Energy And Intensity Of Waves

Question

A point source radiates sound uniformly in all directions, and a detector measures a certain intensity at a distance of 2 m from the source. If the detector is moved to 6 m from the same source, how does the measured intensity change?

RankGuru · Accepted Answer

A point source radiating uniformly spreads its constant power $P$ over the surface of an expanding sphere of area $4\pi r^2$, so the intensity obeys the inverse-square law $I = P/(4\pi r^2)$. Because the total radiated power stays fixed while the spherical area keeps growing, the intensity therefore scales as $1/r^2$. Moving from $r_1 = 2	ext{ m}$ to $r_2 = 6	ext{ m}$ triples the distance, so the ratio of intensities is $\frac{I_2}{I_1} = \left(\frac{r_1}{r_2}ight)^2 = \left(\frac{2}{6}ight)^2 = \frac{1}{9}$. The option of one-third wrongly assumes intensity falls linearly with distance rather than with its square. The option of one-sixth mixes the distance ratio with the area factor incorrectly. The option of no change confuses total radiated power, which is indeed constant, with intensity, which is power per unit area and clearly decreases as the wavefront spreads. This applies the NCERT relation between wave intensity and distance for a spherical wave. As a plausibility check, energy conservation requires the fixed power to thin out over a larger sphere, and since area grows as the square of the radius, the nine-fold drop in intensity is exactly expected.

Energy And Intensity Of Waves

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