Question

An astronomical telescope has the objective lens and the eyepiece of focal lengths $80\text{cm}$ and $5\text{cm}$ respectively. Find its magnifying power, if the final image is formed at a distance of $50\text{cm}$ from the eyepiece.

Answer 1

Given:
$f_o=80\text{cm}$
$f_e=5\text{cm}$
$v_e=-50\text{cm}$

On using lens formula for eyepiece,

$\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}$

$\Rightarrow \frac{1}{(-50)}-\frac{1}{u_e}=\frac{1}{5}$

$\Rightarrow u_e=-4.54\text{cm}$

The general formula for magnifying power of a telescope is,

$m=\frac{f_o}{|u_e|}=\frac{80}{4.54}=17.6$

Hence, option $\left(B\right)$ is the correct answer.