Polarization on Reflection and Total Internal Reflection

Every reflection is a polarizer you did not ask for. Because a boundary treats the two polarization channels — s (field perpendicular to the plane of incidence, TE) and p (field in the plane, TM) — with different amplitude coefficients, any beam that bounces off a surface comes back with its polarization state rearranged. Below the critical angle it acts as a partial polarizer, perfect at Brewster's angle; beyond the critical angle it turns into something subtler — a lossless device that leaves the amplitudes untouched but stamps a controllable phase difference between s and p, turning linear light circular with nothing but a block of glass. This article stands at the meeting point of two owned results it recalls by name. The critical angle $\theta_c = \arcsin(n_2/n_1)$ — the geometry of when light is trapped — belongs to aq-optics-total-internal-reflection and is recalled, not re-established here. That past $\theta_c$ the reflectance is exactly unity in both channels ( $|r_s| = |r_p| = 1$ ), that the transmitted field survives only as an evanescent skin, and that total reflection imprints a polarization-dependent phase, are the electromagnetic content, recalled from aq-em-brewster-angle-total-internal-reflection. What this article owns is the polarization payoff: the reflected degree of polarization below $\theta_c$ (complete at Brewster, recalled from the Brewster article), and above all the differential phase $\delta = \delta_p - \delta_s$ of total internal reflection, together with its flagship application — the Fresnel rhomb, which chains two internal reflections into a broadband quarter-wave plate.

Key takeaways

Reflection is diagonal in the $(p,s)$ basis: it multiplies the p and s components by $r_p$ and $r_s$ , so the reflected polarization state generally differs from the incident one.
Below the critical angle these coefficients are real: reflection partially polarizes the beam, completely at Brewster's angle (recalled from the Brewster article).
Past the critical angle both channels totally reflect, $|r_s| = |r_p| = 1$ (recalled from aq-em-brewster-angle-total-internal-reflection), but with a relative phase $\delta = \delta_p - \delta_s$ that makes the interface a lossless retarder.
The differential phase obeys $\tan(\delta/2) = \cos\theta\,\sqrt{\sin^2\theta - n^2}/\sin^2\theta$ with $n = n_2/n_1 < 1$ ; for Fresnel's glass ( $n_1 = 1.51$ ) it peaks at $45.9°$ near $51.3°$ .
Two total internal reflections at $\delta = 45°$ each (at $48.6°$ or $54.6°$ ) sum to $90°$ — a Fresnel rhomb, an achromatic quarter-wave plate that turns linear light circular.
Key idea: below the critical angle reflection is a partial polarizer (perfect at Brewster); above it, total reflection is a lossless retarder whose differential phase $\delta_p - \delta_s$ — engineered to $45°$ per bounce — builds the Fresnel rhomb.

The core insight

A reflection acts separately on the s and p components of a beam. Below the critical angle the two reflect with unequal real amplitudes, so reflection partially polarizes the light and, at Brewster's angle, polarizes it completely. Above the critical angle both amplitudes have unit magnitude — total reflection — but the p and s waves emerge with different phases, and their difference $\delta = \delta_p - \delta_s$ turns the totally reflecting surface into a lossless wave plate. Setting $\delta = 45°$ per bounce and using two bounces makes a Fresnel rhomb: a broadband quarter-wave retarder.

Where this lives today

The differential-phase side of this article is quietly everywhere polarization must be controlled over a wide band. Fresnel rhombs and their descendants remain the go-to achromatic quarter- and half-wave retarders in broadband, ultrafast, and ultraviolet instruments, where a birefringent plate's sharp $1/\lambda$ chromaticity is a liability; a single glass block delivers a retardance flat across the spectrum. The same TIR phase, generalized, also sets the mode structure of integrated waveguides on every silicon-photonics chip. The polarizing side lives on in ellipsometry, which reads the complex ratio $r_p/r_s$ of a surface to extract film thickness and optical constants, and in polarizing beam-splitter cubes, engineered versions of the single-surface $r_p/r_s$ asymmetry. And the evanescent field of the portrait is now an instrument in its own right, from TIRF microscopy to ATR spectroscopy and frustrated-TIR couplers.

The physical picture

Treat the interface as two independent one-dimensional mirrors — one for the s field, one for the p field — each with its own complex reflectivity. Split any incident beam into these two channels, multiply by $r_s$ and $r_p$ , and recombine; because the channels have different gains and phases, the recombined light is a reshaped ellipse. Below the critical angle the gains differ but the phases are only $0$ or $\pi$ , so reflection partially polarizes the beam (completely at Brewster). Above the critical angle the gains equalize at unity — nothing is lost — but a genuine phase difference opens between the channels, and that is where the interesting engineering lives.

Figure

Three regimes of one glass-to-air interface: below the critical angle reflection partially polarizes, at Brewster it is 100% s-polarized, and past the critical angle total reflection with a differential phase turns linear light elliptical

The third panel is the one this article is built around. Once the incidence passes $\theta_c$ there is no transmitted ray to carry energy away, so both channels reflect completely — yet the s and p waves dive into the evanescent skin and re-emerge with different phases. The reflected beam carries the same two amplitudes it arrived with but a new phase difference between them: send in light linearly polarized at $45°$ to the plane of incidence and it comes back elliptical. Tune the incidence angle and you tune the ellipse — a wave plate made of geometry.

Figure

Differential total-internal-reflection phase for glass: for Fresnel's n = 1.51 the retardance peaks at 45.9° near 51.3°, and the two angles giving exactly 45° per bounce — 48.6° and 54.6° — are the Fresnel-rhomb operating points

The curve is the whole rhomb in one plot. A single bounce peaks at only $45.9°$ near $51.3°$ for Fresnel's $n = 1.51$ glass, so two reflections at $\delta = 45°$ apiece are needed to sum to the $90°$ of a quarter-wave plate. The dashed curve shows the razor-thin margin Fresnel worked with: ordinary $n = 1.50$ glass peaks at just $45.2°$ , its two operating angles nearly merging — a degree of index is the difference between a comfortable design and none at all.

The core equations

Two owned results anchor this article, both recalled by name. The critical angle $\theta_c = \arcsin(n_2/n_1)$ is established in aq-optics-total-internal-reflection and used here without re-derivation; for Fresnel's rhomb glass, $n_1 = 1.51$ into air, it is $\arcsin(1/1.51) = 41.5°$ . And that past $\theta_c$ the reflectance is unity with a polarization-dependent phase is the electromagnetic result of aq-em-brewster-angle-total-internal-reflection. Below $\theta_c$ , reflection partially polarizes an unpolarized beam, and the degree of polarization peaks at Brewster's angle.

P(\theta) = \frac{|R_s - R_p|}{R_s + R_p}, \qquad P(\theta_B) = 1 \qquad (1)

where: $P$ — degree of polarization of the reflected beam for unpolarized input; $R_s, R_p$ — the two power reflectances. Physical reading: reflection is always a partial polarizer, and it becomes a perfect one only where $R_p = 0$ — Brewster's angle, recalled from the companion Brewster article. Above the critical angle both coefficients become complex numbers of unit modulus: no power is transmitted, so all of it reflects, in both channels.

r_s = e^{i\delta_s}, \qquad r_p = e^{i\delta_p}, \qquad |r_s| = |r_p| = 1 \qquad (2)

where: $\delta_s, \delta_p$ — the reflection phases of the two channels past $\theta_c$ . Physical reading (recalled): the transmitted field is evanescent and carries no net power, so reflection is total and lossless — but the physics has moved from the magnitudes, frozen at one, into the phases. The quantity polarization optics cares about is the difference of those two phases — the retardance the interface adds per bounce. Combining the individual channel phases gives a compact closed form, and this is the owned structural result of the article.

\tan\frac{\delta}{2} = \frac{\cos\theta\,\sqrt{\sin^2\theta - n^2}}{\sin^2\theta}, \qquad \delta = \delta_p - \delta_s, \quad n = \frac{n_2}{n_1} < 1 \qquad (3)

where: $\delta$ — differential phase between p and s on one total internal reflection; $\theta$ — internal angle of incidence; $n = n_2/n_1$ — index ratio (below one, dense to rare). Physical reading: $\delta$ vanishes at $\theta_c$ and at grazing and peaks between — the curve in the punchline figure — so a single bounce buys at most about an eighth-wave. A quarter-wave retarder needs $90°$ of retardance, which one bounce cannot supply; two identical bounces can, provided each is tuned to $\delta = 45°$ .

2\delta = 90° \;\Rightarrow\; \delta = 45° \;\text{per bounce} \qquad (4)

where: the two bounces occur inside one prism, the Fresnel rhomb. Physical reading: solving $\delta(\theta) = 45°$ for $n_1 = 1.51$ gives two operating angles, $48.6°$ and $54.6°$ ; a rhomb cut so the beam strikes at either angle twice turns linear light at $45°$ into circular — a quarter-wave plate (recall Wave Plates and Retarders), but achromatic.

Derivation APolarization on reflection below the critical angle

The goal is the reflected degree of polarization as a function of incidence angle, and the demonstration that it climbs from zero at normal incidence to a perfect $1$ at Brewster's angle before falling again — the phenomenology that makes a plain glass plate a partial polarizer and a Brewster-angle plate a perfect one. We work below $\theta_c$ , where the Fresnel coefficients are real, so the only phase changes are sign flips and the physics is entirely in the unequal magnitudes of $R_s$ and $R_p$ . The Brewster null is recalled from the Brewster article; the polarization-versus-angle profile it produces is assembled here.

Step 1. For real coefficients the reflected power in each channel is $R_s = r_s^2$ and $R_p = r_p^2$ . An unpolarized incident beam carries equal s and p power, so the reflected beam's polarization is the imbalance $P(\theta) = |R_s - R_p|/(R_s + R_p)$ . Why unpolarized input: it isolates the interface's own polarizing action from any bias in the source.

Step 2. At normal incidence $R_s = R_p$ (the s/p labels are meaningless there), so $P(0) = 0$ — a head-on reflection does not polarize. Why: with no plane of incidence to break the symmetry, the two channels are physically identical.

Step 3. As $\theta$ grows, $R_p$ falls toward its zero while $R_s$ rises, so the imbalance grows; at Brewster's angle $R_p = 0$ and $P = 1$ exactly — the reflected light is completely s-polarized (recalled from aq-em-brewster-angle-total-internal-reflection and the Brewster article). Why the peak is total: only $R_p$ touches zero, and a numerator with a vanishing counterpart drives $P$ to its ceiling.

Step 4. Past Brewster $R_p$ climbs again, so $P$ falls back below one, and at grazing both reflectances approach unity and $P \to 0$ once more. The full profile is a single hump peaking at Brewster — the reason a stack of plates near $\theta_B$ makes a usable polarizer while a single plate at a random angle barely polarizes. Why this matters downstream: it is the same $R_s/R_p$ asymmetry that TIR converts, above $\theta_c$ , from an amplitude effect into a pure phase effect. *Verified: $P(0) = 0$ , $P(\theta_B) = 1$ at $\theta_B = \arctan(n_2/n_1)$ , and a single-reflection reflected DoP of $0.83$ at $45°$ for air→glass ( $n = 1.51$ ), with $r_s$ and $r_p$ real and of opposite sign there, in verify_polarization-reflection-tir.py.*

Go deeper · Derivation B · 2 minThe differential phase and the Fresnel rhomb

Derivation BThe differential phase and the Fresnel rhomb

The goal is the retardance a single total internal reflection adds between the two channels — the compact law $\tan(\delta/2) = \cos\theta\,\sqrt{\sin^2\theta - n^2}/\sin^2\theta$ — and its flagship use, the Fresnel rhomb, which chains two bounces into a quarter-wave plate. We start from the recalled fact that above $\theta_c$ each Fresnel coefficient is a ratio of complex conjugates, read off and subtract the two channel phases, then invert the result at $\delta = 45°$ to place the rhomb. The critical angle and quarter-wave concept are recalled by name; the differential phase and the rhomb design are owned here.

Step 1. Above $\theta_c$ Snell's law makes $\cos\theta_t$ imaginary, so each Fresnel coefficient takes the form $r = (a - ib)/(a + ib)$ with $a, b$ real (recalled from aq-em-brewster-angle-total-internal-reflection): modulus exactly $1$ , argument $-2\arctan(b/a)$ . For the s-channel $a = \cos\theta$ , $b = \sqrt{\sin^2\theta - n^2}$ ; the p-channel carries an extra $1/n^2$ inside, so $\delta_p \ne \delta_s$ . Subtracting and simplifying gives the closed form (3). Why they differ: the boundary conditions weight the in-plane p-field by an extra index factor, so the channels dip into the evanescent skin differently.

Step 2. Read the shape. At $\theta_c$ the radical vanishes, so $\delta = 0$ ; at grazing the fraction again tends to zero; between them $\delta$ rises to a single maximum. For $n_1 = 1.51$ the peak is $45.9°$ at $\theta = 51.3°$ — a single total internal reflection retards by at most about an eighth of a wave. Why the cap sits just above $45°$ : it is an accident of the index of ordinary glass, and the whole rhomb design lives on the right side of it.

Step 3. Design the rhomb. A quarter-wave plate needs $90°$ , which one bounce cannot supply, so use two bounces at $\delta = 45°$ each. Solving $\delta(\theta) = 45°$ for $n_1 = 1.51$ gives two roots, $48.6°$ and $54.6°$ , both above $\theta_c = 41.5°$ (recalled from aq-optics-total-internal-reflection); Fresnel chose $54.6°$ , where the curve is flatter and the alignment tolerance kinder. Why two solutions: a single hump crosses any level below its peak exactly twice.

Step 4. Cut the prism for two internal reflections at that angle. Each multiplies the p component by $e^{i\delta}$ relative to s, so two bounces give $\mathrm{diag}(e^{i90°}, 1)$ up to a common phase — a quarter-wave plate. Linear light at $45°$ (equal p and s in phase) emerges with the two components a quarter-cycle apart at equal amplitude: circular polarization (recall Wave Plates and Retarders). Because $\delta$ rides only on the mild dispersion of $n$ , the retardance holds across a broad band, unlike a birefringent plate whose retardance scales as $1/\lambda$ . Why this is the payoff: it is the achromatic quarter-wave retarder, still the reference solution in broadband and ultraviolet work. *Verified: the closed form matches the phase of the raw complex Fresnel coefficients to better than $0.1°$ ; $|r_s| = |r_p| = 1$ above $\theta_c$ ; peak $45.9°$ at $51.3°$ ; $\delta = 45°$ at $48.6°$ and $54.6°$ (scipy); two $45°$ bounces on a $45°$ linear Jones vector give equal-amplitude components $90°$ apart — circular — in verify_polarization-reflection-tir.py.*

Limiting cases

Case	Result	Meaning
normal incidence	$r_s = -r_p$ , $P = 0$	the s/p distinction is only a convention; no polarizing action
Brewster's angle (below $\theta_c$ )	$R_p = 0$ , $P = 1$	reflection is a perfect s-polarizer — recalled from the Brewster article
onset of TIR ( $\theta = \theta_c$ )	$\|r_s\| = \|r_p\| = 1$ , $\delta = 0$	total reflection begins; the channels are still in phase
peak retardance ( $n_1 = 1.51$ )	$\delta = 45.9°$ at $51.3°$	the most a single glass bounce can retard — just clears $45°$
rhomb angles ( $n_1 = 1.51$ )	$\delta = 45°$ at $48.6°$ , $54.6°$	two bounces at either angle make a $90°$ quarter-wave plate
grazing ( $\theta \to 90°$ )	$\delta \to 0$	both channels saturate in phase; the retardance closes again

Worked examples

1 · Designing a Fresnel rhomb (confidence builder)

Fresnel's rhomb is a glass prism ( $n = 1.51$ ) cut so a normally entering beam makes two internal reflections. Find the operating angle that gives a quarter-wave of retardance, confirm it lies in the total-reflection range, and state what the device does.

Step 1. Each of the two bounces must contribute $\delta = 45°$ , since $2 \times 45° = 90°$ is a quarter wave. Solving $\tan(\delta/2) = \cos\theta\,\sqrt{\sin^2\theta - n^2}/\sin^2\theta = \tan 22.5° = 0.4142$ for $n = 1/1.51$ gives two roots: $\theta = \mathbf{48.6°}$ and $\theta = \mathbf{54.6°}$ .

Step 2. Check the regime. The critical angle is $\theta_c = \arcsin(1/1.51) = \mathbf{41.5°}$ (recalled from aq-optics-total-internal-reflection), and both operating angles exceed it, so each bounce is a total internal reflection with $|r| = 1$ — lossless. The single-bounce peak is $45.9°$ at $51.3°$ , so $45°$ is comfortably attainable on each side of it.

Step 3. Fresnel used the larger root, $54.6°$ , where the retardance curve is flatter and the design tolerates misalignment better; a parallelogram prism with $54.6°$ base angles sends a normally incident beam through exactly two such reflections.

Step 4. The result. Linear light entering at $45°$ leaves circularly polarized, and because the retardance rides only on the mild dispersion of $n$ it holds across the visible and into the ultraviolet, where birefringent plates fail. Operating angle 54.6° (or 48.6°), two bounces, 90° total: an achromatic quarter-wave plate.

2 · A single reflection at 45°: how polarizing, and what phase jump? (exam level)

An unpolarized beam in air strikes crown glass ( $n = 1.51$ ) at $45°$ incidence — below both Brewster's angle and, being external, any question of TIR. Find the degree of polarization of the reflected light and describe the phase behaviour of the two channels.

Step 1. Refraction: $\sin\theta_t = \sin 45°/1.51 = 0.468$ , so $\theta_t = 27.9°$ . The Fresnel amplitudes are $r_s = (\cos 45° - 1.51\cos\theta_t)/(\cos 45° + 1.51\cos\theta_t) = \mathbf{-0.307}$ and $r_p = (1.51\cos 45° - \cos\theta_t)/(1.51\cos 45° + \cos\theta_t) = \mathbf{+0.094}$ .

Step 2. Reflectances: $R_s = 0.307^2 = 0.0944$ and $R_p = 0.094^2 = 0.0089$ . The reflection is far richer in s than in p because $45°$ sits below Brewster's angle ( $56.5°$ for this glass), where $R_p$ is already small.

Step 3. Degree of polarization: $P = |R_s - R_p|/(R_s + R_p) = 0.0855/0.1033 = \mathbf{0.83}$ . A single glancing reflection off glass already delivers $83\%$ -polarized light — which is why a polarizing filter cuts window and lake glare so effectively even away from the exact Brewster angle.

Step 4. The phases. The coefficients are real here, so the only phase changes are $0$ or $\pi$ : $r_s < 0$ (a $\pi$ flip of the s-wave) while $r_p > 0$ (no flip), because $45°$ is below Brewster's angle where $r_p$ has not yet changed sign. Past Brewster $r_p$ would go negative too, its own $\pi$ jump. Reflected DoP ≈ 0.83; s-wave flips by π, p-wave does not (below Brewster).

Physics of the portrait

The star of the total-reflection regime is not an amplitude but a field that, by the letter of geometrical optics, should not exist at all.

Origins · 1 minevanescent wave and how this was discovered

evanescent wave

the field beyond the totally-reflecting surface · Fresnel's phase-on-reflection → Goos-Hänchen

Geometrical optics says that past the critical angle no light crosses the boundary — yet the electromagnetic boundary conditions forbid the field from simply stopping at the surface. The resolution is the evanescent wave: a real field that clings to the interface, oscillating along it while decaying exponentially into the forbidden medium within a fraction of a wavelength, carrying no net power across. Its fingerprints were seen long before it was understood — Newton reported light crossing the gap between two prisms held in near-contact, a leak classical total reflection could not explain. Augustin-Jean Fresnel, deriving the reflection coefficients in the 1820s, found that total reflection stamps the two polarizations with different phases and turned that discovery into his rhomb. The field's most direct proof came in 1947, when Fritz Goos and Hilda Hänchen measured that a totally reflected beam re-emerges slightly displaced along the surface — the Goos–Hänchen shift — as if it had briefly travelled inside the skin. Today that same field is an instrument, feeding TIRF microscopy, evanescent biosensors, and frustrated-reflection couplers.

History

1704

— Newton reports light crossing the gap between two prisms held in near-contact, an unexplained leak past total reflection

1808

— Malus discovers polarization by reflection

1817

— Fresnel devises his rhomb, turning linear light circular with two internal reflections

1823

— Fresnel derives the differential phase shift that total internal reflection imprints on s and p

1947

— Goos and Hänchen measure the lateral shift of a totally reflected beam, making the evanescent field visible

Going further

Texts. Hecht, Optics, 5th ed., §4.7 (total internal reflection, phase shifts, the Fresnel rhomb) · Born & Wolf, Principles of Optics, 7th ed., §1.5.4 (total reflection phase formulae) · Saleh & Teich, Fundamentals of Photonics, 3rd ed., ch. 6 (polarization on reflection) · Goldstein, Polarized Light, 3rd ed. (reflection retarders, rhomb design).

Provenance. The Fresnel coefficients, Brewster condition, and critical angle $\theta_c = \arcsin(n_2/n_1)$ follow Hecht §4.6–4.7 and Born & Wolf §1.5; the critical-angle derivation is recalled from aq-optics-total-internal-reflection, and the unit-modulus TIR reflectivity, evanescent field, and reflection phases from aq-em-brewster-angle-total-internal-reflection. The differential-phase formula $\tan(\delta/2) = \cos\theta\,\sqrt{\sin^2\theta - n^2}/\sin^2\theta$ and the rhomb design follow Born & Wolf §1.5.4. Index note: this article standardises on Fresnel's rhomb glass $n = 1.51$ (internal), giving $\theta_c = 41.5°$ , a differential-phase peak of $45.9°$ at $51.3°$ , and $\delta = 45°$ at $48.6°$ and $54.6°$ ; ordinary $n = 1.50$ glass peaks at only $45.2°$ , a margin reproduced in verify_polarization-reflection-tir.py. Fresnel's classic rhomb used $\approx 54.6°$ .

Prerequisites. Polarization of Light · Linear Polarization · Circular Polarization · Jones Calculus. Leads to. Wave Plates and Retarders · Stokes Parameters and the Poincaré Sphere · Mueller Calculus · Polarizers and Extinction Ratio.

Why does total internal reflection produce a phase shift at all if no light is transmitted?

Because the field does not stop at the surface — it continues as an evanescent wave, and the boundary conditions that match it to the reflected wave force a phase onto the reflection. Past the critical angle the transmitted cosine turns imaginary, so each Fresnel coefficient becomes a ratio of complex conjugates: magnitude exactly one (total reflection), argument nonzero (the phase). The s and p channels get different phases because the boundary conditions weight the in-plane p-field by an extra index factor. No energy crosses, yet the reflected wave is genuinely retarded — real and measurable, as the Goos–Hänchen shift and the working Fresnel rhomb both attest.

Can one total internal reflection make a quarter-wave plate?

No — a single bounce cannot supply enough retardance. For ordinary glass the differential phase $\delta$ peaks at only about $46°$ ( $45.9°$ at $51.3°$ for $n = 1.51$ ), well short of the $90°$ a quarter-wave plate needs. That is exactly why Fresnel used two reflections at $45°$ each. A single-bounce quarter-wave retarder would need a higher differential-phase peak, meaning a higher index — some infrared designs do this — but for visible-glass indices two bounces is the natural minimum. It is a neat accident that $n \approx 1.5$ glass peaks just above $45°$ , making the two-bounce rhomb possible at all.

What is the difference between the phase jump below the critical angle and the phase shift above it?

Below $\theta_c$ the reflection coefficients are real, so the only phase changes allowed are discrete: $0$ or $\pi$ (a sign flip). The s-wave flips by $\pi$ at every external reflection; the p-wave flips only past Brewster's angle, where $r_p$ changes sign. Above $\theta_c$ the coefficients are complex with unit modulus, and the phase becomes a continuous function of angle, sweeping from $0$ at the critical angle up through the rhomb values and back. That continuous, tunable phase is what makes total reflection a retarder; the discrete $0/\pi$ jumps below only flip signs and partially polarize.

Does the Fresnel rhomb displace the beam, and does that matter?

Yes — reflecting twice inside a solid prism, the beam exits parallel to but laterally offset from the entrance beam, unlike a thin wave plate that leaves the path unchanged. The offset is usually corrected by a second, mirror-image rhomb (a double rhomb) that cancels the displacement and doubles the retardance to a half-wave. The rhomb is also more angle-sensitive than a birefringent plate, since $\delta$ varies with $\theta$ , so it must be used at near-normal entry. These are the price of its great virtue — an achromatic retardance no crystal plate can match across a broad band.

Is the evanescent field the same thing that leaks a totally reflecting prism when it gets wet?

Yes — they are two views of the same field. The evanescent wave extends a fraction of a wavelength beyond the totally reflecting surface, and anything brought into that skin couples to it and spoils the total reflection. A water film or a second prism within the skin lets light cross the gap it classically could not — frustrated total internal reflection — and a droplet or fingerprint on a TIR face bleeds light out. Wetting the face also raises $\theta_c$ , so a design that totally reflected in air may not in contact with water. The field's reality is the whole basis of evanescent sensing.

How does a reflection act on the Poincaré sphere?

Below the critical angle, with unequal real coefficients, a reflection changes the two amplitudes (partially polarizing) and can add a $0$ or $\pi$ phase. Above the critical angle it is a pure retarder: the amplitudes are equal, so it only adds the relative phase $\delta$ , which on the Poincaré sphere is a rotation about the $S_1$ axis by $\delta$ , leaving $S_0$ and $S_1$ fixed and mixing $S_2$ and $S_3$ . A $90°$ rotation carries the linear-at- $45°$ point on the equator up to a pole — circular — which is the Fresnel rhomb seen in Stokes language (recall Stokes Parameters and the Poincaré Sphere).