Question

33.2 energy transport and the poynting vector
when the distance between a point source of light and a light meter is reduced from 6.0 m to 2.0 m, the intensity of illumination at the light meter will be the original value multiplied by
○ 1/3
○ 3
○ 1
○ 1/9
○ 9

Explanation:

Step1: Recall the formula for intensity

The intensity \(I\) of light from a point source is inversely proportional to the square of the distance \(r\) from the source, i.e., \(I=\frac{P}{4\pi r^{2}}\), where \(P\) is the power of the source. Let the initial distance be \(r_1 = 6.0\space m\) and the final distance be \(r_2=2.0\space m\). The initial intensity \(I_1=\frac{P}{4\pi r_{1}^{2}}\) and the final intensity \(I_2=\frac{P}{4\pi r_{2}^{2}}\).

Step2: Find the ratio of intensities

Take the ratio \(\frac{I_2}{I_1}=\frac{\frac{P}{4\pi r_{2}^{2}}}{\frac{P}{4\pi r_{1}^{2}}}\). The \(P\) and \(4\pi\) terms cancel out. So \(\frac{I_2}{I_1}=\frac{r_{1}^{2}}{r_{2}^{2}}\).
Substitute \(r_1 = 6\space m\) and \(r_2 = 2\space m\) into the formula: \(\frac{I_2}{I_1}=\frac{6^{2}}{2^{2}}=\frac{36}{4}=9\)

Answer:

9