The Fresnel Equations

Stand at a shop window at dusk. Look straight through it and the glass is nearly invisible — it steals four percent of the light. Look along it at a shallow angle and it has become a mirror: the street lamps blaze in it, the interior vanishes. Same glass, same light; only the angle changed. Somewhere between those extremes, at one particular angle and for one particular polarization, the reflection does not merely weaken — it vanishes exactly. A sheet of glass can be made perfectly invisible to the right kind of light. The previous article assembled oblique incidence end to end: the matched surface phases, the three kinematic laws, and a first pass through the amplitude coefficients themselves — the Fresnel equations, derived there channel by channel in the standard component frame. Why, then, a dedicated Fresnel article? Because four formulas that run every lens coating, polarizing sunglass, and semiconductor-fab ellipsometer on Earth deserve more than one derivation and a first look. What that first pass necessarily left on the table is this article's cargo: a second, unifying route — one impedance-matching frame that produces both channels at a stroke and keeps working when the media are magnetic (where the $n$ -forms silently fail); the index-free trigonometric forms Fresnel actually published; the energy-bookkeeping theorem the previous article used but deferred here by name — why the transmittance must carry its $n\cos\theta$ prefactor, and why $R + T = 1$ closes channel by channel as an algebraic identity; the reconciliation of the sign conventions that make three textbooks disagree about one number; and the coefficients read as a working instrument, from signed-amplitude curves to the ellipsometric ratio $\rho = r_p/r_s$ that certifies every gate oxide in your phone. So: same boundary conditions, deeper toolkit. The two landmark phenomena the coefficients contain — the Brewster zero the previous article exhibited, and total internal reflection — get their own landmark treatment in the next article.

Key takeaways

At an interface, kinematics (matched phases, Snell's law — the previous article's package) fixes the directions; the Fresnel equations fix the amplitudes, separately for the two polarization channels: $s$ (E perpendicular to the plane of incidence) and $p$ (E in the plane).
Both channels come from one matching problem — continuity of tangential $\mathbf{E}$ and tangential $\mathbf{H}$ — and the whole angle dependence enters through effective impedances: $Z/\cos\theta$ for $s$ , $Z\cos\theta$ for $p$ . Reflection is impedance mismatch.
Nonmagnetic forms: $r_s = \dfrac{n_1\cos\theta_I - n_2\cos\theta_T}{n_1\cos\theta_I + n_2\cos\theta_T}$ , $r_p = \dfrac{n_2\cos\theta_I - n_1\cos\theta_T}{n_2\cos\theta_I + n_1\cos\theta_T}$ — note the swapped index roles; that swap is why the channels part company as the angle grows.
Power bookkeeping: $R = |r|^2$ , but $T = \dfrac{n_2\cos\theta_T}{n_1\cos\theta_I}|t|^2$ — the prefactor (impedance × beam-width geometry) is mandatory; only with it does $R + T = 1$ .
The signed amplitude curves (computed for glass): $r_s$ stays negative at every angle — a universal external-reflection phase flip — while $r_p$ changes sign on its way from $+0.2$ to $-1$ , crossing zero at $56.3°$ : the Brewster null, exhibited in the previous article and owned by the next. The sign structure, invisible in reflectance plots, is what interference stacks and ellipsometry actually use.
KEY IDEA: matching tangential E and H across the interface, channel by channel, turns reflection into an impedance-mismatch problem; the resulting four Fresnel coefficients — with the $n\cos\theta$ energy prefactor — are the complete dynamical content of reflection and refraction.

The core insight

Match tangential $\mathbf{E}$ and tangential $\mathbf{H}$ across the interface in each of the two independent polarization channels; the resulting linear equations, closed by Snell's law, deliver the four Fresnel amplitude coefficients — and the contrast between the two channels is the origin of polarization by reflection.

Where this lives today

The Fresnel equations are run, at scale, everywhere a surface meets design intent. Every anti-reflection coating — on eyeglasses, camera lenses, and solar panels (where the 4%-per-surface tax is revenue) — is a stack of thin films whose design tool is these same coefficients cascaded with propagation phases; the transfer-matrix codes that optimize a modern camera lens evaluate (1)–(2) millions of times per design pass. Polarizing beamsplitter cubes in projectors and interferometers are engineered Fresnel stacks; LiDAR return-signal budgets, rendering engines (physically-based rendering computes per-pixel Fresnel terms — the "Fresnel effect" is a named slider in every 3D package), and fiber-connector loss specs all trace to this article's four ratios. The precision frontier is ellipsometry: measure the complex ratio $\rho = r_p/r_s$ versus angle and wavelength, and invert the Fresnel equations to recover film thicknesses and optical constants with sub-nanometer sensitivity — the standard metrology of the semiconductor industry, where every gate-oxide thickness on every wafer is certified by an inversion of (1)–(2) extended to layered, absorbing media. And at the exotic end, the impedance-form coefficients of Derivation A are the design language of metamaterials and transformation optics: engineered $\mu(\omega)$ lets designers match $Z_1 = Z_2$ while steering $n$ , building interfaces that refract strongly yet reflect nothing — cloaking-adjacent technology that is pure Step-3 algebra with unusual inputs.

Origins · 1 minAugustin-Jean Fresnel and how this was discovered

Augustin-Jean Fresnel

1788–1827 · École des Ponts et Chaussées, France

Fresnel was a road engineer who did optics in the evenings and between postings — and, for a while, while suspended from duty for siding with the royalists during Napoleon's Hundred Days. In one decade, working with home-made apparatus and dying slowly of tuberculosis, he rebuilt optics: the rigorous diffraction theory that won the 1819 Academy prize (with the famous vindication of the "absurd" bright spot in a disc's shadow), the transverse-wave hypothesis worked out with Arago's polarization experiments, and in 1823 the reflection coefficients bearing his name — derived from boundary conditions on an elastic ether, thirty-eight years before Maxwell's equations existed. That the electromagnetic theory reproduces his amplitude ratios exactly is one of physics' great continuity stories: Fresnel had the correct boundary structure with the wrong substrate. His day job left a mark too — the stepped Fresnel lens he designed for lighthouses cut the weight of a first-order beacon lens by an order of magnitude, and its descendants live in overhead projectors, VR headsets, and solar concentrators. Few physicists have one formula in every optics code and one lens in every lighthouse.

History

1808

— Malus discovers polarization by reflection, looking at sunset light reflected off the Luxembourg Palace windows through a calcite crystal

1815

— Fresnel begins his diffraction and polarization program with Arago

1821

— Fresnel and Arago establish that light waves are transverse — the prerequisite for two polarization channels

1823

— Fresnel derives the amplitude coefficients from an elastic-ether model: the trigonometric forms

1861

— Maxwell's electromagnetic theory; the same coefficients re-derived from field boundary conditions

1890

s — Drude turns reflection measurements into surface science, the ancestor of ellipsometry

1945

— Rothen names ellipsometry; Fresnel inversion becomes a thickness gauge

1970

s — Transfer-matrix thin-film design industrializes cascaded Fresnel coefficients

The physical picture

Why must there be reflected and transmitted waves at all? Because the incident wave alone cannot satisfy the boundary conditions. Tangential $\mathbf{E}$ and tangential $\mathbf{H}$ must be continuous across a charge-free, current-free interface (the electrodynamic boundary-condition package, recalled by name from Electrodynamic Boundary Conditions), and a single wave arriving from medium 1 presents field values at the surface that the fields in medium 2 cannot match with nothing on their side. Nature's fix is unique: add exactly one reflected wave and one transmitted wave, with amplitudes chosen so the tangential fields stitch together point by point. The Fresnel coefficients are those stitching amplitudes — not new physics, just the boundary conditions cashed. The deepest way to organize the calculation is borrowed from transmission-line engineering. Every medium presents a wave impedance $Z = \sqrt{\mu/\varepsilon}$ — the ratio of transverse $E$ to transverse $H$ a plane wave maintains (vacuum: $Z_0 = 377\ \Omega$ ; the linear-media article introduced it, by name). At normal incidence, reflection measures the mismatch: $r = (Z_2 - Z_1)/(Z_2 + Z_1)$ , and equal impedances mean no echo — the same rule that governs a wave on two joined ropes. Oblique incidence adds one twist: the boundary conditions only see the tangential projections of the fields, so each channel presents an effective impedance that carries its own angle factor. For $s$ -polarization, $\mathbf{E}$ is already tangential but $\mathbf{H}$ tilts, so $Z_s = Z/\cos\theta$ ; for $p$ -polarization, $\mathbf{H}$ is tangential and $\mathbf{E}$ tilts, so $Z_p = Z\cos\theta$ . One mismatch formula, two different angle dependencies — every curve in this article follows from that asymmetry, including the fact that $Z_p$ matching can be achieved at one angle (the $r_p$ zero this article exhibits and the next one explains) while $Z_s$ mismatch only ever grows.

Figure

The two polarization channels at an interface — s keeps E tangential, p keeps H tangential, and each channel matches its own effective impedance

One bookkeeping decision before the algebra, because it saves endless confusion later. The plane of incidence is the plane containing the incident wavevector and the surface normal. Any incident polarization decomposes into a component with $\mathbf{E}$ perpendicular to that plane — called $s$ (from the German senkrecht), TE, or $\perp$ — and a component with $\mathbf{E}$ in the plane — called $p$ (parallel), TM, or $\parallel$ . The interface, being translation-invariant along itself, cannot mix the two: an incident $s$ -wave produces only $s$ reflected and transmitted waves, likewise $p$ . The two channels are solved independently and superposed at the end, which is why two pairs of coefficients exhaust the problem. Everything upstream of this — why the three waves share one frequency and one tangential wavevector, hence one plane of incidence and Snell's law — is the previous article's kinematic package, Reflection and Transmission at Oblique Incidence, used here without re-derivation.

The coefficient package

The results first, then the derivations. For a monochromatic plane wave incident from medium 1 ( $n_1$ , $\mu_1$ ) onto medium 2 ( $n_2$ , $\mu_2$ ), angles linked by $n_1\sin\theta_I = n_2\sin\theta_T$ , the amplitude ratios in the $s$ -channel are

r_s = \frac{n_1\cos\theta_I - n_2\cos\theta_T}{n_1\cos\theta_I + n_2\cos\theta_T}, \qquad t_s = \frac{2\,n_1\cos\theta_I}{n_1\cos\theta_I + n_2\cos\theta_T}, \qquad (1)

where: $r_s = E_R/E_I$ and $t_s = E_T/E_I$ are ratios of electric-field amplitudes (nonmagnetic form, $\mu_1 = \mu_2 = \mu_0$ ; the general- $\mu$ version replaces each $n\cos\theta$ by $\cos\theta/Z$ — Derivation A produces both). Read the structure: each medium contributes its index weighted by its own wave's obliquity. At $\theta_I = 0$ the cosines drop out and $r_s \to (n_1 - n_2)/(n_1 + n_2)$ — exactly the normal-incidence result of Reflection and Transmission at Normal Incidence, as it must be. For air→glass, $r_s$ is negative at every angle: the $s$ -reflection always carries the external-reflection phase flip. The $p$ -channel:

r_p = \frac{n_2\cos\theta_I - n_1\cos\theta_T}{n_2\cos\theta_I + n_1\cos\theta_T}, \qquad t_p = \frac{2\,n_1\cos\theta_I}{n_2\cos\theta_I + n_1\cos\theta_T}, \qquad (2)

where: the index roles are swapped relative to (1) — $n_2$ now multiplies $\cos\theta_I$ — and that swap is the whole story of polarization by reflection. Because $n_2\cos\theta_I$ falls while $n_1\cos\theta_T$ rises (or vice versa) as the angle grows, the numerator of $r_p$ can pass through zero at one angle, whereas (1)'s numerator, with each index locked to its own cosine, never can. Sign convention: (2) follows the Hecht/Jackson choice for the positive sense of the reflected $\mathbf{E}_p$ ; Griffiths' convention flips the sign of $r_p$ (his Eq. 9.109 form) with zero physical consequence — the warning box below inventories the chaos. Substituting Snell's law turns (1)–(2) into the index-free trigonometric forms

r_s = -\,\frac{\sin(\theta_I - \theta_T)}{\sin(\theta_I + \theta_T)}, \qquad r_p = \frac{\tan(\theta_I - \theta_T)}{\tan(\theta_I + \theta_T)}, \qquad (3)

where: all reference to $n_1, n_2$ has been absorbed into the angle pair — Fresnel's own 1823 forms, derived before anyone knew what was waving. They make the two landmarks visible by inspection: $r_p = 0$ exactly where $\tan(\theta_I + \theta_T)$ diverges, i.e. $\theta_I + \theta_T = 90°$ (the Brewster condition — first derived in the oblique-incidence article, owned in landmark depth by the next), and both coefficients $\to -1$ as $\theta_I \to 90°$ (any polished surface becomes a perfect mirror at grazing incidence — the shop-window effect, and the reason X-ray telescopes, which no material reflects at normal incidence, are built as nested grazing-incidence shells). Finally the power ratios:

R = |r|^2, \qquad T = \frac{n_2\cos\theta_T}{n_1\cos\theta_I}\,|t|^2, \qquad R + T = 1 \ \text{(each channel)}, \qquad (4)

where: the reflectance needs no prefactor (same medium, same angle, same beam width as the incident wave), but the transmittance carries $n_2/n_1$ (intensity per unit amplitude scales with the medium's index) and $\cos\theta_T/\cos\theta_I$ (the refracted beam changes width — a bundle of rays occupying interface area $\mathcal{A}$ has cross-section $\mathcal{A}\cos\theta$ ). Derivation C proves the balance; the warning box explains why forgetting the prefactor is the classic exam disaster, including the unsettling correct fact $t > 1$ for glass→air.

Figure

The signed amplitude coefficients for glass, computed — what reflectance plots hide: r_s never changes sign while r_p crosses zero on its way to the grazing limit

Derivation ABoth channels from one impedance frame

The goal is the full coefficient package (1)–(2), derived once, in a form that also covers magnetic media. The strategy is to reduce each channel to the same one-dimensional matching problem: identify which field is purely tangential in that channel, write the two continuity conditions, and recognize that the only channel-dependent input is an effective impedance — $Z/\cos\theta$ for $s$ , $Z\cos\theta$ for $p$ . This route complements the component-by-component treatment of the oblique-incidence article: same boundary conditions, organized so that both polarizations and the magnetic generalization drop out of a single computation. Watch where the two channels' angle factors come from — one geometric projection each, on opposite fields.

Step 1. Set the frame. Interface at $z = 0$ , plane of incidence the $xz$ -plane, all three waves sharing the tangential wavevector $k_x$ and frequency $\omega$ (the kinematic package, assumed). In each medium a plane wave ties its field magnitudes by $H = E/Z$ with $Z = \sqrt{\mu/\varepsilon}$ — for nonmagnetic media $Z = Z_0/n$ , so "large $n$ " and "small $Z$ " are the same statement.

Step 2. $s$ -channel bookkeeping. Here $\mathbf{E} = E\,\hat{\mathbf{y}}$ for all three waves: $E$ is already tangential, so continuity of tangential $\mathbf{E}$ reads, at $z = 0$ with the common phase cancelled,

E_I + E_R = E_T.

$\mathbf{H} = \hat{\mathbf{k}}\times\mathbf{E}/Z$ lies in the $xz$ -plane, tilted by the propagation angle; its tangential ( $x$ ) component picks up one factor of $\cos\theta$ , with the reflected wave's projection reversed (its $k_z$ flipped):

\frac{\cos\theta_I}{Z_1}\,(E_I - E_R) = \frac{\cos\theta_T}{Z_2}\,E_T.

Why the minus sign sits with $E_R$ : reversing $k_z$ while keeping $\mathbf{E}$ along $+\hat{\mathbf{y}}$ reverses the $x$ -component of $\hat{\mathbf{k}}\times\mathbf{E}$ — the reflected wave's $\mathbf{H}$ projection opposes the incident one. Getting this sign wrong flips $r_s$ and wrecks the normal-incidence audit in Step 5.

Step 3. Solve the $s$ pair. Define the effective impedances $Z_1^{(s)} = Z_1/\cos\theta_I$ , $Z_2^{(s)} = Z_2/\cos\theta_T$ — "tangential E per tangential H" for each wave. The two equations become the standard junction problem, giving

r_s = \frac{Z_2^{(s)} - Z_1^{(s)}}{Z_2^{(s)} + Z_1^{(s)}}, \qquad t_s = \frac{2\,Z_2^{(s)}}{Z_2^{(s)} + Z_1^{(s)}}.

Substituting $Z = Z_0/n$ (nonmagnetic) and clearing denominators lands exactly on (1).

Step 4. $p$ -channel bookkeeping. Now $\mathbf{H} = H\,\hat{\mathbf{y}}$ is the purely tangential field, and $\mathbf{E}$ , in the $xz$ -plane, tilts: tangential $E$ carries the $\cos\theta$ . With the Hecht/Jackson sign convention for the reflected $\mathbf{E}_p$ , continuity of tangential $\mathbf{E}$ and tangential $\mathbf{H}$ read

(E_I - E_R)\cos\theta_I = E_T\cos\theta_T, \qquad \frac{1}{Z_1}(E_I + E_R) = \frac{1}{Z_2}\,E_T.

Solving with effective impedances $Z^{(p)} = Z\cos\theta$ :

r_p = \frac{Z_1^{(p)} - Z_2^{(p)}}{Z_1^{(p)} + Z_2^{(p)}}\bigg|_{Z^{(p)} = Z\cos\theta} = \frac{n_2\cos\theta_I - n_1\cos\theta_T}{n_2\cos\theta_I + n_1\cos\theta_T}, \qquad t_p = \frac{2\,n_1\cos\theta_I}{n_2\cos\theta_I + n_1\cos\theta_T},

which is (2). Why the channels differ structurally: each channel projects ONE of the two fields by its $\cos\theta$ — H in the $s$ -case, E in the $p$ -case — so the impedances enter once divided and once multiplied by the cosine. The $p$ -channel's $Z\cos\theta$ is a product of a decreasing and an increasing factor along the Snell trajectory, so the two media's effective impedances can cross — the matched point where $r_p = 0$ . The $s$ -channel's $Z/\cos\theta$ mismatch is monotonic: no crossing, no zero, ever.

Step 5. Audits. Normal incidence ( $\theta_I = \theta_T = 0$ ): both channels collapse to $r = (n_1 - n_2)/(n_1 + n_2)$ , $t = 2n_1/(n_1 + n_2)$ — the polarization distinction dies with the plane of incidence, and the normal-incidence article's results are recovered (the apparent sign disagreement between $r_s$ and $r_p$ at $\theta = 0$ is pure convention: the two channels' reference arrows for reflected $\mathbf{E}$ point oppositely when the plane of incidence degenerates — see the warning box). Magnetic media: nothing in Steps 1–4 assumed $\mu = \mu_0$ ; the $Z$ -forms in Steps 3–4 are the general Fresnel equations, used for microwave ferrites and metamaterial design, where $Z_1 = Z_2$ with $n_1 \neq n_2$ produces the striking "index contrast with zero reflection" that impedance-matched metamaterials exploit. Verified: nonmagnetic reduction reproduces Griffiths Eqs. 9.109 (modulo his $r_p$ sign convention) and Jackson §7.3; the $Z$ -forms match Jackson Eqs. 7.39/7.41 with $\mu \neq \mu_0$ retained; normal-incidence limit matches $(n_1-n_2)/(n_1+n_2)$ ; numerical spot-check against the oblique-incidence article's 45° glass audit: this frame reproduces its $r_s = -0.303$ , $r_p = +0.092$ exactly, as it must — two routes, one answer.

Go deeper · Derivation B · 2 minThe trigonometric forms and what they expose

Derivation BThe trigonometric forms and what they expose

The goal is Fresnel's own index-free forms (3) from the index forms (1)–(2) — three lines of trigonometry each — and then a reading of the two singular features they make obvious: the $p$ -zero and the grazing mirror. This is worth doing honestly once because the trig forms are the ones that carry the physics on their face, and because their derivation is a pure exercise in spending Snell's law.

Step 1. Eliminate the indices. Snell gives $n_1/n_2 = \sin\theta_T/\sin\theta_I$ . Divide numerator and denominator of $r_s$ in (1) by $n_2$ :

r_s = \frac{\sin\theta_T\cos\theta_I - \sin\theta_I\cos\theta_T}{\sin\theta_T\cos\theta_I + \sin\theta_I\cos\theta_T} = \frac{\sin(\theta_T - \theta_I)}{\sin(\theta_T + \theta_I)} = -\,\frac{\sin(\theta_I - \theta_T)}{\sin(\theta_I + \theta_T)},

by the sine addition formulas, read in reverse.

Step 2. Same maneuver for $r_p$ : divide (2) through by $n_2$ and substitute:

r_p = \frac{\sin\theta_I\cos\theta_I - \sin\theta_T\cos\theta_T}{\sin\theta_I\cos\theta_I + \sin\theta_T\cos\theta_T} = \frac{\sin 2\theta_I - \sin 2\theta_T}{\sin 2\theta_I + \sin 2\theta_T} = \frac{\tan(\theta_I - \theta_T)}{\tan(\theta_I + \theta_T)},

using $\sin 2\theta = 2\sin\theta\cos\theta$ and the sum-to-product identities. Why the $p$ -form grew a tangent where the $s$ -form has a sine: the extra $\cos\theta_I\cos\theta_T$ structure from the index swap converts sums of sines into the $\sin\pm$ / $\cos\pm$ ratio — and it is exactly the $\cos(\theta_I + \theta_T)$ lurking in that tangent's denominator that can vanish.

Step 3. Read the $p$ -zero off the form. $r_p = 0$ requires $\tan(\theta_I + \theta_T) \to \infty$ , i.e. $\theta_I + \theta_T = 90°$ (the alternative, $\theta_I = \theta_T$ , means $n_1 = n_2$ : no interface) — note how the trig form delivers in one line what the index form needed squaring and collecting to say. The condition itself is not news: the oblique-incidence article derived the Brewster angle from the vanishing numerator, and the next article owns its geometry, mechanism, and applications outright. What belongs here is the structural reading: the tangent form shows the zero is a property of the angle pair, independent of which side the light comes from — the seed of the internal/external Brewster complementarity the next article proves.

Step 4. Read the grazing limit. As $\theta_I \to 90°$ with $\theta_T$ pinned near its maximum, both (3) forms $\to -1$ : perfect reflection with a phase flip, in both channels, for any index contrast. Every surface — glass, water, asphalt, the page of this book — becomes a mirror at grazing incidence. This is why wet roads blind you at sunset ( $R$ near 1 at grazing, plus the polarization gap making it worse for one polaroid orientation), and why grazing-incidence optics is the only way to focus X-rays, for which $n$ is barely below 1 and normal-incidence reflectances are $\sim 10^{-5}$ . Verified: forms match Griffiths Prob. 9.17/Hecht §4.6 (Fresnel 1823); numerical spot-check at $\theta_I = 45°$ : $\theta_T = 28.13°$ , $-\sin(16.87°)/\sin(73.13°) = -0.303$ ✓ and $\tan(16.87°)/\tan(73.13°) = 0.092$ ✓, matching Derivation A's index-form values exactly.

Go deeper · Derivation C · 2 minEnergy bookkeeping — why the transmittance carries the n cos-theta prefactor

Derivation CEnergy bookkeeping — why the transmittance carries the n cos-theta prefactor

The goal is the power ledger (4): define reflectance and transmittance as energy fluxes through the interface, derive the $\tfrac{n_2\cos\theta_T}{n_1\cos\theta_I}$ prefactor honestly, and close the books with $R + T = 1$ as an algebraic identity. The strategy: energy conservation at a lossless boundary is a statement about the normal component of the Poynting vector integrated over a patch of interface — not about field amplitudes — so both the medium's intensity scale and the beam's obliquity must enter. This is the derivation that inoculates against the single most common Fresnel error.

Step 1. Intensity of a plane wave in a medium of index $n$ (time-averaged, from the wave-energy articles of this chapter, recalled by name): $I = \tfrac{1}{2}\,n\varepsilon_0 c\,|E|^2$ . The energy crossing a unit interface area per second is the normal projection $I\cos\theta$ — a beam of cross-section $w$ illuminates interface area $w/\cos\theta$ .

Step 2. Define the ratios through one interface patch. Reflected beam: same medium, same angle, so both factors cancel:

R = \frac{I_R\cos\theta_R}{I_I\cos\theta_I} = |r|^2.

Transmitted beam: different medium ( $n_2$ in the intensity) and different angle ( $\cos\theta_T$ in the projection):

T = \frac{I_T\cos\theta_T}{I_I\cos\theta_I} = \frac{n_2\cos\theta_T}{n_1\cos\theta_I}\,|t|^2.

Why both factors are physical and independent: $n_2/n_1$ is the impedance effect — for fixed field amplitude, a denser medium carries more power per unit area; $\cos\theta_T/\cos\theta_I$ is pure geometry — refraction toward the normal squeezes the beam into a narrower cross-section, concentrating the same power onto less area. Air→glass at 45°: the beam narrows by $\cos 28.1°/\cos 45° = 1.25$ while the medium factor adds 1.5; the product 1.87 is why $|t|^2 \approx 0.49$ can carry 91% of the power.

Step 3. Close the ledger for the $s$ -channel. Let $a = n_1\cos\theta_I$ , $b = n_2\cos\theta_T$ . From (1), $R_s = \left(\tfrac{a-b}{a+b}\right)^2$ and $T_s = \tfrac{b}{a}\cdot\tfrac{4a^2}{(a+b)^2} = \tfrac{4ab}{(a+b)^2}$ , so

R_s + T_s = \frac{(a-b)^2 + 4ab}{(a+b)^2} = \frac{(a+b)^2}{(a+b)^2} = 1.

Step 4. Close it for the $p$ -channel. With $a' = n_2\cos\theta_I$ , $b' = n_1\cos\theta_T$ , (2) gives $R_p = \left(\tfrac{a'-b'}{a'+b'}\right)^2$ and $T_p = \tfrac{n_2\cos\theta_T}{n_1\cos\theta_I}\cdot\tfrac{4n_1^2\cos^2\theta_I}{(a'+b')^2} = \tfrac{4a'b'}{(a'+b')^2}$ — the same identity closes it: $R_p + T_p = 1$ . Why the balance holds channel by channel and not merely in sum: the interface cannot convert $s$ into $p$ (translation invariance along the surface forbids the mixing), so each channel runs its own sealed energy account. *Verified: reproduces Griffiths Eqs. 9.115–9.116 and Hecht §4.6.2; numerical audit at 45°, glass: $R_s = 0.0920$ , $T_s = 0.9080$ , sum $1.000000$ ; $R_p = 0.0085$ , $T_p = 0.9915$ , sum $1.000000$ (worked example 1). The identity $(a-b)^2 + 4ab = (a+b)^2$ is exact, so the balance is algebraic, not numerical coincidence.*

Limiting cases

Case	Result	Meaning
Normal incidence	$r \to \dfrac{n_1 - n_2}{n_1 + n_2}$ , both channels	polarization distinction dies with the plane of incidence; 4% per glass surface, 2% for water
Grazing incidence	$r_s, r_p \to -1$	every surface a mirror; wet-road glare, X-ray optics
$\theta_I + \theta_T = 90°$	$r_p = 0$ exactly	the $p$ -null at $\arctan(n_2/n_1)$ — exhibited here, owned by the next article
$n_1 > n_2$ , $\theta_I > \arcsin(n_2/n_1)$	$\cos\theta_T$ imaginary, $\vert r\vert = 1$	total internal reflection — machinery in the next article
$n_1 = n_2$	$r = 0$ , $t = 1$	no interface, no echo — the trivial audit
$Z_1 = Z_2$ , $n_1 \neq n_2$ (magnetic)	$r = 0$ at normal incidence	impedance-matched media reflect nothing while still refracting — metamaterial trick
$\theta_I \to 0$ vs $\theta_I$ small	$R$ flat to second order	reflectance barely changes below $\sim 30°$ ; the curves' long plateau
absorbing medium 2	same formulas, $n_2 \to \tilde{n}_2$ complex	$r$ complex, $\vert r\vert < 1$ ; metals article of this chapter cashes it

Common trapAmplitude is not power, and three textbooks mean three sign conventions

Two distinct traps, both fed by the same formulas. First: $r$ and $t$ are amplitude ratios; the power fractions are $R = |r|^2$ but $T = \tfrac{n_2\cos\theta_T}{n_1\cos\theta_I}|t|^2$ — never $|t|^2$ alone. The failure is loudest for internal incidence: glass→air at normal incidence has $t = 2(1.5)/2.5 = 1.2 > 1$ . The transmitted amplitude really is larger than the incident one (nothing forbids it — the boundary condition is a sum rule, $1 + r = t$ with $r = +0.2 > 0$ ), but the transmitted power is $T = \tfrac{1}{1.5}(1.2)^2 = 0.96 < 1$ . Any bookkeeping that squares $t$ and stops has manufactured energy. Second: the sign of $r_p$ is pure convention — Griffiths defines the positive reflected $\mathbf{E}_p$ arrow one way (his $r_p = \tfrac{n_1\cos\theta_T - n_2\cos\theta_I}{n_1\cos\theta_T + n_2\cos\theta_I}$ at his Eq. 9.109 sign), Hecht and Jackson the other (this article's (2)); at normal incidence the two conventions even disagree about whether $r_s$ and $r_p$ are "equal" or "opposite" — while every measurable ( $R$ , $T$ , Brewster angle, ellipsometric ratios) is convention-independent. Check the arrow diagram before comparing coefficients across books, and never splice formulas from two conventions into one calculation: phases in interference stacks and ellipsometry inversions will silently corrupt.

Worked examples

1 · Full audit at 50° — air to water (confidence builder)

Sunlight grazes a lake at $\theta_I = 50°$ ( $n_2 = 1.333$ ). Find all four Fresnel coefficients, verify both energy ledgers, and compute the degree of polarization of the glare. (The glass version of this audit, at 45°, lives in the oblique-incidence article; water at 50° sits three degrees below its Brewster angle of 53.1° — expect drama in the $p$ -channel.)

Step 1. Kinematics: $\sin\theta_T = \sin 50°/1.333 = 0.5746$ , so $\theta_T = 35.08°$ ; $\cos\theta_I = 0.6428$ , $\cos\theta_T = 0.8185$ .

Step 2. $s$ -channel from (1): denominator $= 1\times0.6428 + 1.333\times0.8185 = 1.7339$ . $r_s = (0.6428 - 1.0911)/1.7339 = \mathbf{-0.259}$ ; $t_s = 2\times0.6428/1.7339 = \mathbf{0.741}$ .

Step 3. $p$ -channel from (2): denominator $= 1.333\times0.6428 + 1\times0.8185 = 1.6753$ . $r_p = (0.8569 - 0.8185)/1.6753 = \mathbf{+0.023}$ — barely positive: three degrees shy of its zero crossing.

Step 4. Energy ledgers with prefactor $\tfrac{n_2\cos\theta_T}{n_1\cos\theta_I} = \tfrac{1.0911}{0.6428} = 1.697$ : $R_s = 0.0668$ , $T_s = 1.697\times0.741^2 = 0.9332$ , sum $= \mathbf{1.0000}$ ✓. $R_p = 0.0005$ , $T_p = 0.9995$ , sum $= \mathbf{1.0000}$ ✓ — both ledgers close to four decimals, the Derivation C identity at work.

Step 5. Polarization of the glare. Unpolarized sun is an incoherent 50/50 mix, so the reflected intensities stand in ratio $R_s : R_p = 0.0668 : 0.0005 \approx 130 : 1$ . Degree of polarization $= (R_s - R_p)/(R_s + R_p) = \mathbf{0.98}$ — lake glare at 50° is 98% pure horizontal polarization. This is why polarized sunglasses feel like a magic trick specifically around water at mid-afternoon sun heights: the geometry parks reflections near the water Brewster angle, where the $p$ -channel is essentially extinguished and a vertical-axis polaroid removes almost the entire reflection while passing half the (unpolarized) scene.

2 · A window pane, both surfaces, at 60° (exam level)

A camera looks through a vertical glass pane ( $n = 1.5$ ) at $60°$ incidence. What fraction of $s$ - and of $p$ -polarized light makes it through both surfaces (ignore multiple reflections and absorption)? What does a polarizing filter buy?

Step 1. First surface, from (1)–(2) at $\theta_I = 60°$ : $\sin\theta_T = 0.8660/1.5 = 0.5774$ , $\theta_T = 35.26°$ ; $\cos\theta_I = 0.500$ , $\cos\theta_T = 0.8165$ . $r_s = (0.500 - 1.2247)/(0.500 + 1.2247) = -0.420$ , so $R_s = 0.1766$ . $r_p = (0.750 - 0.8165)/(0.750 + 0.8165) = -0.0424$ , so $R_p = 0.0018$ .

Step 2. Second surface (glass→air at the internal angle $35.26°$ ): by the reversibility of the Fresnel formulas — swap indices and angles in (1)–(2) and the reflectance is unchanged (Stokes relations; algebraically, $R$ depends only on the unordered pair of $n\cos\theta$ values) — each channel sees the same $R$ again: $R_s = 0.1766$ , $R_p = 0.0018$ .

Step 3. Two-surface transmission per channel: $T^{(2)}_s = (1 - 0.1766)^2 = \mathbf{0.678}$ ; $T^{(2)}_p = (1 - 0.0018)^2 = \mathbf{0.996}$ . Unpolarized: $\tfrac{1}{2}(0.678 + 0.996) = \mathbf{0.837}$ — a pane at 60° eats 16% of unpolarized light, almost all of it from the $s$ -channel.

Step 4. The filter's payoff. A polarizer passing only $p$ costs the 50% of unpolarized light in the wrong channel but delivers 99.6% of what remains through the glass — and, more to the point for a photographer, it kills the reflections off the pane (the light the camera sees bouncing off the window toward it is $R_s$ -dominated by two orders of magnitude at this angle: 17.7% vs 0.18%). Shooting through shop windows or water at steep angles with a rotated polarizer is applied Fresnel analysis: rotate until the transmission axis lies in the plane of incidence, and the window's ghost images fade by an order of magnitude. Note the practical asymmetry: near 60° the pane is nearly lossless for $p$ (0.4% total) — this is the principle the next article's Brewster windows push to exactly zero.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §9.3.3 (Eqs. 9.109–9.116) · Jackson, Classical Electrodynamics, 3rd ed., §7.3 · Hecht, Optics, 5th ed., §4.6 · Born & Wolf, Principles of Optics, 7th ed., §1.5 · Zangwill, Modern Electrodynamics, §17.3 · Purcell & Morin, Electricity and Magnetism, 3rd ed., Ch. 9

Provenance. Fresnel amplitude equations and trigonometric forms cross-checked against Griffiths 4th ed. §9.3.3 (Eqs. 9.109–9.118), Jackson 3rd ed. §7.3, and Hecht Optics 5th ed. §4.6. A verifier should: (1) re-derive $r_s,t_s$ from tangential E,H continuity; (2) confirm $R_s+T_s=1$ via the algebraic identity $(a-b)^2+4ab=(a+b)^2$ ; (3) check $\theta_B=\arctan1.5\approx56.3^\circ$ and air–water $R\approx0.02$ numerically; (4) verify the $\theta_I\to0$ limit gives $(n_1-n_2)/(n_1+n_2)$ . Sign convention for $r_p$ follows Hecht/Jackson; Griffiths' convention differs by an overall sign with no physical consequence. Addendum for this edition: derivations reorganized into the unified impedance frame (both channels + magnetic media from one matching problem, general forms checked against Jackson Eqs. 7.39/7.41) plus the trigonometric-form and energy-ledger derivations; scope split with the neighboring articles executed — the oblique-incidence kinematic package (phase matching, Snell) is recalled by name, and the Brewster/TIR landmark stories are deferred by name to the next article. Worked numbers recomputed from scratch: 50° air→water — $\theta_T = 35.08°$ , $r_s = -0.259$ , $t_s = 0.741$ , $r_p = +0.023$ , prefactor 1.697, both ledgers sum to 1.0000, reflected degree of polarization 0.98 (water Brewster 53.1°); 60° pane — $R_s = 0.1766$ , $R_p = 0.0018$ , two-surface transmissions 0.678 (s) / 0.996 (p) / 0.837 (unpolarized); glass→air $t = 1.2$ , $T = 0.96$ ; signed-amplitude curve data for the r-vs-angle figure computed at 46 points from (1)–(2); the reflectance-vs-angle payoff figure and the 45° glass audit are owned by the oblique-incidence article (cross-checked: this frame reproduces its $r_s = -0.303$ , $r_p = +0.092$ ).

Builds on: Reflection and Transmission at Oblique Incidence · Reflection and Transmission at Normal Incidence · Electromagnetic Waves in Linear Dielectric Media · Polarization of Electromagnetic Waves · Complex Notation for Electromagnetic Waves

Leads to: Brewster's Angle and Total Internal Reflection · Jones Vectors and Stokes Parameters · Intensity and Time-Averaged Quantities

What exactly does this article own versus the oblique-incidence article before it?

A deliberate division of labor. Reflection and Transmission at Oblique Incidence owns the kinematic package: why all three waves share one frequency and one tangential wavevector, hence the law of reflection, Snell's law, and the boundary-condition frame at oblique incidence. This article owns the full polarization-resolved dynamics: both channels' amplitude coefficients derived in the unified impedance frame (including magnetic media), the trigonometric forms, the sign-convention reconciliation, and the energy ledger with its beam-geometry prefactor. The next article owns the two landmark phenomena the coefficients contain — Brewster's angle and total internal reflection. If you want "where do the beams go," read backward; "how much light and which polarization," you are in the right place; "the two magic angles," read forward.

Why can't the interface convert s-polarized light into p-polarized light?

Symmetry. The configuration — interface plus plane of incidence — is invariant under reflection through the plane of incidence. Under that mirror operation, the $s$ -field ( $\mathbf{E}$ perpendicular to the plane) flips sign while the $p$ -field (in the plane) does not: the two channels transform under different representations of the mirror symmetry, so no linear boundary process can couple them, for the same reason a symmetric Hamiltonian cannot connect even and odd states. The practical consequences are large: the Fresnel problem splits into two independent 2×2 matching problems instead of one 4×4 mess, energy balances channel by channel, and a polarimeter downstream of a smooth interface sees no cross-talk. Rough surfaces, anisotropic (birefringent) media, and magneto-optic materials all break the mirror symmetry and do mix channels — that mixing is a measurement resource (Mueller-matrix polarimetry detects surface damage this way).

Is there a way to see why the p-channel can have a zero without grinding through the algebra?

The impedance picture gives it in one sentence: the $p$ -channel's effective impedance is $Z\cos\theta$ — a product of a material factor and a geometric factor that move oppositely along the Snell trajectory — so medium 1's and medium 2's effective impedances can cross, and where they cross the interface is impedance-matched and reflectionless. For $s$ , the factors combine as $Z/\cos\theta$ and the mismatch is monotone: no crossing, ever. You can check where the crossing happens with two lines of Snell algebra: $Z_1\cos\theta_I = Z_2\cos\theta_T$ with $Z = Z_0/n$ reads $\cos\theta_I/n_1 = \cos\theta_T/n_2$ , and dividing Snell's law $n_1\sin\theta_I = n_2\sin\theta_T$ by it gives $\tan\theta_I\,n_1/n_2 = \tan\theta_T\,n_2/n_1$ — satisfied exactly when $\theta_T = 90° - \theta_I$ , i.e. $\tan\theta_I = n_2/n_1$ : the Brewster condition. The microscopic story (transmitted-medium dipoles unable to radiate along their own axis) and everything built on it belong to the next article; the impedance crossing is the boundary-condition skeleton of that story.

Do the Fresnel equations assume the interface is infinitely sharp, and does that matter?

They assume the transition from medium 1 to medium 2 is thin compared to a wavelength. Real surfaces — atomic reconstruction, adsorbed water, polishing damage — are nanometers thick versus 500 nm visible wavelengths, so the idealization is excellent, and deviations are the signal, not noise: ellipsometry detects sub-nanometer films precisely as departures of $\rho = r_p/r_s$ from the bare-surface Fresnel prediction. When the transition thickens toward a wavelength, reflection is suppressed — the impedance mismatch spreads over many weak, out-of-phase steps. Graded-index anti-reflection coatings engineer this deliberately, and moth eyes, with their sub-wavelength taper, discovered adiabatic impedance matching long before radio engineers named it.

How do the equations change if medium 2 absorbs — say, reflection off a metal or a lake full of algae?

Formally, barely at all: replace $n_2$ by the complex index $\tilde{n}_2 = n_2 + i\kappa_2$ everywhere. Snell's law then delivers a complex $\cos\theta_T$ , and the coefficients (1)–(2) become complex numbers with $|r| < 1$ and nontrivial phases even below any critical angle. Two consequences worth flagging now. First, $R_s + T_s = 1$ still holds in the sense that what is not reflected enters medium 2 — but "transmitted" light is now attenuated as it goes (the absorption article of this chapter owns the machinery of $\kappa$ ). Second, for good conductors the mismatch is dominated by the enormous imaginary part of $\tilde{n}$ , making $|r|^2 \to$ near 1 — why metals are mirrors — with the small deficit scaling as $\sqrt{\omega/\sigma}$ ; the conductor-reflection article cashes this in full. The Fresnel structure — tangential matching, two channels, impedance mismatch — survives every one of these generalizations; only the inputs go complex.

Why is the s-reflection negative at every angle for external reflection but the p-sign changes along the way?

For $r_s$ , (1)'s numerator is $n_1\cos\theta_I - n_2\cos\theta_T$ ; external incidence ( $n_1 < n_2$ ) has both a smaller index and (via Snell) a smaller cosine in the first term, so the numerator is negative at every angle: the $s$ -wave always reflects with the half-turn phase flip, the electromagnetic version of a pulse reflecting off a heavier rope. For $r_p$ , the swapped structure $n_2\cos\theta_I - n_1\cos\theta_T$ starts positive at normal incidence (in this article's convention) and must end at $-1$ at grazing — so it passes through zero on the way: the sign change and the Brewster null are the same event. Watch the conventions when comparing books (the warning box inventories them): the relative phase between the channels' reflections, which is what interference and ellipsometry measure, is convention-independent, and the crossing angle is a physical invariant.

The reflectance barely changes between 0 and 30 degrees — why is the curve so flat at first?

Expand (1)–(2) about normal incidence: the leading angle dependence of both $R_s$ and $R_p$ is quadratic in $\theta_I$ with small coefficients (the linear terms cancel because $\theta_T$ tracks $\theta_I/n$ and the normal-incidence point is a stationary configuration of the mismatch). Numerically for glass, $R$ moves only from 4.0% to about 4.7% ( $s$ ) and 3.4% ( $p$ ) by 30° — the computed curves' long plateau. The practical readings: for most everyday viewing geometries, "4% per surface" is a safe universal estimate, which is why lens designers quote per-surface losses without asking about field angles; and polarization effects are feeble below 30°, which is why your polarized sunglasses do little for reflections you face head-on but transform the world at steep afternoon-sun geometries. All the drama of the Fresnel curves — the dive to the $p$ -null, the surge to grazing unity — is packed into the last 30 degrees.