Q. A simple telescope, consisting of an objective of focal length $60\ cm$ and a single eye lens of focal length $5\ cm$ is focussed on a distant object is such a way that parallel rays come out from the eye lens. If the object subtends an angle $2{^\circ}$ at the objective, the angular width of the image
A) $10{^\circ}$
B) $24{^\circ}$
C) $50{^\circ}$
D) $1/6{^\circ}$
Explanation
(b) Since $m = \frac{f_{o}}{f_{e}}$
Also $m = \frac{Angle\ subtended\ by\ the\ image}{Angle\ subtended\ by\ the\ object}$
$\therefore\frac{f_{o}}{f_{e}} = \frac{\alpha}{\beta} \Rightarrow \alpha = \frac{f_{o} \times \beta}{f_{e}} = \frac{60 \times 2}{5} = 24{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.