A retarder with retardance and its fast axis along acts as on a Jones vector

PHYSICS · Multi-part · Fermi University Physics · Optics & Photonics · Polarisation and anisotropic media · Polarisation and anisotropic media

Optics and Photonics, Ch Polarisation states, polarisers and the Jones and Stokes calculus

Problem

A retarder with retardance Γ\Gamma and its fast axis along xx acts as J(0)=diag(1,eiΓ)\mathbf{J}(0) = \mathrm{diag}(1, e^{i\Gamma}) on a Jones vector, adding the phase Γ\Gamma to the slower (yy) component. A retarder with its fast axis at angle θ\theta to xx is built by rotating the incoming light into the device's own axes, applying J(0)\mathbf{J}(0), and rotating the result back to the lab axes: J(θ)=R(θ) J(0) R(−θ),R(θ)=(cos⁡θ−sin⁡θsin⁡θcos⁡θ).\mathbf{J}(\theta) = \mathbf{R}(\theta)\,\mathbf{J}(0)\,\mathbf{R}(-\theta), \qquad \mathbf{R}(\theta) = \begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}. Separately, the ideal linear polariser with its axis along xx has Mueller matrix MP(0)=12(1100110000000000).\mathbf{M}_P(0) = \tfrac{1}{2}\begin{pmatrix}1&1&0&0\\1&1&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}. Because S1S_1 and S2S_2 are built from irradiances measured at 0∘,90∘,45∘,135∘0^\circ, 90^\circ, 45^\circ, 135^\circ, turning the physical polariser axis through θ\theta rotates the pair (S1,S2)(S_1, S_2) through 2θ2\theta, leaving S0S_0 and S3S_3 unchanged, so MP(θ)=RS(2θ) MP(0) RS(−2θ).\mathbf{M}_P(\theta) = \mathbf{R}_S(2\theta)\,\mathbf{M}_P(0)\,\mathbf{R}_S(-2\theta).RS(ϕ)=(10000cos⁡ϕ−sin⁡ϕ00sin⁡ϕcos⁡ϕ00001).\mathbf{R}_S(\phi) = \begin{pmatrix}1&0&0&0\\0&\cos\phi&-\sin\phi&0\\0&\sin\phi&\cos\phi&0\\0&0&0&1\end{pmatrix}. ((a)) Carry out the two matrix multiplications for J(θ)\mathbf{J}(\theta) symbolically to obtain a general formula in θ\theta and Γ\Gamma. A quarter-wave retarder (Γ=90∘\Gamma = 90^\circ) has its fast axis at θ=30∘\theta = 30^\circ to xx. Horizontal linearly polarised light, Jones vector (1,0)(1,0), is incident. Find the two output amplitudes and their phase difference, and say whether the output is linear, circular or elliptical. ((b)) Carry out the two matrix multiplications for MP(θ)\mathbf{M}_P(\theta) symbolically to obtain a general formula in θ\theta. Unpolarised light of irradiance 11 (Stokes vector (1,0,0,0)(1,0,0,0)) falls on a polariser with its transmission axis at θ=30∘\theta = 30^\circ to xx. Find the output Stokes vector and its degree of polarisation.

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