A-Level H2 Physics: Superposition of Waves — One Principle Behind Every Interference and Diffraction Question

Why Superposition Costs JC Students Marks

The Superposition of Waves chapter appears in almost every H2 Physics paper — and it is one of the chapters where students who know the formulas still drop marks. Why? Because the standard approach is to memorise three separate sets of formulae: one for Young's double slit, one for diffraction gratings, one for stationary waves. Each feels like a different topic, so each exam question feels new and risky.

The insight that changes everything is this: every single phenomenon in this chapter follows from one physical idea — path difference determines whether waves add up or cancel out. Once you see that, the chapter collapses from three disconnected topics into one principle with three different geometries.

The Unifying Principle: Path Difference

The superposition principle states that when two or more waves overlap, the resultant displacement at any point is the vector sum of the individual displacements. That is not a formula to memorise; it is a physical statement about what waves do.

The consequence you use in every exam question is:

Two waves arrive at a point. If their crests arrive together — they are in phase — they add up. This is constructive interference, and the resultant amplitude doubles.
If one crest arrives with the other's trough — they are in antiphase — they cancel. This is destructive interference, and the resultant amplitude falls to zero.
What determines which case you are in? The path difference: the difference in distance each wave has travelled to reach that point.

Written as rules:

Path difference = nλ → constructive (n = 0, 1, 2, …)
Path difference = (n + ½)λ → destructive

That is the whole chapter. Everything else is geometry.

What Is Coherence — and Why the Exam Tests It

Before interference can occur, the two sources must be coherent: same frequency, and a constant phase relationship between them. In practice this means they must originate from the same source, or have a phase relationship that does not change with time.

This is why a sodium lamp illuminating two slits produces clear fringes, while two separate lamps do not: independent sources have constantly shifting phase relationships, so the interference pattern washes out to uniform brightness.

The exam loves asking: “Why must the sources be coherent?” The mark-worthy answer is not simply “same frequency”. It is: same frequency and a constant phase difference, so that a stable interference pattern can form. Stating only “same frequency” will lose you the second mark.

Young's Double Slit — Path Difference Applied to Two Slits

Setup: two narrow slits separated by distance a, a screen at distance D, monochromatic light of wavelength λ.

At a point P on the screen, displaced a distance x from the centre, the two waves have travelled slightly different distances. For small angles, the path difference is:

Path difference ≈ ax / D

Apply the path-difference rule:

Bright fringe (constructive): ax / D = nλ → x = nλD / a
Dark fringe (destructive): ax / D = (n + ½)λ

Fringe spacing (subtract consecutive bright positions): Δx = λD / a

Notice that you do not need to memorise a separate fringe-spacing formula. You derive it in two lines from the path-difference condition. This is what “not memorising” looks like in practice — you derive the result from the principle, so you can never misremember it.

Worked Example 1

In a Young's double slit experiment, slit separation a = 0.50 mm, screen distance D = 1.8 m, fringe spacing Δx = 3.6 mm. Calculate the wavelength of light used, and comment on whether it is visible.

From Δx = λD / a:

λ = Δx × a / D = (3.6 × 10⁻³)(0.50 × 10⁻³) / 1.8 = 1.0 × 10⁻⁶ m = 1000 nm

Comment: visible light spans roughly 380–700 nm. At 1000 nm this is infrared, so it would not be visible to the naked eye.

That final comment is worth a mark, and most students skip it. Examiners write “comment on your answer” precisely to test this habit.

Effect-of-X Questions: Reason, Do Not Recall

The H2 exam frequently asks: “The screen is moved further away. What happens to the fringe spacing?” Students who memorised a list of effects are in trouble as soon as the wording changes. Students who reason from Δx = λD / a are not.

Increase D (screen further away): D is in the numerator → Δx increases → fringes spread out.
Increase a (wider slit separation): a is in the denominator → Δx decreases → fringes closer together.
Longer wavelength (red vs blue): λ in numerator → Δx larger → red fringes more spread out than blue.
White light: each wavelength gives its own fringe pattern superimposed → the central fringe is white (all wavelengths satisfy zero path difference at x = 0) and outer fringes show colour dispersion, with violet on the inside and red on the outside of each coloured fringe.

The method is always the same: locate the changed variable in λD / a, decide whether it is numerator or denominator, and state the effect. Three seconds per question, no memorisation required.

Diffraction Grating — Many Slits, Sharper Peaks

A diffraction grating has N slits per metre, giving a slit spacing d = 1 / N. Monochromatic light passes through and diffracts. Bright orders appear at angles where all slits reinforce each other simultaneously.

The condition for a principal maximum of order n:

d sin θ = nλ (n = 0, ±1, ±2, …)

The aha moment: this is the same path-difference logic as before. Adjacent slits are d apart. The path difference between the waves from two adjacent slits arriving at angle θ is d sin θ. If that path difference equals nλ, then every pair of adjacent slits is in phase at that angle, so all N slits reinforce simultaneously → a strong, sharp bright line.

This is why grating peaks are far sharper than double-slit fringes: with N slits, any slight deviation from d sin θ = nλ causes partial cancellation from many slit-pairs at once. The peaks are narrow and intense; the regions between them are effectively dark. Exam sketches of intensity patterns must show this: grating peaks are narrow spikes, double-slit fringes are broader and more evenly spread.

Worked Example 2

A diffraction grating has 600 lines per mm. Light of wavelength 589 nm is incident normally. Find (a) the angle of the second-order maximum, and (b) the highest order visible.

Slit spacing: d = 1 / (600 × 10³ m⁻¹) = 1.667 × 10⁻⁶ m

(a) Second order (n = 2):

sin θ = nλ / d = 2(589 × 10⁻⁹) / (1.667 × 10⁻⁶) = 0.7068

θ = 44.9°

(b) Highest order:

Maximum possible is sin θ = 1, so n_max = d / λ = 1.667 × 10⁻⁶ / 589 × 10⁻⁹ = 2.83

Taking the integer part: highest order = 2.

Always verify by checking the next order: at n = 3, sin θ = 3 × (589 × 10⁻⁹) / (1.667 × 10⁻⁶) = 1.06 > 1, which is impossible. Examiners deduct the mark if you state the highest order without confirming that the next order requires sin θ > 1.

Stationary Waves — Interference Between Two Opposing Waves

A stationary (standing) wave forms when two identical waves travel in opposite directions through the same medium. Unlike the previous sections — where waves from two sources travel outward to a screen — here the “two sources” are the original wave and its own reflection.

The result is a wave pattern that does not travel. Some points remain permanently at zero displacement: these are nodes. Others oscillate with maximum amplitude: these are antinodes.

Key relationships — derived rather than memorised:

Node-to-node distance = λ / 2. The stationary wave has the same wavelength as the component waves. A sine wave crosses zero, rises to maximum, returns to zero, falls to minimum, returns to zero — completing one full wavelength between the first and third zero crossings. The node-to-node distance is half that span: λ / 2.
Node to adjacent antinode = λ / 4. Halfway between two adjacent nodes.

Draw this once on a piece of paper — a standing sine wave with nodes and antinodes labelled — and you will never need to memorise the λ / 2 relationship again.

String and Pipe Resonance: Boundary Conditions First

The exam extends stationary waves to vibrating strings and air columns in pipes. The key is to apply boundary conditions before anything else:

Fixed end of a string: must be a node — the string cannot move there.
Open end of a pipe: must be an antinode — the air molecules have maximum freedom of movement (displacement antinode; pressure node).
Closed end of a pipe: must be a node — the air molecules cannot move at the wall (displacement node; pressure antinode).

Allowed resonance modes follow directly from counting how many half-wavelengths fit between the boundary conditions:

String of length L, fixed both ends: L = nλ / 2 (n = 1, 2, 3, …) → f_n = nv / 2L → all harmonics present.
Closed pipe (one closed end): L = nλ / 4 (n = 1, 3, 5, …) → odd harmonics only.
Open pipe (both ends open): L = nλ / 2 (n = 1, 2, 3, …) → all harmonics present.

Rather than memorising these three sets of conditions separately, sketch the wave pattern for each situation, mark the boundary nodes and antinodes, and count the half-wavelengths that fit. The formula falls out of the sketch every time.

Worked Example 3

A string of length 0.80 m is fixed at both ends. It vibrates in its third harmonic at wave speed 240 m/s. Find the frequency.

Third harmonic, fixed both ends: L = 3λ / 2

λ = 2L / 3 = 2(0.80) / 3 = 0.533 m

f = v / λ = 240 / 0.533 = 450 Hz

No additional formula is needed beyond v = fλ and the boundary condition. A student who memorised f_n = nv / 2L gets the same numerical answer, but cannot adapt if the question changes to a closed pipe or introduces a different boundary condition mid-question.

All Three Contexts Side by Side

Seeing the three applications together makes their common structure visible:

Young's double slit: 2 coherent slits → screen → condition ax / D = nλ → a travelling intensity pattern on the screen.
Diffraction grating: N coherent slits → condition d sin θ = nλ → sharp, bright orders at specific angles.
Stationary wave: wave + its reflection → condition: nodes at fixed ends, antinodes at open ends → a non-travelling wave with fixed nodes and antinodes.

In every case: identify the geometry, write down the path-difference condition, and solve. The “three different topics” feeling disappears once you see them through this single lens.

Seven Exam Traps That Cost JC Students Marks

Getting the slit spacing wrong in grating questions. d = 1 / N where N is lines per metre. If the question gives lines per mm, convert first: 600 lines/mm = 600,000 lines/m → d = 1.667 × 10⁻⁶ m. A wrong d invalidates every subsequent step.
Defining coherence as “same frequency” only. Full definition: same frequency and constant phase difference. Same amplitude is not required for coherence.
Confusing path difference with phase difference. They are related but not the same. A path difference of λ corresponds to a phase difference of 2π. In general: phase difference = (2π / λ) × path difference. Some questions give phase difference; convert before applying the constructive or destructive criteria.
Forgetting to verify sin θ ≤ 1 for the highest grating order. Always compute sin θ for the next order above your answer and confirm it exceeds 1. This verification is worth its own mark in structured questions.
Claiming nodes have no energy. Nodes have zero displacement and zero kinetic energy, but the string has maximum elastic potential energy at the nodes (the displacement gradient is steepest there). Maximum kinetic energy sits at the antinodes; maximum elastic potential energy at the nodes. Data-response questions on energy distribution in stationary waves exploit exactly this misconception.
Misidentifying displacement vs pressure nodes in pipe questions. A closed end is a displacement node but a pressure antinode (the air is compressed and rarefied most at a wall). Draw displacement and pressure diagrams separately if you are unsure; they are λ / 4 out of phase with each other.
Stating that the central fringe in white-light double-slit is coloured. At the central position, path difference = 0 for every wavelength simultaneously, so every wavelength interferes constructively. The central fringe is white. The coloured dispersion begins in the first-order fringes, with violet closest to centre and red furthest out.

How to Practise This Chapter Effectively

1. Anchor Every Calculation in Path Difference

Before writing any formula, ask: what is the path difference, and what condition does it satisfy? Write that down. The formula is just the algebraic rearrangement of the condition for the specific geometry. Getting into this habit means you can reconstruct any formula from scratch under exam pressure.

2. Derive the Fringe-Spacing Formula Each Time You Use It

Write: “constructive when ax / D = nλ, so x_n = nλD / a, fringe spacing Δx = λD / a.” Do this every time for the first ten questions. You will be faster than students who try to recall the formula directly, and immune to the common sign or variable errors that arise from misremembering.

3. Sketch Before You Calculate for Stationary Wave Questions

Draw the standing wave pattern, mark the boundary conditions, label the nodes and antinodes, and count the half-wavelengths. Write L in terms of λ from the sketch. Only then apply v = fλ. Students who skip the sketch miscount the harmonics under time pressure.

4. Build a Variable-Change Reflex

When a question asks “what happens if the wavelength doubles?” or “the grating has fewer lines per mm”, treat it as: locate the variable in the governing equation, decide numerator or denominator, state the directional effect, and quantify it if the question asks. Never rely on a remembered list of effects — the list is too easy to confuse.

5. Past-Paper Pattern Recognition

Singapore H2 Physics past papers show a consistent structure for this topic: a numerical grating or double-slit calculation; a qualitative “explain or describe the effect of” question; and a stationary-wave resonance question involving a string or pipe. Working ten past-paper questions across these three types — rather than doing ten of one type — covers the realistic range of what can appear on your paper.

The Exam-Day Mindset

When you encounter any wave question in the exam, ask yourself one question: what is the path difference, and does it satisfy the constructive or destructive condition?

Everything else — the fringe positions, the diffraction orders, the resonance frequencies — follows from the geometry once that question is answered. If you find yourself reaching for a half-remembered formula, stop. Go back to path difference and derive from there. It costs thirty extra seconds and is far safer than an incorrect recalled formula.

This is the approach we use at Intuitional with our JC H2 Physics students: one principle, understood clearly, applied systematically across every context. The marks follow from the understanding.