Refraction That ClicksEpisode 4 / Reflection Comes from Mismatch

Reflectance is pure impedance mismatch — and total internal reflection is tunnelling, done with light

Reflection Comes
from Mismatch Reflection happens not because matter is there but because there is a mismatch.
It vanishes at Brewster's angle, and it vanishes if you match the index.
And beyond a surface that is totally reflecting, the wave is still there — \(T\propto e^{-2\kappa d}\).

Tools needed: Episode 1's phase, Episode 3's evanescent wave, the wave equation Core of this episode: tunnelling, done with light

We have been talking about refraction, but something else always happens at a boundary: reflection. And there is exactly one satisfying principle here. Reflection happens not because matter is present, but because there is a mismatch. Match the indices exactly and the boundary reflects nothing at all, even though it is still there. Anti-reflection coatings and Brewster's angle are both corollaries of that single line. But the real business of this episode is total internal reflection — light is supposed to transmit nothing, and yet an \(e^{-\kappa z}\) wave is seeping beyond the boundary. Put a second piece of glass right next to it, and light gets through. The transmission is \(T\propto e^{-2\kappa d}\) — the same equation as Episode 1 of the sister series Tunneling That Clicks, with different letters. Tunnelling, done with light. Newton saw it in 1704.

01Reflectance is fixed by mismatch alone

Fresnel reflection at normal incidence

Matching electric and magnetic fields across the boundary gives the amplitude reflection coefficient

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

Air (1) to glass (1.5) gives \(r=-0.2\) and \(R=4\%\). A pane of glass has two surfaces, so it loses about 8%.
If \(n_1=n_2\), then \(r=0\). The boundary is there, and it does not reflect. What causes reflection is not matter but a difference.

About the sign

For \(n_2>n_1\), \(r<0\): the reflected wave is shifted by \(\pi\). The same thing as shaking a rope tied to a wall — the returning pulse is inverted. Going dense to rare (glass to air), it is not.

Anti-reflection coating — split the difference in two A large difference reflects, so insert a step in the middle. With a film of thickness \(\lambda/4n_2\), the reflections off its front and back come back exactly out of phase and cancel.
The perfect-cancellation condition is \(n_{\rm film}=\sqrt{n_{\rm air}n_{\rm glass}}=\sqrt{1.5}=1.225\). The closest usable material is magnesium fluoride, \(n=1.38\), which takes the reflection from 4% down to about 1.4%. The faint purple or green sheen on spectacles and camera lenses is the colour the coating failed to cancel.

02Brewster's angle — a dipole does not radiate along its own axis

Come in at an angle and polarization matters. For light polarized in the plane of incidence (p-polarization), there is an angle at which the reflection drops exactly to zero.

Brewster's angle
$$\tan\theta_B=\frac{n_2}{n_1}\qquad\text{(air to glass at 1.5 gives }\theta_B=56.3^\circ\text{)}$$

Why it vanishes — recall Episode 1. Both the reflected and the transmitted wave are light emitted by oscillating dipoles inside the material. At \(\theta_B\) the refracted and reflected rays are exactly 90° apart, so for p-polarization the dipole's oscillation direction coincides exactly with the reflected ray direction.

And a dipole does not radiate along its own axis (the intensity goes as \(\sin^2\theta\)). There is nobody to emit toward, so the reflected wave cannot exist.

The upshot is that light reflected near \(\theta_B\) is almost purely s-polarized. Glare off a road or a water surface is horizontally polarized for this reason. Polarizing sunglasses are a sheet that cuts exactly that direction.

03Even in total reflection, the wave is on the other side

Going from dense to rare (glass to air) at increasing angle, Snell's law \(n_1\sin\theta_1=n_2\sin\theta_2\) begins to demand \(\sin\theta_2>1\). No such angle exists. That is total internal reflection, with the boundary at

The critical angle, and beyond $$\sin\theta_c=\frac{n_2}{n_1}\qquad\text{(glass to air gives }\theta_c=41.8^\circ\text{)}$$

Beyond \(\theta_c\), the component of the wavevector normal to the boundary becomes imaginary:

$$k_z=\frac{\omega}{c}\sqrt{n_2^2-n_1^2\sin^2\theta}\ \longrightarrow\ i\kappa,\qquad \kappa=\frac{2\pi}{\lambda_0}\sqrt{n_1^2\sin^2\theta-n_2^2}$$

So \(e^{ik_zz}\to e^{-\kappa z}\): not a travelling wave but an exponentially decaying one, called an evanescent wave.
Numbers: at \(\lambda_0=550\) nm, glass (1.5) to air, \(\theta=45^\circ\), the decay length \(1/\kappa\approx\) 248 nm. Less than a wavelength.

"Total" reflection — but is the energy accounted for? Yes. In steady state the evanescent wave carries zero time-averaged energy flux normal to the boundary (the electric and magnetic fields are 90° out of phase). Energy goes and comes back.
So the reflectance is exactly 1. The field is over there, but no energy has crossed — which sets up the next section. Put something there and the story changes.

04Add a second piece of glass and light gets through — optical tunnelling

Bring a second piece of glass within a few hundred nanometres of a totally reflecting surface. Light gets through. Close the gap and more gets through; open it and the transmission falls exponentially.

Core of this episode — the same equation, different letters
Electron tunnelling (Tunneling Ep. 1)Total reflection (here)
The wavewavefunction \(\psi\)electric field \(E\)
Why it is forbidden\(E\(n_1\sin\theta>n_2\) (too steep an angle)
Decay constant\(\kappa=\dfrac{\sqrt{2m(V-E)}}{\hbar}\)\(\kappa=\dfrac{2\pi}{\lambda_0}\sqrt{n_1^2\sin^2\theta-n_2^2}\)
Transmission\(T=\left[1+\dfrac{(k^2+\kappa^2)^2}{4k^2\kappa^2}\sinh^2\kappa d\right]^{-1}\ \xrightarrow{\ \kappa d\gg1\ }\ \propto e^{-2\kappa d}\)  identical
Decay length~0.2 nm (\(V-E=1\) eV)~248 nm (visible, 45°)

These are not merely similar equations. They are the same equation. Both describe a region where the wave equation has \(k^2<0\). The difference is that the decay length is over a thousand times larger for light, which is why you can do the experiment with two slabs of glass held by hand.

Newton saw this in 1704 In Opticks, Newton records pressing two prisms together — light passes through to the far side as they approach contact, even though the surface should be totally reflecting. That is more than 220 years before quantum mechanics.
The modern name is frustrated total internal reflection (FTIR). It has many uses: the optical fingerprint sensor in a smartphone, beam splitters, total-internal-reflection fluorescence microscopy (TIRF, which illuminates only the first 100 nm above an interface), and near-field scanning optical microscopy, which beats the diffraction limit. All of them are instruments for picking up the seeping wave.

05Play with it — opening the gap

Figure: left — the evanescent wave e^(−κz) seeping past a totally reflecting glass surface. Place a second slab at gap d and only the amplitude surviving there crosses. Right — transmission T on a log scale (exact formula). The slope is −2κ — the same shape of plot as Episode 1 of "Tunneling That Clicks"
evanescent wave e^(−κz) transmission T glass

06So the "quantumness" of tunnelling is not in the exponential

The main point of this episode

Electron tunnelling feels strange mostly because of the phrasing "it gets over the wall without enough energy." But look at the table above — exactly the same thing happens with light, and nobody finds that strange.

Because light is a wave to begin with. That a wave goes as \(e^{-\kappa z}\) in a forbidden region is just the solution of a differential equation. There is nothing quantum in the exponential at all.

What is quantum is that the electron is a wave — that one point alone. Grant that, and the rest is the same classical wave physics Newton saw in 1704.
— In this collection's vocabulary, this is the second grade of "similar": different solutions of one framework. The same wave equation, in the region where \(k^2<0\).

◇ ◇ ◇
The honest line

Established: the normal-incidence Fresnel coefficient \(r=(n_1-n_2)/(n_1+n_2)\) and glass's 4% reflection; the anti-reflection condition \(n=\sqrt{n_1n_3}\) at \(\lambda/4\) thickness, and MgF₂ (1.38) taking 4% to about 1.4%; Brewster's angle \(\tan\theta_B=n_2/n_1\) and the dipole-axis explanation; the critical angle \(\sin\theta_c=n_2/n_1\); the evanescent wave \(e^{-\kappa z}\) with \(\kappa=(2\pi/\lambda_0)\sqrt{n_1^2\sin^2\theta-n_2^2}\); zero time-averaged normal energy flux in steady state; frustrated total internal reflection and the fact that the transmission through a symmetric gap has exactly the form of the quantum rectangular barrier; Newton's two-prism observation in Opticks (1704); TIRF, near-field microscopy and optical fingerprint sensors. All standard optics.

Caveats: (1) The transmission in the table and figure is for s-polarization (field perpendicular to the plane of incidence). For p-polarization the prefactor changes, but the \(e^{-2\kappa d}\) exponent is the same. (2) The figure idealises to non-absorbing, non-dispersive, perfectly flat surfaces and a monochromatic plane wave; real glass has surface roughness of a few nanometres, which dominates the near-contact behaviour. (3) "The same equation" means the same mathematical form. Photon tunnelling and electron tunnelling are not the same phenomenon; what is shared is the structure of crossing a region where the wave equation has \(k^2<0\) (the second grade, in the text's phrasing). (4) Brewster's angle gives zero reflection only for p-polarization; s-polarization does not vanish, and magnetic materials (\(\mu\neq1\)) change the condition. (5) "No energy crosses" for the evanescent wave is a statement about the steady-state time average; transiently, energy does flow both ways. (6) The 0.2 nm electron decay length is for \(V-E=1\) eV and varies strongly with barrier height.

Exercises
  1. How can a boundary exist and reflect nothing?
    Show answer
    Because what causes reflection is not matter but a mismatch in refractive index (impedance). \(r=(n_1-n_2)/(n_1+n_2)\) is zero when \(n_1=n_2\); index-matching oil makes a joint in glass disappear. An anti-reflection coating splits one large difference into two small ones that cancel out of phase.
  2. Explain in terms of dipoles why p-polarization does not reflect at Brewster's angle.
    Show answer
    At \(\theta_B\) the refracted and reflected rays are 90° apart, so for p-polarization the dipole's oscillation axis coincides with the reflected direction. A dipole does not radiate along its own axis (intensity \(\propto\sin^2\theta\)), so there is nothing to produce the reflected wave. A direct consequence of Episode 1's "refraction is forward scattering."
  3. In total reflection, why is the reflectance 1 even though there is a field on the far side?
    Show answer
    Because the evanescent wave carries zero time-averaged energy flux normal to the boundary (fields 90° out of phase): energy goes and comes back. Until you put a second slab there, giving the energy somewhere to go — that is FTIR.
  4. How far does the analogy between electron tunnelling and total reflection go?
    Show answer
    The transmission formula is literally identical: \(T=[1+\frac{(k^2+\kappa^2)^2}{4k^2\kappa^2}\sinh^2\kappa d]^{-1}\to\propto e^{-2\kappa d}\). What is shared is the structure of crossing a region where the wave equation has \(k^2<0\) — the second grade of "similar" in this collection's vocabulary. The quantum part of tunnelling is not the exponential; it is only that the electron is a wave. The decay lengths differ by over a thousand (0.2 nm vs 248 nm), which is why light lets you do it by hand.

Episode 4 summaryMismatch makes reflection; seepage makes transmission

Reflectance is fixed by index mismatch alone (\(r=(n_1-n_2)/(n_1+n_2)\); 4% for glass). Zero difference, zero reflection, boundary or not. Anti-reflection coatings split the difference in two and cancel out of phase — \(n=\sqrt{n_1n_3}\) is ideal, MgF₂ gets 4% down to 1.4%. At Brewster's angle (\(\tan\theta_B=n_2/n_1\), 56.3° for glass) p-polarization vanishes because a dipole does not radiate along its own axis — a direct consequence of Episode 1's "refraction is forward scattering."

Past the critical angle you get total internal reflection, but an evanescent wave \(e^{-\kappa z}\) still seeps beyond the boundary (decay length 248 nm for visible light at 45°). In steady state no energy crosses — until you put a second slab there. Then light does get through, with transmission

\(T=\left[1+\frac{(k^2+\kappa^2)^2}{4k^2\kappa^2}\sinh^2\kappa d\right]^{-1}\ \propto\ e^{-2\kappa d}\)

That is Episode 1 of Tunneling That Clicks with different letters. Newton saw it with two prisms in 1704 — 220 years before quantum mechanics. Which licenses the conclusion: the "quantumness" of tunnelling is not in the exponential. What is quantum is only that the electron is a wave.

This document is Episode 4 of the "Refraction That Clicks" series, a reading for physics-loving high-schoolers and undergraduates. The Fresnel coefficients, the anti-reflection condition, Brewster's angle and the dipole-radiation explanation, the critical angle, the evanescent wave and steady-state energy flux, frustrated total internal reflection and the identity of its transmission with the quantum rectangular barrier, Newton's two-prism observation in Opticks (1704), and applications to TIRF, near-field microscopy and fingerprint sensors are all standard optics. That the transmission formula is for s-polarization, that the figure idealises away the few-nanometre surface roughness of real glass, that "the same equation" refers to mathematical form and that photon and electron tunnelling are not the same phenomenon, that only p-polarization vanishes at Brewster's angle, that "no energy crosses" is a steady-state time average, and that the electron decay length depends on barrier height, are all stated in "The honest line" above. — To print, use your browser's Print and "Save as PDF" (sliders freeze and answers are hidden in the print version). Neighbours: Episode 3, When the Refractive Index Is Less Than One / Episode 5, Refraction and Absorption Are One Function / Contents / sister series Tunneling That Clicks, Episode 1.

Print / PDF: ⌘+P (Ctrl+P on Windows). On screen, move the gap d and watch the marker travel along the log plot, landing on a straight line of slope −2κ. Bring the incidence angle toward the critical angle (41.8°) and κ shrinks, so the wave reaches much further. "Show answer" reveals the solutions.