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
Answer & explanation
Correct answer: option 4
If initially the objective ( focal length F) forms the image at distance vo then
$v_0 = \frac{u_0.f_0}{u_0 - f_0} = \frac{3 \times 2}{3 - 2} = 6cm$
$\text{For lenses in contact } \frac{1}{F_0} = \frac{1}{f_1} + \frac{1}{F'_0}$
$\text{If one of the lenses is removed then Focal length of remaining system is } $
$ \frac{1}{F'_0} = \frac{1}{F_0} - \frac{1}{f_1} = \frac{1}{2} - \frac{1}{10}$
$\Rightarrow F'_0 = 2.5 cm$
$\text{This lens will form the image of same object at a distance }$ v0′ $\text{such that}$
$v'_0 = \frac{u_0.F'_0}{u_0 - F'_0} = \frac{3 \times 2.5}{3 - 2.5} = 15 cm$
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.