Question

What is the intensity in watts per meter squared of\(85.0\;{\rm{dB}}\)sound?

Short answer

The intensity is \(3.2 \times {10^{ - 4}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\).

Step-by-step solution

Given Data

The intensity level is \(85.0\;{\rm{dB}}\).

Sound intensity and Loudness

The loudness increases with the increase in sound intensity as it is the ratio of the sound intensity to the threshold.

Calculation of the intensity

The intensity level is given by,

\({\rm{dB}} = 10\log \frac{I}{{{{10}^{ - 12}}}}\)

Plugging the values,

\(\begin{align}85 &= 10\log \frac{I}{{{{10}^{ - 12}}}}\\\frac{{85}}{{10}} &= \log \frac{I}{{{{10}^{ - 12}}}}\\{10^{8.5}} &= \frac{I}{{{{10}^{ - 12}}}}\\I &= {10^{8.5}} \times {10^{ - 12}}\\I &= 3.2 \times {10^{ - 4}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\end{align}\)

Therefore, the intensity is \(3.2 \times {10^{ - 4}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\)