A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm.

Focal length of the objective lens, $f_{\mathrm{o}}=140 \mathrm{~cm}$

Focal length of the eyepiece, fe = 5 cm

Least distance of distinct vision, d = 25 cm

(a) When the telescope is in normal adjustment, its magnifying power is given as:

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

$=\frac{140}{5}=28$

(b) When the final image is formed at d,the magnifying power of the telescope is given as:

$\frac{f_{0}}{f_{e}}\left[1+\frac{f_{e}}{d}\right]$

$=\frac{140}{5}\left[1+\frac{5}{25}\right]$

$=28[1+0.2]$

$=28 \times 1.2=33.6$