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Optical measurement in a curved optical medium with optical birefringence and anisotropic absorption

Yi-Hsin Lin, Yu-Jen Wang, Hung-Chun Lin, Ming-Long Lee, and Po-Lun Chen

Open Access Open Access

Abstract

The planar optical mediums with properties of either birefringence (i.e., waveplates) or anisotropic absorption (i.e., polarizers) are well studied. However, how a beam propagates in a birefringent curved medium with anisotropic absorption, especially for curved-sheet polarizers, still needs to be investigated. In this paper, we study optical wave propagation through a curved-birefringent medium with anisotropic absorption. We built an optical model based on the Mueller matrix to predict the spatial distributions in light intensity and polarization when light propagates in a curved-birefringent medium with anisotropic absorption. To demonstrate how to use the optical model, the experiments based on ellipsometry are also performed. The impact of this study is to analyze the light propagation in birefringent-curved medium with anisotropic absorption, which could affect the performance of curved liquid crystal devices with curved polarizers, such as curved liquid crystal displays (LCDs), flexible LCDs, and flexible LC lenses.

© 2021 Optical Society of America under the terms of the OSA Open Access Publishing Agreement

1. Introduction

Crystal optics dealing with optical wave propagations in birefringent and anisotropic media have been studied theoretically and experimentally [1]. Usually, researchers discuss a wave propagation in optical media with properties of anisotropic absorptions, such as polarizers, in a separate way from a wave propagation in birefringent optical media. However, anisotropic birefringence as well as anisotropic absorption for an optical element is unavoidable in many engineering applications. For example, a polaroid polarizer (sheet polarizer) with anisotropic absorption encounters mechanically-induced-birefringence when a sheet polarizer is attached on a flat or curved substrate under mechanical stresses at high temperature. Typically, a sheet polarizer consists of several layers: a protective polymeric film, two tri-acetate cellulose films (so-called TAC films), a layer of PVA (Polyvinyl alcohol), a layer of adhesive glue, and a release polymeric film. Usually, the protective polymeric film and the release polymeric film are removed from the polarizer in order to attach the polarizer on a substrate, such as a flexible substrate or a glass substrate. During attachment, mechanically-induced-birefringence of the polarizer could affect optical performance in many applications, such as curved liquid crystal displays (LCDs), foldable LCDs, LCDs for vehicle applications, augmented reality and ophthalmic applications [2–3]. As a result, how to predict light propagation in such a birefringent-curved medium with an anisotropic absorption, such as spatial distributions in light intensity and polarization, is necessary. Once we understand the wave propagation in a birefringent-curved medium with anisotropic absorption, we are able to improve or compensate the optical performance of optical systems. In literatures, regarding to curved birefringent media, researchers proposed or demonstrated ray-tracing method for uniaxial optical surfaces [4], and explored different methods of measuring phase retardation and birefringence with bending stress [5–8]. As to curved medium with anisotropic absorption, researchers proposed a method of coupled-wave analysis of a curved wire-grid polarizer [9], but the change of polarization due to bending stress was not considered. To date, studies have yet to report a model describing a birefringent curved medium with anisotropic absorption, especially for sheet polarizers. The properties of output light from curved sheet polarizers, such as polarization state, phase retardation, and intensity, should be considered. In this paper, we investigate wave propagation in a birefringent-curved medium with anisotropic absorption. Based on calculus of Muller matrix (or so-called Stokes calculus method), the optical model for a birefringent-curved medium with anisotropic absorption is built to predict the spatial distributions in light intensity and polarization of an optical element. The impact of this study is to pave a way to analyze the light propagation in birefringent-curved medium with anisotropic absorption, such as curved polarizers, which could affect the performance of curved LCDs, foldable LCDs, curved contact lenses, and curved LC devices with curved polarizers.

2. Optical model

To describe beam propagation through a birefringent medium, Stokes calculus method is commonly exploited [1]. Assume a polarized incident plane wave or light ($\vec{E}$) propagating along z-direction is expressed as:

$$\vec{E} = \hat{x} \cdot {E_x} \cdot {e^{ - j(k \cdot z + {\delta _x})}} + \hat{y} \cdot {E_y} \cdot {e^{ - j(k \cdot z + {\delta _y})}},$$
where Ex and Ey are amplitudes, δx and δy are phases, and k is wave number. A Stoke vector ${\vec{S}_{in}}$ of an incident light consisting of 4 components: S0, S1, S2, and S3 is used to describe polarization properties of a light beam [2–3]. S0 is total intensity (or irradiance) of light. S1 stands for the excess of horizontal polarization (i.e. x-direction) over vertical polarization (i.e. y-direction), S2 is an intensity difference of 45°-linear-polarization minus 135°-linear-polarization, and S3 is an intensity difference of right-circularly-polarization minus left-circularly-polarization. According to Eq. (1), the Stoke vector ${\vec{S}_{in}}$ of incident light could be further expressed as [2–3]:
$${\vec{S}_{in}} = \left( {\begin{array}{c} {{S_0}}\\ {{S_1}}\\ {{S_2}}\\ {{S_3}} \end{array}} \right) = \left( {\begin{array}{c} {E_x^2 + E_y^2}\\ {E_x^2 - E_y^2}\\ {2E_x^2 \cdot E_y^2 \cdot \cos \delta }\\ {2E_x^2 \cdot E_y^2 \cdot \sin \delta } \end{array}} \right),$$
where δ equals to δy-δx. When light propagates in an optical medium, the output and input Stokes parameter (${\vec{S}_{out}}$ and ${\vec{S}_{in}}$) are connected by a Mueller matrix (Ms), as shown in Eq. (3):
$${\vec{S}_{out}} = {M_s} \cdot {\vec{S}_{in}}.$$
In order to analyze optical wave propagation in a birefringent-curved medium with anisotropic absorption, we could simplify the medium as a multilayered structure, as depicted in Fig. 1(a). A birefringent-curved medium with anisotropic absorption could be treated as an equivalent medium consisting of an absorptive layer and a birefringent layer. The birefringent-curved medium is attached on a curved substrate with a refractive index of ng. The refractive index of the absorptive layer is set as na. The birefringent layer is a uniaxial layer with an extraordinary refractive index ne and an ordinary refractive index no. We also add a compensated lens for correcting the propagation direction, which would help to validate with experiments. The radius of curvature of the compensated lens is identical to that of the curved substrate with a refractive index of ng in Fig. 1(a) in order to avoid the divergence of beam which also benefits the light collimation during measurements. T1 to T5 represents five interfaces. The detailed beam propagation between T2 and T3 is illustrated in Fig. 1(b). According to Fig. 1(a), the Mueller matrix of the whole system M is:
$$M = {M_{T5}} \cdot {M_{T4 - T5}} \cdot {M_{T4}} \cdot {M_{T3 - T4}} \cdot {M_{T3}} \cdot {M_{T2 - T3}} \cdot {M_{T2}} \cdot {M_{T1 - T2}} \cdot {M_{T1}},$$
where MTi (i=1∼5) represents the corresponding Mueller matrix of an interface, MTi-Ti+1 (i=1∼4) represents the corresponding Mueller matrix as light propagates from Ti to Ti+1.

 figure: Fig. 1.

Fig. 1. (a) The equivalent optical system of a birefringent-curved medium with anisotropic absorption. A compensated lens is used for light collimation. (b) is the detailed optical layout between T2 and T3 in (a). T1 to T5 represents interfaces. C1 is transmissive axis of the anisotropic absorptive layer. C2 is the optic axis of the birefringent layer. $\vec{k}$ is wave vector of a plane wave. ei and oi stands for e-wave and o-wave at either the absorptive layer (i=1) or the birefringent layer (i=2). p and s mean p-wave and s-wave, respectively. The dash lines in (b) stand for the normal directions of the interfaces.

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When light propagates normally from one isotropic medium to another isotropic medium, a polarization of light remains unchanged. As a result, the Mueller matrix is a unit matrix:

$${M_{T1 - T2}} = {M_{T3 - T4}} = {M_{T4 - T5}} = \left[ {\begin{array}{cccc} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{array}} \right].$$
The isotropic medium between T3 and T4 surface is air. According to the Fresnel equations, the transmittance (A) could be estimated at a normal incidence as $A = 1 - {[({n_t} - {n_i})/({n_t} + {n_i})]^2}$, where ni and nt are refractive indices of medium that light is passing out of and passing into, respectively. Since the optical media that light is passing out of and passing into are the air and the glass substrate, nt = ng = 1.5, and ni = 1. As a result, we could simplify MT1 and MT5 to a transmittance AT1 = AT5 ∼0.96 since only the transmittance changes but the polarization does not change at T1 and T5. Thus, Eq. (4) turns out:
$$M = {A_{T1}} \cdot {A_{T5}} \cdot {M_{T4}} \cdot {M_{T3}} \cdot {M_{T2 - T3}} \cdot {M_{T2}}.$$
Next, we discuss the optical propagation between T2 an T3 as depicted in Fig. 1(b). Here, we start from extended Jones matrix and then convert the extended Jones matrix back to Mueller matrix. Assume the coordinates between T2 and T3 is (x’, y’, z’) which may differ from the coordinate of (x, y, z) as light travels in air (before T1 interface). Both y-z plane and y’-z’ plane are the plane of incidence. When light obliquely impinges into an interface T2 with a wave vector of ${\vec{k}_o}$, the polarization of s-wave (i.e. x-linearly polarized light) is parallel to the polarization of s’-wave (i.e. x’-linearly polarized light). p-wave and p’-wave could be expressed as a function of incident angle in terms of unit vectors of $\hat{y}^{\prime}$ and $\hat{z}^{\prime}$ in Eq. (7).
$$\left\{ \begin{array}{l} \hat{s} = \hat{s}^{\prime} = \hat{x}^{\prime}\\ \hat{p} = \cos {\theta_i} \cdot \hat{y}^{\prime} - \sin {\theta_i} \cdot \hat{z}^{\prime}\\ \hat{p}^{\prime} = \cos {\theta_r} \cdot \hat{y}^{\prime} - \sin {\theta_r} \cdot \hat{z}^{\prime} \end{array} \right.,$$
where θi is incident angle to T2-interface, and θr is the refraction angle of T3-interface (Fig. 1(b)). Two unit vectors, ${\vec{c}_1}$ and ${\vec{c}_2}$, could be expressed as:
$${\vec{c}_1} = \sin [{\theta _{pol}}] \cdot \cos [{\phi _{pol}}] \cdot \hat{x}^{\prime} + \sin [{\theta _{pol}}]\sin [{\phi _{pol}}] \cdot \hat{y}^{\prime} + \cos [{\theta _{pol}}] \cdot \hat{z}^{\prime},$$
$${\vec{c}_2} = \sin [{\theta _c}] \cdot \cos [{\phi _c}] \cdot \hat{x}^{\prime} + \sin [{\theta _c}] \cdot \sin [{\phi _c}] \cdot \hat{y}^{\prime} + \cos [{\theta _c}] \cdot \hat{z}^{\prime},$$
where θpol and ϕpol are the polar and the azimuthal angle of ${\vec{c}_1}$. θc and ϕc are polar and azimuthal angle of ${\vec{c}_2}$. From directions of ${\vec{k}_{o1}}$, ${\vec{k}_{e1}}$, ${\vec{k}_{o2}}$, ${\vec{k}_{e2}}$, ${\vec{c}_1}$, and ${\vec{c}_2}$, we could obtain the polarization directions of o-wave and e-wave:
$$\left\{ \begin{array}{l} {{\hat{o}}_1} = \frac{{{{\vec{k}}_{o1}} \times {{\vec{c}}_1}}}{{|{{{\vec{k}}_{o1}} \times {{\vec{c}}_1}} |}},{{\hat{e}}_1} = \frac{{{{\vec{o}}_1} \times {{\vec{k}}_{e1}}}}{{|{{o_1} \times {k_{e1}}} |}}\\ {{\hat{o}}_2} = \frac{{{{\vec{k}}_{o2}} \times {{\vec{c}}_2}}}{{|{{{\vec{k}}_{o2}} \times {{\vec{c}}_2}} |}},{{\hat{e}}_2} = \frac{{{{\vec{o}}_2} \times {{\vec{k}}_{e2}}}}{{|{{{\vec{o}}_2} \times {{\vec{k}}_{e2}}} |}} \end{array} \right.,$$
where ${\hat{o}_1}$, ${\hat{e}_1}$, ${\hat{o}_2}$, and ${\hat{e}_2}$ are unit vectors standing for polarization directions of o-wave or e-wave in either the anisotropic absorptive layer or the birefringent layer in Fig. 1(b). In Fig. 1(b), θo and θe stands for the refractive angles of o-wave and e-wave when the light is propagating in an optical anisotropic medium. The curved substrate is usually a glass substrate with refractive index of ng ∼1.5. The anisotropic absorptive layer is a polarizer (no1 = ne1 ∼1.5). The birefringence layer is a wave plate and we assume that ne2 > no2 ∼1.5. Since ng ∼ no1 = ne1, we do not consider the refraction of light at T2-interface, and that indicates θo∼θi (in Fig. 1(b)). Thereafter, the incident angle to the birefringence layer is also θo, and ${\vec{k}_{o1}} = {\vec{k}_{o2}}\sim {\vec{k}_o}$ in Eq. (10). Assume ne2 - no2 << no2, θe∼θo. As a result, Eq. (10) could be expressed as:
$$\left\{ \begin{array}{l} {{\hat{o}}_1} = \frac{{{{\vec{k}}_0} \times {{\vec{c}}_1}}}{{|{{{\vec{k}}_0} \times {{\vec{c}}_1}} |}},{{\hat{\rm e}}_1} = \frac{{{{\hat{o}}_1} \times {{\vec{k}}_0}}}{{|{{{\hat{o}}_1} \times {{\vec{k}}_0}} |}}\\ {{\hat{o}}_2} = \frac{{{{\vec{k}}_0} \times {{\vec{c}}_2}}}{{|{{{\vec{k}}_0} \times {{\vec{c}}_2}} |}},{{\hat{\rm e}}_2} = \frac{{{{\hat{o}}_2} \times {{\vec{k}}_0}}}{{|{{{\hat{o}}_2} \times {{\vec{k}}_0}} |}} \end{array} \right..$$
The relation between linear polarizations (Es, Ep) before T2-interface and (Es’, Ep’) after T3-interface could be written according to the extended Jones matrix method:
$$\left( {\begin{array}{cc} {{E_s}^{\prime}}\\ {{E_p}^{\prime}} \end{array}} \right) = {J_{T3}} \cdot {J_{T2 - T3}} \cdot {J_{T2}}\left( {\begin{array}{cc} {{E_s}}\\ {{E_p}} \end{array}} \right),$$
where JT2, JT3, and JT2-T3 are extended Jones matrices for two interfaces T2 and T3, and between those two interfaces, respectively. Although e-wave and o-wave light might split up, the splitting angle of two waves is small (<1 degree) in fact, and the detective area (∼1 cm) of a detector is larger than two slightly splitting light spots. Thus, we could assume the measured light intensity is a sum of two light intensities: ${|{{E_s}^{\prime}} |^2} + {|{{E_p}^{\prime}} |^2}$. Under a small angle approximation, we could ignore the reflection between the absorptive layer and the birefringent layer. Equation (12) could be further extended as:
$$\begin{array}{l} \left( {\begin{array}{@{}cc@{}} {{E_s}^{\prime}}\\ {{E_p}^{\prime}} \end{array}} \right) = \left[ {\begin{array}{@{}cccc@{}} {\hat{s}^{\prime} \cdot {{\hat{e}}_2} \cdot {t_s}^{\prime}}&{\hat{s}^{\prime} \cdot {{\hat{o}}_2} \cdot {t_s}^{\prime}}\\ {\hat{p}^{\prime} \cdot {{\hat{e}}_2} \cdot {t_p}^{\prime}}&{\hat{p}^{\prime} \cdot {{\hat{o}}_2} \cdot {t_p}^{\prime}} \end{array}} \right] \cdot \left[ {\begin{array}{@{}cc@{}} {{e^{ - j \cdot {k_{ez}} \cdot d}}}&0\\ 0&{{e^{ - j \cdot {k_{oz \cdot }}d}}} \end{array}} \right] \cdot \left[ {\begin{array}{@{}cccc@{}} {{{\hat{e}}_1} \cdot {{\hat{e}}_2}}&{{{\hat{o}}_1} \cdot {{\hat{e}}_2}}\\ {{{\hat{e}}_1} \cdot {{\hat{o}}_2}}&{{{\hat{o}}_1} \cdot {{\hat{o}}_2}} \end{array}} \right] \cdot \left[ {\begin{array}{@{}cccc@{}} {{t_e}}&0\\ 0&{{t_o}} \end{array}} \right]\\ \cdot \left[ {\begin{array}{@{}cccc@{}} {{{\hat{e}}_1} \cdot \hat{s} \cdot {t_s}}&{{{\hat{e}}_1} \cdot \hat{p} \cdot {t_p}}\\ {{{\hat{o}}_1} \cdot \hat{s} \cdot {t_s}}&{{{\hat{o}}_1} \cdot \hat{p} \cdot {t_p}} \end{array}} \right] \cdot \left( {\begin{array}{@{}c@{}} {{E_s}}\\ {{E_p}} \end{array}} \right), \end{array}$$
where te and to are absorption coefficients as the oscillation direction of the linear polarization is parallel and perpendicular to the transmissive axis in the absorptive layer, respectively [2–3]. koz and kez are wave numbers of o-wave and e-wave in the birefringent layer [1,10]. ts and tp are transmission coefficients for s-wave and p-wave when light propagates through T2-interface. ts’ and tp’ are transmission coefficients for s’-wave and p’-wave when light propagates through T3-interface. d is the thickness of the birefringent layer, and λ is the wavelength. In Eq. (13), ts, tp, ts’, and tp’ are listed in Eq. (14).
$${t_s} = \frac{{2{n_g} \cdot \cos {\theta _i}}}{{{n_g} \cdot \cos {\theta _i} + {n_o} \cdot \cos {\theta _o}}},$$
$${t_p} = \frac{{2{n_g} \cdot \cos {\theta _i}}}{{{n_g} \cdot \cos {\theta _o} + {n_o} \cdot \cos {\theta _i}}},$$
$${t_s}^{\prime} = \frac{{2 \cdot {n_o} \cdot \cos {\theta _i}}}{{{n_o} \cdot \cos {\theta _i} + \cos {\theta _r}}},$$
$${t_p}^{\prime} = \frac{{2 \cdot {n_e} \cdot \cos {\theta _i}}}{{{n_o} \cdot \cos {\theta _r} + \cos {\theta _i}}},$$
In Eq. (14)-(17), ts = tp = 1 due to ng = no and θi = θo. kez and koz in Eq. (13) could be solved from Eq. (18).
$${k_{ez}} = \frac{{v + \sqrt {{v^2} - 4 \cdot u \cdot w} }}{{2 \cdot u}},$$
$${k_{oz}} = k \cdot \cos {\theta _i},$$
where $w = \frac{{{k_d}^2{{\sin }^2}{\theta _c} + {k_{eb}}^2}}{{{n_e}^2}} + \frac{{{k_d}^2\sin {\theta _c}^2}}{{{n_o}^2}} - {\left( {\frac{{2\pi }}{\lambda }} \right)^2}$, $u = \sin {({\theta _c})^2}/{n_e}^2 + \cos {({\theta _c})^2}/{n_o}^2$, $v = {k_d} \cdot \sin(2{\theta _c}) \cdot ({1/{n_e}^2 - 1/{n_o}^2} )$, ${k_d} = \frac{{2\pi }}{\lambda }\sin {\theta _i}\sin {\phi _c}$, and ${k_{eb}} = \frac{{2\pi }}{\lambda }\sin {\theta _i}\cos {\phi _c}$.

As a result, Eq. (13) could be calculated and then expressed as a general form.

$$\left[ {\begin{array}{c} {{E_s}^{\prime}}\\ {{E_p}^{\prime}} \end{array}} \right] = \left[ {\begin{array}{cc} {{J_{11}}}&{{J_{12}}}\\ {{J_{21}}}&{{J_{22}}} \end{array}} \right] \cdot \left[ {\begin{array}{c} {{E_s}}\\ {{E_p}} \end{array}} \right].$$
By means of a conversion from extended Jones Matrix to Muller matrix as well as a consideration of modified term ${{{\cos {\theta _r}} / {{n_i} \cdot \cos \theta }}_i}$ due to a variation of the projected area from oblique incidence, ${M_{T3}} \cdot {M_{T2 - T3}} \cdot {M_{T2}}$ is [2–3]:
$$\begin{array}{l} {M_{T3}} \cdot {M_{T2 - T3}} \cdot {M_{T2}}\\ = \frac{{\cos {\theta _r}}}{{{n_i} \cdot \cos {\theta _i}}} \cdot \left( {\begin{array}{@{}cccc@{}} 1&0&0&1\\ 1&0&0&{ - 1}\\ 0&1&1&0\\ 0&i&{ - i}&0 \end{array}} \right) \cdot \left( {\begin{array}{@{}cccc@{}} {{J_{11}}{J_{11}}^\ast }&{{J_{11}}{J_{12}}^\ast }&{{J_{12}}{J_{11}}^\ast }&{{J_{12}}{J_{12}}^\ast }\\ {{J_{11}}{J_{21}}^\ast }&{{J_{11}}{J_{22}}^\ast }&{{J_{12}}{J_{21}}^\ast }&{{J_{12}}{J_{22}}^\ast }\\ {{J_{21}}{J_{11}}^\ast }&{{J_{21}}{J_{12}}^\ast }&{{J_{22}}{J_{11}}^\ast }&{{J_{22}}{J_{12}}^\ast }\\ {{J_{21}}{J_{21}}^\ast }&{{J_{21}}{J_{22}}^\ast }&{{J_{22}}{J_{21}}^\ast }&{{J_{22}}{J_{22}}^\ast } \end{array}} \right) \cdot \frac{1}{2}\left( {\begin{array}{@{}cccc@{}} 1&1&0&0\\ 0&0&1&{ - i}\\ 0&0&1&i\\ 1&{ - 1}&0&0 \end{array}} \right). \end{array}$$
Before we combine Eq. (21) with Eq. (6), a coordinate transformation (i.e. a rotational matrix MR) is needed in order to transform x-y-z coordinate to x’-y’-z’ coordinate.
$$\begin{array}{l} M{^{\prime}_{T2 - T3}} = {M_{T3}} \cdot {M_{T2 - T3}} \cdot {M_{T2}} \cdot {M_R}(\phi )\\ = {M_{T3}} \cdot {M_{T2 - T3}} \cdot {M_{T2}} \cdot \left[ {\begin{array}{cccc} 1&0&0&0\\ 0&{\cos (2\phi )}&{\sin (2\phi )}&0\\ 0&{ - \sin (2\phi )}&{\cos (2\phi )}&0\\ 0&0&0&1 \end{array}} \right], \end{array}$$
where $\phi$ is the rotational angle between x-y-z coordinate and x’-y’-z’ coordinate, and $\phi$ depends on the location of the wave and radius of curvature of the substrate. Thereafter, Eq. (6) turns out:
$$M = {A_{T1}} \cdot {A_{T5}} \cdot {M_{T4}} \cdot M{^{\prime}_{T2 - T3}}.$$
In Eq. (23), MT4 is still unknown. MT4 indicates the effects of refraction and reflection from the compensated lens. We assume that the incident angle at the T4-interface is ${\theta _i}^{\prime}$ and refraction angle is ${\theta _r}^{\prime}$. By considering interfacial refraction and reflection using Fresnel’s equations [1,10], the calculated Mueller matrix MT4 is [10–12]:
$${M_{T4}} = \left[ {\begin{array}{cccc} {\frac{{{\tau_s} + {\tau_p}}}{2}}&{\frac{{{\tau_s} - {\tau_p}}}{2}}&0&0\\ {\frac{{{\tau_s} - {\tau_p}}}{2}}&{\frac{{{\tau_s} + {\tau_p}}}{2}}&0&0\\ 0&0&{{{({{\tau_s}{\tau_p}} )}^{1/2}}}&0\\ 0&0&0&{{{({{\tau_s}{\tau_p}} )}^{1/2}}} \end{array}} \right],$$
where τs and τp are effective transmission coefficients and could be expressed as:
$${\tau _s} = \left( {\frac{{\tan {\theta_i}^{\prime}}}{{\tan {\theta_r}^{\prime}}}} \right){\left( {\frac{{2\sin {\theta_r}^{\prime}\cos {\theta_i}^{\prime}}}{{\sin ({\theta_i}^{\prime} + {\theta_r}^{\prime})}}} \right)^2}.$$
$${\tau _p} = \left( {\frac{{\tan {\theta_i}^{\prime}}}{{\tan {\theta_r}^{\prime}}}} \right){\left( {\frac{{2\sin {\theta_r}^{\prime}\cos {\theta_i}^{\prime}}}{{\sin ({\theta_i}^{\prime} + {\theta_r}^{\prime})\cos ({\theta_i}^{\prime} - {\theta_r}^{\prime})}}} \right)^2}.$$
After the T4-interface, light is collimated, we transform the x’-y’-z’ coordinate to the x-y-z coordinate: $M{^{\prime}_{T4}} = M( - \phi ) \cdot {M_{T4}}$. Equation (22) turns out:
$$M = {A_{T1}} \cdot {A_{T5}} \cdot M{^{\prime}_{T4}} \cdot M{^{\prime}_{T2 - T3}}.$$
Equation (27) is able to describe the optical wave propagation through a birefringent-curved medium with anisotropic absorption and a compensated lens.

From the model described above, we could know the wave propagation at each layer by calculation of Muller matrix from input and output stokes vectors as long as we know the radius of curvature of the substrate and the position of light at the substrate. Moreover, this optical model could help analyzing anisotropic absorption and birefringence of a birefringent-curved media which may be fabricated by stretching or molding under heat and pressure.

To use the model with experiments, first, the Mueller matrix of overall system shown in Fig. 1(a) should be obtained. Next, we need to evaluate the relationship between te and to by assuming MT2-T3 is a polarizer (MT2-T3 = Mpol) and inversing the matrix of Eq. (27) as

$$\begin{array}{l} M( - \phi ) \cdot {M_{pol}} \cdot M(\phi ) = \\ {({{A_{T5}} \cdot M( - \phi ) \cdot {M_{T4}} \cdot {M_{T3}} \cdot M(\phi )} )^{ - 1}} \cdot {M_{\exp }} \cdot {({M( - \phi ) \cdot {M_{T2}} \cdot M(\phi ) \cdot {A_{T1}}} )^{ - 1}}, \end{array}$$
where Mexp is the Mueller matrix of whole system. AT1 and AT5 can be obtained from Fresnel’s equations; MT2, MT3, and MT4 can be calculated using Eq. (24). With Eq. (28), we can get relationship between te and to as ${M_{pol}}(1,1) = ({t_e}^2 + {t_o}^2)/2$. Next, we consider the birefringence part and inveres the Eq. (27) again using Eq. (29).
$${M_{T3}} \cdot {M_{T2 - T3}} \cdot {M_{T2}} = {({{A_{T5}} \cdot M( - \phi ) \cdot {M_{T4}}} )^{ - 1}} \cdot {M_{\exp }} \cdot {({M(\phi ) \cdot {A_{T1}}} )^{ - 1}}.$$
Because we have birefrignce now, MT2 and MT3 can not be calcuated with Eq. (24) any more. At this point, experimental result of Eq. (29) is compared with therotical value stated in Eq. (21) and relationship ${M_{pol}}(1,1) = ({t_e}^2 + {t_o}^2)/2$ with numeriacal approach. In numerical approach, te, to, ${\phi _{pol}}$, ${\phi _{wp}}$, and phase retardation of birefrigent layer should be obtained by fitting.

3. Experimental validation of optical model

To show how to use the optical model in the experiments, we setup an ellipsometry for measurement of Mueller matrix and stokes vectors [3–8], as depicted in Fig. 2. In experiments, we used a laser (unpolarized light, wavelength = 632.8 nm, Newport, Model: N-LHR-151), three sheet polarizers (P1, P2, and P3 in Fig. 3), two quarter-wave plates (Q1 and Q2, Newport 10RP54-1 ¼ waveplate for 400-700 nm) mounted to motorized rotation stages (Newport PR50CC) connected to a 3-axes motion controller (Newport Model ESP301) in order to adjust the fast axes of quarter-wave plates, a beam expander (Newport 10x Beam Expander), two optical power detectors (Newport 918D-SL-0D3R) connected to a dual channel power meter (Newport Model 2936-R) as illustrated in Fig. 2. The ellipsometry consists of two parts: one is polarization state generation (PSG) and the other is polarization state detection (PSD). By means of adjusting the angle (${\alpha _{pol1}}$) of transmissive axis of P1 with respect to x-axis and the angle (${\alpha _Q}$) of the fast axis of Q1 with respect to x-axis, PSG could generate arbitrary polarization state of light. As to PSD, by means of measuring of light intensity at different angle (${\alpha _{pol2}}$) of transmissive axis of P2 with respect to x-axis and different angle ($\alpha {^{\prime}_\textrm{Q}}$) of the fast axis of Q1 with respect to x-axis, PSD could help to obtain Stokes vectors of light. The optical power detector #2 in Fig. 2 is used to monitor the light intensity of input light. We changed the angle of fast axes of two quarter-wave plates pairly ($\alpha {^{\prime}_\textrm{Q}} = 5{\alpha _\textrm{Q}}$) and recorded the corresponding optical powers. Thereafter, we could calaulate Mueller matrix of the sample.

 figure: Fig. 2.

Fig. 2. Experimental setup of the ellipsometry for measurement of Mueller matrix. P1, P2, and P3 are polarizers. Q1 and Q2 are quarter-wave plates. PSG and PSD stand for polarization state generation and polarization state detection, respectively.

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 figure: Fig. 3.

Fig. 3. Measured optical power as a function of αQ without the sample.

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In Fig. 2, the laser emits unpolarized light with a stokes vector of ${\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} \over S} _{i,PSG}}$. ${\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} \over S} _{f,PSG}}$ represents a stokes vector after light passes through P1 and Q1 and could be calculated as

$$\begin{array}{l} {{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} \over S} }_{f,PSG}} = {W_{\textrm{Q}1}} \cdot {W_{P1}} \cdot {{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} \over S} }_{i,PSG}}\\ = \left( {\begin{array}{@{}cccc@{}} 1&0&0&0\\ 0&{{{\cos }^2}2{\alpha_Q} + \cos {\Gamma _Q} \cdot {{\sin }^2}2{\alpha_Q}}&{(1 - \cos {\Gamma _Q}) \cdot \sin 2{\alpha_Q} \cdot \cos 2{\alpha_Q}}&{ - \sin {\Gamma _Q} \cdot \sin 2{\alpha_Q}}\\ 0&{(1 - \cos {\Gamma _Q}) \cdot \sin 2{\alpha_Q} \cdot \cos 2{\alpha_Q}}&{{{\sin }^2}2{\alpha_Q} + \cos {\Gamma _Q} \cdot {{\cos }^2}2{\alpha_Q}}&{\sin {\Gamma _Q} \cdot \cos 2{\alpha_Q}}\\ 0&{\sin {\Gamma _Q} \cdot \sin 2{\alpha_Q}}&{ - \sin {\Gamma _Q} \cdot \cos 2{\alpha_Q}}&{\cos {\Gamma _Q}} \end{array}} \right)\\ \cdot \left( {\begin{array}{@{}cccc@{}} A&{B \cdot \cos 2{\alpha_{pol1}}}&{B \cdot \sin 2{\alpha_{pol1}}}&0\\ {B \cdot \cos 2{\alpha_{pol1}}}&{A \cdot {{\cos }^2}2{\alpha_{pol1}} + C \cdot {{\sin }^2}2{\alpha_{pol1}}}&{(A - C) \cdot \sin 2{\alpha_{pol1}} \cdot \cos 2{\alpha_{pol1}}}&0\\ {B \cdot \sin 2{\alpha_{pol1}}}&{(A - C) \cdot \sin 2{\alpha_{pol1}} \cdot \cos 2{\alpha_{pol1}}}&{A \cdot {{\sin }^2}2{\alpha_{pol1}} + C \cdot {{\cos }^2}2{\alpha_{pol1}}}&0\\ 0&0&0&C \end{array}} \right) \cdot \left[ {\begin{array}{@{}c@{}} 1\\ 0\\ 0\\ 0 \end{array}} \right]. \end{array}$$
In Eq. (30), A, B, and C are:
$$\left\{ \begin{array}{l} A = (t_x^2 + t_y^2)/2\\ B = (t_x^2 - t_y^2)/2\\ C = {t_x} \cdot {t_y} \end{array} \right.,$$
WQ1 and WP1 represent Muller matrices of Q1 and P1, respectively. ${\Gamma _Q}$ is phase retardation of Q1. tx and ty are absorption coefficients when oscillation of linear polarization is parallel to x-axis and y-axis, respectively. We set the transmissive axis of the polarizer P1 parallel to x-axis (i.e. ${\alpha _{pol1}}\textrm{ = 0}$), and the transmissive axis of the polarizer P2 is set to be parallel to y-axis (i.e., ${\alpha _{pol2}}\textrm{ = }\pi \textrm{/2}$). Assume that tx >> ty and tx ∼1. As a result, ${S_{f,PSG}}$ could be obtained from Eq. (30) after putting parameters: ${\alpha _{pol1}} = 0$, ${\Gamma _Q} = \pi /2$ radians, ${t_x} \approx 1$,and ${t_y} \approx 0$.
$${\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} \over S} _{f,PSG}} = \frac{1}{2}\left[ {\begin{array}{c} 1\\ {{{\cos }^2}2{\alpha_Q}}\\ {\cos 2{\alpha_Q} \cdot \sin 2{\alpha_Q}}\\ {\sin 2{\alpha_Q}} \end{array}} \right].$$
According to Eq. (32), arbitrary polarization state could be obtained by means of rotating Q1. After light passes through a sample, a Stokes vector ${\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} \over S} _{f,PSG}}$ changes to another Stokes vector ${\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} \over S} _{i,PSD}}$. After keeping propagating to Q2 and P2, a Stokes vector ${\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} \over S} _{f,PSG}}$ could be calculated as:
$${\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} \over S} _{f,PSD}} = {W_{P2}} \cdot {W_{Q2}} \cdot {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} \over S} _{i,PSD}} = \frac{1}{2}\left( {\begin{array}{cccc} 1&{{{\cos }^2}2\alpha_Q^{\prime}}&{\sin 2\alpha_Q^{\prime} \cdot \cos 2\alpha_Q^{\prime}}&{ - \sin 2\alpha_Q^{\prime}}\\ 1&{{{\cos }^2}2\alpha_Q^{\prime}}&{\sin 2\alpha_Q^{\prime} \cdot \cos 2\alpha_Q^{\prime}}&{ - \sin 2\alpha_Q^{\prime}}\\ 0&0&0&0\\ 0&0&0&0 \end{array}} \right) \cdot \left[ {\begin{array}{c} {{S_0}}\\ {{S_1}}\\ {{S_2}}\\ {{S_3}} \end{array}} \right].$$
In Eq. (27), ${\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} \over S} _{i,PSD}}$ is arbitrary polarization state with Stokes components of S0, S1, S2, S3. Assume the Muller matrix of a sample is Ms. From Eqs. (32)-(33), the output Stokes vector of the ellipsometry ${\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} \over S} _f}$ can be expressed as:

$$\begin{aligned} &{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} \over S} }_f} = \frac{1}{4}\left( {\begin{array}{cccc} 1&{{{\cos }^2}2\alpha {^{\prime}_\textrm{Q}}}&{\sin 2\alpha {^{\prime}_\textrm{Q}}\cos 2\alpha {^{\prime}_\textrm{Q}}}&{ - \sin 2\alpha {^{\prime}_\textrm{Q}}}\\ 1&{{{\cos }^2}2\alpha {^{\prime}_\textrm{Q}}}&{\sin 2\alpha {^{\prime}_\textrm{Q}}\cos 2\alpha _Q^{\prime}}&{ - \sin 2\alpha {^{\prime}_\textrm{Q}}}\\ 0&0&0&0\\ 0&0&0&0 \end{array}} \right) \cdot {\textrm{M}_\textrm{s}} \cdot \left[ {\begin{array}{cccc} 1\\ {{{\cos }^2}2{\alpha _Q}}\\ {\cos 2{\alpha _Q}\sin 2{\alpha _Q}}\\ {\sin 2{\alpha _Q}} \end{array}} \right]\nonumber\\ &\textrm{ = }\frac{1}{4}\left( {\begin{array}{cccc} 1&{{{\cos }^2}2\alpha {^{\prime}_\textrm{Q}}}&{\sin 2\alpha {^{\prime}_\textrm{Q}}\cos 2\alpha {^{\prime}_\textrm{Q}}}&{ - \sin 2\alpha {^{\prime}_\textrm{Q}}}\\ 1&{{{\cos }^2}2\alpha {^{\prime}_\textrm{Q}}}&{\sin 2\alpha {^{\prime}_\textrm{Q}}\cos 2\alpha {^{\prime}_\textrm{Q}}}&{ - \sin 2\alpha {^{\prime}_\textrm{Q}}}\\ 0&0&0&0\\ 0&0&0&0 \end{array}} \right) \cdot \left( {\begin{array}{cccc} {{M_{11}}}&{{M_{12}}}&{{M_{13}}}&{{M_{14}}}\\ {{M_{21}}}&{{M_{22}}}&{{M_{23}}}&{{M_{24}}}\\ {{M_{31}}}&{{M_{32}}}&{{M_{33}}}&{{M_{34}}}\\ {{M_{41}}}&{{M_{42}}}&{{M_{43}}}&{{M_{44}}} \end{array}} \right)\nonumber\\ &\cdot \left[ {\begin{array}{cccc} 1\\ {{{\cos }^2}2{\alpha _Q}}\\ {\cos 2{\alpha _Q}\sin 2{\alpha _Q}}\\ {\sin 2{\alpha _Q}} \end{array}} \right]. \end{aligned}$$

In Eq. (34), the components of ${\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} \over S} _f}$ are basically trigonometric functions which could be expanded as Fourier series. Under change of variables, we set: ${p_1} = 1$, ${p_2} = {\cos ^2}2{\alpha _Q}$, ${p_3} = \cos 2{\alpha _Q} \cdot \sin 2{\alpha _Q}$, ${p_4} = \sin 2{\alpha _Q}$, ${h_1} = 1$, ${h_2} = {\cos ^2}2\alpha {^{\prime}_\textrm{Q}}$, ${h_3} = \sin 2\alpha {^{\prime}_\textrm{Q}} \cdot \cos 2\alpha {^{\prime}_\textrm{Q}}$, ${h_4} ={-} \sin 2\alpha {^{\prime}_\textrm{Q}}$. Eq. (34) could be expressed as:

$${\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} \over S} _f} = \frac{1}{4}\left( {\begin{array}{cccc} {{h_1}}&{{h_2}}&{{h_3}}&{{h_4}}\\ {{h_1}}&{{h_2}}&{{h_3}}&{{h_4}}\\ 0&0&0&0\\ 0&0&0&0 \end{array}} \right)\left( {\begin{array}{cccc} {{M_{11}}}&{{M_{12}}}&{{M_{13}}}&{{M_{14}}}\\ {{M_{21}}}&{{M_{22}}}&{{M_{23}}}&{{M_{24}}}\\ {{M_{31}}}&{{M_{32}}}&{{M_{33}}}&{{M_{34}}}\\ {{M_{41}}}&{{M_{42}}}&{{M_{43}}}&{{M_{44}}} \end{array}} \right)\left[ {\begin{array}{c} {{p_1}}\\ {{p_2}}\\ {{p_3}}\\ {{p_4}} \end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{c} {\sum\limits_{i,j = 1}^4 {{h_i}{m_{ij}}{p_j}} }\\ {\sum\limits_{i,j = 1}^4 {{h_i}{m_{ij}}{p_j}} }\\ 0\\ 0 \end{array}} \right].$$
We then define uij:
$$\sum\limits_{i,j = 1}^4 {{h_i}{p_j}{m_{ij}}} \equiv \sum\limits_{i,j = 1}^4 {{u_{ij}}{m_{ij}}} .$$
From Eqs. (35)– (36), uij is:
$$\begin{aligned}{u_{ij}}& = \frac{1}{4}\left( \begin{array}{@{}cccc@{}}1&{{{\cos }^2}2{\alpha_Q}}&{\sin 2{\alpha_Q} \cdot \cos 2{\alpha_Q}}\\{{{\cos }^2}2\alpha {^{\prime}_\textrm{Q}}}&{{{\cos }^2}2{\alpha_Q} \cdot {{\cos }^2}2\alpha {^{\prime}_\textrm{Q}}}&{\sin 2{\alpha_Q} \cdot \cos 2{\alpha_Q} \cdot {{\cos }^2}2\alpha {^{\prime}_\textrm{Q}}}\\{\sin 2\alpha {^{\prime}_\textrm{Q}} \cdot \cos 2\alpha {^{\prime}_\textrm{Q}}}&{{{\cos }^2}2{\alpha_Q} \cdot \sin 2\alpha {^{\prime}_\textrm{Q}} \cdot \cos 2\alpha {^{\prime}_\textrm{Q}}}&{\sin 2{\alpha_Q} \cdot \cos 2{\alpha_Q} \cdot \sin 2\alpha {^{\prime}_\textrm{Q}} \cdot \cos 2\alpha {^{\prime}_\textrm{Q}}}\\{ - \sin 2\alpha {^{\prime}_\textrm{Q}}}&{ - {{\cos }^2}2{\alpha_Q} \cdot \sin 2\alpha {^{\prime}_\textrm{Q}}}&{\sin 2{\alpha_Q} \cdot \cos 2{\alpha_Q} \cdot \sin 2\alpha_Q^{\prime}}\end{array}\right.\nonumber\\&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\left.\begin{array}{@{}cccc@{}}{\sin 2{\alpha_Q}}\\{\sin 2{\alpha_Q} \cdot {{\cos }^2}2\alpha {^{\prime}_\textrm{Q}}}\\{\sin 2{\alpha_Q} \cdot \sin 2\alpha {^{\prime}_\textrm{Q}} \cdot \cos 2\alpha {^{\prime}_\textrm{Q}}}\\{ - \sin 2{\alpha_Q} \cdot \sin 2\alpha {^{\prime}_\textrm{Q}}} \end{array}\right).\end{aligned}$$
In the ellipsometry system, we actually measure light intensity which is a function of angles of fast axes of two quarter-wave plates in Eq. (37). This also indicates that we could calculate the Mueller matrix of the sample by manipulating angles of two quarter-wave plates. As a result, we further expand each element in Eq. (37) using Fourier series [2,13–14]:
$$\frac{I}{{{I_0}}} = \frac{{{a_0}}}{4} + \frac{1}{4}\sum\limits_{i = 1}^{12} {({a_i} \cdot \cos{\alpha _i} + {b_i} \cdot \sin{\alpha _i})} ,$$
where I is the measured light intensity after light passes through PSD, and ${I_o}$ is maximum light intensity at the measurement. Both of ai and bi are coefficients of Fourier series expansion. In Eq. (38), The relations for ${\alpha _i}$, ${\alpha _Q}$ and $\alpha {^{\prime}_\textrm{Q}}$ are:${\alpha _1} = 2{\alpha _Q}$, ${\alpha _2} = 4{\alpha _Q}$, ${\alpha _3} = 2\alpha {^{\prime}_\textrm{Q}} - 4{\alpha _Q}$, ${\alpha _4} = 2\alpha {^{\prime}_\textrm{Q}} - 2{\alpha _Q}$, ${\alpha _5} = 2\alpha {^{\prime}_\textrm{Q}}$, ${\alpha _6} = 2\alpha {^{\prime}_\textrm{Q}} + 2{\alpha _Q}$, ${\alpha _7} = 2\alpha {^{\prime}_\textrm{Q}} + 4{\alpha _Q}$, ${\alpha _8} = 4\alpha {^{\prime}_\textrm{Q}} - 4{\alpha _Q}$, ${\alpha _9} = 4\alpha {^{\prime}_\textrm{Q}} - 2{\alpha _Q}$, ${\alpha _{10}} = 4\alpha {^{\prime}_\textrm{Q}}$, ${\alpha _{11}} = 4\alpha {^{\prime}_\textrm{Q}} + 2{\alpha _Q}$, and ${\alpha _{12}} = 2\alpha {^{\prime}_\textrm{Q}} + 4{\alpha _Q}$. When $\alpha {^{\prime}_\textrm{Q}} = 5{\alpha _Q}$, Eq. (38) could be simplified to:
$$\frac{I}{{{I_0}}} = \frac{{{a_0}}}{4} + \frac{1}{4}\sum\limits_{k = 1}^{12} {[{a_{2k}} \cdot \cos(2 \cdot {\rm k} \cdot {\alpha _Q}) + {b_{2k}} \cdot \sin(2 \cdot {\rm k} \cdot {\alpha _Q})} ].$$
When we have the measured light intensity I using the photodetector, we can accuquire the Fourier coefficients a2k and b2k through fitting. From Eqs. (36) and (39), the inversion of these relation gives the element of Ms in terms of Fourier coefficients using ai and bi :
$${M_s} = \left[ {\begin{array}{cccc} {({a_0} - {a_2} - {a_{10}} + {a_8} + {a_{12}})}&{2({a_2} - {a_8} - {a_{12}})}&{2({b_2} + {{\rm b}_8} - {b_{12}})}&{({b_1} + {{\rm b}_9} - {b_{11}})}\\ {2({a_{10}} - {a_8} - {a_{12}})}&{4({a_8} + {a_{12}})}&{4({b_{12}} - {b_8})}&{2({b_{11}} - {b_9})}\\ {2({b_{10}} - {b_8} - {b_{12}})}&{4({b_8} + {b_{12}})}&{4({a_8} - {a_{12}})}&{4({a_9} - {a_{11}})}\\ {({b_3} - {b_5} + {b_7})}&{ - 2({b_7} + {b_3})}&{2({a_7} - {a_3})}&{({a_6} - {a_4})} \end{array}} \right].$$
Here, Ms is Mexp in Eq. (28). After we obtain Ms (or Mexp), we could first meausred te and to by giving e-wave and o-wave to targeted sample. Second, ${\phi _{pol}}$, ${\phi _{wp}}$, and phase retardation of birefrigent layer can be further meaused and characterized by data fitting based on Eqs. (38)-(40).

First, we did not place any sample and then we measured optical power in a unit of watt as a function of αQ, as shown in Fig. 3. From Fig. 3, we calculated Mueller matrix of air according to Eqs. (39)-(40). The measured Mueller matrix of air show in Eq. (41):

$${\textrm{M}_{air}} = \left[ {\begin{array}{cccc} {0.99734}&{0.00002}&{\textrm{ - 0}\textrm{.00249}}&{\textrm{ - 0}\textrm{.0008}}\\ {\textrm{ - 0}\textrm{.00008}}&{\textrm{0}\textrm{.99882}}&{\textrm{0}\textrm{.005}}&{\textrm{0}\textrm{.00124}}\\ {\textrm{0}\textrm{.00061}}&{\textrm{ - 0}\textrm{.00753}}&{\textrm{0}\textrm{.99835}}&{\textrm{0}\textrm{.00114}}\\ {\textrm{ - 0}\textrm{.00092}}&{\textrm{0}\textrm{.00409}}&{\textrm{0}\textrm{.00273}}&{\textrm{0}\textrm{.99802}} \end{array}} \right].$$
Theoretically, Mueller matrix of air should be an unit matrix. Mair in Eq. (41) is closed to an unit matrix. As a result, the accuracy of our measurement is around 3 decimal digits. Before using the proposed model for measuring a birefringent optical medium with absorption anisotropic, we verified the accuracy of the proposed method using Muller matrix. First, we used a liquid crystal cell with a tunable phase retardation as one of the testing sample, and the phase retardation was pre-determined by a typical measurement method using a pair of crossed polarizers. The difference (or said variance of measurement using Muller matrix method) of phase retardation between the measured data using Muller matrix method and the pre-determined data is less than 0.0174 radians. Second, the accuracy of absorption coefficients measurement was performed by comparing the measured data of a commercial linear polarizer and its pre-determined spec. The variances of two absorption coefficients tx and ty in Eq. (31) could be less than 0.01 and 0.035, respectively. There is high accuracy of measurement using proposed Muller matrix on birefringence and absorption coefficient.

 figure: Fig. 4.

Fig. 4. (a) Schematic illustration of the preparation of the curved polarizer with stress-induced birefringence. (b) Illustration of the 17 sampling points of measurement for r = 1.2 cm and 2 cm at different $\psi$. The transmissive axis of the polarizer was attached along parallel to x-axis. (c) and (d) are the samples with R = 52 mm and R = 77 mm, respectively. The samples were attached a black paper with 17 punched holes.

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By using the optical model, we developed and the ellipsometry we setup, we could measure the birefringence and anisotropic absorption of a polarizer with stress-induced birefringence. The polarizer, provided by General Interface Solution Holding Ltd., consists of a protective film, antireflection layer, a hard coating layer, one iodine-doped PVA layer, two TAC layers, a layer of adhesive glue, and a release film. After we removed protective film and release film, the polarizer was heat up to around 90 degree Celsius and then was pressed to attach the polarizer on a curved glass substrate with a radius of curvature of 52 mm (R = 52 mm), as illustrated in Fig. 4(a). The aperture size of the curved glass substrate was 5 cm. Similarly, we prepsed another sample with a radius of curvature of 77 mm (R = 77 mm). The whole attachment process was done in General Interface Solution Holding Ltd. After high temperature process and pressure applying process, the polarizer could have induced birefringence and the change of anisotropic absorption. Such induced birefringence might result from the internal structure change of the materials under stress and the change of anisotropic absorption might result from the orientation change of local transmissive axes of the iodine-doped PVA layer under stress [3]. In our experiments, we measured 17 points for r = 1.2 cm and 2 cm at different $\psi$ are measured to demonstrate the concept (Fig. 4(b)). We attached a black paper on each sample and punched 17 small holes to assist observation and measurement. Figures 4(c)–4(d) show two samples under two attached with black paper. We also observed a bubble defect at r = 2 cm and $\psi$ = 270 degree for the sample of R = 52 mm which resulted from attachment process (Fig. 4(c)). This indicates the mechanically-induced birefringence for the sample with small radius of curvature (Fig. 4(b)) is larger and also has obvious certain distribution than the one with large radius of curvature (Fig. 4(d)). The outer region of each samples has larger mechanically-induced birefringence compared to the region in the center of each sample.

 figure: Fig. 5.

Fig. 5. The polarizer attached on a substrate with radius of curvature of 52 mm. (a) Angle of transmissive of axis as an azimuthal angle $\psi$ at r = 1.2 cm and 2 cm. (b) te as an azimuthal angle at r = 0, 1.2 cm, and 2 cm. (c) to as an azimuthal angle $\psi$ at r = 0, 1.2 cm, and 2 cm. (d) Mechanically-induced birefringence as an azimuthal angle $\psi$ at r = 0, 1.2 cm, and 2 cm.

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From Muller matrix and stokes parameters we measured at each location and the optical model we developed, we calculated angle of transmissive of axis, absorption coefficients (to and te), mechanically-induced birefringence (Δn) as a function of $\psi$ for two samples, shown in in Figs. 5(a)-(d) and Figs. 6(a)-(d). From Fig. 5(a), the angle variation is around ±1.5 degrees at r = 2 cm and -0.6∼+0.3 degrees at r = 1.2 cm. The outer region has larger variation in the angle of the transmissive axis than the inner region. From Fig. 6(a), the variation is similar in both of the outer and the inner region (-0.7∼+0.4 degrees). Compared Figs. 5(a) and 6(a), the sample with large R has smaller angle variation. This also means the transmissive axis of the sample with R = 77 mm is almost along x-axis after attaching into a curved substrate with a large radius of curvature. The variation of local transmissive axes are because of uneven stress force and the attached process on a curved substrate. The outer region experiences larger stress force. In addition, the small transmissive axis angle could result in light leakage in LCD [15]. The information we obtained about the spatial distribution of the transmissive axis of the samples could help us to exam the process and then to adjust the mechanically attached process.

 figure: Fig. 6.

Fig. 6. The polarizer attached on a substrate with radius of curvature of 77 mm. (a) Angle of transmissive of axis as an azimuthal $\psi$ angle at r = 1.2 cm and 2 cm. (b) te as an azimuthal angle at r = 0, 1.2 cm, and 2 cm. (c) to as an azimuthal $\psi$ angle at r = 0, 1.2 cm, and 2 cm. (d) Mechanically-induced birefringence as an azimuthal angle $\psi$ at r = 0, 1.2 cm, and 2 cm.

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Figures 5(b), 5(c), 6(b), and 6(c) show absorption coefficients (te and to) as a function of $\psi$. Once again, te and to are absorption coefficients as the oscillation direction of the linear polarization is parallel and perpendicular to the transmissive axis of the polarizer, respectively. When R = 52 mm, te is around 0.002 at r = 0 cm and 0.001∼0.003 at r = 1.2 cm and 2 cm. to is around 0.932 at r = 0 cm, 0.887∼0.930 at r = 1.2 cm, and 0.906∼0.926 at r = 2 cm. When R = 77 mm, te is around 0.002 at r = 0 cm, 0.002∼0.003 at r = 1.2 cm, and 0.00056∼0.0003 at r = 2 cm. to is around 0.937 at r = 0 cm, 0.900∼0.926 at r = 1.2 cm, and 0.926∼0.955 at r = 2 cm. For R = 52 mm and R = 77 mm, the absorption coefficient to at r = 1.2 cm and 2 cm is around 2%-3% variation compared to the one at the central part, but to could change 100%. This also indicates the anisotropic absorption changes under stress process. Figures 5(d) and 6(d) depict mechanically-induced birefringence (Δn) as a function of $\psi$. Δn for both samples at the central region are around zero. For R = 52 mm, Δn changes almost periodically between 0 and $2 \times {10^{ - 5}}$ at r = 1.2 cm, and Δn changes between 0 and $4.5 \times {10^{ - 5}}$ at r = 2 cm. The precocity and zero at $\psi$ = 90, 180, and 270 degrees indicate the distribution of mechanically-induced birefringence exhibits radiative distribution. At r = 2 cm, we observed bubbles at $\psi$ = 270 degree due to attachment process which also affects the measuring value of Δn. For R = 77 mm, Δn is around zero at r = 0 and 1.2 cm, and it changes periodically between 0 and $2 \times {10^{ - 5}}$ at r = 2 cm. In Fig. 6(d), Δn at r = 2 cm is zero at $\psi$ = 90, 180, and 270 degrees. In Figs. 5(d) and 6(d), the absolute value of Δn increases with r. Smaller radius of curvature of the substrate, larger Δn. We hope that the information provided in Figs. 5 and 6 could help manufacturing companies to design compensation films or to adjust the attached process in order to improve performance of the LCD. Also, according to the results, the stress-induced birefringence is uneven due to the three-dimension deformation; therefore, if a curved-polarizer exists, it may experience less stress during thermal attaching.

4. Summary

We built up a model for light propagation in a birefringent-curved medium with anisotropic absorption based on calculus of Muller matrix. The proposed model could help us predict the transmittance, absorption, birefringence, phase retardation, and polarization after beam propagates through the optical medium. The method in this paper could be further extended to multi-layered as well as curved-optical components. As long as we know how beam propagates in the birefringent-curved medium with anisotropic absorption, we could develop the corresponding compensation components or devices to improve the performance in the optical systems. We believe the study could benefit to researchers and optical engineers to know better about optical wave propagation in a birefringent-curved medium with anisotropic absorption and to easily manage the polarization state of output light for further optimizing the performance of optical systems.

Funding

General Interface Solution Holding (GIS) Ltd; Ministry of Science and Technology, Taiwan (110-2112-M-A49-024).

Acknowledgments

The authors are indebted to Mr. Jun-Lin Chen for technical assistance.

Disclosures

YHL: GIS Ltd (F), YJW: GIS Ltd (F), HCL: GIS Ltd (F), MLL: GIS Ltd (F), PLC, GIS Ltd (E).

Data availability

Data underlying the results presented in this paper are not publicly available at this time but may be obtained from the authors upon reasonable request.

References

1. P. Yeh and C. Gu, Optics of Liquid Crystal Display (Wiley, 2010).

2. D. H. Goldstein, Polarized light (CRC, 2003).

3. R. A. Chipman and W. S. T. Lam, Polarized light and optical systems (CRC, 2019)

4. Q. T. Liang and X. D. Zheng, “Ray-tracing calculations for uniaxial optical components with curved surfaces,” Appl. Opt. 30(31), 4521–4525 (1991). [CrossRef]  

5. T. A. El-Dessouki, N. I. Hendawy, and A. A. Zaki, “Measuring birefringence of curved sheet and single crystals by double exposure speckle photography,” Optics and Lasers in Engineering 47(6), 622–628 (2009). [CrossRef]  

6. G. Singh and D. S. Mehta, “Measurement of change in refractive index in polymeric flexible substrates using wide field interferometry and digital fringe analysis,” Appl. Opt. 51(35), 8413–8422 (2012). [CrossRef]  

7. A. Sato, T. Ishinabe, and H. Fujikake, “Evaluation of curving characteristics of flexible liquid crystal displays fabricated using polycarbonate substrates,” Jpn. J. Appl. Phys. 55(1), 011701 (2016). [CrossRef]  

8. S. Honda, T. Ishinabe, Y. Shibata, and H. Fujikake, “Evaluation of phase retardation of curved thin polycarbonate substrates for wide-viewing angle flexible liquid crystal displays,” IEICE Trans. Electron. E100.C(11), 992–997 (2017). [CrossRef]  

9. D. Kim and E. Sim, “Segmented coupled-wave analysis of a curved wire-grid polarizer,” J. Opt. Soc. Am. A 25(3), 558–565 (2008). [CrossRef]  

10. E. Collett, “Mueller-Stokes matrix formulation of Fresnel's equations,” Am. J. Phys. 39(5), 517–528 (1971). [CrossRef]  

11. E. Collett, “The Description of Polarization in Classical Physics,” American Journal of Physics 36(8), 713–725 (1968). [CrossRef]  

12. E. Garcia-Caurel, A. De Martino, J. P. Gaston, and L. Yan, “Application of spectroscopic ellipsometry and Mueller ellipsometry to optical characterization,” Appl. Spectrosc. 67(1), 1–21 (2013). [CrossRef]  

13. R. M. A. Azzam, “Photopolarimetric measurement of the Mueller matrix by Fourier analysis of a single detected signal,” Opt. Lett. 2(6), 148–150 (1978). [CrossRef]  

14. P. S. Hauge, “Mueller matrix ellipsometry with imperfect compensators,” J. Opt. Soc. Am. 68(11), 1519–1528 (1978). [CrossRef]  

15. J. S. Hsu, Y. H. Lin, H. C. Lin, and K. H. Yang, “Thermally induced light leakage in in-plane-switching liquid crystal displays,” J. Appl. Phys. 105(3), 033503 (2009). [CrossRef]  

Data availability

Data underlying the results presented in this paper are not publicly available at this time but may be obtained from the authors upon reasonable request.

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Figures (6)
Fig. 1.
Fig. 1. (a) The equivalent optical system of a birefringent-curved medium with anisotropic absorption. A compensated lens is used for light collimation. (b) is the detailed optical layout between T2 and T3 in (a). T1 to T5 represents interfaces. C1 is transmissive axis of the anisotropic absorptive layer. C2 is the optic axis of the birefringent layer. $\vec{k}$ is wave vector of a plane wave. ei and oi stands for e-wave and o-wave at either the absorptive layer (i=1) or the birefringent layer (i=2). p and s mean p-wave and s-wave, respectively. The dash lines in (b) stand for the normal directions of the interfaces.

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Fig. 2.
Fig. 2. Experimental setup of the ellipsometry for measurement of Mueller matrix. P1, P2, and P3 are polarizers. Q1 and Q2 are quarter-wave plates. PSG and PSD stand for polarization state generation and polarization state detection, respectively.

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Fig. 3.
Fig. 3. Measured optical power as a function of αQ without the sample.

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Fig. 4.
Fig. 4. (a) Schematic illustration of the preparation of the curved polarizer with stress-induced birefringence. (b) Illustration of the 17 sampling points of measurement for r = 1.2 cm and 2 cm at different $\psi$. The transmissive axis of the polarizer was attached along parallel to x-axis. (c) and (d) are the samples with R = 52 mm and R = 77 mm, respectively. The samples were attached a black paper with 17 punched holes.

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Fig. 5.
Fig. 5. The polarizer attached on a substrate with radius of curvature of 52 mm. (a) Angle of transmissive of axis as an azimuthal angle $\psi$ at r = 1.2 cm and 2 cm. (b) te as an azimuthal angle at r = 0, 1.2 cm, and 2 cm. (c) to as an azimuthal angle $\psi$ at r = 0, 1.2 cm, and 2 cm. (d) Mechanically-induced birefringence as an azimuthal angle $\psi$ at r = 0, 1.2 cm, and 2 cm.

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Fig. 6.
Fig. 6. The polarizer attached on a substrate with radius of curvature of 77 mm. (a) Angle of transmissive of axis as an azimuthal $\psi$ angle at r = 1.2 cm and 2 cm. (b) te as an azimuthal angle at r = 0, 1.2 cm, and 2 cm. (c) to as an azimuthal $\psi$ angle at r = 0, 1.2 cm, and 2 cm. (d) Mechanically-induced birefringence as an azimuthal angle $\psi$ at r = 0, 1.2 cm, and 2 cm.

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Equations (41)

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E → = x ^ ⋅ E x ⋅ e − j ( k ⋅ z + δ x ) + y ^ ⋅ E y ⋅ e − j ( k ⋅ z + δ y ) ,
S → i n = ( S 0 S 1 S 2 S 3 ) = ( E x 2 + E y 2 E x 2 − E y 2 2 E x 2 ⋅ E y 2 ⋅ cos ⁡ δ 2 E x 2 ⋅ E y 2 ⋅ sin ⁡ δ ) ,
S → o u t = M s ⋅ S → i n .
M = M T 5 ⋅ M T 4 − T 5 ⋅ M T 4 ⋅ M T 3 − T 4 ⋅ M T 3 ⋅ M T 2 − T 3 ⋅ M T 2 ⋅ M T 1 − T 2 ⋅ M T 1 ,
M T 1 − T 2 = M T 3 − T 4 = M T 4 − T 5 = [ 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 ] .
M = A T 1 ⋅ A T 5 ⋅ M T 4 ⋅ M T 3 ⋅ M T 2 − T 3 ⋅ M T 2 .
{ s ^ = s ^ ′ = x ^ ′ p ^ = cos ⁡ θ i ⋅ y ^ ′ − sin ⁡ θ i ⋅ z ^ ′ p ^ ′ = cos ⁡ θ r ⋅ y ^ ′ − sin ⁡ θ r ⋅ z ^ ′ ,
c → 1 = sin ⁡ [ θ p o l ] ⋅ cos ⁡ [ ϕ p o l ] ⋅ x ^ ′ + sin ⁡ [ θ p o l ] sin ⁡ [ ϕ p o l ] ⋅ y ^ ′ + cos ⁡ [ θ p o l ] ⋅ z ^ ′ ,
c → 2 = sin ⁡ [ θ c ] ⋅ cos ⁡ [ ϕ c ] ⋅ x ^ ′ + sin ⁡ [ θ c ] ⋅ sin ⁡ [ ϕ c ] ⋅ y ^ ′ + cos ⁡ [ θ c ] ⋅ z ^ ′ ,
{ o ^ 1 = k → o 1 × c → 1 | k → o 1 × c → 1 | , e ^ 1 = o → 1 × k → e 1 | o 1 × k e 1 | o ^ 2 = k → o 2 × c → 2 | k → o 2 × c → 2 | , e ^ 2 = o → 2 × k → e 2 | o → 2 × k → e 2 | ,
{ o ^ 1 = k → 0 × c → 1 | k → 0 × c → 1 | , e ^ 1 = o ^ 1 × k → 0 | o ^ 1 × k → 0 | o ^ 2 = k → 0 × c → 2 | k → 0 × c → 2 | , e ^ 2 = o ^ 2 × k → 0 | o ^ 2 × k → 0 | .
( E s ′ E p ′ ) = J T 3 ⋅ J T 2 − T 3 ⋅ J T 2 ( E s E p ) ,
( E s ′ E p ′ ) = [ s ^ ′ ⋅ e ^ 2 ⋅ t s ′ s ^ ′ ⋅ o ^ 2 ⋅ t s ′ p ^ ′ ⋅ e ^ 2 ⋅ t p ′ p ^ ′ ⋅ o ^ 2 ⋅ t p ′ ] ⋅ [ e − j ⋅ k e z ⋅ d 0 0 e − j ⋅ k o z ⋅ d ] ⋅ [ e ^ 1 ⋅ e ^ 2 o ^ 1 ⋅ e ^ 2 e ^ 1 ⋅ o ^ 2 o ^ 1 ⋅ o ^ 2 ] ⋅ [ t e 0 0 t o ] ⋅ [ e ^ 1 ⋅ s ^ ⋅ t s e ^ 1 ⋅ p ^ ⋅ t p o ^ 1 ⋅ s ^ ⋅ t s o ^ 1 ⋅ p ^ ⋅ t p ] ⋅ ( E s E p ) ,
t s = 2 n g ⋅ cos ⁡ θ i n g ⋅ cos ⁡ θ i + n o ⋅ cos ⁡ θ o ,
t p = 2 n g ⋅ cos ⁡ θ i n g ⋅ cos ⁡ θ o + n o ⋅ cos ⁡ θ i ,
t s ′ = 2 ⋅ n o ⋅ cos ⁡ θ i n o ⋅ cos ⁡ θ i + cos ⁡ θ r ,
t p ′ = 2 ⋅ n e ⋅ cos ⁡ θ i n o ⋅ cos ⁡ θ r + cos ⁡ θ i ,
k e z = v + v 2 − 4 ⋅ u ⋅ w 2 ⋅ u ,
k o z = k ⋅ cos ⁡ θ i ,
[ E s ′ E p ′ ] = [ J 11 J 12 J 21 J 22 ] ⋅ [ E s E p ] .
M T 3 ⋅ M T 2 − T 3 ⋅ M T 2 = cos ⁡ θ r n i ⋅ cos ⁡ θ i ⋅ ( 1 0 0 1 1 0 0 − 1 0 1 1 0 0 i − i 0 ) ⋅ ( J 11 J 11 ∗ J 11 J 12 ∗ J 12 J 11 ∗ J 12 J 12 ∗ J 11 J 21 ∗ J 11 J 22 ∗ J 12 J 21 ∗ J 12 J 22 ∗ J 21 J 11 ∗ J 21 J 12 ∗ J 22 J 11 ∗ J 22 J 12 ∗ J 21 J 21 ∗ J 21 J 22 ∗ J 22 J 21 ∗ J 22 J 22 ∗ ) ⋅ 1 2 ( 1 1 0 0 0 0 1 − i 0 0 1 i 1 − 1 0 0 ) .
M T 2 − T 3 ′ = M T 3 ⋅ M T 2 − T 3 ⋅ M T 2 ⋅ M R ( ϕ ) = M T 3 ⋅ M T 2 − T 3 ⋅ M T 2 ⋅ [ 1 0 0 0 0 cos ⁡ ( 2 ϕ ) sin ⁡ ( 2 ϕ ) 0 0 − sin ⁡ ( 2 ϕ ) cos ⁡ ( 2 ϕ ) 0 0 0 0 1 ] ,
M = A T 1 ⋅ A T 5 ⋅ M T 4 ⋅ M T 2 − T 3 ′ .
M T 4 = [ τ s + τ p 2 τ s − τ p 2 0 0 τ s − τ p 2 τ s + τ p 2 0 0 0 0 ( τ s τ p ) 1 / 2 0 0 0 0 ( τ s τ p ) 1 / 2 ] ,
τ s = ( tan ⁡ θ i ′ tan ⁡ θ r ′ ) ( 2 sin ⁡ θ r ′ cos ⁡ θ i ′ sin ⁡ ( θ i ′ + θ r ′ ) ) 2 .
τ p = ( tan ⁡ θ i ′ tan ⁡ θ r ′ ) ( 2 sin ⁡ θ r ′ cos ⁡ θ i ′ sin ⁡ ( θ i ′ + θ r ′ ) cos ⁡ ( θ i ′ − θ r ′ ) ) 2 .
M = A T 1 ⋅ A T 5 ⋅ M T 4 ′ ⋅ M T 2 − T 3 ′ .
M ( − ϕ ) ⋅ M p o l ⋅ M ( ϕ ) = ( A T 5 ⋅ M ( − ϕ ) ⋅ M T 4 ⋅ M T 3 ⋅ M ( ϕ ) ) − 1 ⋅ M exp ⋅ ( M ( − ϕ ) ⋅ M T 2 ⋅ M ( ϕ ) ⋅ A T 1 ) − 1 ,
M T 3 ⋅ M T 2 − T 3 ⋅ M T 2 = ( A T 5 ⋅ M ( − ϕ ) ⋅ M T 4 ) − 1 ⋅ M exp ⋅ ( M ( ϕ ) ⋅ A T 1 ) − 1 .
S ⇀ f , P S G = W Q 1 ⋅ W P 1 ⋅ S ⇀ i , P S G = ( 1 0 0 0 0 cos 2 2 α Q + cos ⁡ Γ Q ⋅ sin 2 2 α Q ( 1 − cos ⁡ Γ Q ) ⋅ sin ⁡ 2 α Q ⋅ cos ⁡ 2 α Q − sin ⁡ Γ Q ⋅ sin ⁡ 2 α Q 0 ( 1 − cos ⁡ Γ Q ) ⋅ sin ⁡ 2 α Q ⋅ cos ⁡ 2 α Q sin 2 2 α Q + cos ⁡ Γ Q ⋅ cos 2 2 α Q sin ⁡ Γ Q ⋅ cos ⁡ 2 α Q 0 sin ⁡ Γ Q ⋅ sin ⁡ 2 α Q − sin ⁡ Γ Q ⋅ cos ⁡ 2 α Q cos ⁡ Γ Q ) ⋅ ( A B ⋅ cos ⁡ 2 α p o l 1 B ⋅ sin ⁡ 2 α p o l 1 0 B ⋅ cos ⁡ 2 α p o l 1 A ⋅ cos 2 2 α p o l 1 + C ⋅ sin 2 2 α p o l 1 ( A − C ) ⋅ sin ⁡ 2 α p o l 1 ⋅ cos ⁡ 2 α p o l 1 0 B ⋅ sin ⁡ 2 α p o l 1 ( A − C ) ⋅ sin ⁡ 2 α p o l 1 ⋅ cos ⁡ 2 α p o l 1 A ⋅ sin 2 2 α p o l 1 + C ⋅ cos 2 2 α p o l 1 0 0 0 0 C ) ⋅ [ 1 0 0 0 ] .
{ A = ( t x 2 + t y 2 ) / 2 B = ( t x 2 − t y 2 ) / 2 C = t x ⋅ t y ,
S ⇀ f , P S G = 1 2 [ 1 cos 2 2 α Q cos ⁡ 2 α Q ⋅ sin ⁡ 2 α Q sin ⁡ 2 α Q ] .
S ⇀ f , P S D = W P 2 ⋅ W Q 2 ⋅ S ⇀ i , P S D = 1 2 ( 1 cos 2 2 α Q ′ sin ⁡ 2 α Q ′ ⋅ cos ⁡ 2 α Q ′ − sin ⁡ 2 α Q ′ 1 cos 2 2 α Q ′ sin ⁡ 2 α Q ′ ⋅ cos ⁡ 2 α Q ′ − sin ⁡ 2 α Q ′ 0 0 0 0 0 0 0 0 ) ⋅ [ S 0 S 1 S 2 S 3 ] .
S ⇀ f = 1 4 ( 1 cos 2 2 α Q ′ sin ⁡ 2 α Q ′ cos ⁡ 2 α Q ′ − sin ⁡ 2 α Q ′ 1 cos 2 2 α Q ′ sin ⁡ 2 α Q ′ cos ⁡ 2 α Q ′ − sin ⁡ 2 α Q ′ 0 0 0 0 0 0 0 0 ) ⋅ M s ⋅ [ 1 cos 2 2 α Q cos ⁡ 2 α Q sin ⁡ 2 α Q sin ⁡ 2 α Q ]  =  1 4 ( 1 cos 2 2 α Q ′ sin ⁡ 2 α Q ′ cos ⁡ 2 α Q ′ − sin ⁡ 2 α Q ′ 1 cos 2 2 α Q ′ sin ⁡ 2 α Q ′ cos ⁡ 2 α Q ′ − sin ⁡ 2 α Q ′ 0 0 0 0 0 0 0 0 ) ⋅ ( M 11 M 12 M 13 M 14 M 21 M 22 M 23 M 24 M 31 M 32 M 33 M 34 M 41 M 42 M 43 M 44 ) ⋅ [ 1 cos 2 2 α Q cos ⁡ 2 α Q sin ⁡ 2 α Q sin ⁡ 2 α Q ] .
S ⇀ f = 1 4 ( h 1 h 2 h 3 h 4 h 1 h 2 h 3 h 4 0 0 0 0 0 0 0 0 ) ( M 11 M 12 M 13 M 14 M 21 M 22 M 23 M 24 M 31 M 32 M 33 M 34 M 41 M 42 M 43 M 44 ) [ p 1 p 2 p 3 p 4 ] = 1 4 [ ∑ i , j = 1 4 h i m i j p j ∑ i , j = 1 4 h i m i j p j 0 0 ] .
∑ i , j = 1 4 h i p j m i j ≡ ∑ i , j = 1 4 u i j m i j .
u i j = 1 4 ( 1 cos 2 2 α Q sin ⁡ 2 α Q ⋅ cos ⁡ 2 α Q cos 2 2 α Q ′ cos 2 2 α Q ⋅ cos 2 2 α Q ′ sin ⁡ 2 α Q ⋅ cos ⁡ 2 α Q ⋅ cos 2 2 α Q ′ sin ⁡ 2 α Q ′ ⋅ cos ⁡ 2 α Q ′ cos 2 2 α Q ⋅ sin ⁡ 2 α Q ′ ⋅ cos ⁡ 2 α Q ′ sin ⁡ 2 α Q ⋅ cos ⁡ 2 α Q ⋅ sin ⁡ 2 α Q ′ ⋅ cos ⁡ 2 α Q ′ − sin ⁡ 2 α Q ′ − cos 2 2 α Q ⋅ sin ⁡ 2 α Q ′ sin ⁡ 2 α Q ⋅ cos ⁡ 2 α Q ⋅ sin ⁡ 2 α Q ′ sin ⁡ 2 α Q sin ⁡ 2 α Q ⋅ cos 2 2 α Q ′ sin ⁡ 2 α Q ⋅ sin ⁡ 2 α Q ′ ⋅ cos ⁡ 2 α Q ′ − sin ⁡ 2 α Q ⋅ sin ⁡ 2 α Q ′ ) .
I I 0 = a 0 4 + 1 4 ∑ i = 1 12 ( a i ⋅ cos ⁡ α i + b i ⋅ sin ⁡ α i ) ,
I I 0 = a 0 4 + 1 4 ∑ k = 1 12 [ a 2 k ⋅ cos ⁡ ( 2 ⋅ k ⋅ α Q ) + b 2 k ⋅ sin ⁡ ( 2 ⋅ k ⋅ α Q ) ] .
M s = [ ( a 0 − a 2 − a 10 + a 8 + a 12 ) 2 ( a 2 − a 8 − a 12 ) 2 ( b 2 + b 8 − b 12 ) ( b 1 + b 9 − b 11 ) 2 ( a 10 − a 8 − a 12 ) 4 ( a 8 + a 12 ) 4 ( b 12 − b 8 ) 2 ( b 11 − b 9 ) 2 ( b 10 − b 8 − b 12 ) 4 ( b 8 + b 12 ) 4 ( a 8 − a 12 ) 4 ( a 9 − a 11 ) ( b 3 − b 5 + b 7 ) − 2 ( b 7 + b 3 ) 2 ( a 7 − a 3 ) ( a 6 − a 4 ) ] .
M a i r = [ 0.99734 0.00002  - 0 .00249  - 0 .0008  - 0 .00008 0 .99882 0 .005 0 .00124 0 .00061  - 0 .00753 0 .99835 0 .00114  - 0 .00092 0 .00409 0 .00273 0 .99802 ] .
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