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

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.