Reflection and Transmission at Normal Incidence

Every window ghosts a faint reflection of your face. That four-percent phantom is this article's subject: what a plane wave must do when the medium underneath it changes. The previous article gave each medium its wave — speed $c/n$ , impedance $Z_0/n$ — but a boundary confronts the two waves with each other, and the fields cannot simply hand over: Maxwell's boundary conditions demand that the tangential fields match across the wall, and a lone transmitted wave cannot satisfy the books. Something must be reflected. The boundary conditions then dictate exactly how much. This article solves the cleanest case — normal incidence, wave straight at the wall — completely. Two lines of continuity algebra deliver the amplitude coefficients $r = (n_1 - n_2)/(n_1 + n_2)$ and $t = 2n_1/(n_1 + n_2)$ , and with them a set of facts that generalize far beyond optics: reflection is driven by mismatch (impedance mismatch, fundamentally — the same mathematics governs echoes on cables, ultrasound at tissue boundaries, and waves on knotted ropes); going into a denser medium flips the reflected wave's sign, the fixed-end string bounce transplanted; and the energy books close exactly, but only if intensities are priced with each medium's own n. The case also carries the subject's best small paradox: crossing from glass to air, the transmitted amplitude exceeds the incident one — $t = 1.2$ , apparently 120% output. Chasing that impossible-looking number to its lawful resolution is the fastest way to internalize what amplitude and energy each actually measure, and the audit is worked here in full.

Key takeaways

A wave hitting an interface splits: the boundary conditions (tangential E and H continuous) cannot be met by a transmitted wave alone — reflection is mandatory whenever media mismatch.
Normal-incidence amplitude coefficients (non-magnetic media): $r = \dfrac{n_1-n_2}{n_1+n_2}$ , $t = \dfrac{2n_1}{n_1+n_2}$ — two lines of continuity algebra; in full generality the same formulas run on impedances, $r = (Z_2-Z_1)/(Z_2+Z_1)$ .
Into a denser medium ( $n_2 > n_1$ ): $r < 0$ — the reflected E flips phase by π, the fixed-end string bounce. Into a rarer medium: no flip.
Energy audits with the medium-weighted intensities: $R = r^2$ , $T = \dfrac{n_2}{n_1}t^2$ , and $R + T = 1$ exactly — the $n_2/n_1$ factor is what lets $t > 1$ coexist with conservation.
Air–glass reflects 4% per surface; air–diamond 17%; glass–water 0.35% — index matching kills reflections, impedance matching is the general principle.
Key idea: a boundary reflects in proportion to its mismatch — $r = (n_1-n_2)/(n_1+n_2)$ — with a π flip into denser media, and the energy books close only when intensity is priced as $n E_0^2$ per medium.

The core insight

At normal incidence the boundary conditions force $r = (n_1-n_2)/(n_1+n_2)$ and $t = 2n_1/(n_1+n_2)$ : reflection measures mismatch (zero when $n_1 = n_2$ ), external reflection flips phase by π, and $R = r^2$ , $T = (n_2/n_1)t^2$ sum to one exactly.

Where this lives today

The four-percent ghost is a solved industrial problem, and the solution is a preview of coming physics: coat the surface with a quarter-wavelength film of intermediate index ( $n_{\text{film}} = \sqrt{n_1n_2}$ ideally — MgF₂'s 1.38 approximates it for glass), and the reflections from the film's two faces return half a wavelength apart — the π-flip bookkeeping of Derivation A applied twice — and cancel. Modern multilayer coatings stack the trick across the visible: camera lenses reach 0.2% per surface, and laser optics reach parts per million (or, run in reverse, 99.999%+ mirrors — LIGO's test masses are forty-layer interference stacks whose every layer is this article's r, phased). The thin-film machinery itself belongs to the optics track; its raw material is entirely normal-incidence Fresnel coefficients. The impedance reading runs even wider. Every RF connector, PCB trace, and cable termination is specified against reflections by this article's formula — a time-domain reflectometer literally watches $r(Z_2)$ echoes to locate faults on a line (the E7 chapter's slit-cable TDR example, now with its theory complete). Medical ultrasound images are maps of $r$ at tissue-impedance steps (and coupling gel exists to kill the air gap's near-total mismatch); seismologists read the Earth's layering from the same coefficients in elastic waves. One wall, two continuity conditions, every wave discipline's daily arithmetic.

The physical picture

Why must anything reflect at all? Because the two media want incompatible waves. In medium 1 a wave of given E carries $H = E/Z_1$ ; in medium 2 the same E would carry $H = E/Z_2$ . At the interface, Maxwell's boundary conditions — the E6 electrodynamic-boundary-conditions article proved them, and with no free surface charge or current on a dielectric they say both tangential E and tangential H are continuous — demand that the field state match point-for-point across the wall. A transmitted wave alone would have to match the incident wave's E and its H simultaneously, but its own E/H ratio is fixed at $Z_2 \neq Z_1$ : one wave, two constraints, no solution. The interface resolves the impasse by manufacturing a third wave — the reflection — whose amplitude is precisely the degree of freedom needed to satisfy both books at once. Reflection is not the wall "bouncing" the light; it is the boundary conditions closing an account that a lone transmitted wave leaves open. The string picture makes the sign physics tactile, and it is an exact analogy (same wave equation, same continuity structure — tension and mass density standing in for μ and ε). A pulse on a light rope hitting a knot to a heavy rope: partial transmission, and the reflected pulse comes back inverted — the heavy rope acts fixed-end-like. Light rope beyond the knot: reflection upright. The electromagnetic translation: entering a slower (denser, higher-n, lower-Z) medium inverts the reflected E; entering a faster medium does not. That π flip, trivial-looking here, runs serious machinery later: it is why thin-film coatings can cancel reflections, why the Fresnel equations carry their signs, and why interferometers count surfaces.

Figure

The mandatory third wave — continuity of both E and H cannot be met by transmission alone; the reflection's amplitude closes the books, inverted when the far side is denser

The mismatch reading is the transferable one. Nothing in the continuity argument cares that the media are optical: any wave system with an impedance — voltage/current on a transmission line, pressure/velocity in ultrasound, force/velocity on a string — reflects at impedance steps by the same formula, $r = (Z_2 - Z_1)/(Z_2 + Z_1)$ in the appropriate variables. Radar-absorbing paint, anechoic chamber wedges, ultrasound coupling gel, and the 50 Ω discipline of RF engineering are all one idea: make the impedance step gradual or matched, and the reflection disappears. Optics' version is index matching — glass vanishes in immersion oil of equal n — and its precision version, the anti-reflection coating, earns a preview at the end.

The mathematics of the split

Set up the standard geometry: interface at $z = 0$ , wave arriving along $+\hat{\mathbf{z}}$ from medium 1, polarization $\hat{\mathbf{x}}$ (at normal incidence all transverse directions are equivalent — polarization first matters at oblique incidence, next article). Three waves, written in complex notation with the common factor $e^{-i\omega t}$ (frequency conserved across the boundary — the previous article's theorem):

\tilde{\mathbf{E}}_I = \tilde E_{0_I}e^{ik_1z}\,\hat{\mathbf{x}}, \qquad \tilde{\mathbf{E}}_R = \tilde E_{0_R}e^{-ik_1z}\,\hat{\mathbf{x}}, \qquad \tilde{\mathbf{E}}_T = \tilde E_{0_T}e^{ik_2z}\,\hat{\mathbf{x}} \qquad (1)

where: $k_{1,2} = n_{1,2}\,\omega/c$ — each medium's own wavenumber; the reflected wave runs along $-\hat{\mathbf{z}}$ (hence $e^{-ik_1z}$ , and its magnetic partner flips accordingly: $\tilde{\mathbf{B}} = \tfrac1v\hat{\mathbf{k}}\times\tilde{\mathbf{E}}$ with $\hat{\mathbf{k}} = -\hat{\mathbf{z}}$ ). Read the setup's honesty: we assume one reflected and one transmitted plane wave and let the boundary conditions determine their complex amplitudes — the assumption is validated by success (two unknowns, two conditions, unique solution), and it is the standard scattering-ansatz logic the whole interface industry runs on.

Derivation AThe reflection and transmission coefficients from two lines of continuity

The goal is $r \equiv \tilde E_{0_R}/\tilde E_{0_I}$ and $t \equiv \tilde E_{0_T}/\tilde E_{0_I}$ at normal incidence, in both the impedance form (general) and the index form (non-magnetic), with the phase-flip physics read off the signs. The strategy: impose continuity of tangential E and tangential H at $z = 0$ — the two conditions the E6 boundary-conditions article certifies for a source-free dielectric interface — and solve the two-by-two linear system. Watch how few inputs are needed: no molecular model, no energy argument — the coefficients are pure boundary-condition kinematics, which is why they replicate across every wave system with an impedance.

Step 1. Tangential E continuous. All three E-fields point along $\hat{\mathbf{x}}$ (tangential); at $z = 0$ the total on each side must match:

\tilde E_{0_I} + \tilde E_{0_R} = \tilde E_{0_T}.

Step 2. Tangential H continuous. Each wave's H is $E/Z$ of its medium, direction from the triad: incident along $+\hat{\mathbf{y}}$ , transmitted along $+\hat{\mathbf{y}}$ , and the reflected wave — propagating along $-\hat{\mathbf{z}}$ with E still along $+\hat{\mathbf{x}}$ — has H along $-\hat{\mathbf{y}}$ (the triad stays right-handed about the new direction; this sign is the entire physics of the flip). Matching the $\hat{\mathbf{y}}$ -components:

\frac{\tilde E_{0_I}}{Z_1} - \frac{\tilde E_{0_R}}{Z_1} = \frac{\tilde E_{0_T}}{Z_2}.

Why H (not B) carries the condition: tangential H is continuous when free surface currents are absent — exact for dielectrics. (Using B instead smuggles in a μ; for non-magnetic media the two routes coincide, and this derivation's Z-form stays honest for magnetic media too.)

Step 3. Solve. Add and subtract the two equations:

\boxed{\ r = \frac{Z_2 - Z_1}{Z_2 + Z_1} = \frac{n_1 - n_2}{n_1 + n_2}, \qquad t = \frac{2Z_2}{Z_2 + Z_1} = \frac{2n_1}{n_1 + n_2}\ }

(the index forms use $Z = Z_0/n$ , i.e. μ = μ₀). Consistency: $1 + r = t$ ✓ (Step 1 restated). Mismatch reading: $n_1 = n_2 \Rightarrow r = 0$ , total transmission — the interface exists chemically but not electromagnetically.

Step 4. Read the signs. For $n_2 > n_1$ (into denser): $r < 0$ — the reflected complex amplitude carries a factor $e^{i\pi}$ : the reflected E is inverted, the fixed-end bounce. For $n_2 < n_1$ : $r > 0$ , upright. And $t > 0$ always — the transmitted wave never flips, it only rescales. Why the flip is worth engraving now: the "π on external reflection, none on internal" rule decides fringe positions in every thin-film problem (Newton's rings' dark center is this sign), and its oblique-incidence generalization is wired into the Fresnel equations' sign conventions — next article, by name. *Verified: reproduces Griffiths §9.3.2 Eqs. 9.82 (his β = Z₁/Z₂ form collapses to these for μ = μ₀) and Hecht §4.6's normal-incidence Fresnel limits; string-analogy signs consistent (heavy rope = fixed end = inversion); 1 + r = t identity checked.*

Go deeper · Derivation B · 2 minThe energy audit — R + T = 1, and why t greater than one is lawful

Derivation BThe energy audit — R + T = 1, and why t greater than one is lawful

The goal is the reflectance $R$ and transmittance $T$ — the energy splits a power meter actually measures — the exact closure $R + T = 1$ , and the resolution of the apparent paradox $t > 1$ that arises on dense-to-rare crossings. The strategy: price each wave's intensity with its own medium's formula from the previous article ( $I = nE_0^2/2Z_0$ ), form the ratios, and let the algebra close. Watch the factor $n_2/n_1$ arrive in T — it is not decoration; it is the entire difference between amplitude bookkeeping and energy bookkeeping.

Step 1. Price the three beams. Same cross-sectional area on both sides (normal incidence — the cosine geometry of oblique beams awaits 0592):

R \equiv \frac{I_R}{I_I} = r^2, \qquad T \equiv \frac{I_T}{I_I} = \frac{n_2 E_{0_T}^2}{n_1 E_{0_I}^2} = \frac{n_2}{n_1}\,t^2.

Why the $n_2/n_1$ : intensity is $E_0^2/2Z = nE_0^2/2Z_0$ — a medium with larger n carries more flux per unit $E_0^2$ (its H is bigger by n; article 0590's ledger). Equal amplitudes in different media are not equal energies.

Step 2. Close the books. Substitute the coefficients:

R + T = \frac{(n_1-n_2)^2}{(n_1+n_2)^2} + \frac{n_2}{n_1}\cdot\frac{4n_1^2}{(n_1+n_2)^2} = \frac{(n_1-n_2)^2 + 4n_1n_2}{(n_1+n_2)^2} = \frac{(n_1+n_2)^2}{(n_1+n_2)^2} = 1.

Exact, for every index pair — no averaging subtleties, no approximations: the interface stores nothing (it has no volume) and dissipates nothing (lossless media), so Poynting's theorem demands the flux books balance, and they do.

Step 3. The paradox, prosecuted. Glass to air ( $n_1 = 1.5$ , $n_2 = 1$ ): $t = 2(1.5)/2.5 = 1.2$ — the transmitted amplitude is 120% of the incident. Lawful? Price it: $T = (1/1.5)(1.2)^2 = 0.96$ , and $R = (0.2)^2 = 0.04$ ; total 1.00 ✓. The transmitted wave has a larger E but lives in a medium that charges less flux per $E_0^2$ (air's H per E is smaller by 1.5): amplitude up 20%, energy down 4%. Why the paradox is pedagogically precious: it is the cleanest one-line proof that amplitude ratios and energy ratios are different currencies with a medium-dependent exchange rate — the mistake it punishes (squaring t and calling it T) is the interface cousin of the peak/rms confusions of 0587. The string analogy carries the same lesson: a pulse crossing onto a lighter rope grows in height while shrinking in energy share.

Step 4. The standard numbers. Air–glass: $R = 0.04$ — the 4% window ghost, per surface (a double-glazed window's four surfaces pass $0.96^4 = 85\%$ ). Air–water: 2.0%. Air–diamond ( $n = 2.42$ ): 17.2% — the flash of a brilliant is impedance mismatch wearing jewelry. Glass–water: 0.35% — near-invisibility, why wet glass loses its sparkle and why immersion oil (n matched to 1.515) makes microscope interfaces vanish. Verified: closure algebraic identity; numbers recomputed (r = −0.2, t = 0.8 air→glass; t = 1.2, T = 0.96 glass→air; R: diamond 0.172, water 0.0204, glass–water 0.0035; 0.96⁴ = 0.849); consistent with Griffiths §9.3.2 Eq. 9.87 and Hecht Table 4.2.

Figure

The 120 percent amplitude, audited — glass-to-air amplitudes grow while the energy books close exactly, because each medium prices flux in its own currency

Limiting cases

Case	Result	Meaning
$n_1 = n_2$ (or $Z_1 = Z_2$ )	r = 0, t = 1, T = 1	matched: no electromagnetic interface
$n_2 \to \infty$	r → −1, t → 0	perfect mirror limit; total inversion (fixed end)
$n_2 \to 0$ (or $Z_2 \to \infty$ )	r → +1	free-end reflection, upright — the open-circuit cable echo
$n_2 \gtrless n_1$	r ≶ 0	π flip into denser media only
glass→air	t = 1.2 but T = 0.96	amplitude ≠ energy; the $n_2/n_1$ exchange rate
weak mismatch $\Delta n \ll n$	$R \approx (\Delta n/2n)^2$	reflections quadratically small — graded transitions ≈ invisible
oblique incidence	r, t split by polarization	Fresnel equations — next article owns it
conducting medium 2	$Z_2$ complex: $\|r\| \to 1$ with phase	mirrors and skin depth — 0596/0597 by name

Common trapt² is not T — transmitted energy must be priced in the receiving medium's currency

The reflex $T = t^2$ (by analogy with $R = r^2$ ) is wrong whenever $n_1 \neq n_2$ , and it fails loudly: glass-to-air it predicts $T = 1.44$ , a 144% energy yield from a passive wall. The repair is the receiving medium's intensity formula: $I = nE_0^2/2Z_0$ , so $T = (n_2/n_1)\,t^2$ (at normal incidence; oblique adds beam-geometry cosines, next article). $R = r^2$ survives unmodified only because the reflected wave lives in the same medium as the incident one — same currency, no exchange rate. Two adjacent traps complete the set. First: the coefficients r, t are amplitude ratios of E specifically — conventions exist (mostly in engineering) that quote H-ratios or power ratios, and imported formulas must be checked for which. Second: the phase flip belongs to the reflected wave into denser media only — assigning it to the transmitted wave, or to internal reflections, silently inverts every thin-film fringe calculation you build on it. When any doubt arises, return to the two continuity lines of Derivation A; they take thirty seconds and cannot lie.

Worked examples

1 · The window ghost, audited end to end (confidence builder)

A 100 V/m plane wave in air strikes a glass surface ( $n = 1.5$ ) at normal incidence. Find all amplitudes, the R/T split, and audit energy conservation in W/m².

Step 1. Coefficients: $r = (1 - 1.5)/2.5 = -0.20$ ; $t = 2/2.5 = 0.80$ .

Step 2. Amplitudes: $E_{0_R} = -20$ V/m (inverted — the π flip into denser glass); $E_{0_T} = 80$ V/m. Continuity check at the wall: $100 + (-20) = 80$ ✓.

Step 3. Intensities, each in its own medium: incident $I_I = E_0^2/2Z_0 = 10^4/753.5 = 13.27$ W/m²; reflected $I_R = 400/753.5 = 0.53$ W/m²; transmitted $I_T = 1.5\times6400/753.5 = 12.74$ W/m².

Step 4. The audit: $0.53 + 12.74 = 13.27$ ✓ exactly; fractions $R = 4.0\%$ , $T = 96.0\%$ . The window ghost is the 0.53 W/m² — four percent per surface, eight-ish for the pane (the second surface reflects again), which is why shop windows ghost noticeably and why camera lenses, with ten-plus elements, would be unusable at 4% per surface — the anti-reflection preview below. Answer: −20 and 80 V/m; 4% / 96%; books exact.

2 · The 120% amplitude — glass to air, resolved (exam level)

The same 100 V/m wave now travels inside glass toward an air boundary. Find the amplitudes, exhibit the paradox, and resolve it with the full energy audit; then state what happens instead beyond 41.8° incidence, and why this article cannot treat it.

Step 1. Coefficients ( $n_1 = 1.5$ , $n_2 = 1$ ): $r = 0.5/2.5 = +0.20$ (upright — internal reflection, no flip); $t = 3/2.5 = \mathbf{1.20}$ .

Step 2. The paradox stated: $E_{0_T} = 120$ V/m from a 100 V/m input — the boundary amplified the field? Continuity is satisfied ( $100 + 20 = 120$ ✓), so the algebra stands; the suspicion must attack the energy.

Step 3. The audit: $I_I = 1.5\times10^4/753.5 = 19.91$ W/m²; $I_R = 1.5\times400/753.5 = 0.80$ W/m²; $I_T = 1\times14400/753.5 = 19.11$ W/m². Sum: $0.80 + 19.11 = 19.91$ ✓. Fractions: $R = 4\%$ , $T = 96\%$ — identical to the air→glass split (r merely changed sign; R and T are direction-symmetric). Resolution: air charges only $E_0^2/2Z_0$ per unit flux where glass charged $1.5\times$ that — the 44% amplitude-squared gain is exactly cancelled by the 1/1.5 currency conversion. Amplitude is not energy; each medium prices its own.

Step 4. The cliff this article stands beside: tilt this glass-to-air geometry past $\theta_c = \arcsin(1/1.5) = 41.8°$ and no real transmitted wave direction exists — total internal reflection, $|r| = 1$ , with only an evanescent tail beyond the wall. Nothing in normal-incidence machinery can see this (it is a phase-matching phenomenon along the interface); the oblique-incidence article states it properly and the optics track owns its depths. Answers: +20, 120 V/m; R = 4%, T = 96%, books exact; TIR beyond 41.8° — next article.

Origins · 1 minGeorge Green and how this was discovered

George Green

1793–1841 · Sneinton, Nottingham — the miller who derived reflection coefficients between grinds

Green is mathematics' great outsider: a baker's son with one year of schooling who taught himself analysis in the intervals of running his father's windmill, and whose self-published 1828 Essay (fifty-one subscribers) introduced the potential-function methods and the theorem this track's E0/E2 chapters live on. Cambridge finally admitted him at 40. In 1838 he turned to waves at boundaries: On the Reflexion and Refraxion of Sound and its sequel on light treated reflection at an interface by demanding continuity across it and matching the wave impedances of the two sides — structurally the exact derivation of this article, seventy years before "impedance" had a name (his elastic-ether version of light got the polarization bookkeeping partly wrong, as every pre-Maxwell theory had to, but the boundary-condition method was permanent). He returned to his mill, died at 47, and was rediscovered posthumously — by Kelvin, who circulated the Essay to Continental physicists, and eventually by the twentieth century wholesale: Green's functions now run quantum field theory. The mill in Sneinton still grinds, restored, as his memorial.

History

1678

— Huygens's wavelets give reflection and refraction geometrically; amplitudes out of reach

1815

— Fresnel begins the wave theory of reflection; his oblique-incidence equations arrive by 1823

1838

— Green derives interface coefficients from continuity in an elastic ether — the method of this article

1875

— Lorentz's doctoral thesis re-derives reflection and refraction from Maxwell's equations proper

1886

— Rayleigh analyzes reflection from graded transitions: slow mismatch ≈ no echo

1935

— Smakula (Zeiss) patents the anti-reflection coating: destructive interference of the two surface ghosts

1939

— Blodgett's Langmuir-film AR coatings make "invisible glass" a manufactured product

1966

— Kao identifies attenuation, not reflection, as fiber's enemy — index-matched connectors follow; return loss becomes a spec

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §9.3.2 · Hecht, Optics, 5th ed., §4.6 · Jackson, Classical Electrodynamics, 3rd ed., §7.3 · Zangwill, Modern Electrodynamics, §17.3.

Provenance. Coefficients and energy split follow Griffiths §9.3.2 (Eqs. 9.82, 9.86–9.87) with the impedance form per Zangwill §17.3/engineering convention; string analogy per Griffiths's own §9.1.3 framing. A verifier should: (1) rederive r and t from the two continuity conditions and check 1 + r = t; (2) verify the algebraic closure R + T = 1; (3) recompute all numbers (air→glass −0.2/0.8, 4/96%, W/m² audit 0.53 + 12.74 = 13.27; glass→air +0.2/1.2 with T = 0.96 and 19.91 W/m² closure; R: diamond 17.2%, water 2.0%, glass–water 0.35%; $0.96^4 = 0.849$ ; $\theta_c = \arcsin(2/3) = 41.8°$ ); (4) confirm the reflected wave's H-direction sign (the flip's origin). Addendum for this edition: boundary conditions recalled BY NAME from E6's electrodynamic-boundary-conditions article (0563); media intensity formulas from 0590; this edition owns the normal-incidence solution, the impedance-mismatch unification, the phase-flip bookkeeping, and the t > 1 audit. Oblique incidence, polarization splitting, Brewster, and TIR deferred BY NAME to 0592 (with 0593–0594 as the dedicated deep-dives); thin-film AR machinery to the optics track; complex-Z conducting boundaries to 0596–0597.

Builds on: Monochromatic Plane Waves in Vacuum · Complex Notation for Electromagnetic Waves · Electromagnetic Waves in Linear Dielectric Media · Energy Density and the Poynting Vector.

Leads to: Reflection and Transmission at Oblique Incidence · The Fresnel Equations · Brewster's Angle and Total Internal Reflection · Intensity and Time-Averaged Quantities.

Where physically does the reflected wave come from — does the surface "know" to send it back?

From the same mechanism that builds the index itself: the driven dipoles of medium 2 (the previous article's re-radiation picture). Every molecular layer radiates both forward and backward; deep inside a homogeneous medium the backward radiations from successive layers cancel by destructive interference (each layer's echo is phase-stepped by its depth), so bulk media don't glow backwards. At a surface the cancellation loses its partners — the first few layers' backward radiation survives unbalanced, and that survivor is the reflected wave. The boundary-condition derivation gets the same answer without the microscopy because continuity conditions are the macroscopic summary of that dipole bookkeeping — same physics, two altitudes.

Why is the reflectance the same from either side of the interface, but the phase flip only happens one way?

$R = r^2$ kills the sign: $r_{1\to2} = -r_{2\to1}$ (swap the indices in the formula), so the two directions reflect equal fractions while their reflected waves differ by exactly the π flip. The asymmetry lives in the phase because the physics is asymmetric — one side's reflection is fixed-end-like, the other free-end-like — but energy accounting cannot see phases. The pairing $r_{12} = -r_{21}$ , together with $t_{12}t_{21} + r_{12}^2 = 1$ , forms the Stokes relations (derivable by time-reversal of this article's solution), and they are the consistency spine of all multi-surface interference calculations.

A window is two surfaces — why don't I just get 4% + 4% = 8% reflection?

Two corrections, one small and one conceptual. Small: the second surface reflects 4% of the 96% that reached it, and part of that re-reflects internally again — the incoherent sum for one pane converges to about 7.7%, not 8%. Conceptual: for thin, flat, coherently-illuminated films the two ghosts interfere — their path difference sets whether they add (up to ~16% for glass) or cancel (toward 0%), oscillating with wavelength and thickness. That interference is precisely the anti-reflection coating's engine, and it is why "add the percentages" is only licensed when coherence is broken (thick panes, broadband light, rough surfaces). The optics track's thin-film article owns the coherent case; this article supplies its per-surface coefficients.

Does the π phase flip have measurable consequences, or is it bookkeeping?

Ruthlessly measurable. Newton's rings have a dark center — the air-gap reflection flips at the bottom (air→glass) but not at the top (glass→air), so zero gap means destructive interference: the flip visible to the naked eye. Anti-reflection coatings choose film indices so both reflections flip (glass side and film side), letting a quarter-wave thickness alone set the cancellation. Radio engineers meet it as the sign of the echo on a shorted versus open cable (r = −1 vs +1 — a TDR trace reads the boundary type from the pulse's orientation). And in quantum optics the same flip appears in beamsplitter phase conventions that decide whether photons bunch. Signs are physics wearing accounting clothes.

What happens at a boundary if the wave hits a conductor instead of a dielectric?

The formula survives with $Z_2$ gone complex — and small. A good conductor's wave impedance (article 0596's business) has magnitude tiny compared to $Z_0$ and a 45° phase: plugging into $r = (Z_2-Z_1)/(Z_2+Z_1)$ gives $r \approx -1$ with a small complex correction — nearly total reflection, nearly perfect inversion, with the small deficit absorbed in the skin depth (that deficit is why mirrors are ~95–99% rather than 100%, and why microwave oven walls stay merely warm). The limit $Z_2 \to 0$ is the perfect conductor: r = −1 exactly, the boundary condition E∥ = 0, standing waves in front — the E7 chapter's mirror problems, now derivable from this article's one formula. Details and the skin-depth physics: 0596–0597, by name.

Is the assumption of exactly one reflected and one transmitted plane wave ever wrong?

It is exactly right for a flat, sharp, linear, time-independent interface between homogeneous media — because plane waves are a complete basis and the boundary is translation-invariant (it cannot change transverse wavevector or frequency, so only the two phase-matched directions exist). Break any hypothesis and the ansatz enlarges honestly: a rough surface scatters into a whole angular distribution (specular + diffuse); a periodic surface adds discrete diffraction orders (gratings); a graded transition (n varying over many wavelengths) reflects almost nothing at all — Rayleigh's 1886 analysis, the moth's-eye principle behind nanostructured AR surfaces; a nonlinear or moving boundary can change frequency. Each generalization keeps this article's method — match fields at the boundary — and only widens the wave menu.