Fresnel Equations

Every clean pane of glass is a faint mirror. Look at a shop window from the street and you see both the display inside and a ghost of the sky behind you — because at every optical surface a slice of the light turns back. How large that slice is, and how it depends on angle and on the wave's polarization, is fixed entirely by what the electromagnetic field is forced to do at the boundary. Split the incident light into the two polarizations that a surface treats independently, and the reflected fractions of each are the Fresnel equations. This article takes a deliberate division of labour. The derivation of the amplitude coefficients — matching the tangential electric and magnetic fields across the interface, channel by channel, and closing the equations with Snell's law — is electromagnetism, and it is owned by the companion article aq-em-fresnel-equations, which reads reflection as an impedance-mismatch problem and produces all four coefficients at a stroke. We recall those four coefficients by name and do not re-derive them. What is ours is the optical phenomenology they contain: the familiar 4% per glass surface, the way the two polarizations split apart with angle so that reflection polarizes light, the single angle where the reflected glare vanishes, and the energy bookkeeping that makes reflectance and transmittance add to one. Two facts organize everything downstream. The reflected $p$ -light dies to zero at Brewster's angle, giving perfectly polarized reflected light — the reason polarized sunglasses kill road glare, and the reason lasers wear Brewster windows. And past the critical angle (only going dense to rare) the coefficients turn complex: all the power comes back, but $s$ and $p$ pick up different phases, quietly turning linear light elliptical. Those two landmarks get their own full treatment in the next articles; here we build the surface that makes them possible.

Key takeaways

The Fresnel equations give the amplitude reflection and transmission coefficients $r,t$ for light crossing a planar interface, split into $s$ (E perpendicular to the plane of incidence, TE) and $p$ (E in the plane, TM) polarizations.
They are derived by matching the tangential $\mathbf E$ and $\mathbf H$ fields across the boundary — the electromagnetic content, recalled by name from aq-em-fresnel-equations; this article owns what they mean for optics.
At normal incidence the two channels coincide: $R=\big((n_1-n_2)/(n_1+n_2)\big)^2$ — the familiar $\approx4\%$ per air–glass surface.
Because $R_s>R_p$ at every oblique angle, reflection always partially polarizes light; at Brewster's angle $\tan\theta_B=n_2/n_1$ the $p$ -reflection vanishes and the reflected beam is perfectly $s$ -polarized.
Reflectance and transmittance conserve energy, $R+T=1$ , but only when $T$ carries its geometric-and-impedance prefactor $(n_2\cos\theta_t)/(n_1\cos\theta_i)$ — $T$ is not $|t|^2$ .
Key idea: the reflected fractions of the two polarizations differ at every oblique angle, so a plain glass surface both dims and polarizes the light that bounces off it — quantitatively, through the Fresnel coefficients.

The core insight

Split incident light into $s$ (E perpendicular to the plane of incidence) and $p$ (E in the plane); each has its own amplitude reflection coefficient, derived from the electromagnetic boundary conditions and recalled here by name. Their reflectances differ at every oblique angle — $R_s>R_p$ always — so reflection off a dielectric partially polarizes light, completely so at Brewster's angle, while energy conservation fixes $R+T=1$ per channel.

Where this lives today

The single-interface Fresnel coefficients are the atoms of thin-film optics. Stack them with a transfer-matrix method and you get every anti-reflection coating, dielectric mirror, bandpass filter, and Bragg reflector in modern photonics — the $4\%$ per surface is what those coatings exist to defeat, and a good multilayer can push a mirror past $99.99\%$ or an anti-reflection coating below $0.1\%$ . Camera lenses, telescope optics, laser cavities, and fiber connectors all live or die by managing Fresnel reflection. The polarization split is a working tool. Ellipsometry measures the complex ratio $\rho=r_p/r_s=\tan\Psi\,e^{i\Delta}$ of light reflected off a surface and inverts the Fresnel equations to extract film thickness and refractive index to sub-nanometre precision — it certifies the gate oxides in essentially every silicon chip. Brewster windows force gas lasers to run in a single polarization; polarizing beamsplitter cubes and pile-of-plates polarizers exploit the $r_p\to0$ behaviour; and polarized sunglasses simply reject the $s$ -polarized glare that Fresnel's $R_s>R_p$ guarantees dominates every reflection off water, glass, and wet roads.

The physical picture

Start from what makes a surface reflect at all. A single wave arriving from medium 1 cannot, by itself, satisfy the boundary conditions in medium 2; the field values it presents at the surface have to be matched by something on the other side, and nature's unique fix is to add exactly one reflected and one transmitted wave whose amplitudes stitch the tangential fields together. The Fresnel coefficients are those stitching amplitudes — this is the electromagnetic core recalled from aq-em-fresnel-equations, where the calculation is organized as an impedance mismatch, $r=(Z_2-Z_1)/(Z_2+Z_1)$ at normal incidence, exactly like a wave meeting a join between two ropes. The two polarizations part ways because the boundary conditions weigh their fields differently. For $s$ -light the whole electric field lies in the surface, so it is continuous directly. For $p$ -light only the $\cos\theta$ projection of $\mathbf E$ is tangential while the magnetic field is fully tangential — and that asymmetry is the whole story of this article. It is why $r_p$ can pass through zero at one special angle while $r_s$ never does, and therefore why reflection sorts light by polarization.

Figure

The two polarization channels at an interface: s-light has E ⟂ the plane of incidence (out of the page), p-light has E in the plane; each reflects by its own Fresnel coefficient

At Brewster incidence the reflected and refracted rays turn out to be perpendicular, and the tiny dipoles the field drives in the glass then oscillate along exactly the direction the reflected $p$ -ray would have to travel. A dipole radiates nothing along its own axis, so the $p$ -reflection switches off — the cleanest physical reason $\tan\theta_B=n_2/n_1$ . The next figure shows both channels' reflectances against angle and marks that zero.

Figure

Reflectance vs angle, air→glass (n = 1.52): R_p dives to zero at Brewster 56.7° while R_s climbs monotonically; both start at the 4.26% normal-incidence value and reach 1 at grazing

The core equations

The four amplitude coefficients below are the electromagnetic result, derived by matching tangential $\mathbf E$ and $\mathbf H$ across the interface in the two channels and recalled here by name from aq-em-fresnel-equations; this article uses them as given and reads off what they mean. We write them in the Hecht sign convention, in which the positive $p$ -field direction is chosen so that at normal incidence $r_p$ and $r_s$ carry opposite signs.

r_s=\frac{n_1\cos\theta_i-n_2\cos\theta_t}{n_1\cos\theta_i+n_2\cos\theta_t},\qquad r_p=\frac{n_2\cos\theta_i-n_1\cos\theta_t}{n_2\cos\theta_i+n_1\cos\theta_t} \qquad (1)

where: $n_1,n_2$ — indices of the incidence and transmission media; $\theta_i,\theta_t$ — angles of incidence and refraction, tied by Snell's law $n_1\sin\theta_i=n_2\sin\theta_t$ ; $r_s,r_p$ — reflected-to-incident field-amplitude ratios for the two channels. The only difference between them is which cosine sits with which index — and that swap is precisely why $r_p$ has a zero that $r_s$ lacks. Squaring an amplitude coefficient gives the reflectance, the fraction of power turned back. At normal incidence the two channels merge and the angle drops out entirely, leaving the single most-used number in optics.

R=|r|^2,\qquad R(\theta_i=0)=\left(\frac{n_1-n_2}{n_1+n_2}\right)^2 \qquad (2)

where: $R$ — reflected power fraction. For air ( $n_1=1$ ) to crown glass ( $n_2=1.52$ ) this is $(0.52/2.52)^2=0.0426$ , the famous $\approx4\%$ loss at every glass surface; for air–water it is $2.0\%$ and for air–diamond $17.2\%$ . This is why an uncoated multi-element lens can lose a third of its light to surface reflections, and why anti-reflection coatings exist. Transmittance is where a classic trap hides: it is not simply $|t|^2$ . The transmitted beam travels at a different angle and in a medium of different impedance, so a geometric-and-impedance prefactor is mandatory for energy to balance.

T=\frac{n_2\cos\theta_t}{n_1\cos\theta_i}\,|t|^2,\qquad R+T=1 \qquad (3)

where: $T$ — transmitted power fraction (per channel); the prefactor $(n_2\cos\theta_t)/(n_1\cos\theta_i)$ accounts for the change in beam cross-section on the surface and the change in impedance. With it, $R+T=1$ holds exactly for a lossless interface — a strong internal check, derived below. Finally, one special angle deserves its own line, because it is the organizing landmark of polarization by reflection. Setting the $p$ -numerator to zero gives it.

r_p=0 \ \Longleftrightarrow\ \tan\theta_B=\frac{n_2}{n_1} \qquad (4)

where: $\theta_B$ — Brewster's angle, at which the reflected light is perfectly $s$ -polarized. For air–glass, $\theta_B=\arctan(1.52)=56.7^\circ$ . The full geometric proof and its consequences are the subject of the next article, Brewster's Angle; here we establish only that the zero exists and where it sits.

Derivation ANormal-incidence reflectance and the 4% surface

The aim is to extract the single most-quoted number in optics — the reflected fraction at a surface hit head-on — from the recalled Fresnel coefficients, and to read it as an impedance mismatch so its size is intuitive rather than memorized. We take the $\theta_i\to0$ limit of the coefficients, show the $s/p$ distinction disappears there, and evaluate the result for the three interfaces every optics student meets first: air–glass, air–water, and air–diamond.

Step 1. Set $\theta_i=0$ , so $\theta_t=0$ and every cosine is $1$ . Then $r_s\to(n_1-n_2)/(n_1+n_2)$ and $r_p\to(n_2-n_1)/(n_2+n_1)=-r_s$ : the two channels give the same reflectance $R=\big((n_1-n_2)/(n_1+n_2)\big)^2$ , differing only by a sign that has no plane of incidence to distinguish it. Why: at normal incidence there is no plane of incidence, so " $s$ " and " $p$ " are meaningless labels — the physics cannot depend on a distinction that does not exist.

Step 2. Read it as impedance. With $Z=Z_0/n$ , the coefficient is $r=(Z_2-Z_1)/(Z_2+Z_1)$ : reflection measures how badly the two media's wave impedances mismatch, exactly as a pulse partly reflects at a join between a light and a heavy rope. Why: this reframes "why 4%" — glass is only a mild impedance step from air, so only a mild echo returns; a bigger index jump means a bigger mismatch and a brighter reflection.

Step 3. Evaluate. Air–glass ( $n_2=1.52$ ): $R=(0.52/2.52)^2=0.0426=4.26\%$ . Air–water ( $1.33$ ): $R=(0.33/2.33)^2=2.0\%$ . Air–diamond ( $2.417$ ): $R=(1.417/3.417)^2=17.2\%$ . Why: the monotonic climb with index contrast is exactly why diamond looks so much brighter at its faces than glass, and why high-index camera lenses are the ones that most need coating.

Step 4. Stack surfaces. An uncoated element has two surfaces, so it transmits $(1-R)^2$ ; a six-element lens has twelve air–glass surfaces and transmits $(1-0.0426)^{12}\approx0.59$ — a $41\%$ loss to reflection alone. Why: this multiplicative loss, not absorption, is the historical reason anti-reflection coatings transformed photography and why "coated optics" became a selling point. *Verified: $R=((n_1-n_2)/(n_1+n_2))^2$ gives $4.26\%$ , $2.0\%$ , $17.2\%$ for the three interfaces, and $r_s=-r_p$ at normal incidence, in verify_fresnel-equations.py.*

Go deeper · Derivation B · 2 minWhy reflection polarizes light, and where it does so completely

Derivation BWhy reflection polarizes light, and where it does so completely

Here we prove the optical headline of the Fresnel equations: a plain dielectric surface does not merely dim the reflected light, it sorts it by polarization, because the $s$ channel always reflects more strongly than the $p$ channel. The goal is to establish the strict inequality $R_s>R_p$ at every oblique angle, locate the single angle where $R_p$ hits zero, and show that the reflected beam's degree of polarization therefore rises to a perfect $1$ at Brewster's angle.

Step 1. Compare numerators. Write $a=n_1\cos\theta_i$ , $b=n_2\cos\theta_t$ , $c=n_2\cos\theta_i$ , $d=n_1\cos\theta_t$ ; then $r_s=(a-b)/(a+b)$ and $r_p=(c-d)/(c+d)$ . For $n_2>n_1$ and any oblique angle, algebra (or the impedance forms $Z_s=Z/\cos\theta$ , $Z_p=Z\cos\theta$ ) gives $|r_s|>|r_p|$ , so $R_s>R_p$ . Why: the $s$ channel's mismatch only ever grows with angle, while the $p$ channel's can be tuned away — the asymmetry the physical picture traced to which field is fully tangential.

Step 2. Find the $p$ -zero. $r_p=0$ needs $n_2\cos\theta_i=n_1\cos\theta_t$ ; combined with Snell's law this forces $\theta_i+\theta_t=90^\circ$ , i.e. $\tan\theta_B=n_2/n_1$ . Why: the reflected and refracted rays are then perpendicular, and the medium's driven dipoles — which point along the refracted-then-turned direction — cannot radiate toward the reflected ray, so the $p$ -reflection is switched off. The full argument belongs to the next article.

Step 3. Compute the reflected polarization. Unpolarized input ( $R_s$ and $R_p$ equal weight) reflects with intensities in the ratio $R_s:R_p$ , giving a reflected degree of polarization $P_{\text{refl}}=(R_s-R_p)/(R_s+R_p)$ . At $\theta_B$ , $R_p=0$ , so $P_{\text{refl}}=1$ — perfectly polarized. Why: this is the quantitative content of "glare is polarized" — the reflected light is $s$ -polarized (horizontal off a horizontal surface), which is exactly the component vertical-axis sunglasses reject.

Step 4. Away from Brewster the polarization is partial but real: at $45^\circ$ air–glass ( $n_2=1.52$ ), $R_s\approx0.097$ and $R_p\approx0.009$ give $P_{\text{refl}}\approx0.82$ . Why: reflection polarizes over a broad range of angles, not just at the exact Brewster angle, which is why anti-glare filters work across a whole scene rather than at one perfect tilt. *Verified: $R_s>R_p$ for all $0<\theta_i<90^\circ$ (air–glass), the symbolic zero $r_p=0\Rightarrow\tan\theta_B=n_2/n_1$ , $\theta_B=56.66^\circ$ with $\theta_B+\theta_t=90^\circ$ , and $P_{\text{refl}}\to1$ at Brewster, in verify_fresnel-equations.py.*

Go deeper · Derivation C · 1 minEnergy conservation R + T = 1 and the transmittance prefactor

Derivation CEnergy conservation R + T = 1 and the transmittance prefactor

This derivation settles the most common Fresnel mistake — writing $T=|t|^2$ — by showing that only with the geometric-and-impedance prefactor does power balance. The goal is to prove $R_s+T_s=1$ as an exact algebraic identity for a lossless interface, which simultaneously confirms the coefficients are internally consistent and explains where the seemingly fussy $(n_2\cos\theta_t)/(n_1\cos\theta_i)$ factor comes from physically.

Step 1. Power on the surface. The power a beam delivers to the interface is its intensity times the projected area, $P\propto n|E|^2\cos\theta$ . The transmitted-to-incident power ratio is therefore $T=\dfrac{n_2\cos\theta_t}{n_1\cos\theta_i}|t|^2$ , not $|t|^2$ — the extra factor is the change in both the projected beam footprint and the impedance. Why: the transmitted beam is wider or narrower on the surface and lives in a different medium, so field-squared alone cannot be the power ratio.

Step 2. Insert the $s$ coefficients with $a=n_1\cos\theta_i$ , $b=n_2\cos\theta_t$ : $R_s=\big(\tfrac{a-b}{a+b}\big)^2$ and, using $t_s=2a/(a+b)$ , $T_s=\dfrac{b}{a}\dfrac{4a^2}{(a+b)^2}=\dfrac{4ab}{(a+b)^2}$ . Why: the prefactor $b/a$ is exactly what converts the field ratio $|t_s|^2$ into a power ratio.

Step 3. Add them: $R_s+T_s=\dfrac{(a-b)^2+4ab}{(a+b)^2}=\dfrac{(a+b)^2}{(a+b)^2}=1$ , using the identity $(a-b)^2+4ab=(a+b)^2$ . Why: energy is conserved exactly, at every angle, for a lossless dielectric — a hard check that both the coefficients and the $T$ prefactor are right.

Step 4. The same holds channel by channel for $p$ . Any leftover after $R+T$ would signal absorption or an error; a lossless surface has none. Why: this is the routine consistency test used when deriving multilayer-coating transfer matrices, where a spurious net gain or loss immediately flags a sign or prefactor slip. *Verified: the symbolic identity $(a-b)^2+4ab=(a+b)^2\Rightarrow R_s+T_s=1$ , and numerically $R+T=1$ for both channels at $45^\circ$ air–glass, in verify_fresnel-equations.py.*

Limiting cases

Case	Result	Meaning
Normal incidence	$r_s=-r_p=\dfrac{n_1-n_2}{n_1+n_2},\ R=\big(\tfrac{n_1-n_2}{n_1+n_2}\big)^2$	$s/p$ distinction gone; air–glass gives $R=4.26\%$ ( $n_2=1.52$ )
Brewster's angle	$\tan\theta_B=n_2/n_1\Rightarrow r_p=0$	reflected light purely $s$ -polarized; air–glass $\theta_B=56.7^\circ$
Critical angle / TIR ( $n_1>n_2$ )	$\sin\theta_c=n_2/n_1,\ \lvert r_s\rvert=\lvert r_p\rvert=1$	coefficients unimodular complex; all power reflected with a polarization-dependent phase
Grazing incidence	$\theta_i\to90^\circ:\ r_s,r_p\to-1,\ R\to1$	any surface becomes a near-perfect mirror at grazing, for both polarizations

Common trap“The transmittance is just |t|², so R = 1 − |t|² for the field coefficient”

No — and this is the single most common Fresnel slip. The transmitted beam changes direction and enters a medium of different impedance, so its power fraction carries a mandatory prefactor: $T=\dfrac{n_2\cos\theta_t}{n_1\cos\theta_i}\,|t|^2$ . Only with that factor does $R+T=1$ ; using $|t|^2$ alone breaks energy conservation, sometimes visibly — going glass→air ( $n_1=1.52\to n_2=1$ ) at normal incidence the transmitted amplitude actually exceeds the incident one, $t_s=2(1.52)/2.52=1.21$ , so $|t_s|^2\approx1.46>1$ , which is nonsense as a power fraction until the prefactor $n_2/n_1=1/1.52\approx0.66$ brings it back to $T\approx0.96<1$ (and $R=4.26\%$ closes the ledger). A second, quieter trap: sign conventions for the positive $p$ -field direction differ between textbooks, flipping the sign of $r_p$ — Hecht and Born & Wolf disagree on that sign — so always check the convention before comparing two formulas or two people's Brewster-side algebra. The reflectance $R_p=|r_p|^2$ is convention-independent, which is why it is the safe thing to quote.

Worked examples

1 · Reflection loss of an uncoated lens (confidence builder)

A camera lens element is made of crown glass, $n=1.52$ , used in air. What fraction of light is reflected at one face at normal incidence, and how much does a full six-element (twelve-surface) uncoated lens lose to reflection?

Step 1. One surface: $R=\big(\tfrac{1-1.52}{1+1.52}\big)^2=\big(\tfrac{-0.52}{2.52}\big)^2=(0.2063)^2=\mathbf{0.0426}$ , i.e. $4.26\%$ reflected, $95.74\%$ transmitted.

Step 2. Two surfaces of one element: transmission $(1-R)^2=(0.9574)^2=0.9166$ , so a single uncoated element already loses about $8.3\%$ .

Step 3. Twelve surfaces: $(1-R)^{12}=(0.9574)^{12}=0.593$ . The uncoated six-element lens transmits only about $59\%$ of the light — it throws away $41\%$ to reflection alone, before any absorption.

Step 4. This is the whole case for anti-reflection coatings: a single quarter-wave $\mathrm{MgF_2}$ layer cuts each surface's $4.26\%$ to about $1.3\%$ , lifting the twelve-surface transmission from $59\%$ to about $86\%$ — and multilayer coatings do far better. One face reflects $4.26\%$ ; the uncoated twelve-surface lens transmits $\approx59\%$ .

2 · Reflectance of both polarizations at 45° and the reflected polarization (exam level)

Unpolarized light hits an air–glass surface ( $n=1.52$ ) at $\theta_i=45^\circ$ . Find $R_s$ and $R_p$ , and the degree of polarization of the reflected beam. Compare with Brewster incidence.

Step 1. Refraction angle: $\sin\theta_t=\sin45^\circ/1.52=0.7071/1.52=0.4652$ , so $\theta_t=27.73^\circ$ , $\cos\theta_t=0.8850$ ; and $\cos\theta_i=0.7071$ .

Step 2. $s$ channel: $r_s=\dfrac{1(0.7071)-1.52(0.8850)}{1(0.7071)+1.52(0.8850)}=\dfrac{0.7071-1.3452}{0.7071+1.3452}=\dfrac{-0.6381}{2.0523}=-0.3109$ , so $R_s=0.0967$ .

Step 3. $p$ channel: $r_p=\dfrac{1.52(0.7071)-1(0.8850)}{1.52(0.7071)+1(0.8850)}=\dfrac{1.0748-0.8850}{1.0748+0.8850}=\dfrac{0.1898}{1.9598}=0.0969$ , so $R_p=0.0094$ .

Step 4. Reflected degree of polarization: $P_{\text{refl}}=\dfrac{R_s-R_p}{R_s+R_p}=\dfrac{0.0967-0.0094}{0.0967+0.0094}=\dfrac{0.0873}{0.1061}=0.82$ . At Brewster ( $56.7^\circ$ ) instead, $R_p=0$ and $P_{\text{refl}}=1$ : the reflected light is then perfectly $s$ -polarized. At $45^\circ$ , $R_s\approx9.7\%$ , $R_p\approx0.9\%$ , reflected DoP $\approx0.82$ ; at Brewster it reaches 1.

Physics of the portrait

The equations carry Augustin-Jean Fresnel's name because he wrote them down — remarkably, from a mechanical picture of light as vibrations in an elastic solid, years before Maxwell showed light was electromagnetic and the same formulas fell out of the boundary conditions.

Origins · 1 minfresnel equations and how this was discovered

fresnel equations

the amplitude laws of reflection · Fresnel 1821–1823, from an elastic-solid theory of light

The Fresnel equations are the amplitude reflection and transmission coefficients Augustin-Jean Fresnel (1788–1827) derived around 1821–1823 and presented to the Académie des Sciences. What makes them astonishing is their provenance: Fresnel obtained them from a mechanical theory in which light was a transverse vibration of a hypothetical elastic aether, decades before James Clerk Maxwell's 1860s electromagnetism gave the modern derivation from the continuity of tangential $\mathbf E$ and $\mathbf H$ — which is the route owned by the companion article aq-em-fresnel-equations. That two utterly different pictures of light yield the identical coefficients is one of the quiet triumphs of nineteenth-century optics. Fresnel, a French civil engineer who did physics in his spare hours and died of tuberculosis at 39, had already, with François Arago, established that two perpendicular polarizations do not interfere (the Fresnel–Arago laws of 1819) and built the biprism and the stepped lens that still bears his name. Étienne-Louis Malus had discovered polarization by reflection in 1808, watching sunset light glance off the windows of the Luxembourg Palace; David Brewster pinned the polarizing angle to $\tan\theta_B=n$ in 1815. Fresnel's coefficients explained why — and quantified every reflection from a puddle to a lens coating.

History

1808

— Malus discovers polarization by reflection in light glancing off the Luxembourg Palace windows

1815

— Brewster finds the polarizing angle obeys tan θ_B = n

1819

— Fresnel and Arago publish the laws of interference of polarized light

1823

— Fresnel presents the amplitude reflection and transmission coefficients to the Académie

1875

— Lorentz derives the Fresnel coefficients from Maxwell's electromagnetic theory in his doctoral thesis

Going further

Texts. Hecht, Optics, 5th ed., §4.6 (Fresnel Equations) · Born & Wolf, Principles of Optics, 7th ed., §1.5 · Saleh & Teich, Fundamentals of Photonics, 3rd ed., §6.2 · Jackson, Classical Electrodynamics, 3rd ed., §7.3 · Fowles, Introduction to Modern Optics, §2.7.

Provenance. The coefficient formulas are quoted in the Hecht sign convention (Hecht Eqs. 4.42–4.49) and their electromagnetic derivation is recalled by name from the companion article aq-em-fresnel-equations (matching tangential $\mathbf E,\mathbf H$ channel by channel, the impedance-mismatch view). verify_fresnel-equations.py confirms: (1) normal-incidence $R=((n_1-n_2)/(n_1+n_2))^2$ gives $4.26\%$ / $2.0\%$ / $17.2\%$ for glass/water/diamond and $r_s=-r_p$ ; (2) $R_s>R_p$ for all oblique angles (air–glass); (3) the symbolic zero $r_p=0\Rightarrow\tan\theta_B=n_2/n_1$ , with $\theta_B=56.66^\circ$ and $\theta_B+\theta_t=90^\circ$ ; (4) the energy identity $(a-b)^2+4ab=(a+b)^2\Rightarrow R_s+T_s=1$ , and numerically $R+T=1$ at $45^\circ$ ; (5) the reflected DoP $\to1$ at Brewster. Sign convention for $r_p$ follows Hecht; Born & Wolf use the opposite sign — the reflectance $R_p=|r_p|^2$ is convention-independent.

Prerequisites. Polarization of Light · Linear Polarization · Malus's Law. Leads to. Brewster's Angle · Polarization on Reflection and Total Internal Reflection · Jones Calculus · Wave Plates and Retarders.

Why does r_s never vanish while r_p does at Brewster's angle?

It comes down to which cosine sits with which index. The $s$ -numerator is $n_1\cos\theta_i-n_2\cos\theta_t$ ; because $\cos\theta$ grows as the index falls and shrinks as it rises, these two terms move in the same direction and can only cancel if $n_1=n_2$ (no interface at all). The $p$ -numerator is $n_2\cos\theta_i-n_1\cos\theta_t$ — the indices are swapped between the cosines, so as the angle changes the two terms can cross and cancel at a real angle. That crossing is Brewster's angle, and it exists precisely because the $p$ -channel weighs the fields with the opposite index pairing.

Which polarization is the glare off a road, and why do sunglasses cut it?

Glare reflected from a horizontal surface — a wet road, a lake, a car hood — is predominantly $s$ -polarized, meaning its electric field is horizontal (perpendicular to the vertical plane of incidence). The Fresnel equations guarantee this because $R_s>R_p$ at every oblique angle, and near Brewster the reflected light is almost entirely $s$ . Polarized sunglasses have a transmission axis oriented vertically, so they absorb the horizontal $s$ -glare while passing the vertically polarized light from the scene. That is why tilting your head sideways while wearing them briefly restores the glare — you have rotated the transmission axis toward horizontal.

Does light get a phase shift when it reflects, and does it depend on polarization?

Yes. For a lossless dielectric below the critical angle the coefficients are real, so the only "phase shift" is a sign: reflecting off a higher-index medium (external reflection) flips the field by $\pi$ in the standard convention, while internal reflection off a lower-index medium below Brewster does not. Above the critical angle the coefficients become complex and $s$ and $p$ acquire different, continuously varying phases — a genuine polarization-dependent retardance that turns linear light elliptical, while the transmitted field becomes an evanescent wave that clings to the interface and carries no net power across it. That differential phase is the operating principle of the Fresnel rhomb and is treated fully in the article on polarization at total internal reflection.

Do the Fresnel equations apply to metals and absorbing materials?

Yes, with a complex refractive index $\tilde n=n+i\kappa$ . The same formulas hold, but now the coefficients are complex at all angles, so reflection off a metal imposes a phase shift and there is no true Brewster zero — only a shallow "pseudo-Brewster" minimum in $R_p$ . This is why a polished metal mirror reflects strongly for both polarizations (high $n$ and $\kappa$ push $R$ toward 1) and why metals do not make good polarizers by reflection the way a dielectric surface does. Ellipsometry of metals inverts exactly these complex-index Fresnel equations to measure $n$ and $\kappa$ .

Why is the normal-incidence reflection the same for a beam going glass→air as air→glass?

Because $R=\big((n_1-n_2)/(n_1+n_2)\big)^2$ depends only on the difference of the indices, squared — swapping $n_1$ and $n_2$ flips the sign of $r$ but leaves $R$ identical. So $4.26\%$ reflects at an air–glass surface whichever way the light crosses it. What differs is the phase: the external reflection (air→glass) carries a $\pi$ flip while the internal one (glass→air) does not, a distinction that matters enormously for thin-film interference even though the reflected power is the same both ways.

How does a single anti-reflection coating actually cancel the 4% reflection?

A quarter-wave layer of index $n_c$ between air and glass creates two reflections — one at the air–coating surface, one at the coating–glass surface — that are equal in amplitude and half a wavelength out of step, so they interfere destructively. Making the two amplitudes equal requires $n_c=\sqrt{n_1 n_2}$ ; for air and crown glass that is $\sqrt{1.52}\approx1.23$ . Magnesium fluoride ( $n\approx1.38$ ) is the usual practical compromise, cutting a surface's reflectance from $4.26\%$ to about $1.3\%$ at the design wavelength. It is the Fresnel coefficients of the two interfaces, combined with a $\lambda/4$ path, that make the cancellation work.