The diameter of the moon is 3.5 times 10^3 te | vedclass

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(B) The angular diameter of the moon as seen from the earth is $\alpha = \frac{	ext{diameter}}{	ext{distance}} = \frac{3.5 	imes 10^3 	ext{ km}}{3

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(B) The angular diameter of the moon as seen from the earth is $\alpha = \frac{	ext{diameter}}{	ext{distance}} = \frac{3.5 	imes 10^3 	ext{ km}}{3.8 	imes 10^5 	ext{ km}} = \frac{3.5}{380} 	ext{ rad} \approx 0.00921 	ext{ rad}$.The magnification of the telescope is given by $|m| = \frac{f_o}{f_e} = \frac{400 	ext{ cm}}{10 	ext{ cm}} = 40$.The angle subtended by the final image at the eye is $\beta = |m| 	imes \alpha$.Substituting the values: $\beta = 40 	imes \left( \frac{3.5}{380} ight) 	ext{ rad} = \frac{140}{380} 	ext{ rad} \approx 0.368 	ext{ rad}$.Converting radians to degrees: $\beta (	ext{in degrees}) = 0.368 	imes \frac{180}{\pi} \approx 0.368 	imes 57.3^o \approx 21.1^o$.Rounding to the nearest given option,the value is approximately $20^o$.

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