# A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm.

Question:

A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

Solution:

Focal length of the objective lens, fo = 144 cm

Focal length of the eyepiece, fe = 6.0 cm

The magnifying power of the telescope is given as:

$m=\frac{f_{\mathrm{o}}}{f_{\mathrm{e}}}$

$=\frac{144}{6}=24$

The separation between the objective lens and the eyepiece is calculated as:

$f_{\mathrm{o}}+f_{\mathrm{e}}$

$=144+6=150 \mathrm{~cm}$

Hence, the magnifying power of the telescope is 24 and the separation between the objective lens and the eyepiece is 150 cm.