Wave Propagation, Diffraction & Gratings

One dimension gave us superposition and standing waves. Three dimensions give us the rest of wave physics: a wave now has a direction, and directions can be added. Everything follows — two plane waves crossing at an angle print a fringe pattern in space; a wave squeezed through a gap fans out by an angle $\lambda/d$; a row of scatterers throws light into a discrete set of beams. Those three facts are the entire experimental toolkit of the next two courses, and the last one is how humanity first caught an electron behaving like a wave. A cat's whiskers, incidentally, are a diffraction problem: they must out-span the gap to be worth anything.

Learning Objectives

After this lesson you will be able to:

1. Write the three-dimensional wave equation with the Laplacian, and verify by direct differentiation that the plane wave $\exp[i(\mathbf k\cdot\mathbf r - \omega t)]$ solves it exactly when $k = \omega/c$.
2. Interpret the wavevector $\mathbf k$ geometrically — direction of travel, wavefronts perpendicular to it, spacing $\lambda = 2\pi/k$ — and superpose two plane waves to get a spatial fringe pattern.
3. Apply the Huygens–Fresnel construction: rebuild a propagating wavefront from secondary point sources, and explain why a finite aperture cannot re-emit a perfectly plane wave.
4. Estimate the diffraction angle $\theta \sim \lambda/d$ for an aperture or source of width $d$, and use the inverse relation between aperture size and angular spread.
5. Derive the grating equation $\sin\theta_m = m\lambda/s$ for equally spaced scatterers, identify diffraction orders, and predict how many orders exist for a given $s/\lambda$.

Intuition

In one dimension a wave has two choices: left or right. In three it has a whole sphere of them, and the bookkeeping device for "which way" is the wavevector $\mathbf k$ — a vector pointing along propagation whose length is the wavenumber $2\pi/\lambda$. Everything in this lesson is a consequence of being allowed to add plane waves with different $\mathbf k$.

Add two, and where their crests coincide you get bright bands, where crest meets trough you get dark ones — interference is no longer a pattern in time at one point but a standing pattern painted across space. Add a continuum of them, radiating from every point of a slit, and you get diffraction: the wave spreads into the geometric shadow, because a finite piece of a wavefront simply does not contain the ingredients to keep marching in a perfectly straight line. Add the waves scattered by a regular row of atoms, and almost every direction cancels — except the special ones where every scatterer's contribution arrives in step, which come out as sharp beams at angles set by $\lambda$ and the spacing.

That last mechanism deserves flagging now: it is how X-ray crystallography reads molecules, and in 1927 it is how a nickel crystal proved that electrons have wavelengths (P.3.2). Diffraction is the instrument that detects waviness. Build it here, classically, and it is ready when matter turns out to need it.

Theory

The wave equation in three dimensions

The 1D wave equation of P.1.1 generalizes by replacing $\partial^2/\partial x^2$ with the Laplacian $\nabla^2$, the sum of second derivatives along all three axes. For a scalar field $\phi(\mathbf r, t)$ — a pressure in air, a component of an electric field, later a wavefunction — propagating at speed $c$:

Equivalently $\nabla^2 = \nabla\cdot\nabla$, the divergence of the gradient, with

Writing it as $\nabla\cdot\nabla$ is not decoration: it is what makes the plane-wave check below two lines of vector algebra instead of nine partial derivatives. The equation is still linear and still second order in time, so both structural lessons of P.1.1 survive intact — solutions superpose, and initial value plus initial velocity fix the future.

Plane-wave solutions

A monochromatic plane wave — one frequency, one direction — is

Claim: it solves the 3D wave equation precisely when $k \equiv |\mathbf k| = \omega/c$.

Proof. The gradient brings down $i\mathbf k$, because $\mathbf k\cdot\mathbf r = k_xx + k_yy + k_zz$ is linear in each coordinate:

Taking the divergence of that vector field brings down a second factor of $i\mathbf k$, dotted into the first:

The time derivative is the familiar one, $\partial_t^2\phi = -\omega^2\phi$. Substituting both:

Note what the derivation did not require: any relation among $k_x, k_y, k_z$ individually. Every direction of $\mathbf k$ works, provided the length is $\omega/c$ — the allowed wavevectors of a given frequency fill a sphere of radius $\omega/c$ in $\mathbf k$-space. That sphere is the reason a single frequency can produce two-dimensional patterns at all.

Three readings of $\mathbf k$ worth memorizing:

• Direction. Surfaces of constant phase — wavefronts — are the planes $\mathbf k\cdot\mathbf r = \text{const}$, which are perpendicular to $\mathbf k$; hence "plane" wave. They advance along $\hat{\mathbf k}$ at the phase speed $\omega/k = c$.
• Magnitude. Along $\hat{\mathbf k}$ the phase advances by $2\pi$ over a distance $2\pi/k$, so $\lambda = 2\pi/k$ exactly as in 1D.
• Components. $k_x$ is the wavenumber seen along $x$: a wave travelling at angle $\theta$ to the $z$-axis has $k_z = k\cos\theta$, $k_x = k\sin\theta$, so it prints a stripe pattern along $x$ of period $2\pi/k_x = \lambda/\sin\theta$ — always longer than $\lambda$. This projection is the whole of the grating equation, met later in this lesson.

Caution. $\lambda = 2\pi/k$ measures the spacing of wavefronts along $\mathbf k$, never along an arbitrary axis. A wave crossing a screen at a shallow angle paints stripes far wider than its own wavelength — the same projection that lets a grating with $s \gg \lambda$ still diffract, and the same effect that makes a "wavelength read off the screen" meaningless unless you also state the angle.

Two plane waves: interference in space

Because the wave equation is linear, the sum of two plane-wave solutions with the same $\omega$ but different directions $\mathbf k_1, \mathbf k_2$ (equal lengths $\omega/c$, different orientations) is again a solution. Factor out the mean phase, $\mathbf k_\pm = \tfrac12(\mathbf k_1 \pm \mathbf k_2)$:

Read it as an ordinary travelling wave — direction $\mathbf k_+$, the bisector — multiplied by a stationary spatial envelope $2\cos(\mathbf k_-\cdot\mathbf r)$ that does not move at all. The intensity $\propto|\phi|^2 = 4\cos^2(\mathbf k_-\cdot\mathbf r)$ is a set of straight, motionless bright bands perpendicular to $\mathbf k_-$, spaced by

where $\alpha$ is the angle between $\mathbf k_1$ and $\mathbf k_2$ (using $|\mathbf k_1 - \mathbf k_2| = 2k\sin(\alpha/2)$ for equal lengths). Nearly parallel beams give coarse fringes; beams meeting head-on ($\alpha = \pi$) give $\Lambda = \lambda/2$ — the 3D version of the standing wave of P.1.1, whose nodes sit half a wavelength apart. This is the operating principle of holography, of laser interference lithography, and of the optical lattices that hold cold atoms in place for quantum simulation.

Point sources and the Huygens–Fresnel construction

The other elementary solution is the spherical wave radiated by a point source at the origin,

whose wavefronts are concentric spheres (circles, in a 2D cross-section) expanding at $c$. The $1/r$ is energy conservation, not decoration: intensity $\propto |\phi|^2 = A^2/r^2$ spread over a sphere of area $4\pi r^2$ carries constant total power. Far from the source, a small patch of a huge sphere is indistinguishable from a plane — which is why a distant star delivers plane waves.

Huygens' principle (1678), sharpened by Fresnel: every point of a wavefront acts as a source of secondary spherical wavelets, and the field at a later time is their superposition. It is a computational recipe, and its power is best seen by watching it converge. Illuminate a slit of width $d$ with a plane wave from the left and rebuild the transmitted field from $N$ equally spaced point sources across the opening:

• $N = 3$. A lumpy mess. Three circles cannot imitate a wavefront; the result is a three-source interference pattern, not a beam.
• $N = 5$. Recognizably wrong near the slit, but already plausible far away — the wavelets' common tangent begins to look like a straight front.
• $N = 9$. The right-hand edges of the circles visibly line up along a straight line. Plane wavefronts emerge on the far side, as though the incident wave had simply marched through the gap.
• $N = 17$. Converged, for the purpose of drawing pictures. Adding more sources changes nothing visible.

The moral is that a plane wave "continuing in a straight line" is not a primitive fact — it is the collective result of infinitely many wavelets whose sideways contributions cancel. And that immediately tells you what a finite aperture must do: with the wavefront clipped at the edges, the cancellation is incomplete, so the transmitted beam cannot stay perfectly collimated. It spreads. The rest of the lesson is the size of that spread.

flowchart LR
    A["Plane wave<br/>hits aperture of width d"] --> B["Huygens: every point in the gap<br/>re-radiates a spherical wavelet"]
    B --> C["Wavelets superpose"]
    C --> D["Interior: sideways contributions<br/>cancel → beam goes straight"]
    C --> E["Edges: cancellation incomplete<br/>(no neighbours beyond the rim)"]
    E --> F["Beam fans out by θ ~ λ/d<br/>= diffraction"]
    D --> F

The diffraction angle $\theta \sim \lambda/d$

A wave leaving a source or aperture of transverse width $d$ emerges as a beam that spreads into a cone of half-angle

— the single most useful estimate in wave physics. Three complementary derivations, all worth having:

1. Cancellation (the textbook argument). Pair each Huygens source in the top half of the slit with its partner $d/2$ below. In the direction $\theta$, that pair has a path difference $(d/2)\sin\theta$. Every pair cancels when this equals $\lambda/2$, i.e. at $\sin\theta_1 = \lambda/d$ — the first zero of the single-slit pattern. Between $-\theta_1$ and $+\theta_1$ sits the central lobe containing most of the energy, so its half-width $\lambda/d$ is the diffraction angle. (The full pattern is $I(\theta) \propto [\sin\beta/\beta]^2$ with $\beta = \pi d\sin\theta/\lambda$, the classic $\mathrm{sinc}^2$ envelope of P.3.2.)

2. Transverse wavevector. Confining the wave to width $d$ means the field is nonzero only over an interval $\Delta x \approx d$. Fourier analysis (0.3.2) then requires a spread of transverse wavenumbers $\Delta k_x \gtrsim 2\pi/d$. Since a component $k_x$ corresponds to propagation at angle $\sin\theta = k_x/k$, the angular spread is $\Delta\theta \approx \Delta k_x/k = \lambda/d$. Same answer, and this one shows why: diffraction is the uncertainty principle of the Fourier transform, no quantum mechanics required. When $\hbar$ arrives, multiply both sides by it and this becomes $\Delta x\,\Delta p_x \gtrsim \hbar$ (P.4.3) — the same theorem, re-labelled.

3. Dimensional analysis. The only length scales are $\lambda$ and $d$; an angle must be their ratio. This is why the estimate deserves the "$\sim$": prefactors depend on aperture shape (a circular aperture gives $\theta = 1.22\lambda/D$, the Rayleigh criterion), but the scaling never changes.

The relation is inverse, and the numbers are worth feeling:

Larger aperture, smaller angle; smaller aperture, larger angle. Squeeze a beam and it fights back. This single inequality sets the resolution limit of every microscope and telescope, the minimum spot size of a lithography stepper, and the divergence of every laser and antenna beam ever built.

Diffraction from periodic structures

Now replace one aperture with a row of $N$ identical scatterers spaced by $s$, illuminated by a monochromatic plane wave. Each scatterer radiates a spherical wavelet; their sets of concentric circles interfere. In a general direction the $N$ contributions carry unrelated phases and cancel; the survivors are the directions where every scatterer arrives in step.

Take a direction at angle $\theta$ to the incident beam (which arrives along the normal). The path difference between wavelets from adjacent scatterers is $s\sin\theta$ — read straight off the right triangle whose hypotenuse is $s$. Constructive interference for all $N$ at once requires that this be a whole number of wavelengths:

the grating equation. The integer $m$ labels the diffraction order:

• $m = 0$ — straight ahead, $\theta_0 = 0$. Zero path difference for any spacing or wavelength: the zeroth order always exists and is not dispersive.
• $m = \pm1$ — the first orders, one deflected up and one down, at $\sin\theta_{\pm1} = \pm\lambda/s$. (With normal incidence the geometry is symmetric, so the sign is bookkeeping, not physics.)
• $m = \pm2, \pm3,\dots$ — successively higher orders, each needing an extra wavelength of path per scatterer pair.

Since $|\sin\theta_m| \le 1$, order $m$ exists only if $|m| \le s/\lambda$: the number of observable orders is set by the spacing measured in wavelengths.

Two consequences run in opposite directions, and confusing them is the classic error:

• More scatterers ⇒ sharper beams. With $N$ sources the intensity is $I \propto [\sin(N\delta/2)/\sin(\delta/2)]^2$, $\delta = 2\pi s\sin\theta/\lambda$: the maxima keep their positions but their width shrinks as $1/N$. Two slits give broad $\cos^2$ fringes (P.3.2); a few thousand give the needle-sharp spots that make a grating a precision spectrometer.
• Wider spacing ⇒ more, but closer-packed, beams. Increasing $s$ decreases every $\sin\theta_m = m\lambda/s$, crowding the orders toward the axis while admitting more of them. At $s = 1.5\lambda$ only $m = 0, \pm1$ survive, widely splayed; by $s = 4\lambda$ there are nine orders packed into the same angular range.

The wavelength dependence is what makes gratings instruments: $\theta_m$ depends on $\lambda$, so white light is fanned into a spectrum, and measuring $\theta_m$ measures $\lambda$. Point the same geometry the other way — known $\lambda$, unknown spacing — and it measures $s$. That inversion is X-ray crystallography, and it is exactly the experiment Davisson and Germer performed with electrons in 1927, using a nickel crystal's atomic rows as the grating: their scattered peak at $50^\circ$ from $d = 0.215\ \mathrm{nm}$ rows measured an electron wavelength of $0.165\ \mathrm{nm}$, against de Broglie's predicted $0.167\ \mathrm{nm}$ (P.3.2). A classical instrument, built entirely from the mathematics of this section, is what proved matter is wavelike.

Worked Examples

Example 1 — Two crossed laser beams write an optical lattice

Two beams of a He–Ne laser, $\lambda = 632.8\ \mathrm{nm}$, cross at $\alpha = 30^\circ$. What is the fringe spacing, and what does it become for counter-propagating beams?

Using $\Lambda = \lambda/[2\sin(\alpha/2)]$ with $\alpha/2 = 15^\circ$, $\sin 15^\circ = 0.2588$:

Counter-propagating ($\alpha = 180^\circ$): $\Lambda = \lambda/2 = 316\ \mathrm{nm}$, the tightest spacing two beams of this colour can produce — and the standard geometry for trapping cold atoms one per well. Note the fringes are stationary even though both beams travel at $c$: the envelope $\cos(\mathbf k_-\cdot\mathbf r)$ carries no time dependence.

Example 2 — Reading a grating, then reading a crystal

(a) A grating with 600 lines/mm ($s = 1/600\ \mathrm{mm} = 1.667\ \mathrm{\mu m}$) is illuminated at normal incidence by $\lambda = 550\ \mathrm{nm}$. Which orders exist, and where?

$\lambda/s = 550/1667 = 0.330$. So $m_{\max} = \lfloor 1/0.330\rfloor = 3$, and

plus their negatives and $\theta_0 = 0$: seven beams, with the third order grazing the plane of the grating. Red light at $700\ \mathrm{nm}$ has $\lambda/s = 0.420$, so only orders $0, \pm1, \pm2$ survive and each sits at a larger angle than green's — the fan of a spectrum.

(b) Same equation, unknown spacing. First-order diffraction of $\lambda = 0.154\ \mathrm{nm}$ X-rays (Cu K$\alpha$) is observed at $\theta_1 = 21.0^\circ$. Then

an atomic-scale spacing measured with nothing but an angle and a wavelength. Note the requirement $s > \lambda$ for any first order to exist: this is why you cannot see atoms with visible light — $550\ \mathrm{nm}$ against a $0.43\ \mathrm{nm}$ lattice gives $\lambda/s = 1279 \gg 1$, no orders beyond $m = 0$, no information. Probing a structure requires a wavelength no larger than the structure.

Hands-on (Python)

import numpy as np
import matplotlib.pyplot as plt

# --- 1. Two plane waves crossing: stationary fringes in space ---------------
lam   = 1.0                                     # wavelength (arbitrary units)
k     = 2*np.pi/lam
alpha = np.deg2rad(30.0)                        # full angle between the beams

# Both wavevectors have length k; they straddle the z-axis by +/- alpha/2.
k1 = k*np.array([ np.sin(alpha/2), np.cos(alpha/2)])
k2 = k*np.array([-np.sin(alpha/2), np.cos(alpha/2)])

x = np.linspace(-6, 6, 600); z = np.linspace(0, 8, 400)
X, Z = np.meshgrid(x, z)
phi = np.exp(1j*(k1[0]*X + k1[1]*Z)) + np.exp(1j*(k2[0]*X + k2[1]*Z))   # t = 0

plt.imshow(np.abs(phi)**2, extent=[x[0], x[-1], z[0], z[-1]],
           origin="lower", aspect="auto", cmap="viridis")
plt.xlabel("x"); plt.ylabel("z"); plt.title("Two crossed plane waves")
plt.colorbar(label=r"$|\phi|^2$"); plt.show()

print("predicted fringe spacing:", lam/(2*np.sin(alpha/2)))   # 1.932
# Expected: vertical bright bands, uniformly spaced by ~1.93 lambda, running
# along z. They do not move with t -- only the carrier does.

# --- 2. Huygens convergence: N point sources filling one aperture ----------
lam, kk = 1.0, 2*np.pi                          # lambda = 1
d = 6.0*lam                                     # aperture width
x = np.linspace(-12, 12, 500); z = np.linspace(0.15, 16, 400)
X, Z = np.meshgrid(x, z)

fig, axes = plt.subplots(1, 4, figsize=(14, 3.6), sharey=True)
for ax, N in zip(axes, (3, 5, 9, 17)):
    ys = np.linspace(-d/2, d/2, N)              # source positions in the gap
    field = np.zeros_like(X, dtype=complex)
    for y0 in ys:                               # each source: circular wavelet
        r = np.hypot(X - y0, Z)
        field += np.exp(1j*kk*r)/np.sqrt(r)     # 2D wavelet falls as 1/sqrt(r)
    ax.imshow(field.real, extent=[x[0], x[-1], z[0], z[-1]],
              origin="lower", aspect="auto", cmap="RdBu")
    ax.set_title(f"N = {N}"); ax.set_xlabel("x")
axes[0].set_ylabel("z (propagation)")
plt.tight_layout(); plt.show()
# Expected: N=3 lumpy interference; N=9 straight wavefronts appear beyond the
# gap; N=17 converged. In every panel the beam fans out past |x| ~ z*lam/d.

# --- 3. Single slit and N-slit grating: envelope, orders, sharpening -------
def slit_pattern(sin_t, d_over_lam):
    """sinc^2 envelope of one slit of width d."""
    beta = np.pi*d_over_lam*sin_t
    return np.sinc(beta/np.pi)**2               # np.sinc(u) = sin(pi u)/(pi u)

def grating_pattern(sin_t, s_over_lam, N):
    """N equally spaced scatterers: [sin(N delta/2)/(N sin(delta/2))]^2."""
    delta = 2*np.pi*s_over_lam*sin_t
    num, den = np.sin(N*delta/2), np.sin(delta/2)
    out = np.divide(num, den, out=np.full_like(num, N), where=np.abs(den) > 1e-12)
    return (out/N)**2

sin_t = np.linspace(-1, 1, 4000)

# (a) diffraction angle scales as lambda/d
plt.figure(figsize=(7, 4))
for d_lam in (1.0, 2.0, 4.0):
    plt.plot(np.degrees(np.arcsin(sin_t)), slit_pattern(sin_t, d_lam),
             label=f"d = {d_lam:.0f}$\\lambda$  (first zero at "
                   f"{np.degrees(np.arcsin(min(1/d_lam, 1))):.0f}°)")
plt.xlabel("θ (degrees)"); plt.ylabel("I / I(0)"); plt.legend(); plt.show()
# Expected: first zeros at 90°, 30°, 14.5° -- halving lambda/d halves the angle.

# (b) grating orders and the 1/N sharpening
s_lam = 2.5
plt.figure(figsize=(7, 4))
for N in (2, 5, 20):
    plt.plot(np.degrees(np.arcsin(sin_t)), grating_pattern(sin_t, s_lam, N),
             lw=1.0, label=f"N = {N}")
for m in range(-int(s_lam), int(s_lam) + 1):    # predicted order positions
    plt.axvline(np.degrees(np.arcsin(m/s_lam)), color="gray", ls=":", lw=0.7)
plt.xlabel("θ (degrees)"); plt.ylabel("I / I(0)")
plt.title(f"s = {s_lam}$\\lambda$: orders m = 0, ±1, ±2"); plt.legend(); plt.show()
# Expected: peaks land exactly on the dotted lines predicted by sin(theta)=m*lam/s
# (0°, ±23.6°, ±53.1°); N=2 gives broad cos^2 humps, N=20 gives sharp spikes at
# the SAME angles -- N sets the width, s sets the positions.

for s_lam in (1.5, 2.0, 2.5, 3.0, 4.0):
    print(f"s = {s_lam}λ -> {2*int(s_lam) + 1} orders, "
          f"first at {np.degrees(np.arcsin(1/s_lam)):5.1f}°")
# s = 1.5λ -> 3 orders, first at  41.8°
# s = 2.0λ -> 5 orders, first at  30.0°
# s = 2.5λ -> 5 orders, first at  23.6°
# s = 3.0λ -> 7 orders, first at  19.5°
# s = 4.0λ -> 9 orders, first at  14.5°   <- wider spacing: more, tighter beams

Exercises

E1 (easy). A plane wave of wavelength $\lambda = 500\ \mathrm{nm}$ travels in the $xz$-plane at $30^\circ$ from the $z$-axis. Write $\mathbf k$, and find the spacing of the stripes it prints on the plane $z = 0$.

E2 (easy). Show by direct substitution that $\phi = A\cos(\mathbf k\cdot\mathbf r - \omega t)$ solves the 3D wave equation under the same condition $k = \omega/c$, and explain why the complex form was worth using anyway.

E3 (medium). A laser of wavelength $\lambda = 1.0\ \mathrm{\mu m}$ leaves an aperture of diameter $D = 1.0\ \mathrm{mm}$. Estimate the beam diameter after $1\ \mathrm{km}$, and after $3.8\times10^5\ \mathrm{km}$ (the Moon). Then explain why a bigger telescope sees finer detail.

E4 (medium). Show that for a grating with $s < \lambda$ only the zeroth order exists, and give the physical reading. Then find how large $s/\lambda$ must be to produce at least 5 beams.

E5 (hard). Derive the $N$-slit intensity $I \propto [\sin(N\delta/2)/\sin(\delta/2)]^2$ with $\delta = 2\pi s\sin\theta/\lambda$, then show its principal maxima occur exactly at the grating angles and that their angular width scales as $1/N$.

Checkpoint

1. Write the 3D wave equation and show that $\exp[i(\mathbf k\cdot\mathbf r - \omega t)]$ solves it. What condition on $\mathbf k$ is required, and what is not constrained?
2. Two plane waves of the same frequency cross at angle $\alpha$. Describe the resulting pattern and give its spacing. Does it move?
3. State Huygens' principle. Why does it predict that a wave leaving a finite aperture cannot stay perfectly collimated?
4. Estimate the diffraction angle of an aperture of width $d$, and state which way the relation runs. What is the Fourier-analysis reading of the same estimate?
5. Write the grating equation, define the diffraction order, and say how many orders exist for $s = 3.4\lambda$.

Further Reading

• [Hec] Hecht, Optics, Ch. 9–10 — interference, Huygens–Fresnel, single-slit and grating diffraction; the definitive classical treatment of everything in this lesson.
• [ER] Eisberg & Resnick, §3-1 — the same diffraction mathematics turned into the detector that revealed matter waves.
• [Gri] Griffiths & Schroeter, §1.6 — how the Fourier reading of $\theta \sim \lambda/d$ becomes the uncertainty principle.

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