3.10.2.2
Intensity and the decibel scale
Medical Physics | AQA A-Level Physics
Key Definition
Intensity level: A logarithmic measure of sound intensity relative to the threshold of hearing. Measured in decibels (dB). $$\text{Intensity level (dB)} = 10 \log_{10}\left(\frac{I}{I_0}\right)$$ where $I_0 = 1 \times 10^{-12}$ W m$^{-2}$.
- IntensityThe power transmitted per unit area perpendicular to the direction of wave travel. $I = P/A$. Measured in W m⁻². For a point source, $I = P/(4\pi r^2)$. is power per unit area: $I = \frac{P}{A}$. For a point source radiating uniformly: $I = \frac{P}{4\pi r^2}$.
- The decibel scale is logarithmic because the ear responds to a huge range of intensities (from $10^{-12}$ to about $1$ W m$^{-2}$). A linear scale would be impractical.
- A doubling of intensity corresponds to an increase of roughly 3 dB (since $10 \log_{10}(2) \approx 3$).
- A tenfold increase in intensity gives an increase of exactly 10 dB.
- Equal loudness curvesGraphs showing the intensity level required at each frequency for the sound to be perceived as equally loud. They are U-shaped, with the minimum around 3000 Hz, showing the ear is most sensitive at this frequency. are U-shaped plots that show the intensity level needed at each frequency to sound equally loud. They have a minimum around 3000 Hz.
- Loudness is measured in phonsThe unit of loudness. The loudness in phons of a sound equals the intensity level in dB of a 1 kHz tone that is perceived as equally loud.. A sound has a loudness of $n$ phons if it sounds as loud as a 1 kHz tone at $n$ dB.
- The dBA scaleA frequency-weighted decibel scale that adjusts readings to match the ear's frequency response. It peaks around 3 kHz and equals the standard dB scale at 1 kHz. adjusts measurements to match the ear's sensitivity curve. It peaks around 3 kHz and is identical to the standard dB scale at 1 kHz.
Worked Example
A sound has an intensity of $5.0 \times 10^{-7}$ W m$^{-2}$. Calculate its intensity level in dB.
Show Solution
1
Write the intensity level equation
$$\text{IL} = 10 \log_{10}\left(\frac{I}{I_0}\right)$$
2
Substitute values
$$\text{IL} = 10 \log_{10}\left(\frac{5.0 \times 10^{-7}}{1.0 \times 10^{-12}}\right) = 10 \log_{10}(5.0 \times 10^5)$$
3
Evaluate
$$\text{IL} = 10 \times 5.70 = 57 \text{ dB}$$
Answer
Intensity level $= 57$ dB.