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Learn7408Paper 3B · Option: Medical physicsPhysics of the earHearing and intensity level
Physics of the ear · Subtopic 1 of 1

Hearing and Intensity Level

The outer ear collects sound; the eardrum and ossicles (middle ear) transmit it to the oval window, increasing the pressure about 20 times; the cochlea converts vibrations into nerve signals. Hearing is roughly logarithmic, so loudness is measured by intensity level =10log⁡II0=10\log\frac I{I_0} dB, with I0=1.0×10−12I_0=1.0\times10^{-12} W m⁻². The dBA scale weights frequencies to match the ear's response.

AQA 7408 Paper 3B≈ 35 minRevision 1 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Describe the parts of the ear and how sound reaches the cochlea.
  • Use intensity level =10log⁡II0=10\log\frac I{I_0}.
  • Interpret equal-loudness curves and hearing-loss graphs; explain the dBA scale.
Pause and think

Ten people talking sounds louder than one, but not ten times louder. Why?

Section 2 of 14

Core idea

Start hereFrom whisper to jet engine

The loudest sound you can bear carries about a million million times more power per square metre than the faintest you can hear. To cope, the ear responds to ratios rather than differences, which is why we measure sound in decibels.

The big idea

IL=10log⁡II0IL=10\log\frac I{I_0} dB. Each ×10 in intensity adds 1010 dB. The middle ear raises pressure about 20 times to get sound into the fluid-filled cochlea.

1. How the ear works

Pressure gain of the middle ear
  1. Fout=1.5FinF_{\text{out}}=1.5F_{\text{in}}
    The ossicles act as a lever.
  2. Awindow=Adrum15A_{\text{window}}=\frac{A_{\text{drum}}}{15}
    The oval window is much smaller.
  3. poutpin=1.5×15≈22\frac{p_{\text{out}}}{p_{\text{in}}}=1.5\times15\approx22
    p=FAp=\frac FA.
Without this, most sound would reflect off the fluid in the cochlea.

2. Decibels

Worked exampleTwo machines

Each machine gives 8080 dB.

Goal
Level with both.
Strategy
Intensities add.
  1. I1=10−12×108=10−4 W m−2I_1=10^{-12}\times10^8=10^{-4}\text{ W m}^{-2}
    One machine.
  2. I=2×10−4I=2\times10^{-4}
    Add intensities.
  3. IL=10log⁡(2×108)=83 dBIL=10\log(2\times10^8)=83\text{ dB}
    Back to decibels.
Answer8383 dB.
✓ CheckDoubling adds 10log⁡2=310\log2=3 dB ✓.
Path of sound

Sound travels through air, bone and fluid.

Key takeawayEqual loudness The ear is most sensitive around 33 kHz; low and very high frequencies need more intensity to sound equally loud, hence the dBA weighting.
Common misconceptionAdding decibels Never add dB values directly; convert to intensities first.
Pause and think

Why do rock musicians often lose hearing near 44 kHz?

Summary

  • Outer, middle, inner ear.
  • IL=10log⁡II0IL=10\log\frac I{I_0}.
  • dBA matches ear response.
Key words
Intensity level
Loudness on a log scale, in dB.
Threshold of hearing
I0=10−12I_0=10^{-12} W m⁻² at 1 kHz.
Ossicles
Middle-ear bones that transmit vibration.
HistoryBell and the bel. The bel is named after Alexander Graham Bell; a decibel is a tenth of a bel.
Fun factTiny bones. The stirrup (stapes) is the smallest bone in your body, about 33 mm long.
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