Answer to Question #102792 in Optics for sridhar

Question

The focal lengths of objective lens and eye lens of a Gallelian Telescope are respectively 30 cm and 3.0 cm. Telescope produces virtual, erect image of an object situated far away from it at least distance of distinct vision from the eye lens. In this condition, the Magnifying Power of the Gallelian Telescope should be :

Answer 1

In Astronomical Telescope -

$m=\dfrac{f_0}{f_e}\times(1-\dfrac{f_e}{D})$

 where

$f_0$  is focal length of objective

$f_e$ is focal length of eyepiece

substituting the values of $f_0\ \& \ f_e$ in the above formula then we will get

$m=\dfrac{30}{3}\times(1-\dfrac{3}{25})$

$m=8.8$