Question

33.2 energy transport and the poynting vector
when the intensity at a point is increased by a factor of four, the distance between the source and the measurement will be the original distance at which the original intensity was observed multiplied by
○ 1/4
○ 1/2
○ 1
○ 2
○ 4

Explanation:

Step1: Recall the intensity - distance relationship

The intensity \(I\) of a wave (assuming spherical wavefronts) 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. So, \(I_1r_1^{2}=I_2r_2^{2}\).

Step2: Substitute the given values

Let the initial intensity be \(I_1\) and distance \(r_1\), and the final intensity \(I_2 = 4I_1\) and distance \(r_2\). Substituting into \(I_1r_1^{2}=I_2r_2^{2}\), we get \(I_1r_1^{2}=4I_1r_2^{2}\).

Step3: Solve for \(r_2\)

Cancel out \(I_1\) (since \(I_1
eq0\)) from both sides of the equation \(I_1r_1^{2}=4I_1r_2^{2}\). We have \(r_1^{2}=4r_2^{2}\), then \(r_2^{2}=\frac{r_1^{2}}{4}\), and \(r_2=\frac{r_1}{2}\).

Answer:

B. \(1/2\)