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}}=2.5cm $

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.



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