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, $f_o=+400cm$
and $f_e=+10cm$
$\therefore$ $M=-\left(\frac{400}{10}\right)=-40$
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)
$=(40)\left[\frac{3.5\times {10}^6}{3.8\times {10}^8}\right]\left[\frac{180{}^{\circ }}{\pi }\right]\approx 20{}^{\circ }$