A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5.0 cm. The magnifying power of the telescope is

30

The correct answer is Option (1) → 30

For a telescope in normal adjustment (final image at infinity):

Angular magnification $M = \frac{f_{\text{objective}}}{f_{\text{eyepiece}}}$

Given: $f_{\text{objective}} = 150\ \text{cm}$, $f_{\text{eyepiece}} = 5\ \text{cm}$

$M = \frac{150}{5} = 30$

Magnifying power = 30