Question

The magnifying power of the telescope is found to be $9$ and the separation between the lenses is $20\text{cm}$ for relaxed eye. The difference in focal lengths of the two lenses of the telescope $\left(\text{in cm}\right)$ is (Integer only)

Answer 1

The magnification of the telescope in relaxed eye is given by,

$M=\frac{f_o}{f_e}=9$

$\Rightarrow f_o=9f_e---\left(1\right)$

Separation between lenses in relaxed eye is given by,

$d=f_o+f_e=20\text{cm}---\left(2\right)$

Substituting the value of $f_o$ from (1) in (2), we get,

$9f_e+f_e=20$

$\Rightarrow 10f_e=20$

$\Rightarrow f_e=\frac{20}{10}=2\text{cm}$

From $\left(1\right)$,

$f_o=9f_e=9\times 2=18\text{cm}$

The difference in focal lengths of the telescope

$\Delta f=f_o-f_e=18-2=16\text{cm}$

Hence, $16$ is the correct answer.