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Polarization State Analyzer

Physics Optics • Diffraction and Polarization

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Identify whether a polarization state is linear, circular, or elliptical from its azimuth \(\psi\) and ellipticity angle \(\chi\), then compute the transmitted intensity through a linear polarizer at angle \(\theta\). The calculator also previews the normalized Stokes parameters and the polarization ellipse.

Inputs
This calculator uses the normalized fully polarized Stokes description \(s_1=\cos 2\chi \cos 2\psi\), \(s_2=\cos 2\chi \sin 2\psi\), \(s_3=\sin 2\chi\), and the analyzer law \(I(\theta)=\dfrac{I_0}{2}\bigl[1+\cos 2\chi \cos 2(\theta-\psi)\bigr]\). For \(\chi=0\), this reduces to Malus’ law \(I=I_0\cos^2(\theta-\psi)\).
Animation
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Interactive polarization-state preview
The left panel shows the incident polarization ellipse. The right panel shows the analyzer axis and the transmitted field component. The inset summarizes the normalized Stokes parameters.
Drag to pan. Use the mouse wheel to zoom. Fit view restores the default framing. Positive \(\chi\) is treated here as left-handed and negative \(\chi\) as right-handed.
Enter values and click “Calculate”.

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Frequently Asked Questions

How does this calculator decide whether the state is linear, circular, or elliptical?

It uses the ellipticity angle χ. When χ = 0° the state is linear, when |χ| = 45° the state is circular, and when 0° < |χ| < 45° the state is elliptical.

When does the formula reduce to Malus' law?

It reduces to Malus' law when the incident state is linear, meaning χ = 0°. Then the transmitted intensity becomes I = I0 cos²(θ - ψ).

Why does circularly polarized light always give I = I0/2 through a linear polarizer?

Because circular polarization contains equal components along every linear direction, so an ideal linear polarizer always transmits exactly one half of the incident intensity.

Why can elliptical polarization not usually be extinguished by one linear polarizer?

Because elliptical polarization has two unequal orthogonal components with a phase difference, so rotating a single linear polarizer changes the transmitted intensity but generally cannot make it exactly zero.