Physics 9702 · Paper 1

Intensity & Amplitude

Intensity is the power a wave delivers per unit area. It is how we measure “how much” a wave carries.
Louder sound and brighter light mean higher intensity, and intensity depends on the square of the amplitude.

Intensity: power spread over an area

I = \dfrac{P}{A}

Symbols

$I$ = intensity (W m⁻²)
$P$ = power carried by the wave (W)
$A$ = area the power passes through, at right angles to the wave (m²)

A wave from a small source spreads over a bigger and bigger area as it travels, so the power per unit area gets smaller and the intensity decreases.
For a point source the wave spreads over a sphere of area $4\pi r^2$ , so $I \propto 1/r^2$ : double the distance, quarter the intensity.

Light spreading from a small source onto a near small square and a far larger square, the same power covering a bigger area further away. — Further away the same power is spread over a larger area, so the intensity drops.

Intensity depends on amplitude squared

Intensity is proportional to the square of the amplitude: $I \propto A^2$ .
So doubling the amplitude does not double the intensity; it makes it four times as big.

Worked example

Doubling the amplitude

A wave's amplitude is doubled while nothing else changes. What happens to its intensity?

$I \propto A^2$ , so $\dfrac{I_2}{I_1} = \left(\dfrac{A_2}{A_1}\right)^2 = 2^2 = 4$ .

Answer

I \text{ becomes } 4\times \text{ larger}

Common mistake

It is amplitude squared, not amplitude. Trebling the amplitude multiplies the intensity by 9, not 3.

Your turn— tap to reveal the worked answer (9702/22/M/J/25 Q3(d))

Two waves of the same type: wave X has amplitude 10 cm, wave Y has amplitude 20 cm. Find the ratio I_X / I_Y of their intensities. (9702/22/M/J/25 Q3(d))

Answer: $\dfrac{I_X}{I_Y} = \left(\dfrac{10}{20}\right)^2 = 0.25$ , so 1 : 4.

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