A) 15o
B) 20o
C) 30o
D) 35o
Correct Answer: B
Solution :
\[|m|\ =\frac{{{f}_{o}}}{{{f}_{e}}}=\frac{400}{10}=40\] Angle subtented by moon on the objective of telescope \[\propto \ =\frac{3.5\times {{10}^{3}}}{3.8\times {{10}^{3}}}=\frac{3.5}{3.8}\times {{10}^{-2}}rad\] Also \[|m|\ =\frac{\beta }{\alpha }\Rightarrow \]Angular size of final image \[\beta =|m|\ \times \alpha \]\[=40\times \frac{3.5}{3.8}\times {{10}^{-2}}\] = 0.36 rad \[=0.3\times \frac{180}{\pi }\approx {{21}^{o}}\]
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