Q. An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eye-piece is 36 cm and the final image is formed at infinity. The focal length $f_{0}$ of the objective and the focal length $f_{e}$ of the eye-piece are
A) $f_{0} = 45\ cm\ and\ f_{e} = - 9\ cm$
B) $f_{0} = - 7.2\ cm\ and\ f_{e} = 5\ cm$
C) $f_{0} = 50\ cm\ and\ f_{e} = 10\ cm$
D) $f_{0} = 30\ cm\ and\ f_{e} = 6\ cm$
Explanation
(d) $In\ this\ case\ |m|\ = \ \frac{f_{0}}{f_{e}}\ = \ 5\ \ \ .........(i)$
And length of telescope
$= \ f_{0}\ + \ f_{e}\ = \ 36\ \ \ \ \ \ \ \ \ \ \ ...........(ii)$
Solving eqs (i) and (ii), we get
$f_{e} = 6\ cm,\ f_{0}\ \ = \ 30\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.