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Question

(C) Let the height of the satellite be $h = 500 \, km = 500 	imes 10^3 \, m = 5 	imes 10^5 \, m$.The focal length of the camera lens is $f = 50 \, c

Vedclass · Accepted Answer

$10^{12}$

Vedclass · Answer

(C) Let the height of the satellite be $h = 500 \, km = 500 	imes 10^3 \, m = 5 	imes 10^5 \, m$.The focal length of the camera lens is $f = 50 \, cm = 0.5 \, m$.Let $d_1$ be the diameter of the camera screen and $d_2$ be the diameter of the area observed on the Earth's surface.From the geometry of the lens,the angular field of view is the same for both the screen and the terrestrial area,so $	heta_1 = 	heta_2$.Using similar triangles,we have $\frac{d_1}{f} = \frac{d_2}{h}$,which implies $\frac{d_2}{d_1} = \frac{h}{f}$.The ratio of the terrestrial area $A_0$ to the screen area $A$ is given by the square of the ratio of their linear dimensions:$\frac{A_0}{A} = \frac{(\pi d_2^2 / 4)}{(\pi d_1^2 / 4)} = \left( \frac{d_2}{d_1} ight)^2 = \left( \frac{h}{f} ight)^2$.Substituting the values:$\frac{A_0}{A} = \left( \frac{500 	imes 10^3 \, m}{50 	imes 10^{-2} \, m} ight)^2 = \left( \frac{5 	imes 10^5}{5 	imes 10^{-1}} ight)^2 = (10^6)^2 = 10^{12}$.Thus,the terrestrial area observed is $10^{12} A$.

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