Question

The minimum intensity of audibility of sound is ${10}^{-12}$ watt/$m^2$. If the intensity of sound is ${10}^{-9}$ watt/$m^2$, then calculate the intensity level of this sound in decibels.

Answer 1

Given,

$I_0={10}^{-12}watt/m^2$

$I={10}^{-9}watt/m^2$

The intensity level of the sound in decibels is given by

$B=10log\left(I/I_0\right)$

$B=10log\left({10}^{-9}/{10}^{-12}\right)=10log\left(1000\right)$

$B=10log\left({10}^3\right)dB$

$B=10\times 3log\left(10\right)$

$B=30\times \left(1\right)=30dB$