# A faint sound with an intensity of 10^{-9}\ \mathrm{W/m^2} is measured by an intensity-level...

## Question:

A faint sound with an intensity of {eq}10^{-9}\ \mathrm{W/m^2} {/eq} is measured by an intensity-level meter. What will the reading be in {eq}\mathrm{dB} {/eq}?

## Sound Intensity:

First we need to understand the sound intensity to solve this problem. Suppose {eq}I {/eq} is the sound intensity level in watt per meter and {eq}I_0 {/eq} is a reference intensity, then the sound intensity level in decibels is expressed as: {eq}\displaystyle dB = 10 \log \frac{I}{I_0} {/eq} where {eq}I_0 = 10^{- 12} \ \rm W / m^2 {/eq}.

Become a Study.com member to unlock this answer!

Given:

• The sound intensity is: {eq}I = 10^{- 3} \ \rm W / m^2 {/eq}.

We will compute the sound intensity in decibels. As we know the formula for...