Q. An astronomical telescope has a converging eye-piece of focal length 5 cm and objective of focal length 80 cm. When the final image is formed at the least distance of distinct vision (25 cm), the separation between the two lenses is
A) 75.0 cm
B) 80.0 cm
C) 84.2 cm
D) 85.0 cm
Explanation
(c) Separation $= f_{0} + \frac{f_{e}D}{f_{e} + D}$
$= 80 + \frac{5 \times 25}{5 + 25} = 80 + \frac{125}{30}$
$= 84.16\ cm = 84.2\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.