C.3 Wave phenomena

IB Physics HL· Theme C: Wave Behaviour, C.3 Wave phenomena· 20 min read

1. Standing Waves★★☆☆☆⏱ 5 min

📘 Definition

Standing Wave

also called stationary 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.

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: $λ_{n} = \frac{2 L}{n}$ , $n = 1, 2, 3...$
Pipe of length $L$ closed at one end: $λ_{n} = \frac{4 L}{n}$ , $n = 1, 3, 5...$ (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 $v = f λ$ :
4
$v = 70 \times 4.8 = 336 \approx 340 \text{ m s}^{-1} \text{ (2 s.f.)}$

2. Single-slit Diffraction★★★☆☆⏱ 5 min

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.

📘 Definition

Single-slit Diffraction Minima

The condition for the first minimum (edge of the central maximum) is $b sin θ = λ$ , where $b$ is slit width, $θ$ is the angular position of the minimum, and $λ$ is wavelength. For small angles $sin θ \approx \frac{y}{D}$ 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 ( $2 y = 2 D θ$ ):
6
$2y = 2 \times 2.5 \times 5.5 \times 10^{-3} = 0.028 \text{ m} = 2.8 \text{ cm (2 s.f.)}$

3. Two-source Interference★★★☆☆⏱ 5 min

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.

📘 Definition

Two-source Interference Conditions

For coherent sources separated by distance $d$ : constructive interference (bright fringe) when path difference = $nλ$ , $n = 0, 1, 2...$ ; destructive interference (dark fringe) when path difference = $(n + \frac{1}{2}) λ$ . Fringe separation (distance between adjacent bright fringes) is $Δ s = \frac{λ D}{d}$ .

📐 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
$\Delta s = \frac{\lambda D}{d} = \frac{600 \times 10^{-9} \times 3.0}{0.2 \times 10^{-3}} = 9.0 \times 10^{-3} \text{ m} = 9.0 \text{ mm}$

4. Rayleigh Criterion for Resolution★★★★☆⏱ 5 min

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.

📘 Definition

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 $θ$ (in radians) is $θ = 1.22 \frac{λ}{b}$ , 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
$\theta = 1.22 \frac{550 \times 10^{-9}}{2 \times 10^{-3}} \approx 3.4 \times 10^{-4} \text{ radians}$

5. Common Pitfalls

Wrong move:

Using the 1.22 factor for single slit resolution

Why:

The 1.22 factor is only required for circular apertures, not rectangular single slits

Correct move:

Use $θ = \frac{λ}{b}$ without the 1.22 factor for non-circular apertures

Wrong move:

Forgetting to double the distance to get the full width of the central maximum in single slit diffraction

Why:

The formula gives the distance from the central maximum to the first minimum, not between the two outer minima

Correct move:

Multiply the distance from centre to first minimum by 2 to get the full central width

Wrong move:

Using the open pipe wavelength formula $λ = 2 L / n$ for closed pipes

Why:

Closed pipes have a node at the closed end and antinode at the open end, so only odd harmonics exist

Correct move:

For closed pipes, use $λ = 4 L / n$ where $n = 1, 3, 5...$

Wrong move:

Mixing up $b$ (slit width) and $d$ (slit separation) between single and double slit formulas

Why:

Students often swap the variables, leading to incorrect calculations

Correct move:

Single slit width = $b$ , two-slit separation = $d$ ; always confirm which variable you need for the formula

Wrong move:

Claiming standing waves transfer energy along the medium

Why:

Standing waves are stationary, so no net energy propagation occurs

Correct move:

Energy is stored between nodes and does not travel along the standing wave

6. Quick Reference Cheatsheet

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

7. Frequently Asked

Is the Rayleigh criterion formula given in the data booklet?

Yes, but you must remember that the 1.22 factor only applies to circular apertures (eyes, telescopes, cameras) and not rectangular single slits.

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.

2025 · Paper 1
Rayleigh criterion calculation
2024 · Paper 2
Standing waves in closed pipes
2023 · Paper 1
Single slit diffraction width
2022 · Paper 2
Two-source interference problem

Going deeper

syllabusOfficial IB Physics 2025 SyllabusConfirm latest syllabus updates

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.