**What would be the fractional impact on lowering the sound intensity (in W/m^2 if the level of sound intensity (in dB) is reduced by 40 dB by the installation of unique windows with sound-reflecting properties?****What would be the change in the sound intensity level (in dB) if the intensity is reduced by half?**

The aim of this question is to find the impact of **sound intensity **(in $\dfrac{W}{m^2}$) by reducing the **sound intensity level **(in $dB$). The basic concept behind this article is **Sound Intensity** and **Sound Intensity Level**.

**Sound Intensity** is defined as the energy or power that exists in a **sound wave per unit area**. It is a **vector quantity** whose direction is **perpendicular to the surface area**. As **sound intensity** is the power of sound waves, hence, it is represented by the **SI unit** of **Watt per square meter** $(\dfrac{W}{m^2})$ and expressed as follows:

\[Sound\ Intensity\ I=pv\]

Where:

$p$ is the **sound pressure**

$v$ is the **particle velocity**

**Sound intensity level (SIL)** is the ratio of the **loudness** of the given **intensity** of a sound to the **standard intensity**. It is represented by the SI unit of **Decibels** $(dB)$ and expressed as follows:

\[Sound\ Intensity\ Level\ SIL\ (dB)=\ 10\log_{10}{\left(\frac{I}{I_0}\right)}\]

Where:

$I$ is the **sound intensity** of a given sound

$I_0$ is the **reference sound intensity**

$I_0$ **Reference sound intensity** is generally defined as **standard sound level measurement** corresponding to hearing by a human ear having a **standard threshold** at $1000$ $Hz$

\[I_0=\ {10}^{-12}\ \frac{W}{m^2}\]

## Expert Answer

Given that:

\[Sound\ Intensity\ Level\ SIL\ (dB)\ =\ 40\ dB\]

**Part-1 Solution**

We will substitute the value of given $SIL$ and **Reference sound intensity** $I_0$ in equation of $SIL$:

\[Sound\ Intensity\ Level\ SIL\ (dB)\ =\ 10\log_{10}{\left(\frac{I}{I_0}\right)}\]

\[40\ dB\ =\ 10\log_{10}{\left(\frac{I}{{10}^{-12}}\right)}\]

\[\log_{10}{\left(\frac{I}{{10}^{-12}}\right)}\ =\ \frac{40}{10}\ =\ 4\]

By applying** log formula**:

\[\log_a{b=x}\ \Rightarrow\ a^x=b\]

\[\frac{I}{{10}^{-12}}\ =\ {10}^4\]

\[I\ =\ {10}^4\times{10}^{-12}\]

\[I\ =\ {10}^{-8}\ \frac{W}{m^2}\]

**Part-2 Solution**

Given that:

**Intensity** $I$ is **reduced by half**.

\[Intensity\ =\ \frac{1}{2}I\]

We know that:

\[Sound\ Intensity\ Level\ SIL\ (dB)\ =\ 10\log_{10}{\left(\frac{I}{I_0}\right)}\]

Substituting the value of $I$ and $I_0$ in above equation:

\[SIL\ (dB)\ =\ 10\log_{10}{\left(\frac{I}{{2\ timesI}_0}\right)}\]

\[SIL\ (dB)\ =\ 10\log_{10}{\left(\frac{{10}^{-8}}{2\times{10}^{-12}}\right)}\]

\[SIL\ (dB)\ =\ 10\log_{10}{\left(\frac{{10}^4}{2}\right)}\]

\[SIL\ (dB)\ =\ 10\log_{10}{\left(5000\right)}\]

\[SIL\ (dB)\ =\ 36.989\ dB\]

## Numerical Result

If the level of **sound intensity** (in $dB) is reduced by $40$ $dB$, the **sound intensity** will be:

\[I\ =\ {10}^{-8}\ \frac{W}{m^2}\]

If the **intensity** is **reduced by half**, the **sound intensity level** (in $dB$) will be:

\[SIL\ (dB)\ =\ 36.989\ dB\]

## Example

What would be the fractional impact on lowering the **sound intensity** (in $\dfrac{W}{m^2}$) if the **level of sound intensity** (in $dB$) is reduced by $10$ $dB$?

**Solution**

Given that:

\[Sound\ Intensity\ Level\ SIL\ (dB)\ =\ 10\ dB\]

We will substitute the value of given $SIL$ value and **Reference sound intensity** $I_0$ in equation of $SIL$

\[Sound\ Intensity\ Level\ SIL\ (dB)\ =\ 10\log_{10}{\left(\frac{I}{I_0}\right)}\]

\[40\ dB\ =\ 10\log_{10}{\left(\frac{I}{{10}^{-12}}\right)}\]

\[\log_{10}{\left(\frac{I}{{10}^{-12}}\right)}\ =\ \frac{10}{10}\ =\ 1\]

By applying** log formula**:

\[\log_a{b=x}\ \Rightarrow\ a^x=b\]

\[\frac{I}{{10}^{-12}}\ =\ 10\]

\[I\ =\ 10\times{10}^{-12}\]

\[I\ =\ {10}^{-11}\ \frac{W}{m^2}\]