Optics 076 · Polarization, anisotropy, and modulation

Waveplate & Poincaré Sphere Lab

Compose three ideal waveplates: true Jones order, fast-axis geometry, a rotatable Poincaré sphere, equal-scale field ellipses and analyzer power.

Output normalized Stokes

\mathbf s

$0.50$

Analyzer / incident power

I_D/I_0

$50\%$

Reversed-order angular separation

\gamma_{\rm reverse}

$0.00\pi$

Model regime $\text{valid model regime}$

Validated scientific apparatus

Waveplate & Poincaré Sphere Lab

Plate A retardance

\delta_A

0.50\pi\,\mathrm{rad}

Plate A fast axis from u toward v

\alpha_A

0^{\circ}

Plate B retardance

\delta_B

1\pi\,\mathrm{rad}

Plate B fast axis

\alpha_B

22.50^{\circ}

Plate C retardance

\delta_C

0.50\pi\,\mathrm{rad}

Plate C fast axis

\alpha_C

0^{\circ}

Number of plates: A, then B, then C

N

1

Order: zero forward, one reversed

j_O

0

Input ellipse azimuth

\psi_0

45^{\circ}

Input signed ellipticity angle

\chi_0

0^{\circ}

Analyzer axis from u toward v

\beta

0^{\circ}

One quarter-wave plate makes a circle. A half-wave plate rotates a line. Combine three plates, then reverse them: the same settings need not reach the same state. Follow every transformation on the sphere and test it with the physical analyzer below.

Pure monochromatic states, lossless effective retarders and ideal analyzer. Schematic fixed plate thickness; no material dispersion, walk-off or measured Stokes reconstruction.

Physics tutorial

Audit the physics of Waveplate & Poincaré Sphere Lab

BackgroundA waveplate does not remove a polarization component. It changes the phase between two orthogonal components: $\mathbf J=\mathbf f\mathbf f^{\mathsf T}+e^{i\delta}\mathbf g\mathbf g^{\mathsf T}$ . The fast axis has the smaller phase index. With our negative-time phasor convention, the slow component gains positive relative phase. A diagonal linear input therefore becomes circular after a quarter-wave plate whose fast axis is the first transverse axis.

Why it mattersHow does each retarder rotate a polarization state along a precise path on the Poincaré sphere?

Start with the essentials

Focus question: How does each retarder rotate a polarization state along a precise path on the Poincaré sphere?
One-sentence intuition: One linear retarder is a rigid rotation of the normalized Stokes vector: $\mathbf s_{\rm out}=\mathcal R_{\mathbf n}(\delta)\mathbf s_{\rm in}$ . Its rotation axis lies on the equator at twice the physical fast-axis angle. The colored path follows accumulated retardance through that plate. Successive rotations generally do not commute: reversing the physical plate order can change the final state without changing any plate setting.

Core mathematical model

Define the axes before the matrix

\begin{aligned}\mathbf f&=(\cos\alpha,\sin\alpha)^{\mathsf T}\\\mathbf g&=(-\sin\alpha,\cos\alpha)^{\mathsf T}\\\delta&=\frac{2\pi(n_s-n_f)d}{\lambda_0}\\\mathbf E(x,t)&=\Re\{\mathbf J_E e^{i(kx-\omega t)}\}\end{aligned}

The first vector is fast, the second slow. The global fast-axis phase is omitted because this Lab observes polarization and intensity, not interference between separate arms. Effective retardance is the control; material dispersion and actual optical thickness are not inferred from the drawing.

Jones spinor to the sphere

\begin{aligned}s_1&=|u|^2-|v|^2\\s_2&=2\Re(u^*v)\\s_3&=2\Im(u^*v)\\\mathbf n&=(\cos2\alpha,\sin2\alpha,0)\\|u|^2+|v|^2&=1\end{aligned}

With this sign convention a positive plate retardance rotates Stokes by the positive right-hand angle about the doubled fast-axis vector. Rotation paths need not be great circles. Every ideal retarder preserves power and the unit sphere.

A detector tests projection, not phase

\begin{aligned}\mathbf J_{\rm total}&=\mathbf J_C\mathbf J_B\mathbf J_A\\\frac{I_D(\beta)}{I_0}&=\frac{1+s_1\cos2\beta+s_2\sin2\beta}{2}\\\gamma_{\rm reverse}&=\arccos(\mathbf s\cdot\mathbf s_{\rm reverse})\end{aligned}

The rightmost plate acts first. The analyzer formula is independently checked against time-integrated electric-field projection. It contains no circular Stokes term: opposite circular states have identical linear-analyzer scans. The reversal angle compares two predictions, not an acquired measurement.

Common difficulties

Every quarter-wave plate makes circular light?

Typical misconceptionQuarter-wave retardance is sufficient regardless of the incoming polarization and fast-axis angle.

Better mental modelCircular output requires equal field amplitudes in the two eigenaxes and a quadrature phase difference. Linear input aligned with either axis remains linear. A circular input becomes linear. The analyzer scan alone cannot distinguish the two signs of circular polarization.

Run the experiment

01
Scene 1: Quarter-wave conversion
Begin with quarter-wave conversion. The final point reaches a pole, the field ellipse becomes a circle, and every analyzer angle transmits half the power.
What to observe: The sphere is a state-space analysis, not a lens or detector in the beam. Its poles encode signed circular content; antipodal states are orthogonal polarization states, not opposite ray directions.
02
Scene 2: Half-wave rotation
Turn plate A fast axis to the input azimuth. The circle collapses to a line because no slow-axis component is excited.
What to observe: Our basis follows a right-handed frame: propagation along positive X, first field axis along positive Y and second along positive Z. Angles rotate from the first field axis toward the second. The ellipse plot places those axes rightward and upward.
03
Scene 3: Arbitrary retarder sequence
Choose the half-wave preset. Rotate the fast axis and watch the linear output turn twice as far; power before the analyzer remains unchanged.
What to observe: Positive circular Stokes means that, at a fixed point, the electric field turns from the first axis toward the second as time increases. We use this operational sign instead of an ambiguous right/left label. Ellipse azimuth is undefined at a circular pole.