Wave Intensity

CIE A-Level Physics· Unit 7: Waves· 15 min read

1. Definition and Intensity-Amplitude Relationship★★☆☆☆⏱ 5 min

📘 Definition

Wave Intensity

$I$

Power transmitted by a wave per unit cross-sectional area perpendicular to the direction of wave travel, measured in watts per square metre ( $Wm^{- 2}$ )

Example:

Sunlight reaching Earth's surface has an intensity of approximately 1360 $Wm^{- 2}$

The energy carried by a wave is proportional to the square of its displacement amplitude. Since intensity is energy per unit area per unit time, this gives the key proportional relationship:

I \propto A^2

This means if amplitude doubles, intensity increases by a factor of $2^{2} = 4$ . We can write a ratio for two waves of the same type:

📐 Worked Example

A wave has amplitude $2.0 cm$ and intensity $12 Wm^{- 2}$ . What is the intensity when the amplitude is increased to $3.0 cm$ ?

1
Start with the proportionality ratio for intensity and amplitude:
$\frac{I_1}{I_2} = \left(\frac{A_1}{A_2}\right)^2$
2
Rearrange to solve for the new intensity $I_{2}$ , then substitute known values:
$I_2 = I_1 \left(\frac{A_2}{A_1}\right)^2 = 12 \times \left(\frac{3.0}{2.0}\right)^2$
3
Calculate the final result:
$I_2 = 12 \times 2.25 = 27 \, \text{Wm}^{-2}$

Exam tip:

Always remember it is intensity proportional to amplitude squared, not amplitude. This is the most common error in this topic.

2. Inverse Square Law for Point Sources★★★☆☆⏱ 6 min

📘 Definition

Inverse Square Law

$I \propto 1/ r^{2}$

For an isotropic point source emitting constant power uniformly in all directions, intensity is inversely proportional to the square of the distance from the source

Example:

If distance from a light bulb quadruples, intensity drops to 1/16 of its original value

Power from a point source spreads uniformly over the surface of an expanding sphere of radius $r$ . The surface area of a sphere is $4 π r^{2}$ , so intensity is calculated as:

I = \frac{P}{A} = \frac{P}{4\pi r^2}

Since total power $P$ is constant for a stable source, $I \propto 1/ r^{2}$ , giving the ratio relation:

\frac{I_1}{I_2} = \left(\frac{r_2}{r_1}\right)^2

📐 Worked Example

A small speaker emits sound uniformly in all directions. At 2.0 m from the speaker, intensity is $1.6 \times 1 0^{- 4} Wm^{- 2}$ . What is the intensity at 8.0 m from the speaker?

1
Apply the inverse square law ratio:
$\frac{I_1}{I_2} = \left(\frac{r_2}{r_1}\right)^2$
2
Rearrange for the new intensity $I_{2}$ :
$I_2 = I_1 \left(\frac{r_1}{r_2}\right)^2$
3
Substitute the known values:
$I_2 = 1.6 \times 10^{-4} \times \left(\frac{2.0}{8.0}\right)^2 = 1.6 \times 10^{-4} \times \frac{1}{16}$
4
Final result:
$I_2 = 1.0 \times 10^{-5} \, \text{Wm}^{-2}$

3. Combined Intensity Problems★★★★☆⏱ 4 min

Most structured exam questions require combining both proportionalities: intensity depends on both amplitude squared and inverse square of distance. The combined relation is:

I \propto \frac{A^2}{r^2}

This is used when both amplitude and distance change, or when comparing intensity from two different sources. The ratio form for two sources is:

\frac{I_1}{I_2} = \left(\frac{A_1}{A_2}\right)^2 \left(\frac{r_2}{r_1}\right)^2

📐 Worked Example

Source A emits sound with amplitude $A$ at distance $r$ , with intensity $I$ . Source B emits twice the amplitude of A, at 3 times the distance from an observer. What is the intensity of B in terms of $I$ ?

1
Write the proportionality for both sources, with $k$ as a constant:
$I_A = k \frac{A_A^2}{r_A^2}, \quad I_B = k \frac{A_B^2}{r_B^2}$
2
Divide to eliminate the constant $k$ :
$\frac{I_B}{I_A} = \left(\frac{A_B}{A_A}\right)^2 \left(\frac{r_A}{r_B}\right)^2$
3
Substitute $A_{B} = 2 A_{A}$ , $r_{B} = 3 r_{A}$ , $I_{A} = I$ :
$I_B = I \times (2)^2 \times \left(\frac{1}{3}\right)^2 = I \times 4 \times \frac{1}{9}$
4
Final result:
$I_B = \frac{4}{9}I$

Exam tip:

Always label which quantity corresponds to which source when calculating ratios to avoid swapping distances or amplitudes by mistake.

4. Common Pitfalls

Wrong move:

Stating $I \propto A$ instead of $I \propto A^{2}$

Why:

This is the most common mistake from misremembering the relationship. Wave energy depends on the square of amplitude, so intensity does too.

Correct move:

Memorise that intensity is proportional to the square of wave amplitude: doubling amplitude quadruples intensity.

Wrong move:

Applying inverse square law to plane waves or non-point sources

Why:

The inverse square law only applies when power spreads over an expanding spherical surface.

Correct move:

Only use $I \propto 1/ r^{2}$ when the question confirms you have an isotropic point source.

Wrong move:

Swapping the distance ratio: writing $I_{1} / I_{2} = (r_{1} / r_{2})^{2}$

Why:

Intensity decreases as distance increases, so the ratio is inverted.

Correct move:

Remember: further distance = lower intensity, so $\frac{I _{1}}{I _{2}} = (\frac{r _{2}}{r _{1}})^{2}$ .

Wrong move:

Forgetting the $1/ r^{2}$ factor when only amplitude change is highlighted

Why:

Exam questions often include hidden distance changes to test if you recall both relationships.

Correct move:

Always check if the question mentions a change in distance from the source, and include the inverse square factor.

5. Quick Reference Cheatsheet

Relationship	Formula	Use Case
Intensity vs Amplitude	$I \propto A^{2}$ , $\frac{I _{1}}{I _{2}} = (\frac{A _{1}}{A _{2}})^{2}$	Change in wave amplitude
Inverse Square Law	$I \propto 1/ r^{2}$ , $\frac{I _{1}}{I _{2}} = (\frac{r _{2}}{r _{1}})^{2}$	Point source, changing distance
Combined Relation	$I \propto \frac{A ^{2}}{r ^{2}}$ , $\frac{I _{1}}{I _{2}} = (\frac{A _{1}}{A _{2}})^{2} (\frac{r _{2}}{r _{1}})^{2}$	Both amplitude and distance change
Definition of Intensity	$I = \frac{P}{A}$ , units $Wm^{- 2}$	Calculate intensity from power/area

6. Frequently Asked

Does the inverse square law apply to all waves?

No, the inverse square law only applies to spherical waves from an isotropic point source, where power spreads uniformly over a spherical surface. It does not apply to plane waves (e.g., low-divergence laser beams) or 1-dimensional waves like waves on a string.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 1
Intensity ratio calculation
2022 · 2
Inverse square law problem

Going deeper

syllabusCIE 9702 Official SyllabusCheck page 36 for wave intensity learning outcomes

What's Next

Mastering wave intensity gives you easy marks in CIE A-Level Physics, as intensity ratio questions appear regularly in both multiple choice and structured papers. This concept is foundational for understanding a range of later topics, including standing waves, diffraction of light, electromagnetic radiation, and the photoelectric effect. The proportional relationships you learned here are also used in quantum physics to relate photon flux to radiation intensity. Next, you can progress to the next sub-topic in the waves unit, or build on this knowledge for electromagnetic waves later in your course.