Question

The magnifying power of a telescope is $9$. When it is adjusted for parallel rays, the distance between the objective and the eye-piece is found to be $20\text{cm}$. The focal lengths of the lenses are

Answer 1

Given,

$m=9,L=20\text{cm}$

Let, the focal lengths of the objective and eye-piece are $f_0$ and $f_e$ respectively.

We know,
$m=\frac{f_0}{f_e}=9$

$\Rightarrow f_0=9f_e....\left(i\right)$

and $L=f_0+f_e=20....\left(ii\right)$

Solving equations $\left(i\right)$ and $\left(ii\right)$, we get

$f_0=18\text{cm}$ and $f_e=2\text{cm}$

Hence, $\left(\text{A}\right)$ is the correct answer.

Key Concept: Magnifying power of a telescope is the ration of focal length of the objective to that of the eye piece, when the final image formed at infinity.