Question

The diameter of the 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 image of the moon on the eye will be approximately

• A. ${15}^{\circ }$
• B. ${20}^{\circ }$
• C. ${30}^{\circ }$
• D. ${35}^{\circ }$

Answer 1

The correct option is B ${20}^{\circ }$

The angle subtended by the moon on the objective of telescope $\alpha =\frac{3.5\times {10}^3}{3.8\times {10}^5}=\frac{3.5}{3.8}\times {10}^{-2}rad$.
Let, the angle subtended by the image of the moon on the eye be $\beta$. Also, the angle subtended by the moon on the earth, as well as the eye will be very nearly the same.
Therefore $m=\frac{f_0}{f_e}=\frac{\beta }{\alpha }\Rightarrow \frac{400}{10}=\frac{\beta }{\alpha }\Rightarrow \beta =40\alpha \Rightarrow \beta =40\times \frac{3.5\times {10}^3}{3.8\times {10}^5}\times \frac{180}{\pi }={20}^{\circ }$ (approx).