Q: The focal lengths of the objective and the eyepiece of a compound microscope are 2.0 cm and 3.0 cm respectively. The distance between the objective and the eyepiece is 15.0 cm. The final image formed by the eyepiece is at infinity. Find the distance of object and image produced by the objective, from the objective lens.
Sol: As final image is at infinity, the distance of intermediate image from eye lens ue will be given by
$\large \frac{1}{\infty} – \frac{1}{u_e} = \frac{1}{f_e} $
i.e. ue = -fe = -3 cm
and as the distance between the lenses is 15.0 cm, the distance of intermediate image (formed by objective) from the objective will be
v = L-ue = L-fe = 15-3 = 12 cm
and if u is the distance of object from objective,
$\large \frac{1}{12} – \frac{1}{u} = \frac{1}{2} $
i.e. u = – 24 cm
So object is at a distance of 2.4 cm in front of objective.