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

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(B) The angle subtended by the moon at the objective lens is $\alpha = \frac{	ext{diameter}}{	ext{distance}} = \frac{3.5 	imes 10^3}{3.8 	imes 10^

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(B) The angle subtended by the moon at the objective lens is $\alpha = \frac{	ext{diameter}}{	ext{distance}} = \frac{3.5 	imes 10^3}{3.8 	imes 10^5} \, rad$. The magnification of the telescope is given by $m = \frac{f_o}{f_e} = \frac{\beta}{\alpha}$,where $\beta$ is the angle subtended by the image at the eye. Given $f_o = 4 \, m = 400 \, cm$ and $f_e = 10 \, cm$,we have $m = \frac{400}{10} = 40$. Thus,$\beta = 40 	imes \alpha = 40 	imes \frac{3.5 	imes 10^3}{3.8 	imes 10^5} \, rad$. Converting the angle to degrees: $\beta = 40 	imes \frac{3.5}{380} 	imes \frac{180}{\pi} \approx 21.1^\circ$. Rounding to the nearest provided option,the answer is $20^\circ$.

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