Question

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 in such a way that parallel rays comes out from the eye lens. If the object subtends an angle $2^{\circ}$ at the objective, the angular width of the image is

• A. $10^{\circ}$
• B. $24^{\circ}$
• C. $50^{\circ}$
• D. $\frac{1}{6}^{\circ}$

Answer 1

Answer: (B)

$24^{\circ}$

Answer 2

Magnification, $m=\frac{f_{ o }}{f_{e}}$ Also $m=\frac{\text { Angle subtended by the image }}{\text { Angle subtended by the object }}$ $\therefore \frac{f_{o}}{f_{e}}=\frac{\alpha}{\beta}$ $\Rightarrow \alpha=\frac{f_{o} \times \beta}{f_{e}}$ $=\frac{60 \times 2}{5}$ $=24^{\circ}$