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(D) The solar radiation received per unit area (intensity $I$) follows the inverse square law: $I \propto \frac{1}{r^2}$.Let $r_1$ be the distance of

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$1.5 	imes 10^{-11}$

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(D) The solar radiation received per unit area (intensity $I$) follows the inverse square law: $I \propto \frac{1}{r^2}$.Let $r_1$ be the distance of the Sun from the Earth,$r_1 = 1.5 	imes 10^{11} 	ext{ m}$.Let $r_2$ be the distance of Alpha Centauri,$r_2 = 4 	ext{ light-years} = 4 	imes 9.46 	imes 10^{15} 	ext{ m}$.The ratio of the intensity received at the new distance $(I_2)$ to the original intensity $(I_1)$ is given by:$\frac{I_2}{I_1} = \frac{r_1^2}{r_2^2} = \left( \frac{1.5 	imes 10^{11}}{4 	imes 9.46 	imes 10^{15}} ight)^2$.Calculating the value:$\frac{I_2}{I_1} = \left( \frac{1.5}{37.84} 	imes 10^{-4} ight)^2 \approx (0.0396 	imes 10^{-4})^2 \approx (3.96 	imes 10^{-6})^2 \approx 1.57 	imes 10^{-11}$.Thus,the factor by which the radiation decreases is approximately $1.5 	imes 10^{-11}$.

Suppose the Sun was located at the position occupied by the nearest star,say,Alpha Centauri,$4$ light-years away. By what factor would the solar radiation received per second per unit area decrease?

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