Diffraction

When a wave passes through a gap or around an obstacle, it spreads out. This spreading is called diffraction, and it is one of the most compelling demonstrations that light behaves as a wave. While single-slit diffraction produces a broad central maximum flanked by weaker subsidiary maxima, diffraction gratings produce sharp, well-separated maxima that are immensely useful in spectroscopy.

What Is Diffraction?

Diffraction is the spreading of a wave as it passes through a gap (aperture) or around an obstacle. All waves diffract, but the effect is most noticeable when the size of the gap or obstacle is comparable to the wavelength of the wave.

If the gap is much larger than the wavelength, the wave passes through with very little spreading — diffraction is negligible.
If the gap is comparable to the wavelength, the wave spreads significantly after passing through.
If the gap is much smaller than the wavelength, the wave spreads out almost uniformly in all directions, like a circular wavefront emerging from a point source.

This is why sound (wavelength ~1 m) diffracts easily around doorways and corners, while light (wavelength ~500 nm) does not noticeably diffract around everyday objects — the objects are millions of wavelengths across.

Single-Slit Diffraction

When monochromatic light passes through a single narrow slit, a characteristic diffraction pattern appears on a screen behind it. The pattern consists of:

A central maximum (bright band) that is wide and bright
Subsidiary maxima on either side that are progressively narrower and dimmer
Minima (dark bands) between the maxima

The central maximum is twice as wide as any of the subsidiary maxima and contains most of the light's intensity (about 84% of the total).

Effect of Slit Width

Narrower slit: The central maximum becomes wider and dimmer. The wave spreads out more because the gap is closer in size to the wavelength. The subsidiary maxima also spread out.
Wider slit: The central maximum becomes narrower and brighter. Less diffraction occurs. In the extreme (very wide slit), you simply see a bright patch the same width as the slit — geometric optics.

Effect of Wavelength

Longer wavelength (e.g. red light): The central maximum is wider. More spreading occurs.
Shorter wavelength (e.g. blue light): The central maximum is narrower. Less spreading.

Change	Effect on single-slit pattern
Decrease slit width	Central maximum widens and dims
Increase slit width	Central maximum narrows and brightens
Increase wavelength	Central maximum widens
Decrease wavelength	Central maximum narrows

Diffraction Gratings

A diffraction grating is a plate with a large number of equally spaced, parallel slits. A typical grating might have 300 to 600 slits per millimetre (sometimes expressed as "lines per mm"). When monochromatic light passes through a grating, it produces a pattern of sharp, bright maxima (called orders) at specific angles.

The key equation for a diffraction grating is:

$d \sin\theta = n\lambda$

where:

d = the slit spacing (distance between adjacent slits), in metres
θ = the angle between the central maximum (n = 0) and the nth order maximum
n = the order number (0, 1, 2, 3, ...)
λ = the wavelength of the light

If the grating has N lines per metre, then d = 1/N. If the grating has N lines per millimetre, then d = 1/(1000N) metres.

graph LR
    A["Monochromatic\nlight source"] --> B["Diffraction\ngrating"]
    B --> C["n=0 (central maximum)\nθ = 0°"]
    B --> D["n=1 (1st order)\nθ₁ = sin⁻¹(λ/d)"]
    B --> E["n=2 (2nd order)\nθ₂ = sin⁻¹(2λ/d)"]
    B --> F["n=3 (if sin θ ≤ 1)\nθ₃ = sin⁻¹(3λ/d)"]

Worked Examples: Diffraction Grating Calculations

Worked example 1: A diffraction grating has 500 lines per mm. Monochromatic light of wavelength 589 nm is shone through it. Calculate the angle of the first-order maximum.

First find d: d = 1/(500 × 10³) = 1/(5.0 × 10⁵) = 2.0 × 10⁻⁶ m

d sin θ = nλ

sin θ₁ = nλ/d = (1 × 589 × 10⁻⁹) / (2.0 × 10⁻⁶) = 0.2945

θ₁ = sin⁻¹(0.2945) = 17.1°

Worked example 2: Using the same grating, find the maximum order visible.

The maximum order occurs when sin θ = 1 (θ = 90°):

n_max = d/λ = 2.0 × 10⁻⁶ / 589 × 10⁻⁹ = 3.40

Since n must be a whole number, the maximum order is n = 3. You would see orders 0, ±1, ±2, and ±3 — a total of 7 maxima.

Worked example 3: A grating produces a second-order maximum at 32.0° for a particular wavelength. The grating has 600 lines per mm. Calculate the wavelength.

d = 1/(600 × 10³) = 1.667 × 10⁻⁶ m

λ = d sin θ / n = (1.667 × 10⁻⁶ × sin 32.0°) / 2 = (1.667 × 10⁻⁶ × 0.5299) / 2

λ = 8.833 × 10⁻⁷ / 2 = 4.42 × 10⁻⁷ m = 442 nm

This is violet/blue light.

Worked example 4: A diffraction grating with 300 lines per mm is illuminated with red light of wavelength 650 nm. Calculate the angular separation between the first-order and second-order maxima.

d = 1/(300 × 10³) = 3.333 × 10⁻⁶ m

sin θ₁ = λ/d = 650 × 10⁻⁹ / 3.333 × 10⁻⁶ = 0.1950 → θ₁ = 11.25°

sin θ₂ = 2λ/d = 2 × 650 × 10⁻⁹ / 3.333 × 10⁻⁶ = 0.3900 → θ₂ = 22.95°

Angular separation = 22.95° − 11.25° = 11.7°

Why Gratings Give Sharp Maxima

A double slit produces broad fringes because there are only two sources interfering. A grating with hundreds of slits produces much sharper maxima because:

At the exact angle satisfying d sin θ = nλ, all slits contribute waves that are in phase — the amplitude is N times that of a single slit (where N is the number of slits).
At angles very slightly away from this, the waves from the many slits quickly get out of step and cancel — destructive interference occurs rapidly.

The more slits the grating has, the sharper and brighter the maxima, and the narrower the dark regions between them. This is why gratings are superior to double slits for precise wavelength measurements.

Diffraction Gratings and Spectra

When white light passes through a diffraction grating, each wavelength is diffracted to a different angle (since θ depends on λ). The result is a spectrum in each order.

The zero-order maximum (n = 0) is white — all wavelengths converge at θ = 0°
Each higher order shows a spectrum with violet on the inside (closer to n = 0) and red on the outside (since red has the longest wavelength and sin θ = nλ/d is largest for red)
Higher orders spread the spectrum over a wider angle, giving better resolution but dimmer light

If several spectral lines are present (e.g., from a hydrogen discharge tube), the grating separates them cleanly, making it an essential tool in spectroscopy.

Comparing Double-Slit and Diffraction Grating Patterns

Feature	Double slit	Diffraction grating
Number of sources	2	Hundreds to thousands
Maxima sharpness	Broad, fuzzy fringes	Sharp, bright maxima
Angular positions	Same formula: d sin θ = nλ	Same formula: d sin θ = nλ
Intensity of maxima	Relatively dim	Very bright (N² times single slit)
Use in spectroscopy	Poor — fringes too broad	Excellent — separates wavelengths cleanly
Central maximum	Broad	Sharp

The Single-Slit Envelope

In a double-slit or multi-slit experiment, the individual slit width also matters. Each slit produces its own single-slit diffraction pattern, and the multi-slit interference fringes are modulated by this single-slit envelope. If an interference maximum falls at the same angle as a single-slit minimum, that order is "missing" — it has zero intensity. This explains the observed pattern of missing orders in some experiments.

Practical Applications of Diffraction

CD/DVD/Blu-ray surfaces: The closely spaced tracks on an optical disc act as a reflection diffraction grating, producing rainbow colours when white light is reflected.

X-ray crystallography: X-rays have wavelengths comparable to atomic spacing (~0.1 nm). When X-rays pass through a crystal, the regular lattice of atoms acts as a 3D diffraction grating, producing a pattern that reveals the crystal structure. Bragg's law, nλ = 2d sin θ, is used. This technique revealed the structure of DNA.

Spectrometers: Diffraction gratings are used in spectrometers to measure the wavelengths present in a light source, enabling identification of elements (from emission spectra) and analysis of stars.

Common exam mistake: Students often confuse the slit separation d (in the grating equation d sin θ = nλ) with the slit width. The slit separation is the distance from the centre of one slit to the centre of the next. The slit width affects the single-slit diffraction envelope but does not appear in the grating equation.

Another common error: Forgetting to convert lines per mm to slit spacing in metres. If a grating has 500 lines/mm, then d = 1/(500 × 10³) = 2.0 × 10⁻⁶ m, not 1/500 = 0.002 m.

A-Level Deep Dive: Diffraction

Spec mapping

Edexcel 9PH0 specification Topic 5 — Waves and the Particle Nature of Light, in its sub-strands on diffraction, requires students to recall that waves spread on passing through a gap or around an obstacle (most markedly when the gap is comparable to the wavelength); describe the single-slit pattern produced by monochromatic light (central maximum, subsidiary maxima, minima); use the grating equation $d\sin\theta = n\lambda$ ; and discuss how grating spectra are formed (refer to the official Pearson Edexcel specification document for exact wording). The grating equation is examined synoptically with the photon model ( $E = hc/\lambda$ ) when grating-measured wavelengths are converted to photon energies for emission-spectrum questions, and with wave-particle duality via electron diffraction. Diffraction threads forward into Topic 12 (turning points / nuclear) for X-ray crystallography and de Broglie wavelengths. The Edexcel data-and-formulae booklet lists $d\sin\theta = n\lambda$ but does not list any single-slit minima formula or grating resolving-power expression.

Worked example with full mark scheme

Question (8 marks):

A diffraction grating with $400$ lines per mm is illuminated at normal incidence by monochromatic light of wavelength $589\ \text{nm}$ (a sodium D-line).

(a) Calculate the slit spacing $d$ of the grating. (1)

(b) Calculate the angles, measured from the straight-through direction, of the first three orders of maxima. (4)

(c) Determine the highest order of maximum that can be observed with this grating and wavelength, and state the total number of bright maxima observed (counting orders on both sides of the central maximum). (3)

Solution with mark scheme:

(a) Step 1 — convert lines per mm to slit spacing in metres.

$d = \frac{1}{400 \times 10^3\ \text{m}^{-1}} = 2.50 \times 10^{-6}\ \text{m}$

B1 — correct value $d = 2.50 \times 10^{-6}\ \text{m}$ (or $2.5\ \mu\text{m}$ ). Common slip: writing $d = 1/400 = 0.0025\ \text{m}$ , forgetting the per-mm to per-metre conversion ( $\times 10^3$ ). That single slip propagates as a factor of 1000 through every subsequent angle calculation.

(b) Step 1 — apply the grating equation to each order.

The grating equation is $d\sin\theta = n\lambda$ , so $\sin\theta_n = n\lambda/d$ .

M1 — selecting and rearranging $d\sin\theta = n\lambda$ to find $\sin\theta_n = n\lambda/d$ .

For $n = 1$ :

$\sin\theta_1 = \frac{1 \times 589 \times 10^{-9}}{2.50 \times 10^{-6}} = 0.2356 \implies \theta_1 = 13.6°$

A1 — $\theta_1 = 13.6°$ (accept $13.6°$ to $13.7°$ ).

For $n = 2$ :

$\sin\theta_2 = \frac{2 \times 589 \times 10^{-9}}{2.50 \times 10^{-6}} = 0.4712 \implies \theta_2 = 28.1°$

A1 — $\theta_2 = 28.1°$ .

For $n = 3$ :

$\sin\theta_3 = \frac{3 \times 589 \times 10^{-9}}{2.50 \times 10^{-6}} = 0.7068 \implies \theta_3 = 44.97° \approx 45.0°$

A1 — $\theta_3 = 45.0°$ (accept $44.9°$ to $45.0°$ ).

The maximum order occurs when $\sin\theta = 1$ (i.e. $\theta = 90°$ , the refracted ray skims the grating plane). Setting $\sin\theta = 1$ in $d\sin\theta = n\lambda$ :

$n_{\max} = \frac{d}{\lambda} = \frac{2.50 \times 10^{-6}}{589 \times 10^{-9}} = 4.244$

M1 — recognising $\sin\theta \leq 1$ as the binding constraint and using $n_{\max} = d/\lambda$ .

Step 2 — round down to the nearest integer.

Since $n$ must be a non-negative integer and $n_{\max} = 4.244$ , the highest observable order is $n = 4$ .

A1 — $n_{\max} = 4$ . Note: rounding up to $n = 5$ would require $\sin\theta_5 = 5 \times 589/2500 = 1.178 > 1$ , which is geometrically impossible.

Step 3 — count total maxima.

Orders $n = 0, \pm 1, \pm 2, \pm 3, \pm 4$ are all visible. That gives $1 + 2 \times 4 = 9$ maxima in total.

A1 — total number of maxima = $9$ .

Total: 8 marks (B1 M1 A3 M1 A2).

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A student investigates a diffraction grating with monochromatic light from a laser. The light strikes the grating at normal incidence and produces a series of bright spots on a screen $1.50\ \text{m}$ from the grating. The first-order spots are observed $0.255\ \text{m}$ from the central maximum; the second-order spots are observed $0.555\ \text{m}$ from the central maximum.

(a) Use the first-order data to calculate the wavelength of the laser light, given that the grating has $300$ lines per mm. (3)

(b) Without recalculating, explain whether the second-order spot position is consistent with the first-order data. Support your reasoning quantitatively. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO2) — recognising that the angle to the first-order spot is $\tan\theta_1 = 0.255/1.50 = 0.170$ , hence $\theta_1 = 9.65°$ , and that this small-angle context still requires $\sin\theta$ rather than $\tan\theta$ in $d\sin\theta = n\lambda$ .
M1 (AO1) — substituting into $\lambda = d\sin\theta_1/n$ with $d = 1/(300 \times 10^3) = 3.333 \times 10^{-6}\ \text{m}$ and $\sin\theta_1 = 0.1675$ .
A1 (AO1) — $\lambda = 5.58 \times 10^{-7}\ \text{m}$ (or $558\ \text{nm}$ , accept $555\ \text{nm}$ to $560\ \text{nm}$ ). This is green light, consistent with a common diode-pumped laser.

(b)

M1 (AO2) — checking by predicting the second-order angle from the first-order wavelength: $\sin\theta_2 = 2\lambda/d = 0.3349$ , so $\theta_2 = 19.57°$ .
M1 (AO3) — converting predicted angle to predicted screen distance: $x_2 = L\tan\theta_2 = 1.50 \times \tan 19.57° = 0.533\ \text{m}$ .
A1 (AO3) — comparing with the measured $0.555\ \text{m}$ : agreement is to within $\sim 4\%$ , with the candidate stating that this is consistent within experimental uncertainty (small differences could arise from non-perfect normal incidence, finite spot size, screen-distance measurement, or grating manufacturing tolerance).

Total: 6 marks split AO1 = 2, AO2 = 2, AO3 = 2. This question is typical of the Edexcel pattern of testing whether candidates can use the grating equation forwards (find $\lambda$ ) and backwards (predict $x_2$ and compare). The crucial AO2 step is resisting the temptation to use $\tan\theta \approx \sin\theta$ inside the grating equation — that small-angle shortcut is valid for double-slit fringe spacing $w = \lambda L/s$ but fails for diffraction-grating angles, which are generally not small.

Synoptic links

Connects to:

Topic 5, superposition and interference: the grating equation $d\sin\theta = n\lambda$ is identical in form to the double-slit constructive condition, because both describe the angle at which path differences from successive sources are integer multiples of $\lambda$ . The grating sharpens the maxima by replacing two coherent sources with $N \gg 2$ — full reinforcement at prescribed angles, near-total cancellation elsewhere.