Optics 088 · Waveguides, structured light, and modern optics

Boundary Wave Tunnel

How can light still get through a narrow gap when it should be totally internally reflected? Past the critical angle, the light beyond the surface does not travel away; it becomes an evanescent field that fades exponentially into the gap. If a second medium lies within a few decay lengths, that field couples into it and some light gets through again. This is frustrated total internal reflection, and the transmission falls quickly as the gap widens.

A 3D cutaway of two prisms shows fields found by matching complex Maxwell boundary conditions. Change the air gap, compare the s and p fields and their different decay depths, check that the Poynting flux is conserved and freeze field samples.

Transmitted normal power

T

$0.50$

Reflected normal power

R

$50\%$

Lossless power balance

R+T

$0.00\pi$

Model regime $\text{valid model regime}$

Model checked against analytic results

Boundary Wave Tunnel

Actual normal air gap; paired prisms only

d

250\,\mathrm{nm}

Internal incidence angle from the boundary normal

\theta

50^\circ

Vacuum wavelength; fixed assumed indices

\lambda_0

633\,\mathrm{nm}

Polarization: zero s, one p

q

0

Second prism: zero removed, one present

m

0

Distance of the ideal field sample from the first boundary

z_p

150\,\mathrm{nm}

The field can reach into the air without carrying power away from the surface. Bring a second prism close and the boundary conditions change, so power can cross the gap. Move it away again, compare s and p polarization, then sample the field with the cursor. The cursor reads the field without disturbing it, which a real probe could not do.

The prisms are modeled as infinite, lossless, parallel half-spaces with assumed fixed indices and ideal internal ports. The finite cutaway, the slowed phase motion and the cursor that does not disturb the field are schematic. Scattering by a real probe, a finite beam and the outer prism faces are not modeled, and the Lab says nothing about tunnelling time.

Physics tutorial

Audit the physics of Boundary Wave Tunnel

BackgroundBeyond the critical angle at a glass–air interface, the wavevector component normal to the surface becomes imaginary: $\kappa=k_0\sqrt{n_g^2\sin^2\theta-1}$ . The air then holds a nonzero evanescent electric field, but a single decaying wave carries no time-averaged power across the surface. Bringing in a second glass half-space changes the whole boundary-value problem.

Why it mattersHow can light still get through a narrow gap when it should be totally internally reflected?

Start with the essentials

Focus question: How can light still get through a narrow gap when it should be totally internally reflected?
One-sentence intuition: Transmission has to be computed from the electric and magnetic fields together, not read off from how bright the field looks: $\langle S_z\rangle=\tfrac12\operatorname{Re}(\mathbf E\times\mathbf H^*)_z$ . With two boundaries, the solution contains both evanescent terms and can carry power across the air gap, even though no geometric ray crosses it.

Core mathematical model

Coordinates, indices and decay conventions

k_x=k_0n_g\sin\theta,\quad k_{z,a}^2=k_0^2-k_x^2,\quad\theta_c=\arcsin(1/n_g)

The normal axis Z points from the first glass into the air. X lies along the surface in the plane of incidence, and nothing varies along Y. The angle is measured from the normal, not from the surface. Both glasses have an assumed fixed index of 1.515, and air has index one. The time dependence is an exponential with negative frequency. Below the critical angle the wave in air propagates instead of decaying exponentially.

Penetration depth for amplitude and for intensity

E(z)\propto e^{-\kappa z},\quad |E(z)|^2\propto e^{-2\kappa z},\quad\delta_E=\kappa^{-1},\quad\delta_I=(2\kappa)^{-1}

These depths describe how far the field reaches into the air behind a single totally reflecting interface. The intensity depth is half the amplitude depth. They do not give the exact power transmittance of two prisms, and the field between two prisms need not follow a single decaying exponential. At or below the critical angle there is no finite evanescent decay depth.

Both polarizations satisfy Maxwell boundary conditions

\begin{pmatrix}U\\V\end{pmatrix}_{z+d}=\begin{pmatrix}\cos\delta&i\sin\delta/Y\\iY\sin\delta&\cos\delta\end{pmatrix}\begin{pmatrix}U\\V\end{pmatrix}_{z},\quad Y=w\frac{k_z}{k_0}

For s polarization, U is the electric field component along the invariant direction and w is one. For p polarization, U is the magnetic field along that direction, scaled by the vacuum impedance, and w is the inverse relative permittivity. Its partner V is the weighted derivative divided by the imaginary vacuum wavenumber. Both U and V are continuous across each interface. Taking the sine-over-wavenumber limit analytically keeps the calculation finite at the critical angle.

Power at the input and output ports

T=|t|^2,\quad R=|r|^2,\quad R+T=1,\quad S_z/S_{z,\rm in}=\frac{\operatorname{Re}(UV^*)}{Y_g}

With two prisms, both ports sit in the same glass, so the squared magnitude of the scalar amplitude equals the power fraction. For a single interface, transmission also includes the real admittance ratio of air. A lone evanescent term has companion fields a quarter cycle out of phase, so its normal flux is zero. Balanced power does not mean the fields vanish.

Common difficulties

Can the brightness of an evanescent field be read as transmission?

Typical misconceptionThe field intensity decays exponentially, so the light that gets through two prisms must follow exactly that exponential.

Better mental modelNo. Transmission depends on both interfaces, on polarization and on how the fields match coherently at each boundary. Behind a single interface, the tail carries no power outward, even where its electric intensity is bright. The exponential decay for a thick gap is an asymptotic limit; it does not replace the full solution for a thin gap.

Run the experiment

01
Scene 1: Single-interface TIR
Start with the single boundary. Compare the nonzero electric field in the air with the normal energy flux, which is zero. Move the ideal sampling cursor: neither output power should change.
What to observe: With zero gap, all the light goes through and none reflects. Beyond the critical angle, a thick gap approaches total internal reflection at a single interface. Below the critical angle, the same solver shows lossless Fabry–Perot fringes.
02
Scene 2: Ideal near-field sample
Bring in the second prism and narrow the gap. Compare the s and p transmission curves, the electric field matched at the boundaries and the normal Poynting component, which is conserved.
What to observe: The field in the air looks brighter or dimmer with polarization because the full electric field has different components in each case. For p polarization the scalar variable is magnetic, so its squared magnitude is not the electric intensity.
03
Scene 3: Two-boundary tunnelling
Go below the critical angle, then set the gap to zero. You will see ordinary multiple-reflection resonances, and at zero gap the transmission of uniform glass. Freeze a complex scan before changing any control.
What to observe: A real near-field probe disturbs the boundary problem. The ideal cursor here reports a field value without taking any energy away, so it cannot predict the power a probe collects, its scattering or its back-action.

Boundary Wave Tunnel

Audit the physics of Boundary Wave Tunnel

Core mathematical model

Coordinates, indices and decay conventions

Penetration depth for amplitude and for intensity

Both polarizations satisfy Maxwell boundary conditions

Power at the input and output ports

Common difficulties

Can the brightness of an evanescent field be read as transmission?

Run the experiment

Scene 1: Single-interface TIR

Scene 2: Ideal near-field sample

Scene 3: Two-boundary tunnelling

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