Q: The focal length of the objective of an astronomical telescope is 75 cm and that of the eye-piece is 5 cm. If the final image is formed at the least distance of distinct vision from the eye, calculate the magnifying power of the telescope.

Solution : Here , f_{0} = 75 cm ; f_{e} = 5 cm

We know, D = 25 cm

$\large M = – \frac{f_o}{f_e}(1 + \frac{f_e}{D}) $

$\large M = – \frac{75}{5}(1 + \frac{5}{25}) $

M = = – 15 × 1.2

M = – 18