What are path difference conditions?

For two coherent sources, the type of interference at a point depends on the path difference x (the difference in distances travelled by the two waves):

What is the double-slit experiment?

Two slits a distance a apart, illuminated by light of wavelength λ, produce fringes on a screen a distance D away. The fringe separation is:

In a double-slit experiment with slit separation 0.30\ mm, screen distance 1.5\ m and wavelength 500\ nm, find the fringe separation. [2 marks]

SingaporePhysicsSyllabus dot point

How does the superposition of coherent waves produce predictable patterns of constructive and destructive interference?

State the principle of superposition, explain coherence and path difference, and apply them to two-source interference and the diffraction grating

A focused answer to the H2 Physics learning outcome on superposition and interference. The superposition principle, coherence, path difference conditions for maxima and minima, the double-slit fringe spacing, and the diffraction grating equation.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to state the principle of superposition, explain coherence and path difference, and apply these to two-source (double-slit) interference and the diffraction grating. Interference is the decisive evidence for the wave nature of light and a rich source of quantitative exam questions.

The answer

The principle of superposition

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

y_{\text{total}} = y_1 + y_2 + \dots

Where the waves arrive in phase, they reinforce (constructive interference); where they arrive in antiphase, they cancel (destructive interference).

Coherence and the conditions for interference

A stable interference pattern requires the sources to be coherent: they must have a constant phase difference (and therefore the same frequency). In practice this is achieved by deriving both beams from a single source, as in the double-slit experiment. With incoherent sources the phase relationship fluctuates and no steady pattern forms.

Path difference conditions

For two coherent sources, the type of interference at a point depends on the path difference $\Delta x$ (the difference in distances travelled by the two waves):

Constructive (bright): $\Delta x = m\lambda$ for integer $m$ (in phase).
Destructive (dark): $\Delta x = (m + \tfrac{1}{2})\lambda$ (antiphase).

The double-slit experiment

Two slits a distance $a$ apart, illuminated by light of wavelength $\lambda$ , produce fringes on a screen a distance $D$ away. The fringe separation is:

\Delta y = \frac{\lambda D}{a}

Wider slit separation gives closer fringes; longer wavelength gives wider fringes. Equally spaced bright and dark fringes are the signature of two-source interference.

The diffraction grating

A diffraction grating has many equally spaced slits a distance $d$ apart (the grating spacing). It produces sharp, bright maxima at angles given by:

d\sin\theta = m\lambda

where $m$ is the order of the maximum. The many slits make the maxima much sharper and brighter than the double-slit fringes, so a grating gives precise wavelength measurements. The highest observable order is limited by $\sin\theta \le 1$ .

Worked example

Light of wavelength $500\ \text{nm}$ passes through a diffraction grating with $500$ lines per millimetre. Find the angles of the first- and second-order maxima.

Step 1: Find the grating spacing

d = \frac{1\ \text{mm}}{500} = \frac{1 \times 10^{-3}}{500} = 2.0 \times 10^{-6}\ \text{m}

Step 2: Apply the grating equation for the first order

\sin\theta_1 = \frac{m\lambda}{d} = \frac{1 \times 500 \times 10^{-9}}{2.0 \times 10^{-6}} = 0.25 \implies \theta_1 = 14.5^\circ

Step 3: Apply the grating equation for the second order

\sin\theta_2 = \frac{2 \times 500 \times 10^{-9}}{2.0 \times 10^{-6}} = 0.50 \implies \theta_2 = 30.0^\circ

Step 4: State the results

The first-order maximum is at $14.5^\circ$ and the second-order maximum at $30.0^\circ$ . The orders are not equally spaced in angle, because $\sin\theta$ , not $\theta$ , is proportional to the order.

Common traps

Forgetting that interference needs coherent sources. Two independent bulbs do not interfere; the sources must keep a constant phase difference.

Using the wrong path-difference condition. Bright fringes need $\Delta x = m\lambda$ ; dark fringes need $\Delta x = (m + \tfrac{1}{2})\lambda$ .

Confusing slit separation $a$ (double slit) with grating spacing $d$ (grating). They play different roles; the grating spacing is the reciprocal of lines per unit length.

Assuming grating maxima are equally spaced in angle. It is $\sin\theta$ that is proportional to order $m$ , so the angular spacing widens with order.

Forgetting the limit on the highest order. Since $\sin\theta \le 1$ , the maximum order is $\lfloor d/\lambda \rfloor$ .

Examples in context

Example 1. Measuring a laser's wavelength. Shining a laser through a grating of known spacing and measuring the angle of the first-order maximum lets you find the wavelength from $\lambda = d\sin\theta$ . Because the grating maxima are sharp, this gives a far more precise wavelength than a double-slit measurement, which is why spectrometers use gratings.

Example 2. The colours of a CD. The closely spaced tracks on a CD act as a reflection diffraction grating. White light is split into its component wavelengths at different angles via $d\sin\theta = m\lambda$ , producing the familiar rainbow sheen. Different colours satisfy the grating condition at different angles.

Try this

Q1. State the principle of superposition. [1 mark]

Cue. When waves meet, the resultant displacement is the vector sum of the individual displacements.

Q2. In a double-slit experiment with slit separation $0.30\ \text{mm}$ , screen distance $1.5\ \text{m}$ and wavelength $500\ \text{nm}$ , find the fringe separation. [2 marks]

Cue. $\Delta y = \dfrac{\lambda D}{a} = \dfrac{(500 \times 10^{-9})(1.5)}{0.30 \times 10^{-3}} = 2.5 \times 10^{-3}\ \text{m} = 2.5\ \text{mm}$ .

Q3. Explain why a diffraction grating produces sharper maxima than a double slit. [2 marks]

Cue. Many slits contribute, so light from off-maximum angles cancels more completely, leaving narrow, intense maxima only where all slits are exactly in phase.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksIn a double-slit experiment, light of wavelength

600\ \text{nm}

illuminates two slits

0.50\ \text{mm}

apart. The fringes are observed on a screen

2.0\ \text{m}

away. (a) Find the fringe separation. (b) State the path difference at the third bright fringe from the centre.

Show worked answer →

(a) Fringe separation: $\Delta y = \dfrac{\lambda D}{a} = \dfrac{(600 \times 10^{-9})(2.0)}{0.50 \times 10^{-3}} = \dfrac{1.2 \times 10^{-6}}{5.0 \times 10^{-4}} = 2.4 \times 10^{-3}\ \text{m} = 2.4\ \text{mm}$ .

(b) Bright fringes occur where the path difference is a whole number of wavelengths. The third bright fringe (order $m = 3$ ) has path difference $3\lambda = 3 \times 600\ \text{nm} = 1800\ \text{nm} = 1.8\ \mu\text{m}$ .

Markers reward $\Delta y = \lambda D / a$ with consistent units, and the path difference for the third order as $3\lambda$ .

Original5 marksA diffraction grating has

300

lines per millimetre and is illuminated normally by light of wavelength

589\ \text{nm}

. (a) Find the grating spacing. (b) Find the angle of the first-order maximum. (c) Find the highest order observable.

Show worked answer →

(a) Grating spacing: $d = \dfrac{1}{300\ \text{mm}^{-1}} = \dfrac{1 \times 10^{-3}}{300}\ \text{m} = 3.33 \times 10^{-6}\ \text{m}$ .

(b) Grating equation $d\sin\theta = m\lambda$ with $m = 1$ : $\sin\theta = \dfrac{\lambda}{d} = \dfrac{589 \times 10^{-9}}{3.33 \times 10^{-6}} = 0.1768$ , so $\theta = 10.2^\circ$ .

(c) Highest order: $\sin\theta \le 1$ , so $m \le \dfrac{d}{\lambda} = \dfrac{3.33 \times 10^{-6}}{589 \times 10^{-9}} = 5.66$ . The highest whole order is $m = 5$ .

Markers reward the grating spacing as the reciprocal of lines per metre, the grating equation for the first-order angle, and the maximum order from the condition $\sin\theta \le 1$ rounded down.