image of same object at a distance $ {{{v}’} _{o}} $ such that $ {{{v}’} _{o}}=\frac{u _{o}{{{{F}’}} _{o}}}{u _{o}-{{{{F}’}} _{o}}}=\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.
BETA
