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
Answer & explanation
Correct answer: option 1
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