Stokes Relations and Phase Change on Reflection

Every coated lens you own is a monument to getting these signs right. An anti-reflection coating works only because the designer tracks the sign of $r$ at each interface and arranges the reflected amplitudes to arrive $\pi$ out of phase and cancel; flip one sign and the "anti-reflection" layer becomes a mirror. The same bookkeeping scales up to the dozens of layers in a laser mirror or a dielectric bandpass filter, where the round-trip phase must be accounted to a fraction of a radian. The two-port generalisation is quietly everywhere in precision measurement. A lossless beam splitter must impose a definite phase between its transmitted and reflected outputs — the Stokes relations for a four-terminal device — and that is why the two output ports of a Michelson or Mach–Zehnder interferometer are complementary, why a gravitational-wave detector can hold one port dark and read the signal on the other, and why, in the quantum regime, a beam splitter entangles the photons that meet on it. The humble minus sign Stokes extracted from a soap film is, at bottom, the conservation of probability.

Picture a beam of light striking a glass surface, filmed by a camera. Part reflects, part goes through. Now run the film backwards. The two beams that had left the surface — the reflected one and the transmitted one — are now arriving at it, heading back the way they came. If light in glass truly runs the same forwards and backwards (and in a lossless medium it does, because there is no friction to tell time's direction), then these two incoming beams must do exactly one thing: rebuild the single incident beam we started with, and send nothing off in the wrong direction. Killing that unwanted "wrong-direction" beam is what forces a minus sign into the reflection coefficient. It is a remarkably cheap way to learn something true.

Think of the interface as having two faces — a face looking up into the light medium and a face looking down into the dense one. A wave that reflects off the dense face and a wave that reflects off the light face cannot both stay in step with their incident fields, because if they did, their coherent sum in the reversed movie would produce that forbidden extra beam. Nature's resolution is to make one of the two reflections lag the other by exactly half a wavelength. Which one? The mechanical analogy is exact: a wave pulse running along a light rope toward a junction with a heavier rope reflects inverted — that is external reflection, into the denser (slower) medium, phase $\pi$. Running the other way, toward a lighter rope, the reflected pulse comes back upright — internal reflection, no shift.

Because the whole argument rests on symmetry rather than dynamics, the Stokes relations are not a competing theory of reflection but a consistency check on it. Whatever the detailed Fresnel coefficients turn out to be — for any polarization, at any angle of incidence — they must satisfy $r' = -r$ and $tt' = 1 - r^2$. If a calculation of a coating or a cavity ever violates them, the calculation is wrong.

The two relations are stated first because everything else is their consequence. They connect the four amplitude coefficients of a single interface: reflection and transmission from side 1 ($r,t$) and from side 2 ($r',t'$). Read them as bookkeeping rules that any lossless boundary must obey, fixed by reversibility and nothing else.

where: $r$ — amplitude reflection coefficient for a wave going from medium 1 into medium 2; $r'$ — the same for a wave going from medium 2 into medium 1; $t, t'$ — the corresponding amplitude transmission coefficients. The first relation says the two reflections are equal in magnitude and opposite in sign (a $\pi$ phase difference); the second ties the product of the two transmissions to the reflection. The sign of $r$ carries the physics. At normal incidence the amplitude coefficients take their simplest form, and the sign of $r$ is set purely by which medium is denser. This is the equation that decides "upright or flipped".

where: $n_1$ — index of the incidence medium; $n_2$ — index of the transmission medium. When $n_1 < n_2$ (external reflection, e.g. air into glass) the numerator is negative, so $r < 0$: the reflected field is inverted, the half-wave loss. When $n_1 > n_2$ (internal reflection) $r > 0$: no inversion. Note the tidy identities $t = 1 + r$ and $t' = 1 - r$, from which $tt' = 1 - r^2$ falls out immediately. It is worth being explicit that the amplitude relation $tt' = 1 - r^2$ is not the energy statement $R + T = 1$. Energy transmittance carries an impedance factor that the amplitude relation does not, which is why $t'$ can legitimately exceed one while no energy is manufactured.

where: $R$ — power reflectance; $T$ — power transmittance; $\theta_i, \theta_t$ — incidence and refraction angles. The factor $n_2\cos\theta_t / n_1\cos\theta_i$ converts the squared amplitude into a power ratio by accounting for the change in beam cross-section and wave impedance across the boundary. Its absence from $tt' = 1 - r^2$ is exactly why that relation and energy conservation are different truths.

1 · The amplitudes and the sign at an air–glass surface (confidence builder)

A wave in air ($n_1 = 1$) strikes crown glass ($n_2 = 1.52$) at normal incidence. Find the amplitude reflection and transmission coefficients on both sides, the reflectance, and verify the two Stokes relations numerically.

Step 1. Reflection from air: $r = (n_1 - n_2)/(n_1 + n_2) = (1 - 1.52)/(2.52) = \mathbf{-0.206}$. The minus sign is the half-wave loss — external reflection inverts the field.

Step 2. Reflectance $R = r^2 = (-0.206)^2 = \mathbf{0.0426}$, so about $4.3\%$ of the intensity reflects at each air–glass surface (worth remembering: it is why an uncoated lens flares).

Step 3. Transmission from air $t = 2n_1/(n_1+n_2) = 2/2.52 = 0.794$; reflection from the glass side $r' = (n_2-n_1)/(n_2+n_1) = +0.206$; transmission from glass $t' = 2n_2/(n_1+n_2) = 3.04/2.52 = 1.206$.

Step 4. Check Stokes: $r' = +0.206 = -r$ ✓, and $tt' = 0.794 \times 1.206 = 0.958 = 1 - r^2$ ✓. Note $t' > 1$: an amplitude, not a power — no energy is invented, since $R + T = 0.0426 + 0.9574 = 1$. r = −0.206, R = 4.26%, r′ = +0.206, tt′ = 0.958.

2 · The black film — why a soap bubble goes dark before it bursts (exam level)

A vertical soap film ($n = 1.33$) drains until its top is only a few nanometres thick. Using the half-wave loss, show that the thinnest part reflects almost no light, and estimate the reflectance at $d = 5$ nm and $d = 20$ nm for green light ($\lambda = 550$ nm), against the $4.3\%$ of a single surface.

Step 1. Two beams reflect: from the top (external, air→film, amplitude $r_1 = (1-1.33)/(1+1.33) = -0.142$) and from the bottom (internal, film→air, amplitude $-r_1 = +0.142$, by $r' = -r$). The bottom beam also picks up the geometric phase $\delta_g = (2\pi/\lambda)\,2nd$.

Step 2. Add them: $r_{\text{tot}} = r_1 + (1 - r_1^2)(-r_1)\,e^{i\delta_g}$. As $d \to 0$, $\delta_g \to 0$ and $r_{\text{tot}} \to r_1[1 - (1 - r_1^2)] = r_1^3$, a tiny number. The reflectance collapses to $R(0) = r_1^6 \approx 8\times10^{-6}$ — essentially black.

Step 3. For small $d$ the reflectance grows quadratically: $R \approx r_1^2(1 - r_1^2)^2\,(4\pi n d/\lambda)^2$. At $d = 5$ nm this gives $R \approx \mathbf{0.05\%}$; at $d = 20$ nm, $R \approx \mathbf{0.7\%}$ — still far below the $4.3\%$ of a lone surface.

Step 4. The physics: where the film is thinner than about $\lambda/20$, the geometric path difference is negligible and the half-wave loss alone governs, driving the two reflections toward cancellation. That is the black cap you see at the top of a draining bubble in the seconds before it pops — a direct, everyday sighting of $r' = -r$. R(5 nm) ≈ 0.05%, R(20 nm) ≈ 0.7%, both ≪ 4.3%.

The Stokes relations are usually attributed to a person, but their real author is a principle: the time-reversal invariance — reversibility — of light in a transparent medium. It is worth honouring the idea rather than the man, because the same reversibility argument reaches far beyond this one interface, all the way to the unitarity of quantum scattering.

Texts. Hecht, Optics, 5th ed., §4.3 (Stokes relations, phase change on reflection) and §9.4 (Newton's rings) · Born & Wolf, Principles of Optics, §1.5 (Fresnel formulae and phase) and §7.5 · Loudon, The Quantum Theory of Light, 3rd ed. (beam-splitter relations) · Saleh & Teich, Fundamentals of Photonics, 3rd ed., ch. 6 (reflection and refraction).

Provenance. The reversibility derivation and the sign convention follow Hecht §4.3 and Born & Wolf §1.5; Stokes' original argument is On the perfect blackness of the central spot in Newton's rings (1849). A verifier should reproduce $r' = -r$ and $tt' = 1 - r^2$ from the time-reversal matching equations; confirm the Fresnel s-coefficient obeys $r' = -r$ under $n_1 \leftrightarrow n_2$; check $r \approx -0.206$, $R \approx 4.26\%$ for air→crown-glass ($n = 1.52$); and verify the thin-film reflectance vanishes as $d \to 0$. Index note: this article standardises on crown glass $n = 1.52$ ($r = -0.206$, $R = 4.26\%$); the round $n = 1.5$ used in many texts gives $r = -0.20$, $R = 4.00\%$ — both reproduced in the verify script.

Prerequisites. Superposition Principle of Electromagnetic Waves · Optical Path Difference and Phase Retardation · Two-Wave Interference and the Interference Term. Leads to. Thin-Film Interference · Fringes of Equal Inclination and Equal Thickness · Wavefront-Splitting and Amplitude-Splitting Interferometers.