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

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(A) 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}{1

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$21$

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(A) 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 angle subtended by the moon at the objective is $\alpha = \frac{	ext{Diameter}}{	ext{Distance}} = \frac{3.5 	imes 10^3}{3.8 	imes 10^5} \approx 0.00921 \, 	ext{radians}$.The angular diameter of the image is $\beta = |M| 	imes \alpha = 40 	imes 0.00921 = 0.3684 \, 	ext{radians}$.To convert radians to degrees, multiply by $\frac{180}{\pi}$:$\beta = 0.3684 	imes \frac{180}{3.14159} \approx 21.1^\circ$.Rounding to the nearest integer, the angular diameter is $21^\circ$.

The diameter of the moon is $3.5 \times 10^{3} \, km$ and its distance from the Earth is $3.8 \times 10^{5} \, km$. It is viewed by a telescope with objective and eyepiece focal lengths of $4 \, m$ and $10 \, cm$ respectively. What is the angular diameter of the image of the moon in degrees?

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