A) \[10{}^\circ \]
B) \[20{}^\circ \]
C) \[35{}^\circ \]
D) \[50{}^\circ \]
Correct Answer: B
Solution :
Magnifying power of the telescope, \[M=\frac{-{{f}_{o}}}{{{f}_{e}}}\] Here, \[{{f}_{o}}=+400\text{ }cm\] and \[{{f}_{e}}=+10\text{ }cm\] \[\therefore \] \[M=-\left( \frac{400}{10} \right)=-40\] Angle subtended by the moon on the objective, \[\alpha =\left( \frac{3.5\times {{10}^{6}}}{3.8\times {{10}^{8}}} \right)rad\] But, \[M=\frac{\beta }{\alpha }\Rightarrow \beta =M\alpha \] \[=(40)\left[ \frac{3.5\times {{10}^{6}}}{3.8\times {{10}^{8}}} \right]\](numerically) \[=(40)\left[ \frac{3.5\times {{10}^{6}}}{3.8\times {{10}^{8}}} \right]\left[ \frac{180{}^\circ }{\pi } \right]\approx 20{}^\circ \]
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