# 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](../c03-wave-particle-duality/02-matter-as-waves.md)). 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](01-waves-and-the-wave-equation.md) 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 $c$: $$ \boxed{\;\nabla^2\phi - \frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2} = 0\;}, \qquad \nabla^2 \equiv \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} . $$ Equivalently $\nabla^2 = \nabla\cdot\nabla$, the divergence of the gradient, with $$ \nabla \equiv \hat{\mathbf x}\frac{\partial}{\partial x} + \hat{\mathbf y}\frac{\partial}{\partial y} + \hat{\mathbf z}\frac{\partial}{\partial z} . $$ 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 $$ \phi(\mathbf r, t) = \exp\!\big[i(\mathbf k\cdot\mathbf r - \omega t)\big], \qquad \mathbf r = x\hat{\mathbf x} + y\hat{\mathbf y} + z\hat{\mathbf z}, \qquad \mathbf k = k_x\hat{\mathbf x} + k_y\hat{\mathbf y} + k_z\hat{\mathbf z} . $$ **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: $$ \nabla \exp\big[i(\mathbf k\cdot\mathbf r - \omega t)\big] = i\big(k_x\hat{\mathbf x} + k_y\hat{\mathbf y} + k_z\hat{\mathbf z}\big)\exp\big[i(\mathbf k\cdot\mathbf r - \omega t)\big] = i\mathbf k\,\phi . $$ Taking the divergence of that vector field brings down a second factor of $i\mathbf k$, dotted into the first: $$ \nabla^2\phi = \nabla\cdot(i\mathbf k\,\phi) = (i\mathbf k)\cdot(i\mathbf k)\,\phi = -\big(k_x^2 + k_y^2 + k_z^2\big)\phi = -k^2\phi . $$ The time derivative is the familiar one, $\partial_t^2\phi = -\omega^2\phi$. Substituting both: $$ \nabla^2\phi - \frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2} = \Big({-}k^2 + \frac{\omega^2}{c^2}\Big)\phi = 0 \quad\Longleftrightarrow\quad \boxed{\;k = \frac{\omega}{c}\;}\ \ \blacksquare $$ 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, $\mathbf k_\pm = \tfrac12(\mathbf k_1 \pm \mathbf k_2)$: $$ \phi = e^{i(\mathbf k_1\cdot\mathbf r - \omega t)} + e^{i(\mathbf k_2\cdot\mathbf r - \omega t)} = 2\cos\!\big(\mathbf k_-\!\cdot\mathbf r\big)\; e^{i(\mathbf k_+\cdot\mathbf r - \omega t)} . $$ 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 $$ \Lambda = \frac{\pi}{|\mathbf k_-|} = \frac{\lambda}{2\sin(\alpha/2)} , $$ 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](01-waves-and-the-wave-equation.md), 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, $$ \phi(\mathbf r, t) = \frac{A}{r}\,e^{i(kr - \omega t)}, \qquad r = |\mathbf r| , $$ 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. ```mermaid flowchart LR A["Plane wave
hits aperture of width d"] --> B["Huygens: every point in the gap
re-radiates a spherical wavelet"] B --> C["Wavelets superpose"] C --> D["Interior: sideways contributions
cancel → beam goes straight"] C --> E["Edges: cancellation incomplete
(no neighbours beyond the rim)"] E --> F["Beam fans out by θ ~ λ/d
= 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 $$ \boxed{\;\theta \sim \frac{\lambda}{d}\;} $$ — 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](../c03-wave-particle-duality/02-matter-as-waves.md).) **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](../../term-0-foundations/c03-fourier-methods/02-fourier-series-transforms.md)) 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](../c04-schrodinger-equation/03-matter-waves-uncertainty-momentum.md)) — 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: | Aperture | Behaviour | | --- | --- | | $d = 4\lambda$ | $\theta \approx 0.25$ rad $\approx 14^\circ$ — a recognizable, mildly spreading beam | | $d = 2\lambda$ | $\theta \approx 0.5$ rad $\approx 30^\circ$ — a broad fan | | $d = \lambda$ | $\theta \approx 1$ rad $\approx 57^\circ$ — nearly hemispherical; the slit is now essentially a point source | **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: $$ \boxed{\;s\sin\theta_m = m\lambda \quad\Longleftrightarrow\quad \sin\theta_m = \frac{m\lambda}{s}, \qquad m = 0, \pm1, \pm2, \dots\;} $$ 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. $$ m_{\max} = \left\lfloor \frac{s}{\lambda} \right\rfloor . $$ 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](../c03-wave-particle-duality/02-matter-as-waves.md)); 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](../c03-wave-particle-duality/02-matter-as-waves.md)). 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 $\alpha/2 = 15^\circ$, $\sin 15^\circ = 0.2588$: $$ \Lambda = \frac{632.8\ \mathrm{nm}}{2(0.2588)} = 1.22\ \mathrm{\mu m} . $$ 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 $$ \theta_1 = \arcsin(0.330) = 19.3^\circ,\quad \theta_2 = \arcsin(0.660) = 41.3^\circ,\quad \theta_3 = \arcsin(0.990) = 81.9^\circ , $$ 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 $$ s = \frac{\lambda}{\sin\theta_1} = \frac{0.154}{0.3584} = 0.430\ \mathrm{nm} , $$ 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) ```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. ``` ```python # --- 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. ``` ```python # --- 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 $z = 0$.

Solution

$k = 2\pi/\lambda = 1.257\times10^7\ \mathrm{m^{-1}}$, so $\mathbf k = k(\sin 30^\circ\,\hat{\mathbf x} + \cos 30^\circ\,\hat{\mathbf z}) = (6.28\times10^6\,\hat{\mathbf x} + 1.088\times10^7\,\hat{\mathbf z})\ \mathrm{m^{-1}}$. On $z = 0$ the phase varies as $k_xx$, so the period is $2\pi/k_x = \lambda/\sin 30^\circ = 1000\ \mathrm{nm}$ — twice the wavelength. Stripes on a screen are always *wider* than $\lambda$, by the factor $1/\sin\theta$.

**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.

Solution

$\nabla\phi = -A\mathbf k\sin(\mathbf k\cdot\mathbf r - \omega t)$ and $\nabla^2\phi = -A\,\mathbf k\cdot\mathbf k\,\cos(\cdot) = -k^2\phi$; also $\partial_t^2\phi = -\omega^2\phi$. Substituting gives $(-k^2 + \omega^2/c^2)\phi = 0$, hence $k = \omega/c$ ✓ — unsurprising, since $\cos$ is the real part of the exponential and the equation has real coefficients. The exponential is worth it because differentiation becomes *multiplication* by $i\mathbf k$ (no sign-flipping between sin and cos), and superposing waves becomes adding complex numbers. In quantum mechanics the complex form stops being a convenience and becomes the state itself ([P.4.1](../c04-schrodinger-equation/01-schrodinger-equation-born-rule.md)).

**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.

Solution

$\theta \sim \lambda/D = 10^{-6}/10^{-3} = 10^{-3}$ rad. After $L = 1\ \mathrm{km}$ the beam has grown by roughly $2\theta L = 2\ \mathrm{m}$; after $3.8\times10^5\ \mathrm{km}$, by $2\times10^{-3}\times3.8\times10^{8}\ \mathrm{m} \approx 760\ \mathrm{km}$ — which is why lunar laser-ranging returns a vanishing fraction of the light it sends, and why the Apollo experiments used the largest available launch telescope. Same relation, run backwards, for imaging: a telescope of aperture $D$ cannot resolve angles below $\approx 1.22\lambda/D$, because each point source arrives as a blob of that width. Doubling $D$ halves the blob. Resolution *is* diffraction.

**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.

Solution

Order $m$ requires $\sin\theta_m = m\lambda/s$ with $|\sin\theta_m| \le 1$, i.e. $|m| \le s/\lambda$. If $s < \lambda$ then $s/\lambda < 1$, so no nonzero integer qualifies: only $m = 0$, the undeflected beam. Physically, adjacent scatterers can never accumulate a full wavelength of path difference no matter how far you swing the detector — the structure is too fine for this wavelength to resolve, and the wave passes as if through a uniform medium (the basis of metamaterials and of anti-reflection sub-wavelength textures). For 5 beams we need $m = 0, \pm1, \pm2$, hence $s/\lambda \ge 2$: with $s = 2\lambda$ exactly, $\theta_{\pm2} = \pm90^\circ$ (grazing), so $s/\lambda > 2$ for five beams that are actually observable.

**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 $1/N$.

Solution

Scatterer $n$ (at height $ns$) contributes an extra path $ns\sin\theta$, hence phase $n\delta$. The total amplitude is a geometric series: $$ A = \sum_{n=0}^{N-1}e^{in\delta} = \frac{1 - e^{iN\delta}}{1 - e^{i\delta}} = e^{i(N-1)\delta/2}\,\frac{\sin(N\delta/2)}{\sin(\delta/2)} , $$ using $1 - e^{i\vartheta} = -2i\,e^{i\vartheta/2}\sin(\vartheta/2)$ twice; the modulus squared gives the stated $I$. **Principal maxima:** when $\delta = 2\pi m$ both sine factors vanish, and l'Hôpital (or noting each term is $e^{i2\pi mn} = 1$) gives $|A| = N$, so $I \propto N^2$. The condition $\delta = 2\pi m$ is $2\pi s\sin\theta/\lambda = 2\pi m$, i.e. $\sin\theta_m = m\lambda/s$ ✓ — independent of $N$. **Width:** the nearest zero has $N\delta/2 = \pi m N + \pi$, i.e. $\Delta\delta = 2\pi/N$ from the peak, so $\Delta(\sin\theta) = \lambda/(Ns)$: the peak narrows as $1/N$ while its height grows as $N^2$ — total energy $\propto N$, as conservation demands. Positions from $s$, sharpness from $N$: that is why spectrometers use gratings ruled with $10^4$–$10^5$ lines.

--- ## 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 $s = 3.4\lambda$.

Answers

1. $\nabla^2\phi - c^{-2}\partial_t^2\phi = 0$. Two applications of $\nabla \to i\mathbf k$ give $\nabla^2\phi = -k^2\phi$, and $\partial_t^2\phi = -\omega^2\phi$, so the equation reduces to $-k^2 + \omega^2/c^2 = 0$, i.e. $k = \omega/c$. Only the *length* of $\mathbf k$ is fixed; the direction is completely free, so every wavevector on the sphere of radius $\omega/c$ is a solution. 2. Straight, stationary intensity fringes perpendicular to $\mathbf k_1 - \mathbf k_2$, spaced $\Lambda = \lambda/[2\sin(\alpha/2)]$. The pattern does **not** move: the travelling factor is the carrier $e^{i(\mathbf k_+\cdot\mathbf r - \omega t)}$, while the envelope $\cos(\mathbf k_-\cdot\mathbf r)$ has no time dependence. 3. Every point of a wavefront is a source of secondary spherical wavelets; the field later is their superposition. Straight-line propagation is the *collective* cancellation of the wavelets' sideways contributions, so clipping the wavefront at an aperture's edges removes the neighbours that would have done the cancelling — the beam necessarily spreads. 4. $\theta \sim \lambda/d$: **larger aperture, smaller angle; smaller aperture, larger angle.** In Fourier terms, confining the field to $\Delta x \approx d$ forces a transverse-wavenumber spread $\Delta k_x \gtrsim 2\pi/d$, and $\Delta\theta \approx \Delta k_x/k = \lambda/d$ — diffraction is the Fourier uncertainty relation, the classical ancestor of $\Delta x\,\Delta p \gtrsim \hbar$. 5. $\sin\theta_m = m\lambda/s$; $m$ is the diffraction order, the number of whole wavelengths of path difference between adjacent scatterers. For $s = 3.4\lambda$, $|m| \le 3$, so $m = 0, \pm1, \pm2, \pm3$ — **seven** beams.

--- ## 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. --- ← Prev: [Waves & the Wave Equation](01-waves-and-the-wave-equation.md) · Up: [Pre-Term](../README.md) · Next: [Lagrangian Mechanics](03-lagrangian-mechanics.md) →