# C.3 Wave phenomena

> IB Physics HL · IB Physics HL 2025 Syllabus
> Source: https://www.owlsprep.com/study/ib-physics-hl-u3-c-3-wave-phenomena/

This sub-topic covers standing wave formation, single-slit diffraction, two-source interference, and the Rayleigh criterion for image resolution, core concepts for wave behaviour that are frequently tested in IB HL Physics exams.

**Prerequisites:** [Wave basics and superposition principle](https://www.owlsprep.com/study/ib-physics-hl-u3-c-1-wave-basics/); [Coherence and path difference](https://www.owlsprep.com/study/ib-physics-hl-u3-c-2-superposition/)

## Learning objectives

- Explain standing wave formation from superposition of opposite travelling waves
- Calculate node/antinode positions and frequencies for standing waves in different media
- Solve single-slit diffraction and two-source interference problems
- Apply the Rayleigh criterion to calculate minimum resolvable angular separation

## Standing Waves

**Standing Wave** — A stationary wave pattern formed by the superposition of two identical waves travelling at the same speed in opposite directions. Nodes and antinodes remain in fixed positions, with no net propagation of energy.

*Notation:* also called stationary wave

*Example:* An incident wave reflected off a fixed string end interferes with the original wave to form a standing wave.

Standing wave harmonics depend on boundary conditions: fixed boundaries produce nodes, open boundaries produce antinodes. Key harmonic formulas are:

- String of length $L$ fixed at both ends / pipe open at both ends: $\lambda_n = \frac{2L}{n}$, $n=1,2,3...$
- Pipe of length $L$ closed at one end: $\lambda_n = \frac{4L}{n}$, $n=1,3,5...$ (only odd harmonics)

**Worked example:** A 1.2 m long pipe closed at one end has a fundamental frequency of 70 Hz. Calculate the speed of sound in air.

1. The fundamental frequency is the first harmonic, so $n=1$ for a closed pipe:
2. $$\lambda_1 = \frac{4L}{n} = 4 \times 1.2 = 4.8 \text{ m}$$
3. Use the universal wave equation $v = f\lambda$:
4. $$v = 70 \times 4.8 = 336 \approx 340 \text{ m s}^{-1} \text{ (2 s.f.)}$$

## Single-slit Diffraction

Diffraction describes the spreading of waves when they pass through an aperture. A single narrow slit produces a diffraction pattern with a wide, bright central maximum, and smaller dimmer maxima on either side. Minima occur where destructive interference cancels the wave.

**Single-slit Diffraction Minima** — The condition for the first minimum (edge of the central maximum) is $b\sin\theta = \lambda$, where $b$ is slit width, $\theta$ is the angular position of the minimum, and $\lambda$ is wavelength. For small angles $\sin\theta \approx \frac{y}{D}$ where $y$ is distance from the central maximum on the screen, and $D$ is distance from slit to screen.

**Worked example:** 550 nm monochromatic light is incident on a 0.1 mm wide single slit. The distance from the slit to the screen is 2.5 m. Calculate the width of the central maximum.

1. Convert all values to SI units:
2. $$\lambda = 550 \times 10^{-9} \text{ m}, \quad b = 0.1 \times 10^{-3} \text{ m}, \quad D = 2.5 \text{ m}$$
3. Use small angle approximation to find the angle of the first minimum:
4. $$\theta \approx \frac{\lambda}{b} = \frac{550 \times 10^{-9}}{0.1 \times 10^{-3}} = 5.5 \times 10^{-3} \text{ rad}$$
5. The width of the central maximum is twice the distance from the centre to the first minimum ($2y = 2D\theta$):
6. $$2y = 2 \times 2.5 \times 5.5 \times 10^{-3} = 0.028 \text{ m} = 2.8 \text{ cm (2 s.f.)}$$

## Two-source Interference

Two coherent wave sources produce a stable interference pattern of bright (constructive) and dark (destructive) fringes on a distant screen. The fringe separation depends on wavelength, slit separation, and distance to the screen.

**Two-source Interference Conditions** — For coherent sources separated by distance $d$: constructive interference (bright fringe) when path difference = $n\lambda$, $n=0,1,2...$; destructive interference (dark fringe) when path difference = $(n+\frac{1}{2})\lambda$. Fringe separation (distance between adjacent bright fringes) is $\Delta s = \frac{\lambda D}{d}$.

**Worked example:** Two slits separated by 0.2 mm are illuminated with 600 nm light. The screen is 3.0 m from the slits. Find the separation between adjacent bright fringes.

1. Convert to SI units:
2. $$d = 0.2 \times 10^{-3} \text{ m}, \quad \lambda = 600 \times 10^{-9} \text{ m}, \quad D = 3.0 \text{ m}$$
3. Substitute into the fringe separation formula:
4. $$\Delta s = \frac{\lambda D}{d} = \frac{600 \times 10^{-9} \times 3.0}{0.2 \times 10^{-3}} = 9.0 \times 10^{-3} \text{ m} = 9.0 \text{ mm}$$

> **tip**
>
> Fringe separation increases if wavelength increases, distance to screen increases, or slit separation decreases. This relationship is often tested in Paper 1 multiple choice.

## Rayleigh Criterion for Resolution

Resolution is the ability of an imaging system to distinguish two adjacent point sources as separate objects. If the diffraction patterns of the two sources overlap too much, they appear as a single blurred source.

**Rayleigh Criterion** — Two point sources are just resolvable when the central maximum of the diffraction pattern of one source coincides with the first minimum of the diffraction pattern of the second source. For a circular aperture, the minimum angular separation $\theta$ (in radians) is $\theta = 1.22 \frac{\lambda}{b}$, where $b$ is the aperture diameter.

**Worked example:** The human eye has an aperture diameter of 2 mm in bright light. Estimate the minimum angular separation that can be resolved for 550 nm visible light.

1. Convert values to SI units:
2. $$b = 2 \times 10^{-3} \text{ m}, \quad \lambda = 550 \times 10^{-9} \text{ m}$$
3. Apply the Rayleigh criterion for a circular aperture:
4. $$\theta = 1.22 \frac{550 \times 10^{-9}}{2 \times 10^{-3}} \approx 3.4 \times 10^{-4} \text{ radians}$$

> **warning**
>
> The 1.22 factor is only used for circular apertures. Omit it for rectangular single slit apertures.

## Common pitfalls

- **Wrong:** Using the 1.22 factor for single slit resolution
  - Why it fails: The 1.22 factor is only required for circular apertures, not rectangular single slits
  - Correct: Use $\theta = \frac{\lambda}{b}$ without the 1.22 factor for non-circular apertures
- **Wrong:** Forgetting to double the distance to get the full width of the central maximum in single slit diffraction
  - Why it fails: The formula gives the distance from the central maximum to the first minimum, not between the two outer minima
  - Correct: Multiply the distance from centre to first minimum by 2 to get the full central width
- **Wrong:** Using the open pipe wavelength formula $\lambda = 2L/n$ for closed pipes
  - Why it fails: Closed pipes have a node at the closed end and antinode at the open end, so only odd harmonics exist
  - Correct: For closed pipes, use $\lambda = 4L/n$ where $n = 1, 3, 5...$
- **Wrong:** Mixing up $b$ (slit width) and $d$ (slit separation) between single and double slit formulas
  - Why it fails: Students often swap the variables, leading to incorrect calculations
  - Correct: Single slit width = $b$, two-slit separation = $d$; always confirm which variable you need for the formula
- **Wrong:** Claiming standing waves transfer energy along the medium
  - Why it fails: Standing waves are stationary, so no net energy propagation occurs
  - Correct: Energy is stored between nodes and does not travel along the standing wave

## Cheatsheet

| Concept | Key Formula | Notes |
| --- | --- | --- |
| Standing wave (string/open pipe) | $\lambda_n = 2L/n$, $n=1,2...$ | Fixed ends = nodes |
| Standing wave (closed pipe) | $\lambda_n = 4L/n$, $n=1,3...$ | Only odd harmonics |
| Single slit 1st minimum | $b\sin\theta = \lambda$ | Central width = $2\lambda D / b$ |
| Two-slit fringe separation | $\Delta s = \lambda D / d$ | $d$ = slit separation |
| Rayleigh (circular aperture) | $\theta = 1.22 \lambda / b$ | Omit 1.22 for single slit |

## What's next

This sub-topic builds core wave behaviour concepts that are the foundation for further topics including the Doppler effect and thin film interference, which extend the superposition, path difference and diffraction principles you learned here. Understanding the Rayleigh criterion for resolution is also key for astronomy and imaging technologies that appear in IB Physics HL option topics. Mastering standing waves, interference and diffraction here will make these more advanced follow-on topics much more intuitive and easier to solve problems for.

- [C.4 Standing waves and resonance](https://www.owlsprep.com/study/ib-physics-hl-u3-c-4-standing-waves-and/)
- [C.5 Doppler effect (AHL)](https://www.owlsprep.com/study/ib-physics-hl-u3-c-5-doppler-effect/)
- [C.6 Interference (AHL)](https://www.owlsprep.com/study/ib-physics-hl-u3-c-6-interference/)

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