A double-slit pattern has fringes 1.5\ mm apart with slits 0.50\ mm apart and a screen 1.2\ m away. Find the wavelength. [2 marks]

OCR OCR 2018-style practice: In a double-slit experiment the slit separation is 0.40\ mm and the screen is 1.8\ m away. The bright fringes are 2.7\ mm apart. Calculate the wavelength of the light used.

Use the double-slit fringe equation w = (λ D)/(a), rearranged for wavelength: λ = (wa)/(D). Substitute with consistent units: λ = (2.7 × 10^-3)(0.40 × 10^-3)1.8. Numerator: (2.7 × 10^-3)(0.40 × 10^-3) = 1.08 × 10^-6. So λ = 1.08 × 10^-61.8 = 6.0 × 10^-7\ m. Markers reward rearranging the fringe equation, converting all distances to metres, and the value 6.0 × 10^-7\ m (about 600 nm, orange light).

EnglandPhysicsSyllabus dot point

How do waves combine, and what does interference reveal about the wave nature of light?

Superposition and interference: the principle of superposition, stationary waves on strings and in pipes, coherence and path difference, the two-source (Young's) double-slit experiment, and the diffraction grating equation.

A focused answer to the OCR H556 content on superposition and interference, covering the principle of superposition, stationary waves and harmonics on strings and in pipes, coherence and path difference, the two-source double-slit experiment with the fringe equation, and the diffraction grating equation.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

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

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What this dot point is asking

OCR wants you to apply the principle of superposition, describe stationary waves and their harmonics on strings and in pipes, explain coherence and path difference, account for the two-source (Young's) double-slit pattern and use the fringe equation, and apply the diffraction grating equation.

The answer

The principle of superposition

A stable, observable interference pattern requires coherent sources: sources with the same frequency and a constant phase difference. Light from two separate lamps is incoherent, which is why a single source is split to make the two coherent beams.

Stationary waves

On a string fixed at both ends, the fundamental (first harmonic) has a node at each end and one antinode in the middle, so the string length equals half a wavelength: $L = \frac{\lambda}{2}$ , giving $\lambda = 2L$ and frequency $f_1 = \frac{v}{2L}$ . Higher harmonics occur at integer multiples, $f_n = nf_1$ .

Two-source interference and the double slit

The fringe equation is valid when $D$ is much greater than $a$ , which is why the screen is far from the slits.

The diffraction grating

Wavelength from a string harmonic

A string of length $0.80\ \text{m}$ fixed at both ends vibrates in its third harmonic at a frequency of $375\ \text{Hz}$ . Find the speed of the waves on the string.

step 1: Wavelength of the third harmonic

The third harmonic fits three half-wavelengths in the length: $L = \frac{3\lambda}{2}$ , so $\lambda = \frac{2L}{3} = \frac{2(0.80)}{3} = 0.533\ \text{m}$ .

step 2: Apply the wave equation

$v = f\lambda = 375 \times 0.533 = 200\ \text{m s}^{-1}$ .

step 3: Check

The fundamental would be $f_1 = \frac{v}{2L} = \frac{200}{1.6} = 125\ \text{Hz}$ , and $3 \times 125 = 375\ \text{Hz}$ , confirming the third harmonic.

Examples in context

Diffraction gratings are at the heart of spectrometers, splitting starlight or a flame's emission into its component wavelengths to identify elements. Musical instruments rely on stationary waves: a guitar string and an organ pipe set up harmonics whose frequencies determine the pitch. Noise-cancelling headphones use destructive interference, generating a sound wave in antiphase with the unwanted noise. Anti-reflection coatings on lenses use thin-film interference to cancel reflected light.

Try this

Q1. State the principle of superposition. [2 marks]

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

Q2. A double-slit pattern has fringes $1.5\ \text{mm}$ apart with slits $0.50\ \text{mm}$ apart and a screen $1.2\ \text{m}$ away. Find the wavelength. [2 marks]

Cue. $\lambda = \frac{wa}{D} = \frac{(1.5 \times 10^{-3})(0.50 \times 10^{-3})}{1.2} = 6.3 \times 10^{-7}\ \text{m}$ .

Q3. State the two conditions for two sources to be coherent. [2 marks]

Cue. They must have the same frequency and a constant phase difference.

Exam-style practice questions

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

OCR 20184 marksIn a double-slit experiment the slit separation is

0.40\ \text{mm}

and the screen is

1.8\ \text{m}

away. The bright fringes are

2.7\ \text{mm}

apart. Calculate the wavelength of the light used.

Show worked answer →

Use the double-slit fringe equation $w = \frac{\lambda D}{a}$ , rearranged for wavelength: $\lambda = \frac{wa}{D}$ .

Substitute with consistent units: $\lambda = \frac{(2.7 \times 10^{-3})(0.40 \times 10^{-3})}{1.8}$ .

Numerator: $(2.7 \times 10^{-3})(0.40 \times 10^{-3}) = 1.08 \times 10^{-6}$ . So $\lambda = \frac{1.08 \times 10^{-6}}{1.8} = 6.0 \times 10^{-7}\ \text{m}$ .

Markers reward rearranging the fringe equation, converting all distances to metres, and the value $6.0 \times 10^{-7}\ \text{m}$ (about 600 nm, orange light).

OCR 20224 marksMonochromatic light of wavelength

5.9 \times 10^{-7}\ \text{m}

is shone normally onto a diffraction grating with

300

lines per millimetre. Calculate the angle of the first-order maximum and state the highest order that can be observed.

Show worked answer →

Slit spacing: $d = \frac{1}{300 \times 10^{3}} = 3.33 \times 10^{-6}\ \text{m}$ .

First order ( $n = 1$ ): $\sin\theta = \frac{n\lambda}{d} = \frac{5.9 \times 10^{-7}}{3.33 \times 10^{-6}} = 0.177$ , so $\theta = 10.2^{\circ}$ .

Highest order: $n_{\max} = \frac{d}{\lambda} = \frac{3.33 \times 10^{-6}}{5.9 \times 10^{-7}} = 5.6$ , so $n = 5$ (must be a whole number and $\sin\theta \le 1$ ).

Markers reward finding $d$ from the line density, the first-order angle about $10^{\circ}$ , and rounding the maximum order down to $5$ .

Related dot points

Sources & how we know this

OCR AS and A Level Physics A (H156, H556) specification — OCR (2015)