Q. The focal length of objective and eye lens of an astronomical telescope are respectively 2 m and 5 cm. Final image is format at (1) least distance of distinct vision (2) infinity. Magnifying powers in two cases will be
A) $- 48, - 40$
B) $- 40,\ 48$
C) $- 40, + 48$
D) $- 48 + 40$
Explanation
(a) $(1)M = - \frac{f_{0}}{f_{e}}\left( 1 + \frac{f_{e}}{D} \right)$
$M = - \frac{200}{5}\left( 1 + \frac{5}{25} \right)$
$M = - 40\left( 1 + \frac{1}{5} \right) = - 40 \times \frac{6}{5} = - 48$
(2) $M = \frac{f_{o}}{f_{e}} = - \frac{200}{5} = - 40$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.