Doubling the distance between sound source and a listener will change the intensity, as detected by the listener, by what factor?Select one:a.2b.1/4c.4d.1/2

1. ### Break Down the Problem

We need to determine how the intensity of sound changes when the distance from the sound source to the listener is doubled. The intensity produced by a point source of sound follows the inverse square law.

2. ### Relevant Concepts

According to the inverse square law, the intensity $I$ of sound is inversely proportional to the square of the distance $r$ from the source: $I \propto \frac{1}{r^2}$

3. ### Analysis and Detail

Let's denote the initial distance as $r$ and the initial intensity as $I_1$. When the distance is doubled, the new distance will be $2r$.

Using the inverse square law: $I_1 \propto \frac{1}{r^2}$ $I_2 \propto \frac{1}{(2r)^2} = \frac{1}{4r^2}$

Now, to find the factor of change in intensity, we can express $I_2$ in terms of $I_1$: $\frac{I_2}{I_1} = \frac{\frac{1}{4r^2}}{\frac{1}{r^2}} = \frac{1}{4}$

4. ### Verify and Summarize

Thus, when the distance is doubled, the intensity of the sound decreases by a factor of $\frac{1}{4}$.

Final Answer

The intensity as detected by the listener changes by a factor of 1/4 (option b).