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Polarisation in Detail

13.1 Jones Calculus

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

E=(ExEy)=(AxeiδxAyeiδy)\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)=(cos2θsinθcosθsinθcosθsin2θ)\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)=(1001)\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=MnM2M1Ein\mathbf{E}_{\mathrm{out} = \mathbf{M}_n \cdots \mathbf{M}_2\,\mathbf{M}_1\,\mathbf{E}_{\mathrm{in}}}

13.2 Stokes Parameters

For partially polarised light, the Stokes parameters are:

S0=Ex2+Ey2,S1=Ex2Ey2,S2=2Re(ExEy),S3=2Im(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

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_BThe reflected light for the pp-polarisation has zero amplitude (rp=0r_p = 0). The ss-polarisation is partially reflected with reflectance Rs=((n1cosθin2cosθt)/(n1cosθi+n2cosθt))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=1RsT_s = 1 - R_s. After NN interfaces:

Ip(N)=I0/2,Is(N)=(I0/2)(1Rs)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(1Rs)N1+(1Rs)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 NN \to \infty: P1P \to 1. This is the principle behind “pile-of-plates” polarisers. For glass (n=1.5n = 1.5) at θB56.3°\theta_B \approx 56.3°: Rs((1.5×0.555cosθt)/(1.5×0.555+cosθt))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(1Rs)5)/(1+(1Rs)5)50%P = (1 - (1 - R_s)^5)/(1 + (1 - R_s)^5) \approx 50\%.

\blacksquare