Reflection and Transmission at Oblique Incidence

Tilt the beam and the interface problem blooms. Normal incidence had one geometry and no use for polarization; oblique incidence has angles to determine, a plane of incidence to define, and — the decisive novelty — two polarizations that the boundary treats differently. This article solves the tilted problem completely, and the solution arrives in two installments of very different character. The first installment is pure kinematics, and it is astonishing how much falls out before any amplitude is computed: the requirement that the three waves' phases march in step along the interface forces the reflected and transmitted rays into the plane of incidence, sets the reflection angle equal to the incidence angle, and delivers Snell's law of refraction — the law that took optics two millennia to find, produced here by matching exponents. The second installment is dynamics: imposing the full boundary conditions on amplitudes yields the Fresnel equations, one pair per polarization, derived in full for both. The Fresnel equations repay the labor immediately. They contain Brewster's angle — a specific tilt at which one polarization refuses to reflect at all, turning every puddle, shop window, and lake into a natural polarizer (and explaining, via a one-line dipole argument, why). They quantify the grazing-incidence brightening that makes wet roads glare and lets X-ray telescopes focus photons no mirror could face head-on. And on the dense-to-rare side they run off the edge of Snell's law into total internal reflection — the phenomenon this article states carefully and hands, with its evanescent subtleties, to its own dedicated article.

Key takeaways

Phase matching along the interface — before any amplitude algebra — forces all three wavevectors into the plane of incidence, gives $\theta_r = \theta_i$ , and yields Snell's law $n_1\sin\theta_i = n_2\sin\theta_t$ .
Oblique problems decompose into two independent polarizations: s (E perpendicular to the plane of incidence, "senkrecht") and p (E parallel to it); the boundary conditions couple them to different field components, so they reflect differently.
The Fresnel equations (non-magnetic): $r_s = \dfrac{n_1\cos\theta_i - n_2\cos\theta_t}{n_1\cos\theta_i + n_2\cos\theta_t}$ , $r_p = \dfrac{n_2\cos\theta_i - n_1\cos\theta_t}{n_2\cos\theta_i + n_1\cos\theta_t}$ — both derived in full here.
$r_p$ vanishes at Brewster's angle $\tan\theta_B = n_2/n_1$ (56.3° for glass): reflected light there is purely s-polarized — reflection is a polarizer, and polarized sunglasses are its exploit.
Energy audits need beam-geometry cosines: $T = \dfrac{n_2\cos\theta_t}{n_1\cos\theta_i}\,t^2$ ; then $R + T = 1$ exactly, each polarization separately.
Dense-to-rare beyond $\theta_c = \arcsin(n_2/n_1)$ : Snell has no real $\theta_t$ — total internal reflection, $|r| = 1$ , stated here and owned in depth by the next article.
Key idea: matching phases along the wall gives the three laws of geometric optics for free; matching amplitudes gives Fresnel's equations — with Brewster's polarizing null and total internal reflection inside them.

The core insight

Phase continuity along the interface yields the plane of incidence, $\theta_r = \theta_i$ , and Snell's law before any dynamics; the boundary conditions then give the Fresnel coefficients per polarization, with $r_p = 0$ at Brewster's angle $\tan\theta_B = n_2/n_1$ and total internal reflection past $\theta_c$ on dense-to-rare crossings.

Where this lives today

Fresnel's equations are embedded in more shipped products than perhaps any other formula in this track. Every lens-design and thin-film code evaluates them per surface, per wavelength, per polarization; ellipsometry — the semiconductor industry's workhorse metrology — runs them in reverse, measuring the polarization change on reflection (the $r_p/r_s$ ratio, amplitude and phase) to deduce film thicknesses on wafers to sub-ångström precision. Brewster's null is load-bearing across photonics: laser Brewster windows and Brewster plates enforce polarization; pile-of-plates polarizers stack the effect; photographers tilt polarizers to erase window and water reflections at the 0.83-DOP geometry of Worked Example 2. The grazing limit built X-ray astronomy: at keV energies every material's n dips just below 1 (the dispersion article's regime), so mirrors only work within fractions of a degree of grazing — Chandra's and XMM-Newton's nested paraboloid–hyperboloid shells are Fresnel's $R \to 1$ limit machined to nanometre figure. And the dense-to-rare cliff this article stops at — total internal reflection — carries the internet: every fiber's guidance, every TIR prism in binoculars and rangefinders, every diamond's engineered sparkle ( $\theta_c = 24.4°$ — cutters shape facets so entering light cannot find an exit under the critical angle). The next article opens that cliff's fine print, evanescent tail first.

The physical picture

Start with the picture that makes the kinematics inevitable. The incident wave's crests sweep along the interface like a bar of rain crossing a road: they intersect the wall in moving stripes with a definite spacing (the trace wavelength, $\lambda_1/\sin\theta_i$ ) and a definite sweep speed. Whatever waves the boundary generates — reflected above, transmitted below — are driven by that striped forcing, and a driven linear response must match its driver stripe for stripe, everywhere on the wall, at all times. That single sentence is the whole of geometric optics: matching the stripes' direction puts all rays in one plane; matching their spacing on the reflection side (same medium, same wavelength) forces $\theta_r = \theta_i$ ; matching on the transmission side (compressed wavelength $\lambda_1 n_1/n_2$ ) forces the ray to bend — Snell's law is the transmitted wave tilting exactly enough to keep its shorter crests in step with the driver's stripes.

Figure

Phase matching at the wall — the incident crests stripe the interface, and reflected and transmitted waves must keep step: equal angles above, Snell's bend below

For the amplitudes, the tilt breaks a symmetry normal incidence enjoyed. The plane of incidence — spanned by the incident ray and the surface normal — is now a preferred plane, and the two transverse polarizations relate to it differently: s-polarization holds E perpendicular to that plane (parallel to the interface — German senkrecht), while p-polarization holds E in the plane, tilted partly into and partly along the wall. The boundary conditions treat tangential and normal field components differently, so the two polarizations acquire different reflection coefficients — the interface is birefringent in reflection, splitting what polarization article 0588 treated as a degenerate pair. Any incident polarization decomposes onto s and p (the Jones machinery of 0589, if done systematically), each component obeys its own Fresnel pair, and the outputs recombine — generally into a different polarization state than went in. The p-polarization hides the article's jewel. The transmitted wave's job, microscopically, is done by the driven dipoles of medium 2, all oscillating along the transmitted E — and dipoles do not radiate along their own axis (the sky-polarization geometry of 0588, the radiation chapter's theorem). At one special incidence angle, the direction the reflected ray would have to take lies exactly along the transmitted-field dipole axis: the dipoles cannot supply that ray, and p-reflection dies entirely. Geometry: reflected and transmitted rays perpendicular, $\theta_B + \theta_t = 90°$ , which with Snell gives $\tan\theta_B = n_2/n_1$ — Brewster's angle, where reflection becomes a perfect polarizer.

The kinematics: three laws before any dynamics

Write the three waves with explicit wavevectors, all at the conserved frequency ω (the media article's theorem):

\tilde{\mathbf{E}}_{I,R,T} = \tilde{\mathbf{E}}_{0_{I,R,T}}\,e^{i(\mathbf{k}_{I,R,T}\cdot\mathbf{r} - \omega t)}, \qquad k_I = k_R = \frac{n_1\omega}{c}, \quad k_T = \frac{n_2\omega}{c} \qquad (1)

where: magnitudes are fixed by each medium's dispersion relation; directions are the unknowns. Every boundary condition, whatever field component it constrains, has the structure $(\ )e^{i\mathbf{k}_I\cdot\mathbf{r}} + (\ )e^{i\mathbf{k}_R\cdot\mathbf{r}} = (\ )e^{i\mathbf{k}_T\cdot\mathbf{r}}$ for all points $\mathbf{r}$ on the interface — and no combination of constants can satisfy that identity unless the exponentials agree on the wall. The kinematic content is therefore universal, extracted once:

Derivation APhase matching — the plane of incidence, equal angles, and Snell's law from one identity

The goal is the three laws of geometric optics as consequences of a single requirement: the phases $\mathbf{k}\cdot\mathbf{r}$ of all three waves must agree at every point of the interface (else the boundary conditions, which must hold everywhere on the wall simultaneously, are unsatisfiable). The strategy: put the interface at $z = 0$ , reduce the requirement to the equality of the tangential wavevector components, and read off the geometry. Watch how much arrives with no reference to what the boundary conditions actually say — this derivation would be identical for sound, water waves, or electrons at a potential step, which is why all of them obey Snell-type laws.

Step 1. The matching identity. On $z = 0$ , $\mathbf{r} = (x, y, 0)$ , and agreement of the exponentials for all such $\mathbf{r}$ requires

(k_I)_x = (k_R)_x = (k_T)_x, \qquad (k_I)_y = (k_R)_y = (k_T)_y

— the tangential wavevector is conserved across the boundary. (The normal components are not matched: the wall breaks translation symmetry only in z. Conservation of transverse $\mathbf{k}$ is the interface's version of momentum conservation — the wall can push only along its normal.)

Step 2. The plane of incidence. Orient axes so $\mathbf{k}_I$ lies in the x–z plane: $(k_I)_y = 0$ . Then Step 1 forces $(k_R)_y = (k_T)_y = 0$ : all three wavevectors lie in one plane, the plane of incidence. First law. Why it could not be otherwise: a reflected ray tilted out of the plane would carry a transverse phase variation along y that nothing drives.

Step 3. Equal angles. Writing $(k_I)_x = k_I\sin\theta_i$ and $(k_R)_x = k_R\sin\theta_r$ with $k_I = k_R$ (same medium): $\sin\theta_r = \sin\theta_i$ , so

\theta_r = \theta_i. \qquad \textbf{Second law.}

Step 4. Snell. Matching the transmitted side, $k_I\sin\theta_i = k_T\sin\theta_t$ with the magnitudes of (1):

n_1\sin\theta_i = n_2\sin\theta_t. \qquad \textbf{Third law.}

Why this derivation is the modern one: Snell's law here is not about rays minimizing time (Fermat's principle — an equivalent, deeper variational reading) but about wave crests staying in step; it therefore generalizes verbatim to any interface phenomenon with a conserved tangential wavevector — diffraction gratings (which add reciprocal-lattice vectors to $k_x$ ), photonic crystals, and electron optics. It also exposes the failure mode honestly: for $n_1 > n_2$ and $\sin\theta_i > n_2/n_1$ , no real $\theta_t$ exists — the transmitted tangential wavevector demanded by matching exceeds what medium 2's dispersion can supply with a real ray. That cliff is total internal reflection, stated at this article's end. *Verified: reproduces Griffiths §9.3.3 Eqs. 9.88–9.92 (his "three fundamental laws" from the boundary-exponent argument) and Jackson §7.3; trace-wavelength picture consistent with the hero figure; grating generalization per Hecht §10.2.*

The dynamics: Fresnel's equations, both polarizations

Now the amplitudes. The boundary conditions are the dielectric-interface quartet (E6's boundary-conditions article, source-free form): tangential E and tangential H continuous; normal D and normal B continuous. With the geometry fixed by phase matching, decompose onto s and p and impose the conditions on each. Both derivations follow; μ = μ₀ throughout (the honest general case replaces every n below by $1/Z$ — the impedance form, per the normal-incidence article's Step 2 logic).

Go deeper · Derivation B · 1 minFresnel equations for s-polarization — E parallel to the wall

Derivation BFresnel equations for s-polarization — E parallel to the wall

The goal is $r_s$ and $t_s$ for the polarization with E perpendicular to the plane of incidence — E points along ŷ for all three waves, purely tangential to the interface. The strategy: continuity of tangential E gives one equation; continuity of tangential H (whose in-plane component now carries a $\cos\theta$ projection) gives the second; solve. Watch where the cosines enter — through H's tilt, not through any beam-width argument (those cosines come later, in the energy audit, and confusing the two families is the classic oblique-incidence disaster).

Step 1. Tangential E. All three E-fields are along ŷ, fully tangential:

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

Step 2. Tangential H. Each wave's H has magnitude $E/Z = nE/Z_0$ and lies in the plane of incidence, perpendicular to its ray; its tangential (x) component is $H\cos\theta$ of the respective angle, with signs fixed by the triads ( $\hat{\mathbf{k}}\times\hat{\mathbf{E}}$ direction for each ray — the reflected ray's reversed $k_z$ flips its tangential-H sign, exactly as at normal incidence):

\frac{n_1}{Z_0}\left(\tilde E_{0_I} - \tilde E_{0_R}\right)\cos\theta_i = \frac{n_2}{Z_0}\,\tilde E_{0_T}\cos\theta_t.

Step 3. Solve the pair:

\boxed{\ 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}\ }

with $1 + r_s = t_s$ ✓ (Step 1 restated). Normal-incidence check ( $\theta_i = \theta_t = 0$ ): the previous article's $r = (n_1-n_2)/(n_1+n_2)$ ✓. Why $r_s$ never vanishes (for $n_1 \neq n_2$ ): a zero would need $n_1\cos\theta_i = n_2\cos\theta_t$ while Snell holds, i.e. $n_1^2(1-\sin^2\theta_i)$ ... eliminating $\theta_t$ shows the two conditions are compatible only at $n_1 = n_2$ . There is no s-Brewster: the s-dipoles of the medium oscillate perpendicular to the plane of incidence and radiate happily into every reflected direction — the polarizing null is reserved for p, where geometry can starve the reflected ray. *Verified: reproduces Hecht Eqs. 4.34/4.35 and Jackson Eq. 7.39 (μ = μ₀); Griffiths presents this case as his Problem 9.17 — coefficients match his quoted answer; normal-incidence and grazing ( $r_s \to -1$ ) limits check.*

Go deeper · Derivation C · 2 minFresnel equations for p-polarization — and Brewster's angle inside them

Derivation CFresnel equations for p-polarization — and Brewster's angle inside them

The goal is $r_p$ and $t_p$ for E in the plane of incidence — the case Griffiths works in the main text — and the extraction of Brewster's angle as $r_p$ 's zero, with the dipole reading attached. Now E is tilted: its tangential component carries the $\cos\theta$ , while H (perpendicular to the plane of incidence) is purely tangential. The two conditions swap their cosine burdens relative to the s-case — that swap is the entire origin of the two polarizations' different physics. Sign convention: reference directions chosen so all three E's tilt "forward" (Griffiths Fig. 9.15); at normal incidence this makes $r_p = -r_s$ by pure convention, a notorious trap flagged below.

Step 1. Tangential E (x-components, each $E\cos\theta$ of its own angle, reflected sign per the convention):

\left(\tilde E_{0_I} - \tilde E_{0_R}\right)\cos\theta_i = \tilde E_{0_T}\cos\theta_t.

Step 2. Tangential H (all along ŷ, magnitudes $nE/Z_0$ ):

\frac{n_1}{Z_0}\left(\tilde E_{0_I} + \tilde E_{0_R}\right) = \frac{n_2}{Z_0}\,\tilde E_{0_T}.

(The normal-D condition, $\varepsilon_1$ times the z-components matching, reproduces Step 2 via Snell — the quartet is redundant here, as it must be: two unknowns.)

Step 3. Solve:

\boxed{\ 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}\ }

Normal incidence: $r_p = +(n_2-n_1)/(n_2+n_1)$ — opposite sign to $r_s$ , purely because the p reference arrows rotate with the rays while s arrows are common: same physical flip, different bookkeeping. Never compare $r_s$ and $r_p$ signs across conventions without redrawing the arrows.

Step 4. Brewster's null. Set $r_p = 0$ : $n_2\cos\theta_i = n_1\cos\theta_t$ . Square, use Snell to eliminate $\theta_t$ , and solve: $\sin^2\theta_i = \dfrac{n_2^2(n_2^2 - n_1^2)}{n_2^4 - n_1^4} = \dfrac{n_2^2}{n_1^2+n_2^2}$ , hence

\tan\theta_B = \frac{n_2}{n_1}.

Equivalently $\theta_B + \theta_t = 90°$ (check: $\cos\theta_t = \sin\theta_B$ makes the null condition Snell's law itself). Why the reflected ray starves, in one line: at this geometry the would-be reflected direction is exactly perpendicular to the transmitted ray — i.e. exactly along the axis of the transmitted-medium dipoles that must radiate it, and a dipole emits zero along its axis. The null is radiation-pattern geometry, which is why it is exact, why it happens only for p (s-dipoles point out of the plane, never along the reflected ray), and why at $\theta_B$ an arbitrary incident beam reflects as purely s-polarized light: the interface is a perfect polarization filter for the reflected beam.

Step 5. Energy audit, with the beam cosines. Obliquely, a unit area of wall intercepts beam cross-sections $\propto\cos\theta$ : reflectance is still $R = |r|^2$ (same medium, same angle), but transmittance picks up both the medium factor and the geometry factor:

T = \frac{n_2\cos\theta_t}{n_1\cos\theta_i}\,t^2, \qquad R + T = 1 \ \text{(each polarization separately)}.

Numbers at 45°, air→glass: $\theta_t = 28.13°$ ; $r_s = -0.303$ , $R_s = 9.2\%$ , $T_s = 90.8\%$ ✓; $r_p = +0.092$ , $R_p = 0.85\%$ , $T_p = 99.15\%$ ✓ — already strongly polarizing halfway to Brewster. Verified: reproduces Griffiths §9.3.3 Eqs. 9.109 (Fresnel, p-case), 9.112 (Brewster), 9.115–9.116 (R, T with cosines, closure); Hecht Eqs. 4.38/4.40; dipole argument per Hecht §8.6.1; 45° numbers recomputed (R_s + T_s = 1.0000, R_p + T_p = 1.0000); Brewster algebra checked against $\theta_B = 56.31°$ , where direct evaluation gives $r_p = 0.0000$ .

The full angular story deserves to be seen — the payoff figure of the chapter's interface physics:

Figure

Reflectance versus incidence angle for air-to-glass — the p-curve dives to zero at Brewster's 56.3° while s climbs monotonically; both soar to unity at grazing

Total internal reflection — the statement, and a forward pointer

Run the dense-to-rare case ( $n_1 > n_2$ ) up the angle axis. Snell demands $\sin\theta_t = (n_1/n_2)\sin\theta_i > \sin\theta_i$ — the ray bends away from the normal — and at the critical angle

\theta_c = \arcsin\frac{n_2}{n_1} \qquad (2)

where: $\theta_t$ reaches 90°: the transmitted ray skims the interface. (Glass→air: 41.8°; water→air: 48.6°; diamond→air: 24.4°.) Beyond $\theta_c$ , no real transmitted direction exists — phase matching demands a tangential wavevector larger than medium 2 can carry on any real ray — and the Fresnel machinery returns $|r| = 1$ for both polarizations: total internal reflection, lossless and exact, the physics that traps light in fibers, prisms, and cut diamonds. What survives beyond the wall is not nothing but an evanescent field — exponentially decaying, transporting no average power outward, imposing a polarization-dependent phase on the reflected wave — and every one of those clauses opens machinery (complex $\cos\theta_t$ , penetration depths, frustrated tunneling, the Goos–Hänchen shift) that belongs to the dedicated next article (0594) and the optics track beyond it. This article's contract ends at the statement: past $\theta_c$ , everything comes back.

Limiting cases

Case	Result	Meaning
$\theta_i \to 0$	$r_s \to -r_p \to (n_1-n_2)/(n_1+n_2)$	normal incidence recovered; sign split is convention only
$\theta_i = \theta_B = \arctan(n_2/n_1)$	$r_p = 0$ exactly	Brewster: reflected beam purely s; dipole-axis starvation
$\theta_i \to 90°$ (grazing)	$\|r_s\|, \|r_p\| \to 1$	mirrors for free at glancing angles; X-ray optics live here
$n_1 > n_2$ , $\theta_i \to \theta_c$	$\theta_t \to 90°$ , $T \to 0$	the transmitted ray flattens onto the wall
$\theta_i > \theta_c$	$\|r\| = 1$ , evanescent tail only	total internal reflection — stated here, owned by 0594
$n_1 = n_2$	$r_s = r_p = 0$ at every angle	no interface, no optics
unpolarized input	$R = \tfrac12(R_s + R_p)$	partial polarization on every reflection (Stokes bookkeeping: 0589)
μ ≠ μ₀	replace n by 1/Z throughout	magnetic Brewster angles exist even for s — metamaterial optics

Common trapTwo families of cosines — boundary-condition projections are not beam-geometry factors

Oblique incidence deploys cosines for two unrelated jobs, and merging them is the subject's signature disaster. Family one, inside the Fresnel coefficients: the $\cos\theta_i$ , $\cos\theta_t$ in $r_s$ , $r_p$ come from projecting tilted fields onto the interface (tangential components) — they are boundary-condition bookkeeping, present even for infinitely wide beams. Family two, in the energy audit only: the ratio $\cos\theta_t/\cos\theta_i$ in $T$ prices the change in beam cross-section as the ray bends — pure geometry of tube widths, absent from every amplitude formula. Symptoms of a merge: an "energy-conserving" $R + T$ built as $r^2 + (n_2/n_1)t^2$ that mysteriously fails off-normal (missing family two); or Fresnel coefficients "simplified" by canceling cosines between r and T (destroying family one). Adjacent trap, worth its own sentence: the sign conventions for $r_p$ differ across the literature (Griffiths's arrows give $r_p = +0.2$ air→glass at normal incidence where Verdet-convention texts give −0.2) — physically identical, algebraically opposite; never mix Fresnel formulas from two books in one calculation without redrawing their reference arrows.

Worked examples

1 · Brewster geometry for the lake and the laser (confidence builder)

(a) Find Brewster's angle for air→water ( $n = 1.333$ ) and air→glass ( $n = 1.5$ ), and explain the polarized-sunglasses design. (b) A gas-laser tube seals its ends with windows tilted at Brewster's angle for glass. Why, and what does it cost?

Step 1. (a) $\theta_B = \arctan(1.333) = \mathbf{53.1°}$ for water; $\arctan(1.5) = \mathbf{56.3°}$ for glass — both measured from the vertical normal, so the glare geometry is a sun fairly low over the scene. At (and near) these angles the reflected glare is (almost) purely s-polarized — E horizontal, parallel to the water. Sunglasses therefore mount their polaroid to pass vertical polarization: the horizontal glare component dies by Malus (0588), while the mostly-unpolarized scene loses only half. The physics stack: Brewster selects s → glare is horizontally polarized → crossed analyzer kills it.

Step 2. (b) Inside a laser, light crosses the tube's end windows thousands of times per round trip... per the cavity's many passes. A window at $\theta_B$ transmits p-polarized light with zero reflection loss ( $R_p = 0$ — no coating needed, at any wavelength), while s suffers ~15% reflection loss per pass ( $R_s(\theta_B) = 0.148$ for glass) and is quickly extinguished from the oscillation. Benefit: essentially lossless windows. Cost (or feature): the laser's output is forced to be linearly p-polarized — which is why most gas lasers emit polarized beams. Answers: 53.1°, 56.3°; vertical-pass sunglasses; Brewster windows = lossless for p, and they polarize the laser.

2 · Full Fresnel number-crunch at 45° (exam level)

Unpolarized light strikes an air→glass surface at 45°. Compute both reflectances, verify both energy closures, and give the total reflectance and the degree of polarization of the reflected beam.

Step 1. Kinematics first: $\sin\theta_t = \sin45°/1.5 = 0.4714$ , $\theta_t = 28.13°$ ; $\cos\theta_i = 0.7071$ , $\cos\theta_t = 0.8819$ .

Step 2. s-coefficients: $r_s = (0.7071 - 1.5\times0.8819)/(0.7071 + 1.5\times0.8819) = -0.6158/2.0300 = \mathbf{-0.303}$ ; $t_s = 2(0.7071)/2.0300 = 0.6967$ . Energy: $R_s = 0.0920$ ; $T_s = \dfrac{1.5\times0.8819}{0.7071}(0.6967)^2 = 1.8708\times0.4854 = 0.9080$ ; sum $= 1.0000$ ✓.

Step 3. p-coefficients: $r_p = (1.5\times0.7071 - 0.8819)/(1.5\times0.7071 + 0.8819) = 0.1788/1.9426 = \mathbf{+0.092}$ ; $t_p = 2(0.7071)/1.9426 = 0.7280$ . Energy: $R_p = 0.0085$ ; $T_p = 1.8708\times(0.7280)^2 = 0.9915$ ; sum $= 1.0000$ ✓.

Step 4. The beam bookkeeping. Unpolarized input splits evenly over s and p: total reflectance $R = \tfrac12(0.0920 + 0.0085) = \mathbf{5.0\%}$ (up from 4% at normal incidence — the s-channel is already climbing). Degree of polarization of the reflected beam (Stokes language, 0589): $p = (R_s - R_p)/(R_s + R_p) = 0.0835/0.1005 = \mathbf{0.83}$ — at a mere 45°, eleven degrees shy of Brewster, reflected glare is already 83% polarized, which is why polarized sunglasses work across a broad range of sun angles, not only at the exact Brewster geometry. Answers: $R_s = 9.2\%$ , $R_p = 0.85\%$ , closures exact, $R = 5.0\%$ , reflected DOP = 0.83.

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

Augustin-Jean Fresnel

1788–1827 · Broglie and Paris — the road engineer who gave light its amplitudes

Fresnel built roads for the Corps des Ponts et Chaussées and did optics in the evenings — and, during his political suspension after Napoleon's return in 1815, full-time in a village house, with home-made equipment and his housekeeper's help. In that exile he re-derived diffraction quantitatively (winning the 1819 Academy prize over a corpuscular jury, the famous bright "Poisson spot" confirming his integrals); with Arago's non-interference result and Young's transversality hint, he then rebuilt the wave theory of light as fully transverse — over the objection of nearly every senior physicist, including a reluctant Arago — and from the mechanics of transverse waves at boundaries derived, in 1823, the amplitude coefficients that bear his name: this article's equations, with Brewster's empirical 1815 tangent law falling out as a zero. His sine and tangent forms of $r_s$ , $r_p$ predate Maxwell by four decades; when Lorentz re-derived them electromagnetically in 1875, not a formula changed. Fresnel also engineered his optics into the world's lighthouses — the stepped Fresnel lens multiplied visible range and still ships in every projector and headlamp. He died of tuberculosis at 39, weeks after receiving the Royal Society's Rumford Medal, having remarked that "all the compliments... have never given me so much pleasure as the discovery of a theoretic truth."

History

1621

— Snell finds

\sin\theta_i/\sin\theta_t

constant; Descartes publishes the law in 1637

1690

— Huygens's wavelets construct refraction geometrically; amplitudes remain out of reach

1808

— Malus discovers polarization by reflection in a Paris window's glare

1815

— Brewster's tangent law: the polarizing angle satisfies tan θ_B = n

1823

— Fresnel derives the amplitude coefficients from the transverse-wave theory

1875

— Lorentz re-derives Fresnel's equations from Maxwell's electromagnetism — identical formulas

1929

— Land's Polaroid meets Brewster-angle glare: polarized sunglasses become an industry

1999

— Chandra X-ray Observatory focuses X-rays by grazing-incidence reflection — the R → 1 limit as telescope design

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §9.3.3 · Hecht, Optics, 5th ed., §4.6–4.8 · Jackson, Classical Electrodynamics, 3rd ed., §7.3 · Born & Wolf, Principles of Optics, 7th ed., §1.5.

Provenance. Kinematics and the p-polarization Fresnel derivation follow Griffiths §9.3.3 (Eqs. 9.88–9.116, incl. Brewster 9.112 and the R/T closure 9.115–9.116); the s-case (Griffiths's Problem 9.17) is derived in full following Hecht Eqs. 4.34–4.35/Jackson §7.3, with the sign-convention discrepancy between the two polarizations at normal incidence flagged explicitly. A verifier should: (1) re-extract the three laws from the exponent-matching identity; (2) rederive both Fresnel pairs from tangential E and H continuity and check $1 + r_s = t_s$ and both normal-incidence limits; (3) confirm Brewster algebraically ( $\sin^2\theta_B = n_2^2/(n_1^2+n_2^2) \Leftrightarrow \tan\theta_B = n_2/n_1$ ) and numerically ( $r_p(56.31°) = 0.0000$ air→glass); (4) recompute the 45° table ( $\theta_t = 28.13°$ , $r_s = -0.3033$ , $r_p = +0.0920$ , $R_s + T_s = R_p + T_p = 1.0000$ , mean R = 5.0%, reflected DOP = 0.831) and the anchor angles (θ_B: 53.1° water, 56.3° glass, 67.5° diamond; θ_c: 41.8° glass, 48.6° water, 24.4° diamond). Addendum for this edition (scoping note, per wave brief): this article owns the full oblique package — phase-matching kinematics, both Fresnel derivations, Brewster with the dipole reading, and the cosine-corrected energy audit; total internal reflection is stated at the phenomenon level only (θ_c, |r| = 1, evanescent tail named), with its depths (complex cosθt, penetration depth, Goos–Hänchen, frustrated TIR) deferred BY NAME to Brewster's Angle and Total Internal Reflection (0594) and the optics track's migration; the dedicated Fresnel Equations article (0593) is positioned as this material's deep-dive/applications sequel — division recorded in the lane ledger for the 0593/0594 builders.

Builds on: Monochromatic Plane Waves in Vacuum · Electromagnetic Waves in Linear Dielectric Media · Reflection and Transmission at Normal Incidence · Complex Notation for Electromagnetic Waves.

Leads to: The Fresnel Equations · Brewster's Angle and Total Internal Reflection · Polarization of Electromagnetic Waves.

Why does phase matching give the same laws for sound, light, and electron waves — where's the electromagnetism in Snell's law?

There is none, and that is the point of Derivation A. The three geometric laws use only two ingredients: each medium supports waves with a fixed magnitude of k at given ω, and the flat boundary cannot break translation symmetry along itself (so tangential k is conserved). Any wave system with those properties — acoustic, elastic, quantum-matter, water — obeys the same kinematics with its own dispersion relation supplying the "index." Electromagnetism enters only at the amplitude stage: the Fresnel coefficients encode specifically which field components must be continuous, which is why sound's reflection coefficients differ from light's while their Snell's laws coincide. The division — kinematics universal, dynamics medium-specific — is a template that recurs from crystallography (Bragg) to particle physics (crossing thresholds).

What physically 'chooses' Brewster's angle — why should tan θ_B = n₂/n₁ of all things?

The tangent is the fingerprint of a right angle hiding in the geometry. The null condition is that reflected and transmitted rays be perpendicular: $\theta_B + \theta_t = 90°$ . Feed that to Snell — $n_1\sin\theta_B = n_2\sin\theta_t = n_2\sin(90° - \theta_B) = n_2\cos\theta_B$ — and the tangent law falls out in one line. The right angle matters because of how the reflected light is generated: it is re-radiation by medium-2 dipoles that oscillate along the transmitted E (perpendicular to the transmitted ray, in the plane of incidence for p-light). A dipole radiates nothing along its own axis — and at exactly the perpendicular-rays geometry, the reflected direction is that axis. s-light escapes the trap because its dipoles point out of the plane of incidence, never along any in-plane reflected ray. The null is radiation-pattern geometry, exact at one angle.

Is the transmitted beam at Brewster's angle fully p-polarized, like the reflected one is fully s?

No — and the asymmetry is worth internalizing. The reflected beam at θ_B is purely s because the p-channel's reflection coefficient is exactly zero: nothing p comes back. But the transmitted beam receives all of the p and most of the s ( $T_s(\theta_B) = 85\%$ for glass): it is only mildly polarized toward p. That is why one Brewster surface makes a perfect polarization analyzer in reflection but a poor polarizer in transmission — and why the classical "pile of plates" polarizer stacks many Brewster surfaces, each stripping another 15% of the s, to purify the transmitted beam multiplicatively (ten plates: $0.85^{20} \approx 4\%$ s remaining, counting both faces).

How do polarized sunglasses "know" which axis to block, when glare can come from anywhere?

They exploit a near-universal fact about outdoor glare geometry: the reflecting surfaces that matter — water, roads, car paint, snow crust — are horizontal. For any horizontal surface, the plane of incidence is vertical, so s-polarization (perpendicular to that plane) is horizontal, whatever the compass direction of the glare. Near Brewster incidence the reflection is strongly s — horizontally polarized — so a fixed vertical-pass filter kills glare from every azimuth at once. The design fails exactly where its premise does: vertical reflectors (building glass, car windows seen side-on) produce vertically-polarized glare that sails through, and tilting your head 90° makes lake glare brighter. LCD screens add the modern wrinkle: their emitted polarization axis versus your glasses decides whether your dashboard display survives (some are quarter-wave laminated precisely to defeat this).

At grazing incidence everything reflects — even from glass or water. Why does mismatch stop mattering?

Watch the Fresnel numerators and denominators as $\theta_i \to 90°$ : $\cos\theta_i \to 0$ , so $r_s \to (0 - n_2\cos\theta_t)/(0 + n_2\cos\theta_t) = -1$ and similarly $|r_p| \to 1$ , independent of the indices. Physically: a grazing wave's normal wavevector component $k_z = k\cos\theta_i$ goes to zero — along the normal direction the wave barely oscillates, and even a tiny index step is an enormous relative barrier to so soft an approach (the transmitted channel's phase space, $\cos\theta_t$ -weighted, also pinches shut on the energy side: T → 0). Everyday confirmations: the mirror-like sheen of a wet road toward a low sun, lake reflections at sunset. The engineering jackpot is X-ray optics: at keV energies n differs from 1 by parts in 10⁴ ~ 10⁶ (all materials, per the dispersion article's high-frequency limit), so only grazing geometry reflects usefully — Chandra's mirrors intercept photons at ~0.5° and focus the highest-energy images in astronomy on this FAQ's limit.

Do Fresnel's equations survive when medium 2 is absorbing or conducting?

Formally yes, spectacularly: the same expressions hold with $n_2$ (hence $\cos\theta_t$ , via Snell) promoted to complex values — the boundary-condition derivation never used reality of n. The physics they then describe enlarges: $|r| < 1$ with reflection phases that differ between s and p (the basis of ellipsometry's sensitivity), transmitted waves that are inhomogeneous (planes of constant phase no longer parallel to planes of constant amplitude), and metallic reflectances approaching 1 with characteristic dips. The price is that every intermediate quantity loses its naive geometric reading — "θt" is a complex parameter, not an angle you can draw. The machinery for complex n is built in the absorption/dispersion article (0595) and applied to conductors in 0596–0597; this article's real-n formulas are the template they analytically continue.

Fermat's least-time principle also gives Snell's law. Is that a coincidence, or the same physics?

The same physics at two altitudes. Phase matching says: the fields at the boundary must agree point-by-point, so tangential k is conserved. Fermat says: the ray path between two points makes the travel time stationary. The bridge is stationary phase: a "ray" is where contributions from neighboring paths add coherently, which happens exactly where the total phase is stationary — and stationary total phase across a boundary is the tangential-k matching condition. Fermat is thus the ray-optics shadow of the wave statement, exact in the λ → 0 limit and inherited by quantum mechanics as the stationary-action origin of classical paths. When the next articles bend waves with complex indices or graded media, the phase-matching form is the one that keeps working — variational principles follow the phases, not the other way around.