Question

An astronomical telescope has an angular magnification of 40 times. If the diameter of the objective is 12 cm and the length of the tube is 1m, calculate the focal length of the objective and the diameter of the exit pupil.

Answer 1

In an astronomical telescope with objective lens with focal length $f_o$ and eye piece lens of focal length $f_e$ in the normal mode (where final image is formed at infinite distance for relaxed vision) magnifying power is given by $M = \large\frac{ f_e}{f_o}$ and the tube length (distance between objective and eye piece) is given by 

$L = f_o + f_e$

Given that magnifying power is 40 and L = 1m

$L = f_o + 40f_o=41f_o \to f_o = \frac{L}{41}=\frac{100}{41}=2.44cm$