$A$ small telescope has an objective lens of focal length $140 \; cm$ and an eyepiece of focal length $5.0 \; cm$. What is the magnifying power of the telescope for viewing distant objects when
$(a)$ the telescope is in normal adjustment (i.e.,when the final image is at infinity)?
$(b)$ the final image is formed at the least distance of distinct vision $(25 \; cm)$?

(N/A) Given:
Focal length of the objective lens,$f_{o} = 140 \; cm$
Focal length of the eyepiece,$f_{e} = 5.0 \; cm$
Least distance of distinct vision,$d = 25 \; cm$
$(a)$ When the telescope is in normal adjustment,the final image is at infinity. The magnifying power $m$ is given by:
$m = -\frac{f_{o}}{f_{e}}$
$m = -\frac{140}{5} = -28$
The magnitude of the magnifying power is $28$.
$(b)$ When the final image is formed at the least distance of distinct vision $d$,the magnifying power $m$ is given by:
$m = -\frac{f_{o}}{f_{e}} \left( 1 + \frac{f_{e}}{d} \right)$
$m = -\frac{140}{5} \left( 1 + \frac{5}{25} \right)$
$m = -28 \times (1 + 0.2)$
$m = -28 \times 1.2 = -33.6$
The magnitude of the magnifying power is $33.6$.