Chapter IV: Matrices in Polarization Optics

Source: Gerrard and Burch, Introduction to Matrix Methods in Optics (1975), Chapter IV. Jones vectors are normalized only when an absolute intensity is needed; common phase factors are discarded.

Illustrative problems

Problem 4.1 — Malus’s law and three-polarizer transmission

An ideal polarizer with pass direction \(\mathbf p=(\cos\theta,\sin\theta)^T\) has Jones matrix \(J_p=\mathbf p\mathbf p^T\). Acting on a unit field polarized along \(x\) gives transmitted amplitude \(\cos\theta\), hence

\[\boxed{I=\cos^2\theta}.\]

For initially crossed polarizers with an intermediate polarizer at angle \(\phi\) from the extinction setting, successive projection gives

\[I=\cos^2\phi\sin^2\phi =\boxed{\frac14\sin^2(2\phi)}.\]

The result vanishes with no rotation and peaks at \(1/4\) for the middle polarizer at \(45^\circ\).

Problem 4.2 — Three polaroids illuminated by unpolarized light

The first polarizer transmits half the unpolarized incident intensity. The relative rotations of the next two pass planes are \(12^\circ\) and \(24^\circ\), so repeated use of Malus’s law gives

\[\boxed{\frac{I_{out}}{I_{in}} =\frac12\cos^2(12^\circ)\cos^2(24^\circ) \simeq0.399}.\]

Using the book’s two-decimal trigonometric values gives 0.396. Mueller calculus is required for the first step because the entering beam is unpolarized; after that projection, Jones calculus gives the same result.

Problem 4.3 — Orientation and axes of a polarization ellipse

For \(E_x=H\cos\omega t\) and \(E_y=K\cos(\omega t+\Delta)\), form

\[I=H^2+K^2,\qquad Q=H^2-K^2,\qquad U=2HK\cos\Delta.\]

The ellipse orientation is therefore

\[\boxed{\tan2\alpha=\frac{U}{Q} =\frac{2HK\cos\Delta}{H^2-K^2}}.\]

Its squared semiaxes are the eigenvalues of the real polarization quadratic form:

\[\boxed{a^2,b^2= \frac12\left[I\pm\sqrt{Q^2+U^2}\right]}.\]

The sign of \(V=2HK\sin\Delta\) selects handedness but does not change the axis lengths.

Problem 4.4 — Recovering an ellipse from extinction settings

Multiply the unknown Jones vector by the quarter-wave-plate matrix at \(30^\circ\) and the polarizer projector at \(60^\circ\). Extinction requires both components of the final vector to vanish, so the ratio of the two unknown incident components is fixed. Separating its real and imaginary parts gives the amplitude and phase parameters; substituting them into the ellipse formulas of Problem 4.3 yields

\[\boxed{\text{minor axis at }30^\circ}, \qquad \boxed{a/b=\sqrt3}.\]

Reapplying the plate and analyzer to this recovered Jones vector gives the zero vector, providing a direct check.

Problem 4.5 — Circular light through quarter- and eighth-wave plates

Use a normalized right-circular input and a vertical fast axis. The quarter-wave plate cancels the incident quadrature, leaving equal real components of opposite sign. The output is therefore

\[\boxed{\text{linear polarization at }-45^\circ}.\]

The eighth-wave plate leaves a relative phase of \(135^\circ\). The result is a right-handed ellipse whose quadratic form can be written, after normalization,

\[x^2-\sqrt2xy+y^2=1.\]

Its major axis lies at \(45^\circ\) and its axial ratio is \(\boxed{1+\sqrt2}\). A unitary retarder preserves total intensity in both cases.

Problem 4.6 — Analyzer angle for a phase-shifted equal-amplitude wave

The field has equal component amplitudes and phase difference \(\pi/4\). Projection onto a polarizer at angle \(\theta\) gives

\[I(\theta)=A^2\left[1+ rac{1}{\sqrt2}\sin2\theta\right].\]

Thus the maximum occurs at \(\boxed{\theta=45^\circ}\). With the pass-plane along \(y\), \(I_y=A^2\); therefore

\[\boxed{\frac{I_{max}}{I_y}=1+\frac1{\sqrt2}}.\]

The derivative vanishes at \(45^\circ\), and the negative second derivative confirms a maximum.

Problem 4.7 — Elliptical beam through a linear polarizer

Choose the ellipse axes as coordinates. A right-handed field may be written \((H,iK)^T\); a polarizer at angle \(\alpha\) projects it onto \((\cos\alpha,\sin\alpha)^T\). The projected complex amplitude is \(H\cos\alpha+iK\sin\alpha\), so

\[\boxed{I=H^2\cos^2\alpha+K^2\sin^2\alpha}.\]

The cross term vanishes because the components are in quadrature. The limits \(\alpha=0\) and \(\pi/2\) recover the major- and minor-axis intensities.

Problem 4.8 — Photoelasticity with Jones matrices

Model the stressed specimen as a linear retarder with optic-axis angle \(\alpha\) and retardance \(\delta=csd\), where \(s\) is strain, \(d\) thickness, and \(c\) the strain-optical coefficient. Cascade the entrance polarizer, specimen, and crossed analyzer.

For the book’s polarizers at \(+45^\circ\) and \(-45^\circ\), matrix multiplication gives

\[\boxed{\frac{I}{I_0}=\cos^2(2\alpha) \sin^2\left(\frac\delta2\right)}.\]

For horizontal and vertical crossed polarizers the complementary convention is \(\sin^2(2\alpha)\sin^2(\delta/2)\). Adding mutually perpendicular quarter-wave plates creates a circular polariscope and removes the axis-angle factor:

\[\boxed{\frac{I}{I_0}=\sin^2\left(\frac\delta2\right)}.\]

Zero strain gives extinction, while a half-wave retardance gives the maximum available transmission.