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(D) The intensity $I$ of a point source is inversely proportional to the square of the distance $r$ from the source: $I \propto \frac{1}{r^2}$.Given t

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Decrease by $4\%$.

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(D) The intensity $I$ of a point source is inversely proportional to the square of the distance $r$ from the source: $I \propto \frac{1}{r^2}$.Given that the distance $r$ increases by $2\%$,the new distance $r_2 = r_1 + 0.02r_1 = 1.02r_1$.The ratio of intensities is given by $\frac{I_2}{I_1} = \left( \frac{r_1}{r_2} ight)^2 = \left( \frac{r_1}{1.02r_1} ight)^2 = (1.02)^{-2}$.Using the binomial approximation $(1+x)^n \approx 1+nx$ for small $x$,where $x = 0.02$ and $n = -2$:$I_2 = I_1(1 + 0.02)^{-2} \approx I_1(1 - 2 	imes 0.02) = I_1(1 - 0.04)$.$I_2 = I_1 - 0.04I_1$.This represents a decrease of $4\%$ in intensity.

If the distance between the source and the screen is increased by $2\%$,what will be the effect on the intensity on the screen?

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