$(a)$ $A$ giant refracting telescope at an observatory has an objective lens of focal length $15 \; m$. If an eyepiece of focal length $1.0 \; cm$ is used,what is the angular magnification of the telescope?
$(b)$ If this telescope is used to view the moon,what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is $3.48 \times 10^{6} \; m$,and the radius of the lunar orbit is $3.8 \times 10^{8} \; m$.

(N/A) Given:
Focal length of the objective lens,$f_{o} = 15 \; m = 1500 \; cm$.
Focal length of the eyepiece,$f_{e} = 1.0 \; cm$.
$(a)$ The angular magnification $m$ of a telescope is given by:
$m = \frac{f_{o}}{f_{e}} = \frac{1500 \; cm}{1.0 \; cm} = 1500$.
Hence,the angular magnification of the telescope is $1500$.
$(b)$ Diameter of the moon,$d = 3.48 \times 10^{6} \; m$.
Radius of the lunar orbit,$r = 3.8 \times 10^{8} \; m$.
Let $d'$ be the diameter of the image formed by the objective lens. The angle subtended by the moon at the objective is equal to the angle subtended by the image at the objective.
$\frac{d}{r} = \frac{d'}{f_{o}}$
$d' = \frac{d \times f_{o}}{r} = \frac{3.48 \times 10^{6} \; m \times 15 \; m}{3.8 \times 10^{8} \; m}$
$d' = \frac{3.48 \times 15}{3.8} \times 10^{-2} \; m \approx 13.74 \times 10^{-2} \; m = 13.74 \; cm$.
Thus,the diameter of the image of the moon formed by the objective lens is $13.74 \; cm$.