A small telescope has an objective lens of focal length 120 cm and an eyepiece of focal length of 5.0 cm. The magnifying power of telescope for viewing distant objects in normal adjustment is

24

The correct answer is Option (2) → 24

Objective focal length: $f_0 = 120 \, \text{cm}$

Eyepiece focal length: $f_e = 5.0 \, \text{cm}$

Magnifying power in normal adjustment:

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

$M = \frac{120}{5} = 24$