An astronomical telescope consists of an objective of focal length 50 cm and eyepiece of focal length 2 cm is focused on the moon so that the final image is formed at the least distance of distinct vision (25 cm). Assuming angular diameter of moon as \( (1/2)^\circ \) at the objective, the angular size of image is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 13.5^\circ \)
The moon is far away, so the objective forms its real image in its focal plane. This image acts as the object for the eyepiece, and the angular magnification m is the ratio of the angle subtended by the final image to the angle subtended by the moon at the objective.
The eyepiece forms the final image at the least distance of distinct vision, \( v = -25\,\mathrm{cm} \), with \( f_e = +2\,\mathrm{cm} \). The lens formula gives the object distance \( u_e \) for the eyepiece:
\( \displaystyle \frac{1}{u_e} = \frac{1}{v} - \frac{1}{f_e} = -\frac{1}{25} - \frac{1}{2} = -\frac{27}{50} \Rightarrow |u_e| = \frac{50}{27}\,\mathrm{cm} \)
If the moon subtends \( \theta_o \) at the objective, its image has height \( f_o\theta_o \), and the final image subtends this height over the distance \( |u_e| \) at the eyepiece:
\( \displaystyle m = \frac{\theta_i}{\theta_o} = \frac{f_o}{|u_e|} = \frac{50}{50/27} = 27 \)
\( \displaystyle \theta_i = 27\times\frac{1}{2}^\circ = 13.5^\circ \)
This agrees with the standard result \( m = \frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right) = 25\times1.08 = 27 \) for the final image at the near point D.