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 = fe = +5 cm; v = ve =-25 cm , u = ue = ? substituting in
$\large \frac{1}{f} = \frac{1}{v} – \frac{1}{u} $
$\large \frac{1}{5} = \frac{1}{-25} – \frac{1}{u_e} $
ue = -25/6 cm
object for the eyepiece is to be at a distance of 25/6 cm to its left.
But vo + ue =20 cm where ue = 25/6 cm
vo = 20 – ue = 20-25/6 = 95/6 cm
For the objective, v=vo = +95/6 cm
f = fo =+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.