Physics - Geometrical Optics Question with Solution | TestHub

Given focal length of the objective, $f_o=2\text{m}$.

Length of the telescope under normal adjustment, $L=2.10\text{m}$.

For normal adjustment, the length of the telescope is given by $L=f_o+f_e$.

Substituting the given values:

$2.10=2+f_e$

$f_e=0.10\text{m}$

The magnifying power of the telescope under normal adjustment is given by $M=\frac{f_o}{f_e}$.

$M=\frac{2}{0.10}=20$

Answer: $20$