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(D) The illumination $I$ on the screen is inversely proportional to the square of the distance $r$ between the projector and the screen,given by $I =

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Decreased by $2\%$

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(D) The illumination $I$ on the screen is inversely proportional to the square of the distance $r$ between the projector and the screen,given by $I = \frac{L}{r^2}$,where $L$ is the luminous intensity of the projector.Taking the natural logarithm and differentiating,we get $\ln(I) = \ln(L) - 2\ln(r)$.Differentiating both sides,$\frac{dI}{I} = -2 \frac{dr}{r}$.Given that the distance $r$ is increased by $1\%$,we have $\frac{dr}{r} = 0.01$.Therefore,the percentage change in illumination is $\frac{dI}{I} 	imes 100 = -2 	imes (\frac{dr}{r} 	imes 100) = -2 	imes 1\% = -2\%$.The negative sign indicates a decrease. Thus,the illumination decreases by $2\%$.

In a movie hall,the distance between the projector and the screen is increased by $1\%$. The illumination on the screen is:

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