Optics 075 · Polarization, anisotropy, and modulation

Birefringent Crystal Lab

A finite crystal, tensor eigenmodes and a physical screen: parallel exit rays, true walk-off, equal-scale double images and an ideal biaxial comparison.

Equal-scale image separation

\Delta r

$0.50$

Phase indices, fast / slow

n_1,n_2

$50\%$

Total normal-flux transmission

P_{\rm out}/P_{\rm in}

$0.00\pi$

Model regime $\text{valid model regime}$

Validated scientific apparatus

Birefringent Crystal Lab

Physical crystal thickness

t

25\,\mathrm{mm}

Third principal-axis tilt from beam

\theta

45^{\circ}

Principal frame azimuth from upward

\phi

20^{\circ}

Frame spin about third axis (uniaxial invariant)

\chi

0^{\circ}

Material: zero calcite, one ideal biaxial

j_M

0

Vacuum wavelength (biaxial indices fixed)

\lambda_0

532\,\mathrm{nm}

Analyzer axis from upward toward image right

\alpha

0^{\circ}

Analyzer: zero removed, one inserted

j_A

0

One pattern enters; two equal-sized copies emerge. Rotate the crystal frame and watch only the extraordinary copy move. Insert an analyzer to erase one copy, then align the optic axis and make both coincide. White marks the wave normal; the colored paths follow energy, not an invented second Snell angle.

Normal-incidence geometrical transport, reflectionless interfaces, unpolarized collimated pattern. Manufacturer calcite dispersion; synthetic biaxial indices. No acquired photograph, diffraction or conoscope.

Physics tutorial

Audit the physics of Birefringent Crystal Lab

BackgroundInside an anisotropic crystal, phase and energy need not travel in the same direction. Energy follows $\mathbf S=\mathbf E\times\mathbf H$ . At normal incidence both transmitted wavevectors are normal to the parallel faces. The extraordinary Poynting vector can still tilt. At the second face both outgoing beams recover the original direction, but at different positions.

Why it mattersHow does crystal orientation split one ray into ordinary and extraordinary propagation?

Start with the essentials

Focus question: How does crystal orientation split one ray into ordinary and extraordinary propagation?
One-sentence intuition: The same energy directions determine both the 3D ray exits and every pixel of the screen: $\Delta\mathbf r_j=t\,\frac{\mathbf S_{j,\perp}}{S_{j,x}}$ . Every transverse input point receives the same translation. There is no independent image magnification and no made-up offset. A thicker crystal makes a larger displacement, not a bigger copy.

Core mathematical model

Solve the transverse displacement modes

\begin{aligned}\boldsymbol\eta&=\boldsymbol\varepsilon_r^{-1}\\\mathbf P\boldsymbol\eta\mathbf P\,\mathbf D_j&=n_j^{-2}\mathbf D_j\\\mathbf P&=\mathbf I-\widehat{\mathbf k}\widehat{\mathbf k}^{\mathsf T}\\\mathbf E_j&=\boldsymbol\eta\mathbf D_j\\\mathbf H_j&\propto n_j\widehat{\mathbf k}\times\mathbf E_j\end{aligned}

Nonmagnetic, real symmetric positive dielectric tensor. Displacement eigenvectors lie in the plane normal to the common wavevector. The electric vector may have a longitudinal component; the cross product gives energy direction. Field units here are arbitrary and only directions are used.

Check the uniaxial angle convention

\begin{aligned}\frac{1}{n_{\rm ex}^2(\theta)}&=\frac{\cos^2\theta}{n_o^2}+\frac{\sin^2\theta}{n_e^2}\\n_{\rm ex}(0)&=n_o\\n_{\rm ex}(\pi/2)&=n_e\end{aligned}

The angle is from wave normal to optic axis, not from the plate surface. Along the optic axis the two phase indices agree. Away from it the extraordinary energy direction can differ from the wave normal.

One translation operator feeds both images

\begin{aligned}I_{\rm out}(\mathbf r)&=\sum_{j=1}^{2}w_j I_{\rm in}(\mathbf r-\Delta\mathbf r_j)\\w_j&=\tfrac12\left|\mathbf a\cdot\widehat{\mathbf D}_j\right|^2\\w_1+w_2&=\tfrac12\end{aligned}

Weights shown are for the inserted ideal analyzer. Without it, each is one half and total flux is one. Transverse electric and displacement directions are parallel at normal incidence; reflectionless matching preserves those output axes. Spatial translations are shared by screen texture, enlarged data and ray exits.

Common difficulties

Two refractive indices mean two exit angles?

Typical misconceptionEven at normal incidence, the outgoing beams should keep diverging because their refractive indices differ.

Better mental modelTangential wavevector continuity fixes both internal wave normals here. The separation accumulates through energy walk-off inside the crystal, not different Snell angles in air. Plane-parallel faces restore parallel outgoing directions. A Wollaston prism has different geometry and is not this apparatus.

Run the experiment

01
Scene 1: Double-image calcite
Begin with calcite and no analyzer. Compare the same-scale input and screen. Rotate the azimuth; one copy remains fixed while the extraordinary copy travels around it.
What to observe: Source and screen are forward geometric intensity maps, not acquired camera photographs. The input is an assumed collimated pattern, not a diffuse object being imaged by an unspecified lens.
02
Scene 2: Along the optic axis
Pass the cyan channel, then the amber channel. The corresponding physical ray and copy survive; total power is halved in either case because the input is unpolarized.
What to observe: Both figures use the same physical field, linear normalization and sRGB display mapping. The white source pattern contains two mutually incoherent polarization channels. A phase difference alone cannot generate interference in this experiment.
03
Scene 3: Ideal biaxial comparison
Choose along the optic axis: the copies merge. In the ideal biaxial comparison both energy paths walk off. Spin the frame about its third principal axis to leave the principal plane; this spin has no effect on uniaxial calcite. The third principal axis is not either optic axis.
What to observe: The ellipse is the true-ratio section of the index ellipsoid, not the wavevector surface and not an image reconstructed from detector data. Enlarged ray widths and axis markers are visual guides.