A compound microscope has an objective and eye-piece as thin lenses of focal lengths 1 cm and 5 cm respectively. The distance between the objective and the eye-piece is 20 cm. The distance at which the object must be placed infront of the objective if the final image is located at 25 cm from the eye-piece, it numerically

Question

Dr. Abubacker · Accepted Answer

(c) For the eye-piece
$v_{e} = - 25\ cm,f_{e} = 5\ cm$
$\frac{1}{- 25} - \frac{1}{u_{e}} = \frac{1}{5}$
Or $\frac{1}{u_{e}} = - \frac{1}{25} - \frac{1}{5}\ \ $or$\ \ \frac{1}{u_{e}} = - \frac{- 1 - 5}{25}$
Or $\ u_{e} = - \frac{25}{6}$
Now , $v_{o} = L - \left| u_{e} \right| = 20 - \frac{25}{6}$
$= \frac{120 - 25}{6}cm = \frac{95}{6}cm$
Now,$\ \frac{1}{95/6} - \frac{1}{u_{o}} = \frac{1}{1}\ $or$\ \frac{1}{u_{o}} = \frac{6}{95} - 1$
Or $u_{0} = - \frac{95}{89}$cm or $\left| u_{o} \right| = \frac{95}{89}\ $cm

Explanation

Reviewed by Dr. Abubacker

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