(a)
Hint: \(\alpha =\frac{f_{0}}{f_{e}}\)

Step: Find the angular magnification of a telescope.
Focal length of the objective lens, ${\mathrm{f}}_{\mathrm{o}}$=15 m = 15 x ${10}^2$ cm and focal length of the eyepiece, ${\mathrm{f}}_{\mathrm{e}}$ = 1.0 cm
The angular magnification of a telescope is given by:

$\alpha =\frac{f_{\circ }}{f_e}=\frac{15\times {10}^2}{1.0}=1500$

(b)
Hint: The angle subtended by the diameter of the moon is equal to the angle subtended by the image of the moon.

Step: Find the diameter of the image of the moon.
Let d’ be the diameter of the image of the moon formed by the objective lens.
The angle subtended by the diameter of the moon is equal to the angle subtended by the image of the moon.
\(tan\theta =\frac{d}{r_{0}}=\frac{d^{'}}{f_{0}}\\\Rightarrow \frac{3.48\times10^{6}}{3.8\times10^{8}}=\frac{d^{'}}{15}\\\therefore d^{'}=13.74 cm\)
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