Question

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

• A. ${15}^o$
• B. ${20}^o$
• C. ${30}^o$
• D. ${35}^o$

Answer 1

Answer: (B)

${20}^o$

$|m|=\frac{f_o}{f_e}=\frac{400}{10}=40$
Angle subtended by moon on the objective of telescope
$\propto =\frac{3.5\times {10}^3}{3.8\times {10}^3}=\frac{3.5}{3.8}\times {10}^{-2}ard$
Also $|m|=\frac{\beta }{\alpha }\Rightarrow$ Angular size of final image
$\beta =|m|\times \alpha =40\times \frac{3.5}{3.8}\times {10}^{-2}=0.36rad$
$=0.3\times \frac{180}{\pi }\approx {20}^o$