Let d be thediameter ofmoon᾿simage formed by the objective lens.
Therefore, Angle subtended by the moon at the objective lens
\(\alpha =\frac{\text{diameter}\ \text{of}\ \text{moon}}{Radius\ \text{of}\ \text{lunar}\ \text{orbit}} = \frac{3.48\times10^6}{3.8\times10^8}\ ..(1)\)
Similarly, the angle subtended by moon᾿s image (formed by the objective) at the objective
\(\alpha = \frac{\text{diameter} \ \text{of}\ \text{moon's}\ \text{image}}{f_0} = \frac{d}{15}\ ..(2)\)
Comparing equations (1) and (2) we have
\(\frac{d}{15} = \frac{3.48\times10^6}{3.8\times10^8}\)
\(d = \frac{3.48\times10^6}{3.8\times10^8}\times 15 = 0.137m = 13.7cm\)
