Optics 080 · Polarization, anisotropy, and modulation

TIR Phase & Fresnel Rhomb

How can total internal reflection, which returns all the light, act as a waveplate that works across many wavelengths? Total internal reflection sends all the power back, but it delays the two polarizations by different amounts. A Fresnel rhomb uses two such reflections, which add up to a quarter-wave retardance. That delay changes only slightly with wavelength. A crystal waveplate is different: its delay, counted in waves, drifts as the wavelength changes.

Trace two reflections through exact rhomb geometry and see the complex Fresnel phase each one adds, using N-BK7 dispersion. Polarization is drawn at equal scale, and analyzer records can be frozen.

Total relative retardance

\Delta

$0.50$

Detector / prepared input power

P_D/P_0

$50\%$

Glass–air critical angle

\theta_c

$0.00\pi$

Model regime $\text{valid model regime}$

Model checked against analytic results

TIR Phase & Fresnel Rhomb

Internal angle from face normal; recuts the solid

\theta

55.35^\circ

Rhombs in series; two reflections each

N

1

Vacuum wavelength

\lambda_0

550\,\mathrm{nm}

Input angle from in-plane p toward s

\alpha

45^\circ

Analyzer angle from output p toward s

\beta

0^\circ

Inspected reflection on first rhomb

j

1

A block of clear glass with no birefringence turns linear polarization nearly circular. Follow the two total internal reflections, add a second rhomb to get back a rotated linear state, then record an analyzer scan. Recut the glass and watch the geometry, the phase and the signal change together.

Total internal reflection at the flat faces is lossless. The uncoated entry and exit faces at normal incidence lose some light, and only the direct path is kept. Field symbols are enlarged. Ghost reflections, coatings and finite-beam shifts are not modeled. The constant-birefringence reference is an ideal retarder, not quartz.

Physics tutorial

Audit the physics of TIR Phase & Fresnel Rhomb

BackgroundA block of transparent isotropic glass can change polarization without any crystal birefringence. At each total reflection the two polarizations pick up different complex reflection phases, so $\Delta=2N(\phi_s-\phi_p)$ . Each rhomb here has two parallel reflecting faces, and light enters and leaves at normal incidence. Cutting the glass differently changes both the ray path and the retardance.

Why it mattersHow can total internal reflection, which returns all the light, act as a waveplate that works across many wavelengths?

Start with the essentials

Focus question: How can total internal reflection, which returns all the light, act as a waveplate that works across many wavelengths?
One-sentence intuition: Total reflection keeps all of the reflected power but not the phase: $|r_s|=|r_p|=1,\qquad\phi_s\ne\phi_p$ . Two reflections can turn equal in-plane and out-of-plane components into nearly circular polarization, and a second identical rhomb turns that back into linear polarization at a rotated angle. The entrance and exit faces still reflect a little light at normal incidence, so the main beam loses some power to Fresnel reflection.

Core mathematical model

Complex reflection coefficients and the sign convention

\begin{aligned}m&=n_{\rm air}/n_g\\q&=\sqrt{\sin^2\theta-m^2}\\r_s&=\frac{\cos\theta-iq}{\cos\theta+iq}\\r_p&=\frac{m^2\cos\theta-iq}{m^2\cos\theta+iq}\end{aligned}

The time dependence is a negative imaginary exponential. The out-of-plane basis vector is fixed along positive Z, and the in-plane basis vector is its cross product with the direction of each segment. After two reflections the outgoing basis is the same as the incoming one. The angle is measured from the normal of the face.

Output polarization and the power in the main beam

\begin{aligned}\mathbf E&=\begin{pmatrix}r_p^{2N}\cos\alpha\\r_s^{2N}\sin\alpha\end{pmatrix}\\T_0&=\frac{4n_g}{(1+n_g)^2}\\\frac{P_D}{P_0}&=T_0^{2N}|\mathbf a^\mathsf T\mathbf E|^2\\\mathbf a&=(\cos\beta,\sin\beta)^\mathsf T\end{aligned}

Input power is counted after the light has been prepared in a pure linear polarization. Every normal crossing between air and glass transmits the fraction of power given here. The small beams reflected at the entrance and exit faces are dropped, and their further reflections are not added back as ghost light. The ideal analyzer absorbs the rest of the rejected power.

How far the evanescent field reaches

\begin{aligned}\theta_c&=\arcsin(1/n_g)\\d_{\rm ev}&=\frac{\lambda_0}{2\pi\sqrt{n_g^2\sin^2\theta-1}}\\|E(z)|&\propto e^{-z/d_{\rm ev}}\end{aligned}

The depth is measured along the outward normal into the air. It is the distance over which the amplitude falls by a factor of e; the intensity depth is half as large. A near field exists even though the normal energy flux is zero. The model leaves out a second nearby surface, frustrated total reflection and the shifts of a finite beam.

What a linear analyzer scan can tell you

\begin{aligned}I(\beta)&=A+B\cos2\beta+C\sin2\beta\\P_{\rm lin}&=\frac{\sqrt{B^2+C^2}}{A}\\s_3&=2\operatorname{Im}(E_p^*E_s)\end{aligned}

The fit coefficients come from the stored intensities through discrete Fourier sums. The circular Stokes value shown elsewhere is the model's reference value; it is not reconstructed from this scan. Its sign follows the basis defined above, and the Lab avoids left and right labels, which are ambiguous.

Common difficulties

Does total reflection mean no field outside?

Typical misconceptionIf no ray is transmitted, there is no field outside the glass, and a flat analyzer scan tells me which way the light is circularly polarized.

Better mental modelThere is an evanescent field outside that decays exponentially away from the surface, but it carries no net energy flow along the normal into the far field. Its decay length is computed below. A rotating linear analyzer measures only the two linear Stokes components; the sign of circular polarization needs a different analyzer. The records here are simulated expected values, not counted photons.

Run the experiment

01
Scene 1: One rhomb: near circular
Start with one rhomb and solve for its design angle at 550 nm. Compare the polarization ellipses at the input, after the first reflection and at the exit, and look at both reflection points.
What to observe: The light enters along positive X, the plane of incidence is XY, and the perpendicular axis is positive Z. The local in-plane basis turns with the reflected beam. The gold loops are enlarged cross-sections of the electric field, not photon paths.
02
Scene 2: Two rhombs: rotated linear
Add the second rhomb, rotate the analyzer and acquire a scan. Then change the controls: the acquired points and the settings they were taken with stay frozen.
What to observe: The output beam is parallel to the input but shifted sideways. The glass, the reflection points and the sensor share one millimetre coordinate system. The second rhomb adds two more normal-incidence faces, which lowers the throughput. The optical path ends at the opaque detector.
03
Scene 3: Near critical: not quarter wave
Choose the near-critical preset and tune the wavelength. Drag across the angle and spectrum plots, and compare the rhomb stack with the ideal reference wave plate.
What to observe: The rhomb retardance depends on wavelength only through the dispersion of the glass index, with no explicit inverse-wavelength factor. That makes it broadband, but not perfectly achromatic. The comparison wave plate has a constant birefringence and does not model any named crystal.