Q: A microscope consists of two convex 20 cm apart. Where must the object be placed so that the final virtual image is at a distance of 25 cm from the eye ?

(f_o = 2 cm , f_e = 5 cm)

Sol: For the eyepiece, focal length

f = f_{e} = +5 cm; v = v_{e} =-25 cm , u = u_{e} = ? substituting in

$\large \frac{1}{f} = \frac{1}{v} – \frac{1}{u} $

$\large \frac{1}{5} = \frac{1}{-25} – \frac{1}{u_e} $

u_{e} = -25/6 cm

object for the eyepiece is to be at a distance of 25/6 cm to its left.

But v_{o} + u_{e} =20 cm where u_{e} = 25/6 cm

v_{o} = 20 – u_{e} = 20-25/6 = 95/6 cm

For the objective, v=v_{o} = +95/6 cm

f = f_{o} =+2 cm; u = ?

$\large \frac{1}{f} = \frac{1}{v} – \frac{1}{u} $

$\large \frac{1}{2} = \frac{6}{95} – \frac{1}{u} $

u = -190/83 = -2.29 cm

The object is to be placed at a distance of 2.29 cm to the left side of the objective.