---
title: "Waveplate & Poincaré Sphere Lab Interactive Lab | Sandbox Physics"
description: "How does a waveplate move a polarization state on the Poincaré sphere?"
canonical_url: "https://sandboxphysics.com/labs/waveplate-poincare-sphere-lab/"
language: "en"
page_type: "lab"
alternate_languages:
  zh-CN: "https://sandboxphysics.com/zh/labs/waveplate-poincare-sphere-lab/"
last_modified: "2026-10-27"
---

Optics 076 · Polarization, anisotropy, and modulation

# Waveplate & Poincaré Sphere Lab

**How does a waveplate move a polarization state on the Poincaré sphere?** An ideal waveplate turns the polarization point rigidly around an axis on the sphere's equator. The axis sits at twice the fast-axis angle, and the turn equals the plate's retardance. A quarter-wave plate takes light polarized at 45 degrees to its axis into circular; a half-wave plate mirrors the polarization about its fast axis. Order matters, because turns about different axes do not commute.

Stack three ideal waveplates, applied in the correct Jones-matrix order, and set each fast axis. Watch the state move on a rotatable Poincaré sphere, compare field ellipses drawn at equal scale and read the analyzer power.

> Interactive simulation: open this page in a web browser to change the parameters and run the model.

The light is fully polarized and monochromatic, the plates are lossless effective retarders, and the analyzer is ideal. Plate thickness is fixed and schematic. Material dispersion, walk-off and reconstructing Stokes parameters from measured data are not included.

Physics tutorial

## Audit the physics of Waveplate & Poincaré Sphere Lab

**Background** A 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 is the one with the smaller phase index. With the time dependence of the phasor written as a negative exponent, the slow component gains a positive phase relative to the fast one. So a quarter-wave plate with its fast axis along the first transverse axis turns diagonal linear light into circular light.

**Why it matters** How does a waveplate move a polarization state on the Poincaré sphere?

Start with the essentials

- **Focus question**: How does a waveplate move a polarization state on the Poincaré sphere?
- **One-sentence intuition**: A single linear retarder rigidly rotates the normalized Stokes vector: $\mathbf s_{\rm out}=\mathcal R_{\mathbf n}(\delta)\mathbf s_{\rm in}$. The rotation axis lies on the equator, at twice the angle of the physical fast axis. The colored path traces the retardance as it builds up inside the plate. Successive rotations generally do not commute, so reversing the order of the plates can change the final state even though no plate setting changes.

### 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 the fast axis and the second the slow axis. The overall phase along the fast axis is dropped, because this Lab looks at polarization and intensity, not at interference between separate arms. You set the effective retardance directly; material dispersion and the real 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 retardance rotates the Stokes vector by a positive right-handed angle about the doubled fast-axis vector. The paths need not be great circles. Every ideal retarder keeps the power unchanged and keeps the state on 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 in the product acts first. The analyzer formula is checked independently against a time integral of the projected electric field. It has no circular Stokes term, so opposite circular states give identical scans with a linear analyzer. The reversal angle compares two predictions; it is not a recorded measurement.

### Common difficulties

#### Every quarter-wave plate makes circular light?

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

**Better mental model** Circular output needs equal field amplitudes along the two eigenaxes and a quarter-cycle phase difference between them. Linear light aligned with either axis stays linear. Circular input comes out linear. A scan with a linear analyzer alone cannot tell the two signs of circular polarization apart.

### Run the experiment

1. 01

   #### Scene 1: Quarter-wave conversion

   Begin with quarter-wave conversion. The end point reaches a pole, the field ellipse becomes a circle, and the analyzer passes half the power at every angle.

   **What to observe:** The sphere is a map of polarization states, not a lens or detector in the beam. Its poles stand for circular polarization of each sign. Opposite points on the sphere are orthogonal polarization states, not light travelling in opposite directions.

2. 02

   #### Scene 2: Half-wave rotation

   Turn the fast axis of plate A to the input azimuth. The circle collapses to a line, because the light no longer has a component along the slow axis.

   **What to observe:** The axes form a right-handed frame: light travels along positive X, the first field axis points along positive Y and the second along positive Z. Angles are measured from the first field axis toward the second. The ellipse plot draws these two axes to the right and upward.

3. 03

   #### Scene 3: Arbitrary retarder sequence

   Choose the half-wave preset. Rotate the fast axis and watch the linear output turn through twice the angle, while the power before the analyzer stays the same.

   **What to observe:** A positive circular Stokes value means that, at a fixed point, the electric field turns from the first axis toward the second as time goes on. This operational sign is used instead of right or left labels, which are ambiguous. At a circular pole the ellipse azimuth is undefined.

Physical conclusion

Predict quarter-wave and half-wave conversions, find the doubled rotation axis and explain why plate order matters. Keep state space, real space and measured power apart while you do it.

Waves & optics · Starter path

## From one ray to interference measurement

Start with refraction, find where the ray picture stops working, then use interference to measure tiny distances.

[Start here → **Optical Boundary Explorer** Turn the incident angle and watch reflected and transmitted power trade off. Then switch to the critical-angle scene. Observe · 11 min](https://sandboxphysics.com/labs/optical-boundary-explorer/)

1. [1 Optical Boundary Explorer](https://sandboxphysics.com/labs/optical-boundary-explorer/)
2. [2 Rainbow Caustic Observatory](https://sandboxphysics.com/labs/rainbow-laboratory/)
3. [3 Camera Obscura](https://sandboxphysics.com/labs/camera-obscura/)
4. [4 Coherent Wave Composer](https://sandboxphysics.com/labs/coherent-wave-composer/)
5. [5 Young Double-Slit Bench](https://sandboxphysics.com/labs/young-double-slit-bench/)
6. [6 Michelson Precision Lab](https://sandboxphysics.com/labs/michelson-precision-lab/)

[View the full path →](https://sandboxphysics.com/topics/waves-and-optics/#starter-path)