The intensity of illumination on a surface from a source of light varies inversely as the square of the distance of the surface from the source. The effect of moving a piece of paper 3 times as far from the source is to

### Concept & Inverse Square Law The inverse square law states that a specified physical quantity (like illumination intensity) is inversely proportional to the square of the distance from the source. $$I \propto \frac{1}{d^2}$$ ### Step-by-Step Solution * **Initial State**: Let the original distance be $d$ and the original intensity be $I$. Thus, $I = \frac{k}{d^2}$, where $k$ is a constant. * **New State**: The piece of paper is moved $3$ times as far, meaning the new distance $d_{new} = 3d$. * **Calculate New Intensity**: $I_{new} = \frac{k}{(3d)^2}$ $I_{new} = \frac{k}{9d^2}$ $I_{new} = \frac{1}{9} \times \left(\frac{k}{d^2}\right)$ $I_{new} = \frac{I}{9}$ * The new intensity is the original intensity divided by 9. ### Exam Strategy & Shortcut If the distance changes by a factor of $n$, the intensity changes by a factor of $\frac{1}{n^2}$. Here, distance increases by a factor of $3$, so intensity is multiplied by $\frac{1}{3^2} = \frac{1}{9}$, which means dividing by 9. ### Common Pitfall Forgetting to square the distance factor. Many students will mistakenly think that moving it $3$ times as far will divide the intensity by $3$ (Option b). ### Final Answer Therefore, the correct answer is **divide the intensity by 9**.