Question

A small telescope has an objective lens of focal length $150cm$ and eyepiece of focal length $5cm$. What is the magnifying power of the telescope for viewing distant objects in normal adjustment?
If this telescope is used to view a $100m$ tall tower $3$ km away, what is the height of the image of the tower formed by the objective lens?

Answer 1

Given, Focal length of objective, $f_o=150cm$

Focal length of eye-piece, $f_e=5cm$

Height of tower, $H=100m$

Distance of tower, $u=3km$

Magnification of telescope is given by:

$m=-\frac{f_o}{f_e}(1+\frac{f_e}{D})$

$m=-\frac{150}{5}(1+\frac{5}{25})=-36$

Also, $m=\frac{tan\beta }{tan\alpha }$

$tan\alpha =H/u=100/3000=1/30$

$tan\beta =-\frac{36}{30}$

$tan\beta =\frac{H^{\prime }}{D}$

Height of the image of the tower, $H^{\prime }=-\frac{-36\times 25}{30}=-30cm$