Virtual laboratory/Physics/Waves and acoustics/Sound intensity level — decibels

Sound intensity level — decibels

Increase the sound intensity and watch the level in decibels L=10·log(I/I₀). The scale is logarithmic: every tenfold increase in intensity is only +10 dB. That is why a whisper and a concert are separated by a huge difference in intensity, but only a small one — in decibels.

Sound intensity level — decibels
Physics · Waves and acoustics · Sound intensity level (FIZ.FALE.PP.07)
Sound intensity level LL60 dB
Examples:
The sound intensity II increases 10× — the level LL rises by:
level L
60 dB
intensity I
1.0·10⁻⁶ W/m²
ratio I/I₀
×1.0·10⁶
like
conversation
🎯 Challenge: set the level in decibels
streak: 0

I will give you an intensity ratio I/I₀ — use the slider to set the level L that matches it. L = 10·log₁₀(I/I₀).

What just happened?

The intensity rose 10×, but the level rose only by 10 dB. The ear and the decibel scale respond to sound logarithmically — multiplying the intensity turns into adding decibels.

Sound intensity level:

L=10 log⁡10 ⁣II0[dB]L = 10\,\log_{10}\!\dfrac{I}{I_0}\quad[\text{dB}]

— where II is the sound intensity (W/m²), and I0=10−12 W/m2I_0 = 10^{-12}\ \text{W/m}^2 is the threshold of hearing. Conversely: I=I0⋅10L/10I = I_0\cdot 10^{L/10}.

What does it mean in practice?

Every 10× greater intensity is a constant +10 dB, regardless of the starting point:

10×I⇒+10 dB,100×I⇒+20 dB,1000×I⇒+30 dB10\times I \Rightarrow +10\,\text{dB},\qquad 100\times I \Rightarrow +20\,\text{dB},\qquad 1000\times I \Rightarrow +30\,\text{dB}

Why a logarithmic scale?

From a whisper to the threshold of pain the intensity grows over 101310^{13} times. On a linear scale (the purple bar in the panel) almost everything "sticks" to zero. The logarithm squeezes this huge range into a convenient 0–140 dB.

Why you need this for the exam:

Acoustics problems rely on L=10log⁡10(I/I0)L = 10\log_{10}(I/I_0). The trick: compute the intensity ratio I/I0I/I_0, take its base-10 logarithm and multiply by 10. Level difference: ΔL=10log⁡10(I2/I1)\Delta L = 10\log_{10}(I_2/I_1).

Remember: decibels are a logarithm — 10× the intensity = +10 dB, 100× = +20 dB. Multiplying the intensity means adding decibels. L = 10·log₁₀(I/I₀).

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Theory & formulas

The decibel scale

L=10log⁡10II0L = 10\log_{10}\dfrac{I}{I_0} The scale is logarithmic — matched to the enormous range of intensities heard by the ear.

Rules of thumb

  • 10× greater intensity → +10 dB (because log⁡1010=1\log_{10}10 = 1),
  • 2× greater intensity → +3 dB (10log⁡102≈310\log_{10}2\approx3).

Notation: LL — intensity level (dB), II — sound intensity, I0=10−12 W/m2I_0=10^{-12}\ \mathrm{W/m^2} — threshold of hearing.

Frequently asked questions

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Why is a 10× greater intensity only +10 dB?

Because we compute the level logarithmically: L=10·log₁₀(I/I₀); the log of 10 is 1, so a tenfold increase in intensity gives +10 dB.

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Why a logarithmic scale?

The ear responds to an enormous range of intensities — the logarithm brings them down to a convenient scale that matches perceived loudness.

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How many dB is a doubling of intensity?

About +3 dB (10·log2≈3).

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