## Abstract

Materials showing birefringence and polarization selective absorption (dichroism) affect the polarization state of incoming light in a peculiar way, quite different from the one exhibited by phase retarders like waveplates. In this paper, we report on the characterization of a Polymer LIquid CRYstal Polymer Slices (POLICRYPS) diffraction grating used as a dichroic phase retarder; the dichroic behaviour of the grating is due to the polarization-dependent diffraction efficiency. Experimental data are validated with a theoretical model based on the Jones matrix formalism, while the grating behavior is modeled by means of the dichroic matrix. In this way, the birefringence of the analyzed structure is easily obtained. For comparison purposes, also two systems different from POLICRYPS have been fabricated and tested.

© 2010 OSA

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### Equations (11)

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(1)
$$L=\left(\begin{array}{cc}H& 0\\ 0& V\end{array}\right)$$
(2)
$$A\left(\beta \right)=\left(\begin{array}{cc}\mathrm{cos}\beta & -\mathrm{sin}\beta \\ \mathrm{sin}\beta & \mathrm{cos}\beta \end{array}\right)\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}\mathrm{cos}\beta & \mathrm{sin}\beta \\ -\mathrm{sin}\beta & \mathrm{cos}\beta \end{array}\right)=\left(\begin{array}{cc}{\mathrm{sin}}^{2}\beta & -\mathrm{sin}\beta \mathrm{cos}\beta \\ -\mathrm{sin}\beta \mathrm{cos}\beta & {\mathrm{cos}}^{2}\beta \end{array}\right)$$
(3)
$$M=\left(\begin{array}{cc}{e}^{i\raisebox{1ex}{$\delta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}& 0\\ 0& {e}^{-i\raisebox{1ex}{$\delta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\end{array}\right)$$
(4)
$$\begin{array}{l}{\tilde{E}}_{out}^{}=\left(\begin{array}{cc}{\mathrm{sin}}^{2}\beta & -\mathrm{sin}\beta \mathrm{cos}\beta \\ -\mathrm{sin}\beta \mathrm{cos}\beta & {\mathrm{cos}}^{2}\beta \end{array}\right)\left[\left(\begin{array}{cc}H& 0\\ 0& V\end{array}\right)\left(\begin{array}{cc}{e}^{i\raisebox{1ex}{$\delta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}& 0\\ 0& {e}^{-i\raisebox{1ex}{$\delta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\end{array}\right)\right]\frac{\sqrt{2}}{2}\sqrt{{I}_{inc}}\left(\begin{array}{c}-1\\ 1\end{array}\right)=\\ =\frac{\sqrt{2}}{2}\sqrt{{I}_{inc}}\left(\begin{array}{c}-H{e}^{i\frac{\delta}{2}}{\mathrm{sin}}^{2}\beta -V{e}^{-i\frac{\delta}{2}}\mathrm{sin}\beta \mathrm{cos}\beta \\ H{e}^{i\frac{\delta}{2}}\mathrm{sin}\beta \mathrm{cos}\beta +V{e}^{-i\frac{\delta}{2}}{\mathrm{cos}}^{2}\beta \end{array}\right)\end{array}$$
(5)
$${I}_{out}\left(\beta \right)={\tilde{E}}_{out}\left(\beta \right)\cdot {\tilde{E}}_{out}^{*}\left(\beta \right)=\frac{{I}_{inc}}{2}\left[{H}^{2}{\mathrm{sin}}^{2}\beta +{V}^{2}{\mathrm{cos}}^{2}\beta +HV\mathrm{sin}2\beta \mathrm{cos}\delta \right]$$
(6)
$${I}_{out}\left(\beta =\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)={I}_{inc}\raisebox{1ex}{${H}^{2}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$$
(7)
$${I}_{out}\left(\beta =0\right)={I}_{inc}\raisebox{1ex}{${V}^{2}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$$
(8)
$$H=\sqrt{\frac{2{I}_{out}\left(\beta =\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}{{I}_{inc}}}$$
(9)
$$V=\sqrt{\frac{2{I}_{out}\left(\beta =0\right)}{{I}_{inc}}}$$
(10)
$$\mathrm{cos}\delta =\frac{1}{HV}\left[\frac{2{I}_{out}\left(\beta =\pi /4\right)}{{I}_{inc}}-\frac{{H}^{2}+{V}^{2}}{2}\right]$$
(11)
$$\frac{{H}^{2}}{{V}^{2}}=\frac{{I}_{out}\left(\beta =\frac{\pi}{2}\right)}{{I}_{out}\left(\beta =0\right)}$$