IB SL

Wave Behaviour

How to Use This Guide

Wave Properties — transverse vs longitudinal, amplitude, wavelength, frequency, wave speed, phase difference, intensity
Sound Waves — longitudinal pressure waves, Doppler effect, decibel scale
Light and the EM Spectrum — Snell’s law, total internal reflection, refraction
Superposition and Interference — constructive/destructive interference, double-slit, single-slit diffraction, gratings
Standing Waves — nodes, antinodes, harmonics in strings and pipes

Aligned to IB Physics 2025 syllabus — Theme C: Wave Behaviour (first assessment 2025)

Jump to section: Wave Properties · Sound Waves · Light and EM Spectrum · Superposition and Interference · Standing Waves · Practice Questions

Videos on this page: Watch: Doppler Effect

Section 1: Wave Properties

Transverse and Longitudinal Waves

A wave is a periodic disturbance that transfers energy through a medium (or, for electromagnetic waves, through a vacuum) without transferring matter.

Type	Direction of oscillation	Examples
Transverse	Perpendicular to the direction of wave travel	Light, all EM waves, waves on a string, seismic S-waves
Longitudinal	Parallel to the direction of wave travel	Sound, seismic P-waves, ultrasound

A longitudinal wave consists of alternating compressions (regions of higher pressure/density) and rarefactions (regions of lower pressure/density).

Key Wave Quantities

Symbol	Quantity	Definition	SI Unit
$A$	amplitude	maximum displacement from equilibrium	m
$\lambda$	wavelength	distance between two successive points in phase (e.g., crest to crest)	m
$T$	period	time for one complete oscillation	s
$f$	frequency	number of complete oscillations per unit time	Hz ( $\text{s}^{-1}$ )
$v$	wave speed	speed at which the wave pattern moves	$\text{m s}^{-1}$

Fundamental relationships:

$f = \frac{1}{T} \qquad v = f\lambda$

$v = f\lambda$ is in the data booklet but you must know when to use it. Wave speed $v$ is determined by the medium (not by the source). When a wave passes from one medium to another, the frequency stays constant but the wavelength and speed both change. This is why light bends at an interface (refraction).

Phase Difference

Two points on a wave are in phase if they have identical displacement and velocity at all times (separated by a whole number of wavelengths). Anti-phase (phase difference = $180° = \pi$ rad) means one is at a crest while the other is at a trough.

$\phi = \frac{\Delta x}{\lambda} \times 360° \qquad \text{or} \qquad \phi = \frac{2\pi \Delta x}{\lambda} \text{ rad}$

where $\Delta x$ is the path difference or spatial separation.

Path difference $= n\lambda$ ( $n$ integer) → in phase ( $\phi = 0°, 360°, ...$ )
Path difference $= (n + \tfrac{1}{2})\lambda$ → anti-phase ( $\phi = 180°, 540°, ...$ )

Intensity and the Inverse Square Law

Intensity is the power per unit area of wavefront:

$I = \frac{P}{A} \qquad \text{unit: W m}^{-2}$

For a point source radiating equally in all directions (spreading over the surface of a sphere of area $4\pi r^2$ ):

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

Intensity–amplitude relationship:

$I \propto A^2$

Doubling the amplitude quadruples the intensity. Halving the distance from a source quadruples the intensity.

Worked Example C1 — Inverse square law:

A point source of sound has intensity $I_1 = 4.0 \text{ W m}^{-2}$ at a distance $r_1 = 2.0 \text{ m}$ . Find the intensity at $r_2 = 6.0 \text{ m}$ .

$\frac{I_2}{I_1} = \frac{r_1^2}{r_2^2} = \frac{4.0}{36} = \frac{1}{9}$

$I_2 = \frac{4.0}{9} \approx 0.44 \text{ W m}^{-2}$

Section 2: Sound Waves

Properties of Sound

Sound is a longitudinal mechanical wave — it requires a medium and cannot travel through a vacuum.

Speed of sound in air: approximately $340 \text{ m s}^{-1}$ at room temperature (increases with temperature)
Speed of sound in solids $>$ liquids $>$ gases (denser, stiffer media transmit sound faster)
Sound consists of alternating compressions and rarefactions in the medium
Frequency range for human hearing: approximately 20 Hz to 20,000 Hz

Decibel Scale

The decibel (dB) scale is a logarithmic scale for sound intensity level. You need a qualitative understanding: the scale compresses a very wide range of intensities ( $10^{12}$ :1) into a manageable scale (0–120 dB). Every increase of 10 dB corresponds to a tenfold increase in intensity. The IB SL exam only tests qualitative knowledge here.

The Doppler Effect

The Doppler effect is the change in observed frequency of a wave when the source and observer are in relative motion.

Doppler formula (given in data booklet):

$f' = f_s \frac{v \pm v_o}{v \mp v_s}$

Symbol	Meaning
$f'$	observed frequency
$f_s$	source frequency
$v$	wave speed in the medium
$v_o$	speed of observer
$v_s$	speed of source

Sign rule:

Top sign ( $+$ for observer, $-$ for source): source/observer approaching (observed frequency increases)
Bottom sign ( $-$ for observer, $+$ for source): source/observer receding (observed frequency decreases)

Memory trick: “Observer on top, source on bottom. Moving toward = frequency increases (use signs that increase the fraction).”

Worked Example C2 — Ambulance Doppler shift:

An ambulance siren emits sound at $f_s = 800 \text{ Hz}$ . The ambulance moves toward a stationary observer at $v_s = 30 \text{ m s}^{-1}$ . Speed of sound $v = 340 \text{ m s}^{-1}$ .

(a) Frequency heard as ambulance approaches:

$f' = 800 \times \frac{340 + 0}{340 - 30} = 800 \times \frac{340}{310} = 800 \times 1.097 \approx 878 \text{ Hz}$

(b) Frequency heard as ambulance recedes:

$f' = 800 \times \frac{340}{340 + 30} = 800 \times \frac{340}{370} = 800 \times 0.919 \approx 735 \text{ Hz}$

The observer hears the frequency drop from 878 Hz to 735 Hz as the ambulance passes.

HL The full derivation of the Doppler formula from first principles — considering wavefront compression and expansion geometrically — is an HL topic. At SL, you are expected to apply the formula and explain the effect qualitatively.

Doppler direction errors are very common. Source approaching observer: use $v - v_s$ in the denominator (smaller denominator = larger fraction = higher frequency). Getting the signs backward is the most frequent error. Write out the full formula and substitute carefully rather than trying to memorise direction shortcuts.

Watch: Doppler Effect — Khan Academy

Section 3: Light and the Electromagnetic Spectrum

The EM Spectrum

All electromagnetic waves travel at $c = 3.00 \times 10^8 \text{ m s}^{-1}$ in a vacuum. They differ only in frequency (and therefore wavelength).

EM spectrum order (increasing frequency / decreasing wavelength):

$\text{Radio} \to \text{Microwave} \to \text{Infrared} \to \text{Visible} \to \text{UV} \to \text{X-ray} \to \text{Gamma}$

Mnemonic: Rather Mighty Ignorant Vigilantes Use X-ray Guns

Visible light occupies approximately 400 nm (violet) to 700 nm (red). For visible: ROYGBIV from longest to shortest wavelength (red to violet).

Refraction and Snell’s Law

Refraction is the change in direction of a wave as it passes from one medium to another, caused by the change in wave speed.

Snell’s Law:

$n_1 \sin\theta_1 = n_2 \sin\theta_2$

Refractive index:

$n = \frac{c}{v_{\text{medium}}} = \frac{\text{speed of light in vacuum}}{\text{speed of light in medium}}$

Angles are measured from the normal (perpendicular to the interface), not from the surface.
Light bends toward the normal when entering a denser medium (higher $n$ ), and away from the normal when entering a less dense medium.
For vacuum (or approximately for air): $n = 1.00$ .

Worked Example C3 — Snell’s Law:

A ray of light in air ( $n_1 = 1.00$ ) strikes a glass surface ( $n_2 = 1.50$ ) at an angle of incidence $\theta_1 = 40°$ .

(a) Find the angle of refraction $\theta_2$ .

$n_1 \sin\theta_1 = n_2 \sin\theta_2 \implies \sin\theta_2 = \frac{1.00 \times \sin 40°}{1.50} = \frac{0.643}{1.50} = 0.429$

$\theta_2 = \arcsin(0.429) \approx 25.4°$

The ray bends toward the normal on entering the denser glass — confirmed by $\theta_2 < \theta_1$ .

(b) Calculate the critical angle for this glass–air interface.

At the critical angle, $\theta_2 = 90°$ , so $\sin\theta_c = n_2/n_1 = 1/n$ (for glass to air):

$\sin\theta_c = \frac{n_{\text{air}}}{n_{\text{glass}}} = \frac{1.00}{1.50} = 0.667 \implies \theta_c \approx 41.8°$

Total Internal Reflection

Conditions for TIR:

Light is in the denser medium (higher $n$ ) and trying to enter the less dense medium.
The angle of incidence exceeds the critical angle: $\theta > \theta_c$ .

$\sin\theta_c = \frac{n_2}{n_1} \quad (n_1 > n_2)$

Applications: optical fibres (telecommunications, endoscopes), diamond brilliance (high $n$ → small $\theta_c$ → most light totally internally reflected), periscope prisms.

Optical fibres work via TIR. A glass core ( $n \approx 1.5$ ) surrounded by cladding ( $n$ slightly lower) ensures that light entering at an angle greater than $\theta_c$ is totally internally reflected along the fibre. Signal loss occurs from absorption and scattering, not from TIR itself. This is a common short-answer topic in IB exams.

Section 4: Superposition and Interference

Principle of Superposition

When two or more waves meet at a point, the resultant displacement is the vector sum of the individual displacements at that instant.

Constructive interference: waves arrive in phase (path difference $= n\lambda$ , $n = 0, 1, 2, ...$ )

$\text{Resultant amplitude} = A_1 + A_2 \quad \text{(maximum)}$

Destructive interference: waves arrive anti-phase (path difference $= (n + \tfrac{1}{2})\lambda$ )

$\text{Resultant amplitude} = |A_1 - A_2| \quad \text{(minimum; zero if } A_1 = A_2\text{)}$

For a stable (coherent) interference pattern, the sources must have the same frequency and a constant phase relationship (coherent sources).

Young’s Double-Slit Experiment

Two coherent slits separated by distance $d$ produce alternating bright and dark fringes on a screen at distance $D$ .

Fringe spacing (distance between adjacent bright fringes):

$s = \frac{\lambda D}{d}$

Symbol	Meaning	Relationship
$s$	fringe spacing	increases with $\lambda$ and $D$ ; decreases with $d$
$\lambda$	wavelength of light
$D$	slit-to-screen distance
$d$	slit separation

The formula is in the data booklet.

Worked Example C4 — Double-slit fringe spacing:

Light of wavelength $\lambda = 600 \text{ nm}$ passes through two slits separated by $d = 0.30 \text{ mm}$ . The screen is $D = 2.0 \text{ m}$ away.

$s = \frac{\lambda D}{d} = \frac{600 \times 10^{-9} \times 2.0}{0.30 \times 10^{-3}} = \frac{1.2 \times 10^{-6}}{3.0 \times 10^{-4}} = 4.0 \times 10^{-3} \text{ m} = 4.0 \text{ mm}$

Single-Slit Diffraction

A single slit of width $b$ produces a central bright maximum flanked by dark fringes.

Position of dark fringes in single-slit diffraction:

$\sin\theta = \frac{m\lambda}{b} \quad m = \pm 1, \pm 2, \pm 3, ...$

The central maximum is the widest and brightest.
The first dark fringe occurs at $\sin\theta_1 = \lambda/b$ .
Narrower slit ( $b$ smaller) → wider diffraction pattern (more spreading).

Diffraction Gratings

A diffraction grating has many parallel slits, producing sharper, brighter maxima than a double-slit.

Grating equation (principal maxima):

$n\lambda = d\sin\theta$

where $d$ is the slit spacing ( $d = 1/\text{number of lines per metre}$ ) and $n$ is the order (0, 1, 2, …).

The maximum order is limited by $\sin\theta \leq 1$ , so $n_{\max} = d/\lambda$ .

$n$ in the grating equation is the order number, NOT the number of moles or refractive index. In IB papers all three appear — make sure you know from context which $n$ is which. In the grating equation specifically, $n$ is always the order of the maximum (0 = central, 1 = first order, etc.).

Exam command term “outline”: For interference experiments, a full “outline” answer earns marks for: (1) naming the setup (double-slit, grating), (2) stating the condition for constructive or destructive interference, (3) giving the relevant formula. Always write the formula.

Section 5: Standing Waves

Formation of Standing Waves

A standing (stationary) wave forms when two identical waves travelling in opposite directions superpose. Unlike a travelling wave, a standing wave does not transfer energy along the medium.

Nodes: points of zero displacement (permanent destructive interference)
Antinodes: points of maximum displacement (permanent constructive interference)
Adjacent nodes are separated by $\lambda/2$
The amplitude varies from zero (at nodes) to maximum (at antinodes)
All points between two adjacent nodes oscillate in phase but with different amplitudes

Harmonics in Strings and Pipes

A string fixed at both ends, or an open pipe, forms standing waves with nodes at the fixed ends (or antinodes at both open ends for an open pipe).

Stretched string (or open pipe) — both ends are nodes (or both antinodes for open pipe):

$L = \frac{n\lambda_n}{2} \quad n = 1, 2, 3, ... \qquad f_n = \frac{nv}{2L}$

$n = 1$ : fundamental (1st harmonic), $f_1 = v/(2L)$
$n = 2$ : 2nd harmonic, $f_2 = 2f_1$
$n = 3$ : 3rd harmonic, $f_3 = 3f_1$

Closed pipe (one open end, one closed end) — node at closed end, antinode at open end:

$L = \frac{(2n-1)\lambda_n}{4} \quad n = 1, 2, 3, ... \qquad f_n = \frac{(2n-1)v}{4L}$

Only odd harmonics are present: $f_1$ , $f_3$ , $f_5$ , …

Worked Example C5 — Harmonics in a string:

A guitar string of length $L = 0.64 \text{ m}$ has a wave speed $v = 256 \text{ m s}^{-1}$ .

(a) Find the fundamental frequency.

$f_1 = \frac{v}{2L} = \frac{256}{2 \times 0.64} = \frac{256}{1.28} = 200 \text{ Hz}$

(b) What is the wavelength of the 3rd harmonic?

$\lambda_3 = \frac{2L}{3} = \frac{2 \times 0.64}{3} = \frac{1.28}{3} \approx 0.43 \text{ m}$

(c) Find $f_3$ : $f_3 = 3f_1 = 3 \times 200 = 600 \text{ Hz}$ , or $f_3 = v/\lambda_3 = 256/0.427 \approx 600 \text{ Hz}$ .

Resonance occurs when a system is driven at a frequency that matches a natural frequency (harmonic). At resonance, the amplitude of oscillation is maximum and energy transfer from the driving source is most efficient.

Pipes often catch students out. For a closed pipe, the closed end is always a node and the open end is always an antinode — only odd harmonics can form. If an exam question mentions “a pipe closed at one end” and asks for the next resonant frequency above the fundamental, the answer is $3f_1$ (not $2f_1$ ). Forgetting this costs marks frequently.

Exam-Style Practice Questions

Paper 1 Style (MCQ)

Q1. Light of wavelength $\lambda$ passes through a double slit of separation $d$ . The interference pattern is observed on a screen at distance $D$ . The slit separation is then doubled. What happens to the fringe spacing?

A. It doubles.

B. It is halved.

C. It is unchanged.

D. It increases by a factor of 4.

Answer

B. Fringe spacing $s = \lambda D / d$ . If $d$ doubles, $s$ halves. All other variables ( $\lambda$ , $D$ ) are unchanged.

Q2. Which of the following correctly identifies the conditions needed for total internal reflection?

A. Light travelling from a less dense to a more dense medium at any angle.

B. Light travelling from a more dense to a less dense medium at an angle less than the critical angle.

C. Light travelling from a more dense to a less dense medium at an angle greater than the critical angle.

D. Light travelling along the normal at any interface.

Answer

C. TIR requires light to be in the denser medium (higher refractive index) and the angle of incidence must exceed the critical angle. If the angle equals the critical angle, the refracted ray travels along the boundary ( $\theta_r = 90°$ ). Below the critical angle, partial refraction occurs.

Paper 2 Style (Structured Response)

Q3. A source emits sound at frequency $f_s = 1200 \text{ Hz}$ . The speed of sound in air is $v = 340 \text{ m s}^{-1}$ .

(a) The source moves toward a stationary observer at $v_s = 25 \text{ m s}^{-1}$ . Calculate the frequency heard by the observer. [2]

(b) Calculate the frequency heard when the source moves away from the observer at the same speed. [1]

Mark-scheme answers

(a) Source approaching, observer stationary ( $v_o = 0$ ):

$f' = f_s \times \frac{v}{v - v_s} = 1200 \times \frac{340}{340 - 25} = 1200 \times \frac{340}{315} = 1200 \times 1.079 \approx 1295 \text{ Hz}$

[1 mark for correct formula with correct sign convention; 1 mark for correct answer]

(b) Source receding:

$f' = 1200 \times \frac{340}{340 + 25} = 1200 \times \frac{340}{365} \approx 1117 \text{ Hz}$

[1 mark for correct answer]

(c) When the source approaches, successive wavefronts are emitted closer together in the direction of travel (the wavelength in front of the source is compressed), so the observer receives more wavefronts per second — higher observed frequency. When the source recedes, the wavefronts are stretched out (longer effective wavelength) and the observer receives fewer per second — lower frequency. The drop occurs as the source passes.

[1 mark for compression/stretching of wavefronts; 1 mark for linking to increased/decreased frequency received]

Common Theme C errors that cost marks:

Measuring angles from the surface rather than from the normal in Snell’s Law — always use the angle with the normal (perpendicular to the interface).
Getting Doppler sign conventions wrong — write out the full formula and substitute; do not guess.
Confusing single-slit ( $\sin\theta = m\lambda/b$ ) with grating ( $n\lambda = d\sin\theta$ ) — both look similar but the physics is different.
Forgetting that only odd harmonics exist in a pipe closed at one end.
Stating that total internal reflection occurs when going from low- $n$ to high- $n$ medium — it only occurs going from high to low.
Using the grating formula for a double-slit fringe spacing question — use $s = \lambda D/d$ for fringe spacing; grating formula gives angle of maxima.

May 2026 Exam Predictions

Based on past IB Physics paper patterns, Theme C questions in May 2026 are likely to include:

Paper 1 MCQ: Snell’s Law / critical angle calculation (frequently tested), or a standing wave harmonics question (identify the correct harmonic diagram).
Paper 2 Short Answer: A double-slit or diffraction grating calculation with follow-up “explain the effect of” question (e.g., “explain the effect on the fringe pattern of using light with a longer wavelength”).
Doppler effect: Numerical calculation plus qualitative explanation — both parts appear regularly.
Standing waves in pipes/strings: Sketch the standing wave pattern and calculate frequency or wavelength. Closed-pipe odd-harmonics restriction is a reliable examiner favourite.

Drawing standing wave diagrams: IB examiners award a mark for each node and antinode correctly placed. When sketching, always: (1) mark all nodes with an N and antinodes with an A, (2) ensure nodes are at fixed ends (strings, closed pipe ends), antinodes at free/open ends, (3) draw the envelope of motion (not a snapshot of a single moment). Label the wavelength and length of the medium.