Sound Intensity and the Decibel Scale Revision Notes | A Level AQA Physics
Use intensity level = 10 log(I/I₀), where I₀ = 1.0 × 10⁻¹² W m⁻² is the threshold of hearing. Notes, worked examples and 14 questions answered.
Updated 12 Sept 2026 · Revise Right Team
Sound Intensity and the Decibel Scale is part of Physics of the ear (A-level only) in A Level AQA Physics.
6 min
Reading time
34
Lessons in this topic
A Level AQA
Pathway
Reviewed by Revise Right
Content review
What this lesson covers
Use intensity level = 10 log(I/I₀), where I₀ = 1.0 × 10⁻¹² W m⁻² is the threshold of hearing. Notes, worked examples and 14 questions answered.
Define intensity and state its SI unit, linking it to power per unit area
Apply the intensity level formula using the threshold of hearing to calculate decibel values
Explain the logarithmic nature of the decibel scale and justify why it is used for sound measurement
Compare the dB and dBA scales and evaluate their relative usefulness in different contexts
Key ideas to keep in view
Use this lesson to stay inside Physics of the ear (A-level only) while connecting the detail back to the wider Medical physics (A-level only) topic.
Medical physics (A-level only)
Physics of the ear (A-level only)
A Level AQA Physics
Revision notes
Read the core explanations from Sound Intensity and the Decibel Scale before testing yourself from memory.
1. Why Sound Needs Its Own Scale
A whisper carries a power of around $10^{-12}$ W m$^{-2}$. A jet engine at close range delivers around $10^{2}$ W m$^{-2}$. That is a factor of one hundred trillion between two sounds you might encounter in a single day. No ordinary linear scale handles that range comfortably — which is exactly why physicists use the decibel system.
The decibel scale is logarithmic, meaning each step of 10 dB represents a tenfold change in intensity. This compresses an enormous physical range into a manageable set of numbers between 0 and roughly 140. Understanding why this works — and how to calculate with it — is the core of this lesson.
2. Intensity: The Physical Foundation
Picture a point source of sound radiating equally in all directions. The energy spreads outward over an ever-growing spherical surface. As the radius $r$ doubles, the surface area quadruples ($A = 4\pi r^2$), so the intensity drops to one quarter. This inverse-square relationship means intensity falls off rapidly with distance — a fact that underpins how audiologists and acoustic engineers model hearing damage risk.
At A-level, you need to be precise: intensity is a measure of power per unit area, not energy per unit time per unit area (though those are equivalent). The direction matters too — the area must be perpendicular to the wave's direction of travel. Examiners will penalise vague definitions that omit the perpendicularity condition.
3. The Intensity Level Formula and the Threshold of Hearing
The human ear does not respond linearly to intensity. Doubling the physical intensity of a sound does not produce a sensation of twice the loudness. The ear's response is approximately logarithmic, which is why the decibel scale — also logarithmic — maps so naturally onto perceived loudness.
The threshold of hearing, $I_0 = 1.0 \times 10^{-12}$ W m$^{-2}$, is the internationally agreed reference intensity — the quietest sound a healthy young adult can detect at 1 kHz. When $I = I_0$, the ratio inside the log is exactly 1, and $\log_{10}(1) = 0$, giving an intensity level of 0 dB. This is not silence; it is the boundary of audibility.
Every tenfold increase in intensity adds exactly 10 dB. A sound at $10^{-11}$ W m$^{-2}$ gives 10 dB; at $10^{-10}$ W m$^{-2}$ gives 20 dB; at $10^{-2}$ W m$^{-2}$ gives 100 dB. This pattern is worth memorising as a quick sanity check during calculations.
Worked example: Calculating Intensity Level from a Given Intensity
A sound source produces an intensity of $1.0 \times 10^{-5}$ W m⁻² at a listener's position. Calculate the intensity level in dB. [3 marks]
$$\text{Intensity level} = 10 \log_{10}\left(\frac{I}{I_0}\right)$$ — [M1] Stating the correct formula earns the method mark. Always write it out — examiners award this mark even if your arithmetic later goes wrong.
$$\text{Intensity level} = 10 \log_{10}\left(\frac{1.0 \times 10^{-5}}{1.0 \times 10^{-12}}\right)$$ — [M1] Substituting $I = 1.0 \times 10^{-5}$ W m⁻² and $I_0 = 1.0 \times 10^{-12}$ W m⁻² correctly.
$$\frac{1.0 \times 10^{-5}}{1.0 \times 10^{-12}} = 1.0 \times 10^{7}$$ — Dividing powers of ten: $10^{-5} \div 10^{-12} = 10^{-5-(-12)} = 10^{7}$. The coefficient $1.0 \div 1.0 = 1.0$, so the ratio is $1.0 \times 10^{7}$.
$$\log_{10}(1.0 \times 10^{7}) = 7$$ — $\log_{10}(10^{7}) = 7$ exactly. This is a clean result because the intensity is an exact power of ten multiple of $I_0$.
$$\text{Intensity level} = 10 \times 7 = 70 \text{ dB}$$ — [A1] The final answer with correct unit (dB). A sound at $10^{-5}$ W m⁻² corresponds to 70 dB — roughly the level of a normal conversation.
Answer: Intensity level = 70 dB
Notice the pattern: the exponent in the ratio ($10^7$) multiplied by 10 gives the decibel value directly. This shortcut works whenever the intensity is an exact power-of-ten multiple of $I_0$. For non-integer ratios, you must use your calculator's log function and cannot rely on this shortcut.
4. Working Backwards: Finding Intensity from Intensity Level
Exam questions frequently reverse the calculation — giving you a decibel value and asking for the physical intensity. This requires rearranging the formula, which is where many students lose marks. The key step is isolating the logarithm before applying the inverse operation.
Worked example: Finding Intensity from a Given Intensity Level
A noise measurement instrument records an intensity level of 85 dB. Calculate the intensity of the sound in W m⁻². [3 marks]
$$85 = 10 \log_{10}\left(\frac{I}{I_0}\right)$$
$$\log_{10}\left(\frac{I}{I_0}\right) = \frac{85}{10} = 8.5$$ — [M1] Dividing both sides by 10 isolates the logarithm. This is the step most commonly skipped under exam pressure.
$$\frac{I}{I_0} = 10^{8.5}$$ — If $\log_{10}(x) = 8.5$, then $x = 10^{8.5}$. On a calculator: $10^{8.5} = 3.162 \times 10^{8}$.
$$I = 3.162 \times 10^{8 + (-12)} = 3.162 \times 10^{-4} \approx 3.2 \times 10^{-4} \text{ W m}^{-2}$$ — [A1] Adding the exponents: $8 + (-12) = -4$. Rounding to 2 significant figures matches the precision of the given data. Always include units.
Answer: $I \approx 3.2 \times 10^{-4}$ W m⁻²
5. Relative Intensity Levels Between Two Sounds
A particularly useful application of the decibel formula is comparing two sounds directly, without needing to know their absolute intensities in W m⁻². The difference in intensity levels between two sounds, $L_2 - L_1$, depends only on the ratio $I_2 / I_1$.
This formula follows directly from subtracting one intensity level expression from the other: the $I_0$ terms cancel, leaving the ratio of the two intensities. Recognising this saves time in exam questions that ask how many decibels louder one source is than another.
Worked example: Comparing Two Sound Intensity Levels
Sound A has an intensity of $2.0 \times 10^{-4}$ W m⁻². Sound B has an intensity of $5.0 \times 10^{-3}$ W m⁻². Calculate how many dB louder sound B is than sound A. [3 marks]
$$L_B - L_A = 10 \log_{10}\left(\frac{I_B}{I_A}\right)$$ — [M1] The $I_0$ reference cancels when taking the difference, so we work directly with the ratio of the two intensities.
$$L_B - L_A = 10 \log_{10}\left(\frac{5.0 \times 10^{-3}}{2.0 \times 10^{-4}}\right)$$ — [M1] Substituting $I_B = 5.0 \times 10^{-3}$ W m⁻² and $I_A = 2.0 \times 10^{-4}$ W m⁻².
$$L_B - L_A = 10 \times \log_{10}(25) = 10 \times 1.398 = 13.98 \approx 14 \text{ dB}$$ — [A1] $\log_{10}(25) = 1.398$ (calculator). Multiplying by 10 gives 13.98 dB, rounded to 14 dB to 2 significant figures. Sound B is approximately 14 dB louder than sound A.
Answer: Sound B is approximately 14 dB louder than sound A
6. The dB and dBA Scales: Measurement in Practice
The dB scale as derived from the intensity level formula is physically objective — it reflects the actual power per unit area of the sound wave. However, the human ear does not respond equally to all frequencies. At the same physical intensity, a 1 kHz tone sounds much louder than a 50 Hz tone of identical intensity. The dB scale alone does not capture this perceptual reality.
The dBA scale (A-weighted decibels) addresses this by applying a frequency-dependent correction to the measured intensity level. Frequencies to which the ear is less sensitive — particularly those below about 500 Hz and above about 10 kHz — are attenuated in the dBA reading. The result is a single number that better represents the perceived loudness of a complex sound.
Feature
dB scale
dBA scale
Basis
Physical intensity ratio relative to $I_0$
Frequency-weighted intensity, adjusted for ear sensitivity
Formula used
$10 \log_{10}(I / I_0)$
Same formula, but applied after A-weighting filter
Assessing hearing damage risk and noise nuisance in occupational health
Limitation
Does not reflect perceived loudness at different frequencies
Less useful for pure physics analysis; depends on frequency content of sound
In occupational health and environmental noise regulation, dBA is the standard unit. Legal exposure limits — such as the UK Control of Noise at Work Regulations — are specified in dBA precisely because it correlates better with hearing damage risk than the unweighted dB scale. A sound dominated by low frequencies will read noticeably lower on the dBA scale than on the dB scale.
7. Exam Technique and Common Errors
The decibel calculation is one of the most reliably examined topics in the medical physics option. The marks are predictable, but so are the errors. The guidance below consolidates the most important exam-technique points for this topic.
Returning to where this lesson began: the reason sound needs its own scale is that the ear operates over a range of intensities spanning fourteen orders of magnitude. The decibel system — built on the logarithm of the ratio $I / I_0$ — maps that vast physical range onto a scale from 0 to roughly 140, while the dBA variant adjusts for the ear's frequency-dependent sensitivity to make those numbers clinically and legally meaningful.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-12.
A jet engine at close range delivers around 10² W m⁻², which is 10¹⁴ times the threshold of hearing, giving an intensity level of 10 × 14 = 140 dB. That is the top of the roughly 0–140 dB range the decibel scale covers.
How do you calculate intensity level in decibels?
Use intensity level = 10 log(I/I₀), where I₀ = 1.0 × 10⁻¹² W m⁻² is the threshold of hearing. For I = 1.0 × 10⁻⁵ W m⁻², the ratio is 1.0 × 10⁷, so the intensity level is 10 × 7 = 70 dB. Every tenfold increase in intensity adds exactly 10 dB, and I = I₀ gives 0 dB.
What is the difference between dB and dBA?
The dB scale measures physical intensity relative to I₀, while the dBA scale applies an A-weighting that attenuates frequencies the ear is less sensitive to, particularly those under about 500 Hz and over about 10 kHz. dBA better represents perceived loudness and hearing-damage risk, so UK noise-at-work limits are set in dBA.
What is sound intensity?
Intensity is the power transmitted per unit area perpendicular to the direction of propagation of a wave: I = P/A, measured in W m⁻². For a point source radiating equally in all directions, A = 4πr², so doubling the distance makes the intensity drop to one quarter.
How do you convert decibels to intensity?
Divide the intensity level by 10, raise 10 to that power, then multiply by I₀. For 85 dB, log(I/I₀) = 8.5, so I/I₀ = 10⁸.5 = 3.162 × 10⁸ and I = 3.162 × 10⁸ × 1.0 × 10⁻¹² ≈ 3.2 × 10⁻⁴ W m⁻².
How do you compare the decibel levels of two sounds?
The difference in intensity levels is L₂ - L₁ = 10 log(I₂/I₁), because the I₀ terms cancel. A sound of 5.0 × 10⁻³ W m⁻² compared with one of 2.0 × 10⁻⁴ W m⁻² has an intensity ratio of 25, so it is 10 × log(25) = 10 × 1.398 ≈ 14 dB louder.
What is the formula for intensity?
Intensity is I = P/A, the power transmitted per unit area perpendicular to the direction the wave travels, measured in W m⁻². The area must be perpendicular to the wave's direction of travel, a condition that vague definitions often leave out.
How does sound intensity change with distance?
Sound intensity from a point source follows an inverse-square relationship. The energy spreads over a spherical surface of area 4πr², so when the distance doubles, the area quadruples and the intensity drops to one quarter.
What does 0 dB mean?
An intensity level of 0 dB means the sound's intensity equals the threshold of hearing, 1.0 × 10⁻¹² W m⁻², so the ratio I/I₀ is 1 and log₁₀(1) = 0. It is not silence, but the boundary of audibility for a healthy young adult at 1 kHz.
What does a 10 dB increase mean?
A 10 dB increase means the sound's intensity has risen by a factor of ten. A sound at 10⁻¹¹ W m⁻² gives 10 dB, 10⁻¹⁰ W m⁻² gives 20 dB and 10⁻² W m⁻² gives 100 dB.
What is A-weighting?
A-weighting is a frequency-dependent correction applied to a measured intensity level to produce a dBA reading. Frequencies to which the ear is less sensitive, particularly under about 500 Hz and over about 10 kHz, are attenuated, so the single number better matches perceived loudness.
Why is dBA used for noise at work?
The dBA scale is used for noise at work because it correlates better with hearing damage risk than the unweighted dB scale. Legal exposure limits, such as the UK Control of Noise at Work Regulations, are therefore specified in dBA.
What is the intensity of an 85 dB sound?
An 85 dB sound has an intensity of about 3.2 × 10⁻⁴ W m⁻². Dividing by 10 gives log₁₀(I/I₀) = 8.5, so I/I₀ = 10⁸·⁵ = 3.162 × 10⁸, and multiplying by I₀ = 1.0 × 10⁻¹² W m⁻² gives the intensity.
Does doubling the intensity of a sound double its loudness?
Doubling the physical intensity of a sound does not produce a sensation of twice the loudness. The ear's response is approximately logarithmic, which is why the decibel scale, also logarithmic, maps so naturally onto perceived loudness.