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Q. A telescope has an objective of focal length $50\, cm$ and an eye-piece of focal length $5\, cm$. The least distance of distinct vision is $25\, cm$. The telescope is focussed for distinct vision on a scale $200\, cm$ away. The separation between the objective and the eye-piece is

ManipalManipal 2009Ray Optics and Optical Instruments

Solution:

$\frac{1}{v_{o}}+\frac{1}{200}=\frac{2}{50}$
$\Rightarrow v_{0}=\frac{200}{3}=66.6\, cm$
$\therefore L=v_{o}+f_{e}=66.6+5 \approx 71\, cm$