Question

The diameter of moon is $3.5 \times 10^{3}km$ and its distance from the earth is $3.8 \times 10^{5}km$. If it is seen through a telescope whose focal length for objective and eye lens are 4 m and 10 cm respectively, then the angle subtended by the moon on the eye will be approximately

• A. 15⁰
• B. 20⁰
• C. 30⁰
• D. 35⁰

Answer 1

Answer: (B)

20⁰

Answer 2

The angle subtended by the moon on the objective of telescope $\alpha = \frac{3.5 \times 10^{3}}{3.8 \times 10^{5}} = \frac{3.5}{3.8} \times 10^{- 2}$rad

Also $m = \frac{f_{o}}{f_{e}} = \frac{\beta}{\alpha}$ ⇒ $\frac{400}{10} = \frac{\beta}{\alpha}$ ⇒ $\beta = 40\alpha$

⇒ $\beta = 40 \times \frac{3.5 \times 10^{3}}{3.8 \times 10^{5}} \times \frac{180}{\pi} = 20{^\circ}$