Question

The diameter of the moon is $ 3.5\times {{10}^{3}} $ km and its distance from the earth is $ 3.8\times {{10}^{5}} $ km seen by a telescope having focal lengths of the objective and eye piece as 4 m and 10 cm respectively, the diameter of image of the moon will be approximately

• A. $ 10{}^\circ $
• B. $ 20{}^\circ $
• C. $ 35{}^\circ $
• D. $ 50{}^\circ $

Answer 1

Answer: (B)

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 $