# Wave Intensity

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u7-wave-intensity/

This sub-topic defines wave intensity, explains its proportional relationship to wave amplitude, and covers the inverse square law for intensity from a point source. You will learn to solve common ratio problems that appear regularly in CIE A-Level Physics exams.

**Prerequisites:** [Basic wave properties (amplitude, power, energy)](https://www.owlsprep.com/study/cie-9702-u7-basic-wave-properties/)

## Learning objectives

- Define wave intensity and state its relationship to wave amplitude
- Apply the inverse square law for intensity from a point wave source
- Solve combined problems involving both amplitude and distance changes

## Definition and Intensity-Amplitude Relationship

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

*Notation:* $I$

*Example:* Sunlight reaching Earth's surface has an intensity of approximately 1360 $\text{Wm}^{-2}$

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 $2.0 \, \text{cm}$ and intensity $12 \, \text{Wm}^{-2}$. What is the intensity when the amplitude is increased to $3.0 \, \text{cm}$?

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.

## Inverse Square Law for Point Sources

**Inverse Square Law** — 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

*Notation:* $I \propto 1/r^2$

*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 $4\pi r^2$, 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 $1.6 \times 10^{-4} \, \text{Wm}^{-2}$. 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:

   $$I_2 = 1.0 \times 10^{-5} \, \text{Wm}^{-2}$$

> **info**
>
> The inverse square law only applies when there is no absorption or reflection of the wave by the medium. If absorption is mentioned in the question, account for it separately.

## Combined Intensity Problems

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 $A_B = 2A_A$, $r_B = 3r_A$, $I_A = I$:

   $$I_B = I \times (2)^2 \times \left(\frac{1}{3}\right)^2 = I \times 4 \times \frac{1}{9}$$
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.

## Common pitfalls

- **Wrong:** Stating $I \propto A$ instead of $I \propto A^2$
  - Why it fails: This is the most common mistake from misremembering the relationship. Wave energy depends on the square of amplitude, so intensity does too.
  - Correct: Memorise that intensity is proportional to the square of wave amplitude: doubling amplitude quadruples intensity.
- **Wrong:** Applying inverse square law to plane waves or non-point sources
  - Why it fails: The inverse square law only applies when power spreads over an expanding spherical surface.
  - Correct: Only use $I \propto 1/r^2$ when the question confirms you have an isotropic point source.
- **Wrong:** Swapping the distance ratio: writing $I_1/I_2 = (r_1/r_2)^2$
  - Why it fails: Intensity decreases as distance increases, so the ratio is inverted.
  - Correct: Remember: further distance = lower intensity, so $\frac{I_1}{I_2} = \left(\frac{r_2}{r_1}\right)^2$.
- **Wrong:** Forgetting the $1/r^2$ factor when only amplitude change is highlighted
  - Why it fails: Exam questions often include hidden distance changes to test if you recall both relationships.
  - Correct: Always check if the question mentions a change in distance from the source, and include the inverse square factor.

## Cheatsheet

| Relationship | Formula | Use Case |
| --- | --- | --- |
| Intensity vs Amplitude | $I \propto A^2$, $\frac{I_1}{I_2} = \left(\frac{A_1}{A_2}\right)^2$ | Change in wave amplitude |
| Inverse Square Law | $I \propto 1/r^2$, $\frac{I_1}{I_2} = \left(\frac{r_2}{r_1}\right)^2$ | Point source, changing distance |
| Combined Relation | $I \propto \frac{A^2}{r^2}$, $\frac{I_1}{I_2} = \left(\frac{A_1}{A_2}\right)^2 \left(\frac{r_2}{r_1}\right)^2$ | Both amplitude and distance change |
| Definition of Intensity | $I = \frac{P}{A}$, units $\text{Wm}^{-2}$ | Calculate intensity from power/area |

## 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.

- [Polarisation of Waves](https://www.owlsprep.com/study/cie-9702-u7-polarisation/)
- [Superposition](https://www.owlsprep.com/study/cie-9702-u8-overview/)
- [Principle of superposition](https://www.owlsprep.com/study/cie-9702-u8-principle-of-superposition/)

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/cie-9702-u7-wave-intensity/