An astronomical telescope uses an objective lens of focal length 30 cm. In normal adjustment, distance between the two lenses is 32 cm. The magnifying power of telescope is

15

The correct answer is Option (1) → 15

Distance between the objective and eyepiece lens, $d=32cm$

focal length of the objective lens, $f_0=30cm$

$d=f_0+f_e$

$f_e=d-f_0=32-30=2cm$

$⇒f_e=2cm$

Magnifying Power, $M=\frac{f_0}{f_e}=\frac{30}{2}=15$