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(A) For a point source,the intensity $I$ at a distance $r$ is given by the inverse square law: $I \propto \frac{1}{r^2}$.Let $I_0$ be the intensity at

Vedclass · Accepted Answer

$3$

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(A) For a point source,the intensity $I$ at a distance $r$ is given by the inverse square law: $I \propto \frac{1}{r^2}$.Let $I_0$ be the intensity at $r_0 = 1 	ext{ m}$,where $I_0 = 16 	imes 10^{-8} 	ext{ W m}^{-2}$.The intensity at distance $r$ is $I(r) = \frac{I_0 \cdot r_0^2}{r^2} = \frac{16 	imes 10^{-8} 	imes (1)^2}{r^2} = \frac{16 	imes 10^{-8}}{r^2} 	ext{ W m}^{-2}$.At $r_1 = 2 	ext{ m}$,the intensity is $I_1 = \frac{16 	imes 10^{-8}}{2^2} = \frac{16 	imes 10^{-8}}{4} = 4 	imes 10^{-8} 	ext{ W m}^{-2}$.At $r_2 = 4 	ext{ m}$,the intensity is $I_2 = \frac{16 	imes 10^{-8}}{4^2} = \frac{16 	imes 10^{-8}}{16} = 1 	imes 10^{-8} 	ext{ W m}^{-2}$.The difference in intensity is $|I_1 - I_2| = |4 	imes 10^{-8} - 1 	imes 10^{-8}| = 3 	imes 10^{-8} 	ext{ W m}^{-2}$.Thus,the value is $3$.

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