Q. The diameter of moon is 3.5$\times 10^{3}\ $km and its distance from the earth is $3.8 \times 10^{5}$ km. The focal length of the objective and eye-piece are 4 m and 10 cm respectively. The diameter of the image of the moon will be approximately
A) $2{^\circ}$
B) $21{^\circ}$
C) $40{^\circ}$
D) $50{^\circ}$
Explanation
(b) $\alpha = \frac{3.5 \times 10^{3}}{3.8 \times 10^{5}}rad = \frac{3.5}{3.8 \times 100} \times \frac{180{^\circ}}{\pi}$
$= \frac{3.5 \times 180 \times 7{^\circ}}{38 \times 100 \times 22}$
Also, $\ M = \frac{f_{o}}{f_{e}} = \frac{400}{10} = 40$
$\beta = \frac{40 \times 35 \times 180 \times 7{^\circ}}{35 \times 100 \times 22} = 21.1{^\circ} \approx 21{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.