Apply Wave Model — Apply C.2 Wave model in problem-solving...

SECTION 1

Historical Context & Motivation

For centuries, scientists debated whether light and sound were streams of particles or disturbances travelling through a medium. The wave model gradually won ground as experiments revealed behaviours—such as bending around corners and creating patterns of reinforcement and cancellation—that particles alone could not explain. Understanding this history helps you appreciate why the IB Physics C.2 syllabus asks you to apply the wave model rather than merely describe it: the model's real power lies in solving problems and predicting outcomes.

1678

Huygens' Wave Theory

Christiaan Huygens proposed that light travels as wavefronts, with every point on a wavefront acting as a source of secondary wavelets. This principle remains central to explaining diffraction and refraction today.

1801

Young's Double-Slit Experiment

Thomas Young demonstrated interference by passing light through two closely spaced slits, producing alternating bright and dark fringes that only a wave model could explain.

1864

Maxwell's Electromagnetic Theory

James Clerk Maxwell unified electricity, magnetism, and light into electromagnetic waves, predicting that they all travel at 3.0 × 10⁸ m s⁻¹ in a vacuum.

1900s

Wave–Particle Duality Emerges

Experiments on the photoelectric effect and electron diffraction showed that the wave model alone is incomplete—yet it remains the essential framework for analysing interference, diffraction, and standing waves in the IB course.

The question the C.2 topic poses is practical: given that waves superpose, diffract, and interfere, how do we use these properties to make quantitative predictions and explain observations? The rest of this lesson equips you to do exactly that.

SECTION 2

Core Principles of the Wave Model

Before you can apply the wave model to problems, you need to lock in five foundational ideas. Each one connects directly to the equations and diagrams you will use throughout the IB exam.

Superposition

When two or more waves meet at a point, the resultant displacement equals the algebraic sum of the individual displacements. This principle underlies all interference patterns.

Constructive & Destructive Interference

Constructive interference occurs when waves arrive in phase (path difference = nλ); destructive interference occurs when they are half a wavelength out of phase (path difference = (n + ½)λ).

Diffraction

Waves spread out when they pass through an aperture or around an obstacle. The effect is most pronounced when the gap width is comparable to the wavelength.

Path Difference & Phase Difference

Path difference (Δx) is the extra distance one wave travels compared to another. Phase difference (Δφ) = 2π × Δx / λ, linking geometry to wave behaviour.

Standing Waves

When two waves of equal frequency travel in opposite directions, they form a standing wave with fixed nodes (zero displacement) and antinodes (maximum displacement).

✦ KEY TAKEAWAY

Think of the wave model like a recipe: superposition is the rule for combining ingredients, path difference measures how much of each ingredient you add, and interference is the result—constructive when flavours complement, destructive when they cancel. Master the recipe and you can predict the outcome of any wave problem.

SECTION 3

Visualising Double-Slit Interference

The double-slit experiment is the classic demonstration of the wave model in action. The diagram below shows how two coherent sources (the slits) create overlapping wavefronts that produce an interference pattern on a distant screen.

A plane wave enters from the left and passes through two slits (S₁ and S₂) separated by distance d. Circular wavefronts (cyan from S₁, violet from S₂) overlap. On the distant screen at distance D, bright fringes appear where waves arrive in phase (constructive interference) and dark fringes where they cancel (destructive interference). The angle θ locates each fringe order n.

Notice that the central bright fringe (n = 0) sits directly opposite the midpoint between the slits. Moving away from the centre, the bright fringes become progressively dimmer because the single-slit diffraction envelope modulates the overall intensity. This interplay between interference (from two slits) and diffraction (from each slit's finite width) is precisely the kind of reasoning the IB expects when you apply the wave model to explain observed patterns.

SECTION 4

Mathematical Framework

The wave model becomes a powerful problem-solving tool once you connect the geometry of a setup to a handful of key equations. In the IB, you need to select and manipulate these relationships confidently.

WAVE SPEED EQUATION

v = f × λ

v = wave speed (m s⁻¹), f = frequency (Hz), λ = wavelength (m). This is the starting point for nearly every wave calculation.

DOUBLE-SLIT / DIFFRACTION GRATING (MAXIMA)

d sin θ = nλ

d = slit separation (m), θ = angle to the nth-order maximum, n = order number (0, 1, 2, …), λ = wavelength (m). This equation locates bright fringes. For destructive interference, replace nλ with (n + ½)λ.

FRINGE SPACING (SMALL-ANGLE APPROXIMATION)

s = λD / d

s = fringe spacing on the screen (m), D = distance from slits to screen (m). Valid when θ is small (sin θ ≈ tan θ ≈ θ). This is extremely useful for quick IB calculations.

SINGLE-SLIT DIFFRACTION (FIRST MINIMUM)

b sin θ = λ

b = slit width (m), θ = angle to the first minimum. For higher-order minima: b sin θ = mλ (m = 1, 2, 3, …). A narrower slit produces wider diffraction spreading.

💡 IB Exam Tip

Always check units before substituting values. Convert nm to m (1 nm = 1 × 10⁻⁹ m) and mm to m. Many marks are lost to unit errors, not physics errors.

SECTION 5

Interference & Diffraction Patterns Compared

IB questions frequently ask you to compare or identify different wave patterns. The diagram below places the single-slit diffraction envelope next to the double-slit interference pattern so you can see how they relate. In reality, the double-slit pattern is always contained within the single-slit envelope—this is why some bright fringes appear weaker or even missing.

Top: The single-slit diffraction pattern shows a broad central maximum and weaker side lobes. Bottom: The double-slit interference pattern (solid cyan) consists of equally spaced fringes whose amplitudes are modulated by the single-slit envelope (dashed amber). Notice that higher-order fringes are dimmer because the envelope falls off. If a fringe falls exactly on a diffraction minimum, it is a missing order.

Comparison of single-slit and double-slit patterns

Feature	Single-Slit Diffraction	Double-Slit Interference
Key equation	b sin θ = mλ (minima)	d sin θ = nλ (maxima)
Central maximum width	Twice as wide as side maxima	Same width as other fringes
Effect of narrowing the slit/gap	Pattern widens (more spreading)	Fringe spacing increases
Pattern shape	Smooth envelope, unequal maxima	Equally spaced fringes inside envelope

SECTION 6

Worked Example — Double-Slit Fringe Spacing

A laser of wavelength 632.8 nm illuminates two slits separated by 0.25 mm. The screen is placed 2.0 m from the slits. Calculate the fringe spacing on the screen and determine the angle to the second-order maximum.

Double-Slit Fringe Spacing & Angle

Step 1 — List Known Values & Convert Unitsλ = 632.8 nm = 632.8 × 10⁻⁹ m = 6.328 × 10⁻⁷ m. Slit separation d = 0.25 mm = 2.5 × 10⁻⁴ m. Screen distance D = 2.0 m.

Step 2 — Apply the Fringe Spacing FormulaUsing s = λD / d: s = (6.328 × 10⁻⁷ m × 2.0 m) / (2.5 × 10⁻⁴ m) = 1.2656 × 10⁻⁶ / 2.5 × 10⁻⁴ = 5.06 × 10⁻³ m

s ≈ 5.1 mm

Step 3 — Find the Angle to the Second-Order MaximumUsing d sin θ = nλ with n = 2: sin θ = nλ / d = (2 × 6.328 × 10⁻⁷) / (2.5 × 10⁻⁴) = 1.2656 × 10⁻⁶ / 2.5 × 10⁻⁴ = 5.06 × 10⁻³ θ = sin⁻¹(5.06 × 10⁻³)

θ ≈ 0.29° (or about 5.1 × 10⁻³ rad)

Step 4 — Interpret the ResultsThe fringes are about 5 mm apart—easily visible on a lab screen. The very small angle confirms that the small-angle approximation (sin θ ≈ θ) was valid, since sin θ and θ in radians agree to three significant figures here.

✓ Check Your Work

If you had used the exact formula d sin θ = nλ and then computed the screen position y = D tan θ, you would get the same 5.1 mm spacing. The small-angle approximation saves a step and is accepted on IB exams when θ < about 10°.

SECTION 7

Strengths & Limitations of the Wave Model

The wave model is remarkably powerful, but it has clear boundaries. Understanding both sides is essential for IB Paper 2 and Paper 3 responses, where examiners often ask you to evaluate the suitability of a model.

Strengths vs limitations of the classical wave model

Strengths	Limitations
Accurately predicts interference and diffraction patterns for all wave types (light, sound, water)	Cannot explain the photoelectric effect or blackbody radiation (requires the photon/quantum model)
Allows quantitative calculation of fringe spacing, angles, and wavelengths	Assumes coherent, monochromatic sources; real sources often require more complex treatment
Explains polarisation, providing evidence that light is a transverse wave	Does not describe energy quantisation or particle-like behaviour of photons at very low intensities
Applies equally to mechanical (sound, seismic) and electromagnetic waves	Cannot predict the direction of individual photon detections in the double-slit experiment

✦ KEY TAKEAWAY

Think of the wave model as a detailed map of a city. It shows you streets, distances, and intersections brilliantly—but it can't tell you what's happening inside individual buildings. For that, you need a different map (the quantum model). In the IB, knowing when to use each model is just as important as knowing how to use it.

SECTION 8

Connection to Advanced & HL Topics

The wave model you have mastered in C.2 is the launching pad for several more advanced ideas you may encounter in IB HL or at university. Recognising these connections now will deepen your understanding and help you answer synoptic exam questions.

How C.2 concepts connect to higher-level physics

C.2 Wave Model Concept	Advanced Extension	Where You'll See It
Superposition of two waves	Fourier analysis: any wave shape decomposed into sine components	IB HL Option / University physics
d sin θ = nλ (double slit)	Diffraction grating with N slits; resolving power R = mN	IB C.3 / Spectroscopy
Single-slit diffraction (b sin θ = λ)	Rayleigh criterion for resolution: θ₁ = 1.22 λ / b	IB C.3 / Telescope optics
Standing waves on strings	Quantum mechanical standing waves (particle in a box, electron orbitals)	IB E.1 / University quantum mechanics

The key insight is that the same mathematical machinery—superposition plus geometry—scales beautifully. Whether you are analysing a guitar string, an X-ray crystallography pattern, or an electron's probability wave, you are applying the same wave model you learn in C.2. Building rock-solid skills here gives you a head start on every one of these topics.

SECTION 9

Practice Problems

PROBLEM 1 — CONCEPTUAL

Two loudspeakers emit identical sound waves. A listener standing equidistant from both speakers hears a loud sound. She then moves to a point where she hears almost nothing. Using the wave model, explain why the sound nearly disappears at the new position.

PROBLEM 2 — BASIC CALCULATION

Monochromatic light of wavelength 520 nm passes through a double slit with slit separation 0.40 mm. A screen is placed 1.5 m away. Calculate the fringe spacing on the screen.

PROBLEM 3 — INTERMEDIATE

A single slit of width 0.10 mm is illuminated by light of wavelength 600 nm. Calculate the angular position of the first diffraction minimum. Then determine the width of the central maximum on a screen 3.0 m away.

PROBLEM 4 — APPLIED

A diffraction grating with 600 lines per mm is used to analyse light from a sodium lamp. The yellow doublet of sodium has wavelengths 589.0 nm and 589.6 nm. Calculate the angle of the first-order maximum for 589.0 nm and explain whether the first-order spectrum can resolve the two lines (use the grating's resolving power R = mN, where N is the total number of illuminated slits for a 1.0 cm wide grating).

PROBLEM 5 — CRITICAL THINKING

A student sets up a double-slit experiment and measures 12 bright fringes across the central diffraction maximum. If the slit separation d is six times the slit width b (d = 6b), use the wave model to explain why exactly 11 bright fringes (plus 2 missing orders) appear within the central diffraction envelope. What would happen to the pattern if d were increased to 8b while keeping b constant?

SUMMARY

Lesson Summary

The wave model explains a wide range of phenomena by treating disturbances as oscillations that obey superposition. When waves overlap, they produce constructive interference (path difference = nλ) and destructive interference (path difference = (n + ½)λ). For the double slit, use d sin θ = nλ to locate maxima and s = λD / d for fringe spacing. For single-slit diffraction, the first minimum satisfies b sin θ = λ.

Applying the model means: (1) identifying whether the situation involves interference, diffraction, or standing waves; (2) selecting the correct equation; (3) converting units carefully; and (4) interpreting results physically. Remember the model's limitations—it cannot explain phenomena that require the quantum/photon model. Mastery of C.2 means you can move fluently between diagrams, equations, and physical explanations to tackle any wave-behaviour question the IB throws at you.