When dealing with sound intensity and its relationship to distance, it's essential to understand how sound levels change as you move away from a source. In this scenario, we have a sound source emitting sound uniformly in all directions, and we want to determine how the sound intensity level, measured in decibels, changes when the distance from the source is doubled.

To analyze this, we start by defining two distances: r₁ (the initial distance) and r₂ (the distance after moving away, which is twice r₁). The corresponding sound intensities at these distances are I₁ and I₂. The sound intensity level in decibels is calculated using the formula:

β = 10 \log_{10} \left( \frac{I}{I_0} \right)

where I₀ is the reference intensity. To find the change in sound intensity level, we need to calculate the difference between the two levels:

Δβ = β₁ - β₂

Substituting the formula for β, we have:

Δβ = 10 \log_{10} \left( \frac{I₁}{I_0} \right) - 10 \log_{10} \left( \frac{I₂}{I_0} \right)

Factoring out the common term gives us:

Δβ = 10 \left( \log_{10} \left( \frac{I₁}{I₂} \right) \right)

Using the inverse square law, we know that the intensity ratio can be expressed in terms of the distances:

\frac{I₁}{I₂} = \frac{r₂^2}{r₁^2}

Since r₂ is twice r₁, we can substitute:

\frac{I₁}{I₂} = \frac{(2r₁)^2}{r₁^2} = \frac{4r₁^2}{r₁^2} = 4

Substituting this back into our equation for Δβ gives:

Δβ = 10 \log_{10}(4)

Calculating this yields:

Δβ = 10 \times 2 = 20 \text{ dB}

However, since we are interested in the decrease in sound intensity level when moving away, we note that the sound intensity level decreases by 6 dB for each doubling of distance. Therefore, moving from r₁ to r₂ results in a decrease of 6 dB.

This relationship illustrates that sound intensity diminishes non-linearly with distance, specifically following the inverse square law. Each time the distance is doubled, the sound intensity level decreases by 6 dB, highlighting the significant impact of distance on sound perception.