Question

The magnifying power of an astronomical telescope in normal adjustment is $100.$ The distance between the objective and the eyepiece is $101\text{cm}$. The difference between the focal lengths of lenses will be $\left(\text{in cm}\right)$

Answer 1

In normal adjustment, the object and the final image both are at infinity and the separation between the objective and the eyepiece is,

$l=f_0+f_e.$

Given that,

$f_0+f_e=101\text{cm}---\left(1\right)$

Also, magnification in normal adjustment is given by,

$M=\frac{f_0}{f_e}$

Given that, $M=100$

$\Rightarrow \frac{f_o}{f_e}=100$

$\Rightarrow f_o=100f_e---\left(2\right)$

Substituting $\left(2\right)\text{in}\left(1\right)$

$\Rightarrow 101f_e=101$

$\Rightarrow f_e=1\text{cm}$

$\Rightarrow f_0=100\text{cm}$

$\therefore f_o-f_e=100-1=99\text{cm}$

Hence, $99$ is the correct answer.