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
A) $75\ cm$
B) $60\ cm$
C) $71\ cm$
D) $74\ cm$
Explanation
(c) $f_{o} = 50\ cm,\ f_{e} = 5\ cm,\ D = 25\ cm$ and $u_{o} = 200\ cm$.
Separation between the objective and the eye lens is
$L = \frac{u_{o}f_{o}}{(u_{o} - f_{o})} + \frac{f_{e}D}{(f_{e} + D)} = \frac{200 \times 50}{(200 - 50)} + \frac{5 \times 25}{(5 + 25)} = 71\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.