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Polarisation in Detail

The Jones vector represents the polarisation state of a monochromatic plane wave:

E=(ExEy)=(Ax eiδxAy eiδy)\mathbf{E} = \begin{pmatrix} E_x \\ E_y \end{pmatrix} = \begin{pmatrix} A_x\,e^{i\delta_x} \\ A_y\,e^{i\delta_y} \end{pmatrix}

Optical elements are represented by 2×22 \times 2 matrices:

  • Linear polariser at angle θ\theta: P(θ)=(cos⁡2θsin⁡θcos⁡θsin⁡θcos⁡θsin⁡2θ)\mathbf{P}(\theta) = \begin{pmatrix}\cos^2\theta & \sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin^2\theta\end{pmatrix}

  • Quarter-wave plate (fast axis horizontal, retardation π/2\pi/2): Q=(100eiπ/2)=(100i)\mathbf{Q} = \begin{pmatrix}1 & 0 \\ 0 & e^{i\pi/2}\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & i\end{pmatrix}

  • Half-wave plate (retardation π\pi): H=(100eiπ)=(100−1)\mathbf{H} = \begin{pmatrix}1 & 0 \\ 0 & e^{i\pi}\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}

Theorem 13.1. The output of a sequence of optical elements is the product of their Jones matrices applied to the input Jones vector:

Eout=Mn⋯M2 M1 Ein\mathbf{E}_{\mathrm{out} = \mathbf{M}_n \cdots \mathbf{M}_2\,\mathbf{M}_1\,\mathbf{E}_{\mathrm{in}}}

For partially polarised light, the Stokes parameters are:

S0=∣Ex∣2+∣Ey∣2,S1=∣Ex∣2−∣Ey∣2,S2=2 Re(ExEy∗),S3=2 Im(ExEy∗)S_0 = |E_x|^2 + |E_y|^2, \quad S_1 = |E_x|^2 - |E_y|^2, \quad S_2 = 2\,\mathrm{Re}(E_x E_y^*), \quad S_3 = 2\,\mathrm{Im}(E_x E_y^*)

The degree of polarisation is

P=S12+S22+S32S0P = \frac{\sqrt{S_1^2 + S_2^2 + S_3^2}}{S_0}

For fully polarised light: P=1P = 1. For unpolarised light: S1=S2=S3=0S_1 = S_2 = S_3 = 0.

13.3 Worked Example: Polarisation by Multiple Reflections

Section titled “13.3 Worked Example: Polarisation by Multiple Reflections”

Problem. Unpolarised light is incident on a stack of NN glass plates at the Brewster angle. Find the degree of polarisation of the transmitted light.

Solution

At the Brewster angle θB\theta_B The reflected light for the pp-polarisation has zero amplitude (rp=0r_p = 0). The ss-polarisation is partially reflected with reflectance Rs=((n1cos⁡θi−n2cos⁡θt)/(n1cos⁡θi+n2cos⁡θt))2R_s = ((n_1\cos\theta_i - n_2\cos\theta_t)/(n_1\cos\theta_i + n_2\cos\theta_t))^2.

For one interface, the transmitted pp-intensity is Tp=1T_p = 1 and the transmitted ss-intensity is Ts=1−RsT_s = 1 - R_s. After NN interfaces:

Ip(N)=I0/2,Is(N)=(I0/2)(1−Rs)NI_p^{(N)} = I_0/2, \quad I_s^{(N)} = (I_0/2)(1 - R_s)^N

The degree of polarisation:

P=Ip(N)−Is(N)Ip(N)+Is(N)=1−(1−Rs)N1+(1−Rs)NP = \frac{I_p^{(N)} - I_s^{(N)}}{I_p^{(N)} + I_s^{(N)}} = \frac{1 - (1 - R_s)^N}{1 + (1 - R_s)^N}

For N→∞N \to \infty: P→1P \to 1. This is the principle behind “pile-of-plates” polarisers. For glass (n=1.5n = 1.5) at θB≈56.3°\theta_B \approx 56.3°: Rs≈((1.5×0.555−cos⁡θt)/(1.5×0.555+cos⁡θt))2R_s \approx ((1.5 \times 0.555 - \cos\theta_t)/(1.5 \times 0.555 + \cos\theta_t))^2.

For five plates: P=(1−(1−Rs)5)/(1+(1−Rs)5)≈50%P = (1 - (1 - R_s)^5)/(1 + (1 - R_s)^5) \approx 50\%.

■\blacksquare

13.4 Jones Matrices for Common Optical Elements

Section titled “13.4 Jones Matrices for Common Optical Elements”

A systematic reference for standard Jones matrices:

ElementJones Matrix
Linear polariser (xx-axis)(1000)\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}
Linear polariser (yy-axis)(0001)\begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}
Quarter-wave plate (fast axis xx)(100i)\begin{pmatrix}1 & 0 \\ 0 & i\end{pmatrix}
Half-wave plate (fast axis xx)(100−1)\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}
Rotator (angle θ\theta)(cos⁡θ−sin⁡θsin⁡θcos⁡θ)\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}

The Poincaré sphere provides a geometric representation of polarisation states. The Stokes parameters (S1,S2,S3)(S_1, S_2, S_3) normalised by S0S_0 are the Cartesian coordinates of a point on a unit sphere:

  • North pole (S3=1S_3 = 1): right circular polarisation
  • South pole (S3=−1S_3 = -1): left circular polarisation
  • Equator (S3=0S_3 = 0): linear polarisation (orientation varies with longitude)
  • Intermediate latitudes: elliptical polarisation

13.6 Worked Example: Determining Unknown Polarisation

Section titled “13.6 Worked Example: Determining Unknown Polarisation”

Problem. Unpolarised light passes through a linear polariser at 0∘0^\circ, then a quarter-wave plate with fast axis at 45∘45^\circ, then a linear polariser at 90∘90^\circ. Find the transmitted intensity.

Solution

After first polariser (0∘0^\circ): E1=(10)\mathbf{E}_1 = \begin{pmatrix}1 \\ 0\end{pmatrix} (normalised).

Quarter-wave plate with fast axis at 45∘45^\circ: rotate to fast-axis basis, apply retardation, rotate back:

E2=R(−45∘)(100i)R(45∘)(10)\mathbf{E}_2 = R(-45^\circ)\begin{pmatrix}1 & 0 \\ 0 & i\end{pmatrix}R(45^\circ)\begin{pmatrix}1 \\ 0\end{pmatrix}

where R(θ)=(cos⁡θsin⁡θ−sin⁡θcos⁡θ)R(\theta) = \begin{pmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{pmatrix}.

R(45∘)(10)=12(1−1)R(45^\circ)\begin{pmatrix}1 \\ 0\end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ -1\end{pmatrix}

After retardation: 12(1−i)\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ -i\end{pmatrix}. Rotating back: E2=12(1i)\mathbf{E}_2 = \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ i\end{pmatrix} (right circular).

Second polariser at 90∘90^\circ selects yy-component: Ey=i/2E_y = i/\sqrt{2}.

Transmitted intensity: I=∣Ey∣2=1/2I = |E_y|^2 = 1/2.

■\blacksquare

13.7 Optical Activity and Faraday Rotation

Section titled “13.7 Optical Activity and Faraday Rotation”

Some materials exhibit optical activity: the plane of linear polarisation rotates as light propagates. This arises from circular birefringence, different refractive indices for left and right circularly polarised light.

The rotation angle is ϕ=αL\phi = \alpha L where α\alpha is the specific rotation and LL is the path length.

In the Faraday effect, a magnetic field along the propagation direction induces circular birefringence:

ϕ=VBL\phi = V B L

where VV is the Verdet constant. Faraday rotation is non-reciprocal: reversing the propagation direction doubles the rotation, unlike natural optical activity which cancels upon reflection.

  1. 3D cinema (IMAX): Projectors use orthogonal polarisation states for left and right eye images; polarising glasses separate them.
  2. LCD displays: Liquid crystals rotate the polarisation of light; crossed polarisers convert rotation into intensity modulation.
  3. Stress analysis (photoelasticity): Transparent materials under stress become birefringent; viewing through crossed polarisers reveals stress patterns.
  4. Radar and remote sensing: Polarimetric radar measures the full Stokes vector of backscattered radiation to identify terrain and targets.
  5. Quantum cryptography: The BB84 protocol encodes qubits in the polarisation states of single photons (horizontal/vertical and diagonal/anti-diagonal bases).
  6. Optical isolators: Combining a polariser with a Faraday rotator creates a non-reciprocal device that allows light to pass in one direction only.

Mistake 1: Multiplying Jones matrices in the wrong order Matrix multiplication is not commutative. The correct order for a sequence of optical elements is Mn⋯M2 M1\mathbf{M}_n \cdots \mathbf{M}_2\,\mathbf{M}_1, where M1\mathbf{M}_1 is the first element encountered by the light. Students frequently reverse this order, especially when rotating coordinate systems for wave plates at arbitrary orientations.

Mistake 2: Assuming the Stokes parameters fully determine the Jones vector The Stokes parameters describe the polarisation state including partially polarised light, while the Jones vector only describes fully polarised light. For partially polarised light, there is no Jones vector representation. Students sometimes try to invert the Stokes parameters to get a Jones vector without checking whether P=1P = 1.

Mistake 3: Forgetting that the Poincare sphere represents polarisation states, not field amplitudes The Poincare sphere uses the normalised Stokes parameters (S1/S0,S2/S0,S3/S0)(S_1/S_0, S_2/S_0, S_3/S_0) as coordinates. Points on the sphere surface represent fully polarised states, while interior points represent partially polarised light. Students sometimes confuse the Poincare sphere with a representation of the electric field vector in real space.

  • Polarization: Introduces the fundamental polarisation states, Malus’s law, and birefringence that the Jones calculus and Stokes parameters formalise.

  • Fresnel Equations: The polarisation-dependent reflection and transmission coefficients at interfaces are described by the Fresnel equations.

  • Geometric Optics: Brewster’s angle and polarisation by reflection arise from the boundary conditions in the ray optics limit.

  • Vector Calculus

  • Linear Algebra

  • Quantum Computing

Polarisation describes the direction in which light’s electric field oscillates. Unpolarised light has random orientations, but passing it through a polariser selects one direction. Malus’s law gives the intensity after a second polariser as proportional to the cosine squared of the angle between them. Birefringent materials split light into two polarisation components that travel at different speeds, creating phase shifts. Circular polarisation occurs when two orthogonal linear components have a quarter-cycle phase difference. Polarisation is used in LCD screens, 3D cinema, and stress analysis, where birefringent patterns reveal mechanical strain.