13.1 Jones Calculus The Jones vector represents the polarisation state of a monochromatic plane wave:
E = ( E x E y ) = ( A x e i δ x A y e i δ y ) \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} E = ( E x E y ) = ( A x e i δ x A y e i δ y )
Optical elements are represented by 2 × 2 2 \times 2 2 × 2 matrices:
Linear polariser at angle θ \theta θ : P ( θ ) = ( cos 2 θ sin θ cos θ sin θ cos θ sin 2 θ ) \mathbf{P}(\theta) = \begin{pmatrix}\cos^2\theta & \sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin^2\theta\end{pmatrix} P ( θ ) = ( cos 2 θ sin θ cos θ sin θ cos θ sin 2 θ )
Quarter-wave plate (fast axis horizontal, retardation π / 2 \pi/2 π /2 ): Q = ( 1 0 0 e i π / 2 ) = ( 1 0 0 i ) \mathbf{Q} = \begin{pmatrix}1 & 0 \\ 0 & e^{i\pi/2}\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & i\end{pmatrix} Q = ( 1 0 0 e iπ /2 ) = ( 1 0 0 i )
Half-wave plate (retardation π \pi π ): H = ( 1 0 0 e i π ) = ( 1 0 0 − 1 ) \mathbf{H} = \begin{pmatrix}1 & 0 \\ 0 & e^{i\pi}\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix} H = ( 1 0 0 e iπ ) = ( 1 0 0 − 1 )
Theorem 13.1. The output of a sequence of optical elements is the product of their Jones matrices applied to the input Jones vector:
E o u t = M n ⋯ M 2 M 1 E i n \mathbf{E}_{\mathrm{out} = \mathbf{M}_n \cdots \mathbf{M}_2\,\mathbf{M}_1\,\mathbf{E}_{\mathrm{in}}} E out = M n ⋯ M 2 M 1 E in
13.2 Stokes Parameters For partially polarised light, the Stokes parameters are:
S 0 = ∣ E x ∣ 2 + ∣ E y ∣ 2 , S 1 = ∣ E x ∣ 2 − ∣ E y ∣ 2 , S 2 = 2 R e ( E x E y ∗ ) , S 3 = 2 I m ( E x E y ∗ ) 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^*) S 0 = ∣ E x ∣ 2 + ∣ E y ∣ 2 , S 1 = ∣ E x ∣ 2 − ∣ E y ∣ 2 , S 2 = 2 Re ( E x E y ∗ ) , S 3 = 2 Im ( E x E y ∗ )
The degree of polarisation is
P = S 1 2 + S 2 2 + S 3 2 S 0 P = \frac{\sqrt{S_1^2 + S_2^2 + S_3^2}}{S_0} P = S 0 S 1 2 + S 2 2 + S 3 2
For fully polarised light: P = 1 P = 1 P = 1 . For unpolarised light: S 1 = S 2 = S 3 = 0 S_1 = S_2 = S_3 = 0 S 1 = S 2 = S 3 = 0 .
13.3 Worked Example: Polarisation by Multiple Reflections Problem. Unpolarised light is incident on a stack of N N N glass plates at the Brewster angle. Find the degree of polarisation of the transmitted light.
Solution At the Brewster angle θ B \theta_B θ B The reflected light for the p p p -polarisation has zero amplitude (r p = 0 r_p = 0 r p = 0 ). The s s s -polarisation is partially reflected with reflectance R s = ( ( n 1 cos θ i − n 2 cos θ t ) / ( n 1 cos θ i + n 2 cos θ t ) ) 2 R_s = ((n_1\cos\theta_i - n_2\cos\theta_t)/(n_1\cos\theta_i + n_2\cos\theta_t))^2 R s = (( n 1 cos θ i − n 2 cos θ t ) / ( n 1 cos θ i + n 2 cos θ t ) ) 2 .
For one interface, the transmitted p p p -intensity is T p = 1 T_p = 1 T p = 1 and the transmitted s s s -intensity is T s = 1 − R s T_s = 1 - R_s T s = 1 − R s . After N N N interfaces:
I p ( N ) = I 0 / 2 , I s ( N ) = ( I 0 / 2 ) ( 1 − R s ) N I_p^{(N)} = I_0/2, \quad I_s^{(N)} = (I_0/2)(1 - R_s)^N I p ( N ) = I 0 /2 , I s ( N ) = ( I 0 /2 ) ( 1 − R s ) N
The degree of polarisation:
P = I p ( N ) − I s ( N ) I p ( N ) + I s ( N ) = 1 − ( 1 − R s ) N 1 + ( 1 − R s ) N P = \frac{I_p^{(N)} - I_s^{(N)}}{I_p^{(N)} + I_s^{(N)}} = \frac{1 - (1 - R_s)^N}{1 + (1 - R_s)^N} P = I p ( N ) + I s ( N ) I p ( N ) − I s ( N ) = 1 + ( 1 − R s ) N 1 − ( 1 − R s ) N
For N → ∞ N \to \infty N → ∞ : P → 1 P \to 1 P → 1 . This is the principle behind “pile-of-plates” polarisers. For glass (n = 1.5 n = 1.5 n = 1.5 ) at θ B ≈ 56.3 ° \theta_B \approx 56.3° θ B ≈ 56.3° : R s ≈ ( ( 1.5 × 0.555 − cos θ t ) / ( 1.5 × 0.555 + cos θ t ) ) 2 R_s \approx ((1.5 \times 0.555 - \cos\theta_t)/(1.5 \times 0.555 + \cos\theta_t))^2 R s ≈ (( 1.5 × 0.555 − cos θ t ) / ( 1.5 × 0.555 + cos θ t ) ) 2 .
For five plates: P = ( 1 − ( 1 − R s ) 5 ) / ( 1 + ( 1 − R s ) 5 ) ≈ 50 % P = (1 - (1 - R_s)^5)/(1 + (1 - R_s)^5) \approx 50\% P = ( 1 − ( 1 − R s ) 5 ) / ( 1 + ( 1 − R s ) 5 ) ≈ 50% .
■ \blacksquare ■