Brewster's Angle

Tilt a pane of glass toward the low sun and the glare off it does something quietly remarkable: at one particular tilt the reflected light becomes perfectly polarized — every reflected ray vibrating in the same horizontal plane, the vertical component gone without a trace. That angle is Brewster's angle, and it is the one place in ordinary dielectric optics where a polarization is not merely weakened but removed exactly. It is why polarized sunglasses erase the sheen on a wet road, why a photographer's filter can dissolve the reflection on a shop window, and why every gas laser on a bench wears its end windows at a precise slant. That the reflected parallel-polarized amplitude $r_p$ actually passes through zero is a result of the Fresnel equations — the boundary-condition bookkeeping of Maxwell's fields at the interface — and this article recalls it by name from the electromagnetic companion aq-em-brewster-angle-total-internal-reflection rather than re-deriving it. What optics owns, and what we build here, is the geometry of that zero: the clean law $\tan\theta_B = n_2/n_1$ , its striking signature that the reflected and refracted rays leave exactly perpendicular, and the microscopic picture that makes it unforgettable — reflection is re-radiation by the medium's driven dipoles, and a dipole cannot radiate along its own axis. Point the reflected ray straight down the p-dipole axis and there is no one left to build it. Two lines of trigonometry turn that sentence into the polarizing angle; the rest of the article spends it on the residual reflected polarization, on Brewster windows, and on why the effect cuts one polarization channel while leaving the other whole.

Key takeaways

At Brewster's angle the p-polarized (in-plane, TM) reflection coefficient $r_p$ vanishes, so an unpolarized beam reflects as pure s-polarized light — the maximally polarizing angle of a dielectric.
The condition is $\tan\theta_B = n_2/n_1$ : for air onto crown glass ( $n = 1.52$ ) it is $56.66°$ , for air onto water ( $1.33$ ) it is $53.06°$ .
Its geometric signature is that the reflected and refracted rays are exactly perpendicular, $\theta_B + \theta_t = 90°$ — the fact that makes the law memorable and the dipole picture work.
Microscopically, reflection is re-radiation by the second medium's oscillating dipoles; at $\theta_B$ the p-dipole axis points along the reflected ray, and a dipole radiates nothing along its axis — so $r_p = 0$ . No such alignment is ever possible for s, so there is no s-Brewster angle.
That $r_p = 0$ at Brewster is the Fresnel/boundary-condition result, recalled by name from aq-em-brewster-angle-total-internal-reflection; the geometry and the dipole mechanism are owned here.
Key idea: the p-polarized reflection vanishes precisely when the reflected and refracted rays are perpendicular, which forces $\tan\theta_B = n_2/n_1$ and leaves the reflected beam completely s-polarized.

The core insight

Reflection off a dielectric is a partial polarizer, and at one special incidence — Brewster's angle — it becomes a perfect one. There the reflected and refracted rays are perpendicular, the driven p-dipoles of the second medium are asked to radiate along their own axis and cannot, so the parallel-polarized reflection $r_p$ drops to zero. The surviving reflected light is 100% s-polarized, and the polarizing angle obeys the compact law $\tan\theta_B = n_2/n_1$ .

Where this lives today

Brewster's angle is one of those pieces of physics that is invisible precisely because it works so well. Every gas laser and many solid-state cavities use Brewster-angle windows and Brewster-cut elements to fix the polarization with geometry alone, at zero insertion loss for the wanted mode. Photography and everyday eyewear monetize the same asymmetry: polarizing camera filters and polarized sunglasses reject the strongly s-polarized glare dielectric surfaces throw off near $\theta_B$ , cutting the sheen on water, glass, and wet asphalt. At the research bench the angle becomes a measuring instrument. Ellipsometry parks a beam near the (pseudo-)Brewster angle, where the reflected amplitude and phase swing most steeply, and reads film thickness and refractive index to sub-nanometre precision — the workhorse metrology of semiconductor and coating fabrication. Brewster-angle microscopy sits a beam exactly at the reflection zero of a bare water surface and watches a molecular film light up as anything perturbing the interface spoils the null, imaging Langmuir monolayers a single molecule thick.

The physical picture

Think of reflection not as a mirror bouncing light but as a two-step relay. The incident wave crosses into the second medium and drives its electrons into oscillation; those oscillating charges are tiny antennas, and the reflected wave is nothing but the coherent back-radiation of that whole sheet of driven dipoles. Now recall the one fact everyone remembers about an antenna: it beams broadside and goes silent along its own axis. For light polarized in the plane of incidence — the p, or TM, mode — the dipoles oscillate in that plane, perpendicular to the refracted ray. If the reflected ray's direction lines up with that oscillation axis, the reflected wave has no source. That alignment happens at exactly one incidence — where the reflected and refracted rays stand at a right angle — and it is Brewster's angle.

Figure

Brewster geometry: reflected ray perpendicular to refracted ray, the reflected beam purely s-polarized because the p-dipole axis points straight down the reflected direction

Notice how asymmetric the two polarizations are. The s-mode (electric field perpendicular to the plane of incidence, TE) drives dipoles that oscillate out of the page, broadside to every ray in the diagram; they radiate into the reflected direction at every angle, so the s-reflection never vanishes and there is no Brewster angle for s. Only the p-channel has an axis that can be aimed at the reflected ray, and only at one incidence. So reflection is a partial polarizer at every angle — always richer in s than in p — and a perfect polarizer at Brewster. The next figure makes that quantitative.

Figure

Reflectances for air onto crown glass: R_p dives to zero at 56.66° while R_s climbs, so the reflected degree of polarization reaches 100% exactly at Brewster's angle

Both reflectances start together at the normal-incidence value $R = ((n_1-n_2)/(n_1+n_2))^2 = 4.26\%$ for crown glass, where the s/p distinction is only a label. From there $R_s$ climbs steadily to unity at grazing, while $R_p$ dips, touching zero at $\theta_B = 56.66°$ before it too rises to one. The reflected degree of polarization — the green dashed curve, $\mathrm{DoP} = |R_s - R_p|/(R_s + R_p)$ for unpolarized input — is zero at normal incidence, spikes to exactly $1$ where $R_p$ hits its zero, then falls away. That single spike is Brewster's angle in one glance: the only incidence at which reflection hands back perfectly polarized light.

The core equations

Everything follows from one recalled fact and one geometric substitution. The recalled fact is that the p-polarized amplitude reflection coefficient $r_p$ — derived from the continuity of the tangential fields at the boundary in aq-em-brewster-angle-total-internal-reflection — passes through zero at a single incidence angle. Setting that condition against Snell's law collapses it to the polarizing law. The condition $r_p = 0$ is the electromagnetic input; the compact form below is the optical output.

\tan\theta_B = \frac{n_2}{n_1} \qquad (1)

where: $\theta_B$ — Brewster (polarizing) angle, measured from the surface normal; $n_1$ — index of the incident medium; $n_2$ — index of the transmitting medium. Physical reading: the polarizing angle depends only on the ratio of indices, and on wavelength only through the mild dispersion of $n$ — so a single window tilt works across a broad band. The law carries a geometric twin that is often easier to remember and is the seed of the dipole argument. Feeding $\theta_t = 90° - \theta_B$ into Snell's law reproduces equation (1) exactly, so the vanishing of $r_p$ is equivalent to a right angle between the reflected and refracted rays.

\theta_B + \theta_t = 90° \qquad (2)

where: $\theta_t$ — refraction angle at Brewster incidence. Physical reading: reflected ray ⟂ refracted ray. Because the reflected ray then lies exactly along the p-dipole oscillation axis, no parallel-polarized wave can be radiated into it — the geometry is the mechanism. Brewster's angle zeroes the p-channel but leaves a real s-reflection, and its size sets how bright the perfectly polarized reflection is. Evaluating the Fresnel s-coefficient with $\theta_t = 90° - \theta_B$ gives a compact residual reflectance in the indices alone.

R_s(\theta_B) = \left(\frac{n_2^2 - n_1^2}{n_2^2 + n_1^2}\right)^2 \qquad (3)

where: $R_s(\theta_B)$ — s-power reflectance at Brewster's angle. Physical reading: for air onto crown glass this is $0.157$ , so only about $16\%$ of the s-power (and none of the p) reflects — a pure but dim polarizer, which is why a Brewster surface polarizes weakly. Finally, the polarizing power of the reflection is captured by the degree of polarization of the reflected beam for unpolarized incident light, which peaks at unity wherever $R_p = 0$ .

P = \frac{R_s - R_p}{R_s + R_p} \qquad (4)

where: $P$ — degree of linear polarization of the reflected light; $R_s, R_p$ — the two power reflectances. Physical reading: away from Brewster $R_p \ne 0$ and $P < 1$ ; at $\theta_B$ , $R_p = 0$ and $P = 1$ exactly — the spike in the figure.

Derivation ABrewster's angle two ways, and why they must agree

The goal is the polarizing law $\tan\theta_B = n_2/n_1$ , reached along two independent roads that meet at the same number — an algebraic road starting from the recalled vanishing of $r_p$ , and a geometric-mechanistic road starting from the picture of driven dipoles that never touches a Fresnel coefficient. That two such different starting points land on one law is the strongest evidence the picture is right, and it is the reasoning you can rebuild on a napkin years later. The algebraic input is owned upstream; the geometry and mechanism are owned here.

Step 1. The algebraic road. Recall the parallel-polarization coefficient $r_p = (n_2\cos\theta_i - n_1\cos\theta_t)/(n_2\cos\theta_i + n_1\cos\theta_t)$ — quoted from the companion aq-em-fresnel-equations, which derives all four coefficients from the boundary conditions — together with the fact, recalled from aq-em-brewster-angle-total-internal-reflection, that this coefficient passes through zero at a single incidence: $r_p = 0$ demands $n_2\cos\theta_i = n_1\cos\theta_t$ . Why we recall rather than re-derive: the vanishing numerator is a boundary-condition result of Maxwell's fields, established in the electromagnetic articles; our business is what that zero means geometrically.

Step 2. Pair it with Snell's law $n_1\sin\theta_i = n_2\sin\theta_t$ . Dividing the two relations eliminates the indices and gives $\tan\theta_i = \tan(90° - \theta_t)$ , i.e. $\theta_i + \theta_t = 90°$ ; substituting $\theta_t = 90° - \theta_i$ back into Snell's law yields $n_1\sin\theta_i = n_2\cos\theta_i$ , hence $\tan\theta_B = n_2/n_1$ . Why this works in one line: the two conditions are a cosine-ratio and a sine-ratio of the same two angles, and the only way both hold is with the angles complementary.

Step 3. The geometric-mechanistic road, starting from the picture. Reflection is the coherent back-radiation of the second medium's driven dipoles. For p-polarization the transmitted field, and therefore the dipoles, oscillate in the plane of incidence, perpendicular to the refracted ray. A Hertzian dipole radiates with the pattern $\propto \sin^2\Theta$ , where $\Theta$ is measured from its axis; along the axis ( $\Theta = 0$ ) it radiates nothing. Why this is the whole physics: the reflected wave exists only if the dipoles can radiate into its direction, and they cannot radiate along their own axis.

Step 4. Convert the null to an angle. The reflected wave dies exactly when its direction coincides with the dipole axis — that is, when the reflected ray is perpendicular to the refracted ray, $\theta_B + \theta_t = 90°$ . This is precisely the geometric condition of Step 2, so Snell's law closes it into $\tan\theta_B = n_2/n_1$ once more. The two roads meet: the algebra and the antenna agree to the last digit. Why the s-channel is exempt: s-dipoles oscillate out of the plane of incidence, broadside to every possible reflected direction, so $\Theta$ can never reach $0$ and $r_s$ never vanishes — no s-Brewster angle exists for a nonmagnetic interface. *Verified: both routes reproduced symbolically ( $\theta_B = \arctan(n_2/n_1)$ satisfies $n_1\sin\theta = n_2\cos\theta$ , and $r_p$ vanishes numerically at that angle), giving $56.66°$ for air→glass and $53.06°$ for air→water, with $\theta_B + \theta_t = 90°$ , in verify_brewster-angle.py.*

Go deeper · Derivation B · 2 minThe reflected light is 100% s-polarized at Brewster's angle

Derivation BThe reflected light is 100% s-polarized at Brewster's angle

The goal is to show that the reflected beam is not merely richer in s-polarization at Brewster's angle but entirely s-polarized, whatever the incident state, and to pin down how bright that perfectly polarized reflection is. We compute the surviving s-reflectance from the Fresnel geometry at $\theta_B$ , note the p-reflectance is identically zero, and feed both into the degree of polarization — the crisp statement that reflection at $\theta_B$ is a perfect, if inefficient, polarizer.

Step 1. From $\tan\theta_B = n_2/n_1$ read off $\cos\theta_B = n_1/\sqrt{n_1^2 + n_2^2}$ and $\sin\theta_B = n_2/\sqrt{n_1^2 + n_2^2}$ , and from $\theta_t = 90° - \theta_B$ read $\cos\theta_t = \sin\theta_B$ and $\sin\theta_t = \cos\theta_B$ . Why these substitutions: the Brewster condition fixes both angles in terms of the two indices alone, so every reflectance collapses to a function of $n_1, n_2$ .

Step 2. Insert into the s-coefficient $r_s = (n_1\cos\theta_B - n_2\cos\theta_t)/(n_1\cos\theta_B + n_2\cos\theta_t)$ . With $\cos\theta_t = \sin\theta_B$ the numerator becomes $(n_1^2 - n_2^2)/\sqrt{n_1^2 + n_2^2}$ and the denominator $(n_1^2 + n_2^2)/\sqrt{n_1^2 + n_2^2}$ , so $r_s = (n_1^2 - n_2^2)/(n_1^2 + n_2^2)$ and $R_s(\theta_B) = ((n_2^2 - n_1^2)/(n_2^2 + n_1^2))^2$ . For air→glass this is $0.157$ . Why it is modest: the residual s-reflection at Brewster is a little larger than the normal-incidence value but still a small fraction — a Brewster surface polarizes cleanly but weakly.

Step 3. The p-reflectance is $R_p(\theta_B) = 0$ by Derivation A. Feed both into the degree of polarization $P = (R_s - R_p)/(R_s + R_p) = R_s/R_s = 1$ . Why the input state does not matter: any incident beam splits into s and p; the p-part is not reflected at all, so whatever reflects is pure s, regardless of how the incident light was polarized.

Step 4. Read the consequence for a single surface as a polarizer. It gives perfect purity ( $P = 1$ ) at low efficiency (only $16\%$ of the incident s-power). To make a bright polarizer one either stacks many Brewster surfaces so the transmitted beam is progressively stripped of s, or — the more common trick — uses the losslessly transmitted p-beam directly, which is exactly what a Brewster window does. Why purity and brightness trade off here: the same zero that guarantees perfect polarization also guarantees a small reflected amplitude, because nothing amplifies the surviving s. *Verified: $R_s(\theta_B) = 0.157$ for crown glass and $0.148$ for common glass ( $n = 1.5$ ), $R_p(\theta_B) = 0$ , and $P = 1$ at $\theta_B$ reproduced in verify_brewster-angle.py.*

Limiting cases

Case	Result	Meaning
air → crown glass ( $n_2 = 1.52$ )	$\theta_B = 56.66°$ , $\theta_t = 33.34°$	the standard optical-glass value; $56.66° + 33.34° = 90°$ ⟂
air → water ( $n_2 = 1.33$ )	$\theta_B = 53.06°$	the polarized-glare geometry off lakes, wet roads, and windows
air → diamond ( $n_2 = 2.417$ )	$\theta_B = 67.5°$	high index pushes the polarizing angle toward grazing
matched indices ( $n_2 \to n_1$ )	$\theta_B \to 45°$	reflection vanishes at all angles, but the limiting polarizing direction is $45°$
s-polarization, any interface	no Brewster angle	s-dipoles are broadside to every reflected ray — $r_s$ never zeroes
internal ( $n_1 = 1.52 \to n_2 = 1$ )	$\theta_B = 33.34°$	the complementary internal partner of the external $56.66°$ ( $33.34° + 56.66° = 90°$ )

Common trap“Brewster's angle is the angle of zero reflection.”

It is the angle of zero p-reflection, not zero reflection. An unpolarized beam striking glass at $56.66°$ still reflects — about $16\%$ of its s-power comes back — but that reflected light is now perfectly s-polarized, the p-component entirely absent. "No reflection at Brewster" is true only for a beam that is already pure p-polarized; feed such a beam in and the surface transmits it losslessly, which is the whole point of a Brewster window. Feed in unpolarized or s-light and you get a bright, fully polarized reflection. And the tidy formula assumes a non-magnetic, non-absorbing dielectric: for a metal the complex index prevents an exact zero, leaving only a nonzero pseudo-Brewster minimum with elliptically polarized reflection.

Worked examples

1 · Brewster's angle and the reflected polarization for glass and water (confidence builder)

Find Brewster's angle for light going from air into crown glass ( $n = 1.52$ ) and into water ( $n = 1.33$ ). At the glass value, verify the ray geometry and state the polarization and brightness of the reflected beam.

Step 1. Glass: $\theta_B = \arctan(1.52/1) = \arctan(1.52) = \mathbf{56.66°}$ . Water: $\theta_B = \arctan(1.33) = \mathbf{53.06°}$ . Denser media polarize at steeper angles, closer to grazing.

Step 2. Check the geometry for glass. Snell's law gives $\theta_t = \arcsin(\sin 56.66°/1.52) = \arcsin(0.8355/1.52) = \arcsin(0.5497) = \mathbf{33.34°}$ , and $56.66° + 33.34° = \mathbf{90.0°}$ — the reflected and refracted rays are exactly perpendicular, as the polarizing law demands.

Step 3. The reflected polarization. Since $R_p(\theta_B) = 0$ , the reflected light is $\mathbf{100\%}$ s-polarized regardless of the incident state — an unpolarized beam reflects as pure horizontally-vibrating light. Its brightness is the residual s-reflectance $R_s(\theta_B) = ((1.52^2 - 1)/(1.52^2 + 1))^2 = (1.3104/3.3104)^2 = \mathbf{0.157}$ : about $16\%$ of the incident s-power.

Step 4. The everyday payoff. Glare off a horizontal wet road near $\theta_B$ is strongly horizontally polarized, so a polarizing filter with its axis vertical blocks it — exactly how polarized sunglasses kill road and water glare. θ_B = 56.66° (glass), 53.06° (water); reflected light 100% s-polarized, R_s ≈ 0.157.

2 · Why a gas laser wears its windows at Brewster's angle (exam level)

A He–Ne laser tube is sealed with glass windows ( $n = 1.52$ ) tilted at Brewster's angle. Explain quantitatively why this forces the laser to oscillate in a single linear polarization, given a small-signal round-trip gain of only a few percent.

Step 1. Set the window at $\theta_B = 56.66°$ . For the p-polarization the reflection loss is exactly zero — and, by the complementary-pair geometry, the beam meets the second (internal) face at $33.34°$ , which is that face's own Brewster angle, so it too reflects no p. A Brewster window is therefore lossless for p at both surfaces from a single tilt.

Step 2. For the s-polarization each surface reflects $R_s = 0.157$ out of the beam. One window has two surfaces, so per pass the s-transmission is $(1 - 0.157)^2 = 0.843^2 = \mathbf{0.711}$ — a $28.9\%$ loss every time the s-mode crosses a window.

Step 3. Compare with the gain. The intracavity beam crosses the window twice per round trip, so the s-mode retains $0.711^2 = \mathbf{0.505}$ per round trip (a $\sim50\%$ loss) while the p-mode retains $1.000$ . A He–Ne small-signal gain is only a few percent per round trip, nowhere near enough to overcome a $50\%$ loss. The s-mode cannot reach threshold; only the lossless p-mode lases.

Step 4. The design lesson. The Brewster window does two jobs at once with pure geometry and no coatings: it seals the gas tube with a lossless path for the wanted polarization, and it discriminates so hard against the orthogonal one that the laser self-selects a clean linear output — the same logic that underlies pile-of-plates polarizers and low-loss intracavity optics. p-loss per window = 0, s round-trip retention ≈ 0.505 ⟹ single linear polarization.

Physics of the portrait

Brewster's angle is named for the law, but the law grew from a chance observation of light glancing off a Paris palace at sunset.

Origins · 1 minbrewster's angle and how this was discovered

brewster's angle

the angle of pure reflected polarization · Malus 1808 → David Brewster 1815

The story begins with Étienne-Louis Malus (1775–1812), a French army engineer and veteran of Napoleon's Egyptian campaign. One evening in 1808, idly viewing the setting sun reflected in the windows of the Luxembourg Palace through a crystal of Iceland spar, he saw the reflected image brighten and dim as he rotated the crystal — reflected light, he realized, was polarized, as if it had passed through a calcite prism. It was the discovery that reflection itself sorts light by polarization. Seven years later the Scottish physicist David Brewster (1781–1868) — a prodigious experimentalist, scientific editor, and the inventor of the kaleidoscope — measured how the polarizing angle depended on the reflecting material and distilled it into the law that bears his name: the tangent of the polarizing angle equals the refractive index, $\tan\theta_B = n$ . Brewster worked empirically, decades before Fresnel's wave theory and the dipole picture explained why the p-reflection vanishes — his law a precise, useful fact in search of a mechanism. That mechanism arrived with electromagnetism and completes the concept that still carries his name on every polarizer, laser window, and pair of polarized sunglasses.

History

1808

— Malus discovers polarization by reflection while watching the sunset in the windows of the Luxembourg Palace

1811

— Arago observes chromatic polarization, deepening the study of reflected and transmitted polarization

1815

— Brewster establishes his law, tan θB = n, relating the polarizing angle to refractive index

1819

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

1823

— Fresnel's amplitude reflection coefficients locate the vanishing p-reflection within wave optics

1962

— helium-neon lasers with Brewster-window tubes and external mirrors enforce a single cavity polarization

Going further

Texts. Hecht, Optics, 5th ed., §4.6 (Fresnel equations, Brewster's angle) · Born & Wolf, Principles of Optics, 7th ed., §1.5 (Fresnel formulae, reflected polarization) · Saleh & Teich, Fundamentals of Photonics, 3rd ed., §6.2 (reflection and refraction of polarized light) · Feynman, Lectures on Physics, Vol. I, §33 (polarization and dipole radiation).

Provenance. The Fresnel equations, the Brewster condition, and the dipole-radiation argument follow Hecht Optics §4.6; the trigonometric Fresnel forms and reflectance curves follow Born & Wolf §1.5 and Saleh & Teich §6.2. The polarizing law $\tan\theta_B = n_2/n_1$ and the residual $R_s(\theta_B) = ((n_2^2-n_1^2)/(n_2^2+n_1^2))^2$ are standard and recomputable from Snell's law plus the vanishing of $r_p$ (recalled from aq-em-brewster-angle-total-internal-reflection). Index note (base reconciliation): the base migration quoted the round-number crown glass value $n = 1.5$ ( $\theta_B = 56.31°$ , $R_s = 0.148$ ); this edition standardises on $n = 1.52$ ( $\theta_B = 56.66°$ , $R_s = 0.157$ ) to match the optical-glass value used across the polarization track, and on water $n = 1.33$ ( $53.06°$ ). Every quoted angle and reflectance is reproduced from its stated index in verify_brewster-angle.py so the number always matches the index.

Prerequisites. Polarization of Light · Linear Polarization. Leads to. Polarizers and Extinction Ratio · Jones Calculus · Stokes Parameters and the Poincaré Sphere · Mueller Calculus · Polarization on Reflection and Total Internal Reflection.

Is Brewster's angle measured from the surface or from the normal?

From the normal, like every angle of incidence in optics. Brewster's angle for air onto glass is $56.66°$ from the normal, which is only about $33°$ from the glancing surface — a fairly shallow tilt of the glass toward the beam. This matters in the lab: when you set a plate to its polarizing angle you rotate it until the beam strikes at $56.66°$ to the plate's normal, not to its face. Quoting the complement by mistake ( $33.34°$ ) is a common slip, and it happens to be exactly the refraction angle at Brewster incidence, which makes the error easy to talk yourself into.

Does Brewster's angle depend on the colour of the light?

Only weakly, and only through dispersion. The formula $\tan\theta_B = n_2/n_1$ contains no wavelength; the sole colour dependence enters because the refractive index $n_2$ itself drifts slightly with wavelength. For ordinary glass across the visible, $n$ varies by roughly one percent, so $\theta_B$ shifts by only about half a degree from red to blue. This near-achromaticity is a practical gift: a single Brewster window works well for a broadband beam, and a laser tube sealed at one fixed tilt stays essentially lossless for its wanted polarization across its whole gain bandwidth. Contrast this with a birefringent wave plate, whose retardance scales as $1/\lambda$ and is sharply chromatic.

Why is the reflected beam at Brewster's angle so dim if it is perfectly polarized?

Because the same zero that guarantees purity guarantees weakness. At $\theta_B$ the entire p-channel is transmitted and only the s-channel reflects, and even that s-reflectance is modest, $R_s(\theta_B) \approx 0.157$ for glass. So the reflected beam is $100\%$ polarized but carries only about $16\%$ of the incident s-power (roughly $8\%$ of an unpolarized beam's total). To get a bright polarized beam you use the transmitted p-light, which is nearly full intensity and highly (though not perfectly) polarized, or you stack many surfaces. A single Brewster reflection is the cleanest polarizer you can make and one of the dimmest.

What is the pseudo-Brewster angle, and why do metals have one instead?

A metal has a complex refractive index — it absorbs — and a complex index prevents the p-reflection coefficient from ever reaching an exact zero at a real angle of incidence. Instead $|r_p|$ dips to a nonzero minimum at the pseudo-Brewster angle, and the reflected light there is elliptically rather than perfectly plane-polarized. That minimum's position and depth encode the metal's optical constants, which is exactly what ellipsometry exploits. So "Brewster's angle" in its clean, exact form is a property of transparent dielectrics; for absorbing media it survives only as a softened, information-rich minimum.

How is Brewster's angle related to the critical angle for total internal reflection?

They are different special angles of the same interface. Brewster's angle exists for light crossing in either direction and zeroes the p-reflection; the critical angle exists only from the denser side and marks the onset of total internal reflection. For any interface the internal Brewster angle lies below the critical angle, because $\arctan(n_2/n_1) < \arcsin(n_2/n_1)$ for any index ratio — so climbing in angle from inside the glass you pass the quiet Brewster zero first and hit the total-reflection wall later. The critical angle, the evanescent field, and the polarization phases past it are the subject of the companion article on polarization and total internal reflection.

Can Brewster's angle ever appear for the s-polarization instead of the p?

Not for ordinary materials. In non-magnetic media the s-dipoles oscillate broadside to every possible reflected direction, so there is no angle at which they fall silent and $r_s$ never vanishes. The one loophole is magnetic response: if the two media differ in magnetic permeability, the reflection zero can migrate into the s-channel, because magnetic dipoles then enter the re-radiation budget. Natural materials have negligible magnetism at optical frequencies, so this stayed a curiosity for a century — until engineered metamaterials made the permeability a design variable, and "generalized Brewster" surfaces that null a chosen polarization at a chosen angle became a real metasurface product.