Diameter of the moon is 3.5 times 10^3 , km a | vedclass

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(D) The angular diameter of the moon as seen from the earth is $\alpha = \frac{	ext{Diameter of moon}}{	ext{Distance from earth}} = \frac{3.5 	imes

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(D) The angular diameter of the moon as seen from the earth is $\alpha = \frac{	ext{Diameter of moon}}{	ext{Distance from earth}} = \frac{3.5 	imes 10^3 \, km}{3.8 	imes 10^5 \, km} = \frac{3.5}{380} \approx 0.00921 \, 	ext{radians}$.The magnification of the telescope is given by $M = -\frac{f_o}{f_e}$.Given $f_o = 4 \, m = 400 \, cm$ and $f_e = 10 \, cm$.So, $M = -\frac{400}{10} = -40$.The angular diameter of the image is $\beta = |M| 	imes \alpha$.$\beta = 40 	imes 0.00921 = 0.3684 \, 	ext{radians}$.To convert radians to degrees, multiply by $\frac{180}{\pi}$:$\beta = 0.3684 	imes \frac{180}{3.14159} \approx 21.11^\circ$.Rounding to the nearest integer, the angular diameter is $21^\circ$.

Diameter of the moon is $3.5 \times 10^3 \, km$ and its distance from earth is $3.8 \times 10^5 \, km$. It is seen by a telescope whose objective and eyepiece have focal lengths $4 \, m$ and $10 \, cm$ respectively. What will be the angular diameter of the image of the moon in degrees?

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