9Polarization Ray Tracing Calculus

The objective of polarization ray tracing is to calculate the evolution of the polarization state of rays through an optical system and to determine the polarization properties, such as diattenuation and retardance, associated with these ray paths. By tracing many rays, the polarization aberrations associated with an optical system can be assessed, and the behavior of a particular optical system, associated coating designs, polarization elements, and other components can be compared with the optical system’s polarization specifications. Moreover, the knowledge of polarization aberrations due to coatings, individual surfaces, polarization elements, and other components can also aid in informing the design and defining polarization specifications. Polarization ray tracing enables detailed insights into how individual surfaces or components affect system polarization performance.

Among different polarization ray tracing matrix methods, Jones matrices have been used in optical design for at least 20 years.1,2 Polarization ray tracing, which calculates the Jones matrix associated with an arbitrary ray path through an optical system, was introduced to calculate the polarization aberration function.3 The Jones matrix deals with Jones vectors, which specifically refer to a monochromatic plane wave, describing the electric field and the polarization ellipse with respect to an x–y coordinate system in the transverse plane. If the plane wave is not propagating along the z-axis, then the x–y coordinates are referred to as “local coordinates” associated with a particular transverse plane. However, to use Jones vectors and matrices in optical design for the ray tracing of highly curved beams, local coordinate systems are required for each ray, and each of its ray segments, to define the direction of the Jones vector’s x- and y-components in space, and these local coordinate systems lead to complications due to the intrinsic singularities of local coordinates. In the experience of the authors, working in Jones vector local coordinates leads to a cascade of minor complications, both in handling rays near the singularities and in describing high numerical aperture beams. Such issues are intrinsic to any choice of local coordinates. According to the Winding Number Theorem,4 it is impossible to define a continuous and differentiable vector field constrained to lie on the surface of a sphere over the entire sphere without at least two zeros in the field; a set of latitude vectors or conversely a set of longitude vectors provide two examples, where the zeros occur at the poles. All local coordinate choices have such singularities.

In this chapter, a different polarization ray tracing calculus is presented to solve the problem of singularities in local coordinates. Polarization effects at each ray intercept are described by a three-by-three polarization ray tracing matrix, P, which is a generalized version of a two-by-two Jones matrix. With a three-by-three matrix, arbitrary propagation directions are easily accommodated and the problem of singularities in local coordinates is avoided (see Chapter 11, The Jones Pupil and Local Coordinate Systems). Polarization effects are propagated along ray paths through optical systems by matrix multiplication of the P matrices for each ray intercept. In image space, arrays of P matrices can then be used to determine the resultant polarization state on curved surfaces such as spherical and aberrated wavefronts as a three-dimensional (3D) electric field. 3D polarization ray tracing methods have been mentioned in several manuscripts, but not developed into full mathematical methods.5–7 3D polarization ray tracing algorithms in Refs. 8 and 9 are the basis of this chapter and the polarization ray tracing code, “Polaris-M,”10,11 which was developed at the Polarization Laboratory at the University of Arizona and has now been commercialized by Airy Optics, Inc.

In Section 9.1, the 3D polarization ray tracing matrix is developed. Sections 9.2 and 9.3 further develop the P matrix with examples. Section 9.4 derives an algorithm to calculate the diattenuation of a given P matrix. An interferometer is used as an example in Section 9.5, and the issue of adding P matrices is addressed in Section 9.6. A hollow corner cube retroreflector is analyzed in Section 9.7.

9.1Definition of Polarization Ray Tracing Matrix, P

The polarization ray tracing matrixP characterizes the change in a three-element electric field vector due to interaction with an optical surface, element, a sequence of optical elements, or an entire optical system. Consider the evolution of the polarization state of a ray through an optical system with N interfaces labeled by index q. Assume for the moment that all the materials are isotropic so that polarization changes will occur only at interfaces and not along ray segments. (This restriction will be removed later.) The ray will reflect or refract at interface q = 1, 2, 3, and so on. The ray exits interface q − 1 propagating in direction ${\widehat{\mathbf{k}}}_{q-1}$ with electric field vector Eq−1 and is incident upon interface q. At interface q, the polarization state will be modified, perhaps by a polarization element, a reflection, or a refraction, and the ray will exit interface q with propagation vector ${\widehat{\mathbf{k}}}_q$ and electric field vector Eq. The incident electric field vector Eq−1 and exiting electric field vector Eq are linearly related by the polarization ray tracing matrix for the qth ray intercept Pq, as shown in Figure 9.1,

Figure 9.1For the ray or plane wave incident at the qth optical interface, the matrix Pq relates the incident polarization state, Eq−1, and propagation vector, kq−1, to the exiting polarization state, Eq, and propagation vector, kq.

$${\mathbf{E}}_q=\left(\begin{array}{l}{E}_{x,q}\\ E_{y,q}\\ E_{z,q}\end{array}\right)={\mathbf{P}}_q{\mathbf{E}}_{q-1}=\left(\begin{array}{lll}{p}_{xx,q}& p_{xy,q}& p_{xz,q}\\ p_{yx,q}& p_{yy,q}& p_{yz,q}\\ p_{zx,q}& p_{zy,q}& p_{zz,q}\end{array}\right)\left(\begin{array}{l}{E}_{x,q-1}\\ E_{y,q-1}\\ E_{z,q-1}\end{array}\right).$$9.1

9.1

Similarly, the incident propagation vector${\widehat{\mathbf{k}}}_{q-1}$ and exiting propagation vector ${\widehat{\mathbf{k}}}_q$ are also linearly related by the polarization ray tracing matrix Pq.1

The net polarization effect of a series of isotropic optical elements is represented by cascading the Pq matrices for each ray intercept to yield polarization ray tracing matrix PTotal,

$${\mathbf{P}}_{Total}={\mathbf{P}}_N{\mathbf{P}}_{N-1}\cdots {\mathbf{P}}_q\cdots {\mathbf{P}}_2{\mathbf{P}}_1={\displaystyle \prod _{q=N,-1}^1{\mathbf{P}}_q}.$$9.2

9.2

The polarization ray tracing matrix is a 3D generalization of the Jones matrix and Equation 9.2 has the same form as the Jones matrix product equation, Equation 5.10, from Chapter 5 (Jones Matrices and Polarization Properties).

In a conventional ray trace, to calculate a ray’s contribution to the wavefront aberration, the optical path lengths along all ray segments between the optical system’s entrance pupil and exit pupil are summed (Chapter 10, Optical Ray Tracing). Similarly, the polarization ray tracing matrix describes the polarization-dependent transmission and phase contributions to the optical path length. Polarization-dependent transmission and phase contributions may be due to coated and uncoated interfaces, diffraction gratings, holographic elements, and other polarization effects.

Pq as defined in Equation 9.1 is under-constrained; it doesn’t uniquely define Pq. In Equation 9.1, the transformation of all polarization states can be described as linear combinations of the transformations of any two linearly independent basis vectors, Ea and Eb,

$${{\mathbf{E}}^{\prime }}_a={\mathbf{P}}_q{\mathbf{E}}_a,{{\mathbf{E}}^{\prime }}_b={\mathbf{P}}_q{\mathbf{E}}_b.$$9.3

9.3

The relationship in Equation 9.3 yields six equations, one for each row, but Pq has nine elements. Thus, Equation 9.1 does not fully constrain Pq. To uniquely define Pq, Equation 9.3 is augmented with an additional condition relating the incident and the exiting propagation vectors,

$${\widehat{\mathbf{k}}}_q={\mathbf{P}}_q{\widehat{\mathbf{k}}}_{q-1}.$$9.4

9.4

9.2Formalism of Polarization Ray Tracing Matrix Using Orthogonal Transformation

Ray tracing calculations using the polarization ray tracing calculus involve frequent transformations between the global coordinates where the optical system and each P are defined and the local coordinates where the physics of polarization elements, anisotropic materials, thin film interfaces, diffraction gratings, reflection, refraction, and other phenomena are formulated. Orthogonal transformations between different coordinate systems, such as s–p coordinates, are straightforward and ubiquitous. This section explains the coordinate transformation notation.

Orthogonal matrices, also known as real unitary matrices, describe rotations of orthogonal coordinate systems. In our case, orthogonal matrices transform between a local coordinate basis selected for a calculation at an interface and the global coordinate basis and vice versa. In general, a separate pair of basis vectors is needed before and after the interface due to the change of ray direction.

For reflection and refraction from surfaces, the s- and p-polarization states along with the propagation vector form a natural basis, the $(\widehat{\mathbf{s}},\widehat{\mathbf{p}},\widehat{\mathbf{k}})$ basis, as shown in Figure 9.2. For smooth interfaces of isotropic media, $\widehat{\mathbf{s}}$ and $\widehat{\mathbf{p}}$ are defined to be perpendicular and parallel to the plane of incidence, respectively, and thus are the eigenpolarizations of the Fresnel equations. The surface local coordinates incident on $({\widehat{\mathbf{s}}}_q,{\widehat{\mathbf{p}}}_q,{\widehat{\mathbf{k}}}_{q-1})$ and exiting from $({{\widehat{\mathbf{s}}}^{\prime }}_q,{{\widehat{\mathbf{p}}}^{\prime }}_q,{\widehat{\mathbf{k}}}_q)$ the qth intercept are

Figure 9.2Local coordinate bases for the incident and refracted rays (left) and incident and reflected rays (right).

$${\widehat{s}}_q=\frac{{\widehat{k}}_{q-1}\times {\widehat{\eta }}_q}{|{\widehat{k}}_{q-1}\times {\widehat{\eta }}_q|},{\widehat{p}}_q={\widehat{k}}_{q-1}\times {\widehat{s}}_q,{{\widehat{s}}^{\prime }}_q={\widehat{s}}_q,{{\widehat{p}}^{\prime }}_q={\widehat{k}}_q\times {\widehat{s}}_q,$$9.5

9.5

where $\widehat{\eta }$ is the surface normal of the qth surface. Our convention will direct $\widehat{\eta }$ away from the incident medium into the next medium. There is a special case for normally incident rays where ${\widehat{\mathbf{\eta }}}_q$ is parallel to ${\widehat{\mathbf{k}}}_{q-1}$ such that $({\widehat{\mathbf{\eta }}}_q\times {\widehat{\mathbf{k}}}_{q-1}=0)$ and ${\widehat{\mathbf{s}}}_q$ of Equation 9.5 is undefined. In this case, any local coordinates that form an orthonormal right-handed coordinate system can be used for normal incidence. One simple choice is to reuse ${{\widehat{\mathbf{s}}}^{\prime }}_{q-1}$ from the previous ray intercept and calculate ${\widehat{\mathbf{p}}}_q$,

$${\widehat{\mathbf{s}}}_q={{\widehat{\mathbf{s}}}^{\prime }}_q={{\widehat{\mathbf{s}}}^{\prime }}_{q-1},{\widehat{\mathbf{p}}}_q={\widehat{\mathbf{k}}}_{q-1}\times {\widehat{\mathbf{s}}}_q,{{\widehat{\mathbf{p}}}^{\prime }}_q={\widehat{\mathbf{k}}}_q\times {\widehat{\mathbf{s}}}_q.$$9.6

9.6

Figure 9.2 shows the local coordinate bases for the incident ray along ${\widehat{\mathbf{k}}}_{In}$ at a planar interface with a surface normal $\widehat{\mathbf{\eta }}$. $\left({{\widehat{\mathbf{s}}}^{\prime }}_T,{{\widehat{\mathbf{p}}}^{\prime }}_T,{\widehat{\mathbf{k}}}_T\right)$ are the local coordinate bases for the refracted ray and $\left({{\widehat{\mathbf{s}}}^{\prime }}_R,{{\widehat{\mathbf{p}}}^{\prime }}_R,{\widehat{\mathbf{k}}}_R\right)$ are the local coordinate bases for the reflected ray.

The incident $({\widehat{\mathbf{s}}}_q)$ and exiting $\left({{\widehat{\mathbf{s}}}^{\prime }}_q\right)$ vectors for the qth interface are always the same; thus, ${\widehat{\mathbf{s}}}_q$ is used for both vectors; only the ${\widehat{\mathbf{p}}}_q$ vector changes. The orthogonal matrices are

$$\begin{array}{ll}{\mathbf{O}}_{in,q}^{-1}=\left(\begin{array}{lll}{\widehat{s}}_{x,q}& {\widehat{s}}_{y,q}& {\widehat{s}}_{z,q}\\ {\widehat{p}}_{x,q}& {\widehat{p}}_{y,q}& {\widehat{p}}_{z,q}\\ {\widehat{k}}_{x,q-1}& {\widehat{k}}_{y,q-1}& {\widehat{k}}_{z,q-1}\end{array}\right),& {\mathbf{O}}_{out,q}=\left(\begin{array}{lll}{\widehat{s}}_{x,q}& {{\widehat{p}}^{\prime }}_{x,q}& {\widehat{k}}_{x,q}\\ {\widehat{s}}_{y,q}& {{\widehat{p}}^{\prime }}_{y,q}& {\widehat{k}}_{y,q}\\ {\widehat{s}}_{z,q}& {{\widehat{p}}^{\prime }}_{z,q}& {\widehat{k}}_{z,q}\end{array}\right).\end{array}$$9.7

9.7

${\mathbf{O}}_{in,q}^{-1}$ operates on the incident electric field Eq−1, defined in the global coordinate system, and calculates a projection of Eq−1 onto the incident sp local coordinate system $({\widehat{\mathbf{s}}}_q,{\widehat{\mathbf{p}}}_q,{\widehat{\mathbf{k}}}_{q-1})$; that is,

$${\mathbf{E}}_{sp,q-1}=\left(\begin{array}{l}{E}_{s,q-1}\\ E_{p,q-1}\\ 0\end{array}\right)={\mathbf{O}}_{in,q}^{-1}\left(\begin{array}{l}{E}_{x,q-1}\\ E_{y,q-1}\\ E_{z,q-1}\end{array}\right).$$9.8

9.8

Oout,q rotates the global coordinate system $(\widehat{\mathbf{x}},\widehat{\mathbf{y}},\widehat{\mathbf{z}})$ to the exiting sp′ local coordinate system $\left({\widehat{\mathbf{s}}}_q,{{\widehat{\mathbf{p}}}^{\prime }}_{\mathbf{q}},{\widehat{k}}_q\right)$ and operates on the electric field ${\mathbf{E}}_{s{p}^{\prime },q}=\left(\begin{array}{l}{E}_{s,q}\\ E_{p^{\prime },q}\\ 0\end{array}\right)$, defined in the exiting local coordinate system $\left({\widehat{\mathbf{s}}}_q,{{\widehat{\mathbf{p}}}^{\prime }}_q,{\widehat{\mathbf{k}}}_q\right)$, and calculates the exiting electric Eq in the global coordinate system,

$$\begin{array}{l}{\mathbf{E}}_q={\mathbf{O}}_{out,q}\cdot \left(\begin{array}{l}{E}_{s,q}\\ E_{p^{\prime },q}\\ 0\end{array}\right)=\left(\begin{array}{lll}{\widehat{s}}_{x,q}& {{\widehat{p}}^{\prime }}_{x,q}& {\widehat{k}}_{x,q}\\ {\widehat{s}}_{y,q}& {{\widehat{p}}^{\prime }}_{y,q}& {\widehat{k}}_{y,q}\\ {\widehat{s}}_{z,q}& {{\widehat{p}}^{\prime }}_{z,q}& {\widehat{k}}_{z,q}\end{array}\right)\cdot \left(\begin{array}{l}{E}_{s,q}\\ E_{p^{\prime },q}\\ 0\end{array}\right)=E_{s,q}\left(\begin{array}{l}{\widehat{s}}_{x,q}\\ {\widehat{s}}_{y,q}\\ {\widehat{s}}_{z,q}\end{array}\right)+E_{p^{\prime },q}\left(\begin{array}{l}{{\widehat{p}}^{\prime }}_{x,q}\\ {{\widehat{p}}^{\prime }}_{y,q}\\ {{\widehat{p}}^{\prime }}_{z,q}\end{array}\right)\hfill \\ =E_{s,q}{\widehat{\mathbf{s}}}_q+E_{p^{\prime },q}{{\widehat{\mathbf{p}}}^{\prime }}_q.\hfill \end{array}$$9.9

9.9

Math Tip 9.1Orthogonal Transformation

An orthogonal transformation is a rotation from one coordinate system to another. This is performed with a rotation matrix, a unitary matrix that maps one set of orthonormal coordinates $(\widehat{\mathbf{a}},\widehat{\mathbf{b}},\widehat{\mathbf{c}})$ into another $(\widehat{\mathbf{d}},\widehat{\mathbf{e}},\widehat{\mathbf{f}})$. A rotation matrix R1 that has $(\widehat{\mathbf{a}},\widehat{\mathbf{b}},\widehat{\mathbf{c}})$ as its columns,

$${\mathbf{R}}_1=\left(\begin{array}{lll}{a}_x& b_x& c_x\\ a_y& b_y& c_y\\ a_z& b_z& c_z\end{array}\right),$$9.10

9.10

rotates global coordinates $(\widehat{\mathbf{x}},\widehat{\mathbf{y}},\widehat{\mathbf{z}})$ to $(\widehat{\mathbf{a}},\widehat{\mathbf{b}},\widehat{\mathbf{c}})$,

$$\left(\begin{array}{lll}{a}_x& b_x& c_x\\ a_y& b_y& c_y\\ a_z& b_z& c_z\end{array}\right)\left(\begin{array}{l}1\\ 0\\ 0\end{array}\right)=\left(\begin{array}{l}{a}_x\\ a_y\\ a_z\end{array}\right),\left(\begin{array}{lll}{a}_x& b_x& c_x\\ a_y& b_y& c_y\\ a_z& b_z& c_z\end{array}\right)\left(\begin{array}{l}0\\ 1\\ 0\end{array}\right)=\left(\begin{array}{l}{b}_x\\ b_y\\ b_z\end{array}\right),\left(\begin{array}{lll}{a}_x& b_x& c_x\\ a_y& b_y& c_y\\ a_z& b_z& c_z\end{array}\right)\left(\begin{array}{l}0\\ 0\\ 1\end{array}\right)=\left(\begin{array}{l}{c}_x\\ c_y\\ c_z\end{array}\right).$$9.11

9.11

Similarly, the inverse of R1 rotates $(\widehat{\mathbf{a}},\widehat{\mathbf{b}},\widehat{\mathbf{c}})$ back to $(\widehat{\mathbf{x}},\widehat{\mathbf{y}},\widehat{\mathbf{z}})$. When $(\widehat{\mathbf{a}},\widehat{\mathbf{b}},\widehat{\mathbf{c}})$ are real-valued vectors, the inverse of R1 is also the transpose of R1.

$$\begin{array}{l}{\mathbf{R}}_1^{-1}=\left(\begin{array}{lll}{a}_x& a_y& a_z\\ b_x& b_y& b_z\\ c_x& c_y& c_z\end{array}\right),\hfill \\ \left(\begin{array}{lll}{a}_x& a_y& a_z\\ b_x& b_y& b_z\\ c_x& c_y& c_z\end{array}\right)\left(\begin{array}{l}{a}_x\\ a_y\\ a_z\end{array}\right)=\left(\begin{array}{l}1\\ 0\\ 0\end{array}\right),\left(\begin{array}{lll}{a}_x& a_y& a_z\\ b_x& b_y& b_z\\ c_x& c_y& c_z\end{array}\right)\left(\begin{array}{l}{b}_x\\ b_y\\ b_z\end{array}\right)=\left(\begin{array}{l}0\\ 1\\ 0\end{array}\right),\left(\begin{array}{lll}{a}_x& a_y& a_z\\ b_x& b_y& b_z\\ c_x& c_y& c_z\end{array}\right)\left(\begin{array}{l}{c}_x\\ c_y\\ c_z\end{array}\right)=\left(\begin{array}{l}0\\ 0\\ 1\end{array}\right).\hfill \end{array}$$9.12

9.12

Using Equation 9.10, an orthogonal transformation matrix that rotates $(\widehat{\mathbf{x}},\widehat{\mathbf{y}},\widehat{\mathbf{z}})$ to $(\widehat{\mathbf{d}},\widehat{\mathbf{e}},\widehat{\mathbf{f}})$ can be calculated,

$${\mathbf{R}}_2=\left(\begin{array}{lll}{d}_x& e_x& f_x\\ d_y& e_y& f_y\\ d_z& e_z& f_z\end{array}\right).$$9.13

9.13

Example 9.1Find the Orthogonal Transformation Matrix That Rotates $(\widehat{\mathbf{a}},\widehat{\mathbf{b}},\widehat{\mathbf{c}})$ to $(\widehat{\mathbf{d}},\widehat{\mathbf{e}},\widehat{\mathbf{f}})$

Using  equations  in  Math Tip 0.1, ${\mathbf{R}}_2{\mathbf{R}}_1^{-1}$ rotates $(\widehat{\mathbf{a}},\widehat{\mathbf{b}},\widehat{\mathbf{c}})$ to $(\widehat{\mathbf{d}},\widehat{\mathbf{e}},\widehat{\mathbf{f}})$ since ${\mathbf{R}}_1^{-1}$ transforms $(\widehat{\mathbf{a}},\widehat{\mathbf{b}},\widehat{\mathbf{c}})$ to $(\widehat{\mathbf{x}},\widehat{\mathbf{y}},\widehat{\mathbf{z}})$ and R2 transforms $(\widehat{\mathbf{x}},\widehat{\mathbf{y}},\widehat{\mathbf{z}})$ to $(\widehat{\mathbf{d}},\widehat{\mathbf{e}},\widehat{\mathbf{f}})$. Therefore, ${\mathbf{R}}_2{\mathbf{R}}_1^{-1}\widehat{\mathbf{a}}=\widehat{\mathbf{d}},{\mathbf{R}}_2{\mathbf{R}}_1^{-1}\widehat{\mathbf{b}}=\widehat{\mathbf{e}},{\mathbf{R}}_2{\mathbf{R}}_1^{-1}\widehat{\mathbf{c}}=\widehat{\mathbf{f}}.$

Example 9.2Orthogonal Transformation Matrix for a Flat Glass Plate

A ray propagating along ${\widehat{\mathbf{k}}}_{In}=(0,\text{sin}(\pi /6),\text{cos}(\pi /6))$ refracts into a flat glass plate with $\widehat{\mathbf{\eta }}=(0,0,1)$ and refracts along ${\widehat{\mathbf{k}}}_T=\left(0,\frac{1}{3},\frac{2\sqrt{2}}{3}\right)$, as shown in Figure 9.2 (left). Calculate the orthogonal transformation matrices ${\mathbf{O}}_{in}^{-1}$ and Oout.

First, calculate local coordinate vectors,

$$\begin{array}{l}\widehat{\mathbf{s}}=\frac{{\widehat{\mathbf{k}}}_{In}\times \widehat{\mathbf{\eta }}}{|{\widehat{\mathbf{k}}}_{In}\times \widehat{\mathbf{\eta }}|}=(1,0,0),\hfill \\ \widehat{\mathbf{p}}={\widehat{\mathbf{k}}}_{In}\times \widehat{\mathbf{s}}=\left(0,\frac{\sqrt{3}}{2},-\frac{1}{2}\right),\hfill \\ {\widehat{\mathbf{s}}}^{\prime }=\widehat{\mathbf{s}}=(1,0,0),\hfill \\ {\widehat{\mathbf{p}}}^{\prime }={\widehat{\mathbf{k}}}_T\times {\widehat{\mathbf{s}}}^{\prime }=\left(0,\frac{2\sqrt{2}}{3},-\frac{1}{3}\right).\hfill \end{array}$$

Then, the orthogonal transformation matrices are

$$\begin{array}{l}{\mathbf{O}}_{in}^{-1}=\left(\begin{array}{lll}{\widehat{s}}_x& {\widehat{s}}_y& {\widehat{s}}_z\\ {\widehat{p}}_x& {\widehat{p}}_y& {\widehat{p}}_z\\ {\widehat{k}}_{x,In}& {\widehat{k}}_{y,In}& {\widehat{k}}_{z,In}\end{array}\right)=\left(\begin{array}{lll}1& 0& 0\\ 0& \frac{\sqrt{3}}{2}& \frac{-1}{2}\\ 0& \frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right),\\ O_{out}=\left(\begin{array}{lll}{\widehat{s}}_x& {{\widehat{p}}^{\prime }}_x& {\widehat{k}}_{x,T}\\ {\widehat{s}}_y& {{\widehat{p}}^{\prime }}_y& {\widehat{k}}_{y,T}\\ {\widehat{s}}_z& {{\widehat{p}}^{\prime }}_z& {\widehat{k}}_{z,T}\end{array}\right)=\left(\begin{array}{lll}1& 0& 0\\ 0& \frac{2\sqrt{2}}{3}& \frac{1}{3}\\ 0& \frac{-1}{2}& \frac{2\sqrt{2}}{3}\end{array}\right).\end{array}$$

The physics of reflection and refraction at dielectric, metal, and multilayer coated interfaces is described in terms of the incident $(\widehat{\mathbf{s}},\widehat{\mathbf{p}})$ components. Pq for a refraction or reflection can be derived by using Jt,q and Jr,q, which are defined in a local $\widehat{\mathbf{s}}$ and $\widehat{\mathbf{p}}$ basis, and Equation 9.7,

$${\mathbf{J}}_{t,q}=\left(\begin{array}{lll}{\alpha }_{s,t,q}& 0& 0\\ 0& {\alpha }_{p,t,q}& 0\\ 0& 0& 1\end{array}\right),{\mathbf{J}}_{r,q}=\left(\begin{array}{lll}{\alpha }_{s,r,q}& 0& 0\\ 0& {\alpha }_{p,r,q}& 0\\ 0& 0& 1\end{array}\right).$$9.14

9.14

Note that the Jones matrices for reflection and refraction are diagonal matrices when expressed in sp coordinates. The subscript t indicates refraction, r indicates reflection, s indicates s-polarization, and p indicates p-polarization. αs,t,q and αp,t,q are s- and p-amplitude transmission coefficients and αs,r,q and αp,r,q are s- and p-amplitude reflection coefficients. For an uncoated interface between two isotropic media, the coefficients are calculated from the Fresnel equations. For coated interfaces, the coefficients are calculated from multilayer coating calculations (Chapter 13, Thin Films).12–14 The polarization ray tracing matrices for refraction and reflection, respectively, are

$${\mathbf{P}}_q={\mathbf{O}}_{out,q}{\mathbf{J}}_{t,q}{\mathbf{O}}_{in,q}^{-1}\text{and}{\mathbf{P}}_q={\mathbf{O}}_{out,q}{\mathbf{J}}_{r,q}{\mathbf{O}}_{in,q}^{-1}.$$9.15

9.15

Example 9.3Pq for a Refraction

Using Equations 9.8, 9.9, and 9.15, the exiting electric field Eq in the global coordinate system after a refraction is related to the incident field Eq−1 through a series of coordinate transformations:

$$\begin{array}{l}{\mathbf{E}}_q={\mathbf{P}}_q{\mathbf{E}}_{q-1}\hfill \\ \phantom{E_q}={\mathbf{O}}_{out,q}\left(\begin{array}{lll}{\alpha }_{s,t,q}& 0& 0\\ 0& {\alpha }_{p,t,q}& 0\\ 0& 0& 1\end{array}\right)O_{in,q}^{-1}\left(\begin{array}{l}{E}_{x,q}\\ E_{y,q}\\ E_{z,q}\end{array}\right)\hfill \\ \phantom{E_q}={\mathbf{O}}_{out,q}\left(\begin{array}{lll}{\alpha }_{s,t,q}& 0& 0\\ 0& {\alpha }_{p,t,q}& 0\\ 0& 0& 1\end{array}\right)\left(\begin{array}{l}{E}_{s,q-1}\\ E_{p,q-1}\\ 0\end{array}\right)={\mathbf{O}}_{out,q}\cdot \left(\begin{array}{l}{\alpha }_{s,t,q}{E}_{s,q-1}\\ {\alpha }_{p,t,q}{E}_{p,q-1}\\ 0\end{array}\right)={\mathbf{O}}_{out,q}\cdot \left(\begin{array}{l}{E}_{s,q}\\ E_{p^{\prime },q}\\ 0\end{array}\right)\hfill \\ \phantom{E_q}=\left(\begin{array}{lll}{\widehat{s}}_{x,q}& {{\widehat{p}}^{\prime }}_{x,q}& {\widehat{k}}_{x,q}\\ {\widehat{s}}_{y,q}& {{\widehat{p}}^{\prime }}_{y,q}& {\widehat{k}}_{y,q}\\ {\widehat{s}}_{z,q}& {{\widehat{p}}^{\prime }}_{z,q}& {\widehat{k}}_{z,q}\end{array}\right)\cdot \left(\begin{array}{l}{E}_{s,q}\\ E_{p^{\prime },q}\\ 0\end{array}\right)=E_{s,q}{\widehat{s}}_q+E_{p^{\prime },q}{{\widehat{\mathbf{p}}}^{\prime }}_q.\hfill \end{array}$$

1. ${\mathbf{O}}_{in,q}^{-1}$ projects the incident electric field Eq−1 defined in the global coordinate system onto the incident local coordinate bases $({\widehat{\mathbf{s}}}_q,{\widehat{\mathbf{p}}}_q,{\widehat{\mathbf{k}}}_{q-1})$, resulting in Esp,q−1. Note that Esp,q−1 is a Jones vector defined in $({\widehat{\mathbf{s}}}_q,{\widehat{\mathbf{p}}}_q,{\widehat{\mathbf{k}}}_{q-1})$ with an additional zero.

2. Jt,q (or Jr,q) calculates the exiting Jones vector with an additional zero, Esp′,q, defined in $({\widehat{\mathbf{s}}}_q,{{\widehat{\mathbf{p}}}^{\prime }}_q,{\widehat{\mathbf{k}}}_q)$.

3. Oout,q converts Esp′,q to the exiting electric field Eq in the global coordinate system.

Polarization transformations during reflection and refraction have s- and p-polarizations as eigenpolarizations. Transformations due to diffraction gratings, holograms, sub-wavelength gratings, and other non-isotropic interfaces often couple some s- into p-polarization and vice versa. Thus, the Jones matrices for gratings, holograms, wire grid polarizers, sub-wavelength gratings, and other non-isotropic interfaces can have off-diagonal elements. Therefore, in general, the Pq matrix for a given ray intercept is

$${\mathbf{P}}_q={\mathbf{O}}_{out,q}{\mathbf{J}}_q{\mathbf{O}}_{in,q}^{-1},$$9.16

9.16

where ${\mathbf{J}}_q=\left(\begin{array}{ccc}{j}_{11}& j_{12}& 0\\ j_{21}& j_{22}& 0\\ 0& 0& 1\end{array}\right)$ represents the Jones matrix for the interaction in its arbitrarily chosen local coordinate system.2

For interactions that do not change the ray direction, for example, sheet polarizers and retarders, the surface local coordinates are arbitrarily chosen to be perpendicular to the propagation vector and Equation 9.16 becomes

$${\mathbf{P}}_q={\mathbf{O}}_{in,q}{\mathbf{J}}_q{\mathbf{O}}_{in,q}^{-1}.$$9.17

9.17

9.3Retarder Polarization Ray Tracing Matrix Examples

The matrix P is a polarization ray tracing matrix for a single ray since orthogonal transformation matrices (Equation 9.7) are different for each ray, unless the beam of light is collimated and all surfaces are planar surfaces. The P matrix in Equation 9.16 is dependent not only on the corresponding Jones matrix (optical element) but also on the propagation vector.

Figure 9.3 shows quarter wave linear retarders acting upon beams with different propagation vectors (blue arrows). Each of the three rays is normally incident onto a retarder element. The Jones matrix for a horizontal fast axis quarter wave retarder with $\widehat{\mathbf{\eta }}=(0,0,1)$ is $\left(\begin{array}{cc}{e}^{-i\pi /4}& 0\\ 0& e^{i\pi /4}\end{array}\right)$. The Jones matrix is specified in a symmetric phase convention where the fast axis polarization state is advanced by an eighth of a wave and the slow axis is delayed by an eighth of a wave. The corresponding P matrix for the same quarter wave retarder with $\widehat{\mathbf{\eta }}=(0,1,0)$ or $\widehat{\mathbf{\eta }}=(1,0,0)$ is different. Table 9.1 shows the P matrices for all three cases.

Figure 9.3Quarter wave linear retarders for rays propagating along x-, y-, and z-axes are shown. Blue arrows indicate the propagation vectors and red arrows indicate the fast axis of each retarder.

Table 9.1Polarization Ray Tracing Matrices for Identical Horizontal Fast Axis Linear Quarter Wave Retarders with Normally Incident Rays But Having Three Different Surface Normal Vectors

J	$\widehat{\mathbf{\eta }}$	${\widehat{\mathbf{k}}}_{In}$	${\widehat{\mathbf{k}}}_{Out}$	P
$\left(\begin{array}{cc}{e}^{-i\pi /4}& 0\\ 0& e^{i\pi /4}\end{array}\right)$	$\left(\begin{array}{l}0\\ 0\\ 1\end{array}\right)$	$\left(\begin{array}{l}0\\ 0\\ 1\end{array}\right)$	$\left(\begin{array}{l}0\\ 0\\ 1\end{array}\right)$	$\left(\begin{array}{ccc}{e}^{-i\pi /4}& 0& 0\\ 0& e^{i\pi /4}& 0\\ 0& 0& 1\end{array}\right)$
$\left(\begin{array}{cc}{e}^{-i\pi /4}& 0\\ 0& e^{i\pi /4}\end{array}\right)$	$\left(\begin{array}{l}0\\ 1\\ 0\end{array}\right)$	$\left(\begin{array}{l}0\\ 1\\ 0\end{array}\right)$	$\left(\begin{array}{l}0\\ 1\\ 0\end{array}\right)$	$\left(\begin{array}{ccc}{e}^{-i\pi /4}& 0& 0\\ 0& 1& 0\\ 0& 0& e^{i\pi /4}\end{array}\right)$
$\left(\begin{array}{cc}{e}^{-i\pi /4}& 0\\ 0& e^{i\pi /4}\end{array}\right)$	$\left(\begin{array}{l}1\\ 0\\ 0\end{array}\right)$	$\left(\begin{array}{l}1\\ 0\\ 0\end{array}\right)$	$\left(\begin{array}{l}1\\ 0\\ 0\end{array}\right)$	$\left(\begin{array}{ccc}1& 0& 0\\ 0& e^{-i\pi /4}& 0\\ 0& 0& e^{i\pi /4}\end{array}\right)$

Note that P for a ray propagating along the z-axis is the same as the Jones matrix padded with zeros and a one in the lower right element, but in general, P matrices are different from Jones matrices. The P’s in Table 9.1 relate the phase to the corresponding component of the electric field in global coordinates.

Example 9.4Calculate the P Matrix for a Ray Reflected from a Flat Glass Plate Using Fresnel Reflection Coefficients

A ray propagating along ${\widehat{\mathbf{k}}}_{In}=(0,\text{sin}(\pi /6),\text{cos}(\pi /6))$ reflects from the flat glass plate (n = 1.5) with $\widehat{\mathbf{\eta }}=(0,0,1)$ and refracts along ${\widehat{\mathbf{k}}}_R$, as shown in Figure 9.2 (right). Calculate the P matrix for this reflection.

First, calculate ${\widehat{\mathbf{k}}}_R$ using the law of reflection,

$${\widehat{\mathbf{k}}}_R=(0,\text{sin}(\pi /6),-\text{cos}(\pi /6)).$$

Then, calculate local coordinate vectors,

$$\begin{array}{l}\widehat{\mathbf{s}}=\frac{{\widehat{\mathbf{k}}}_{In}\times \widehat{\mathbf{\eta }}}{|{\widehat{\mathbf{k}}}_{In}\times \widehat{\mathbf{\eta }}|}=(1,0,0),\hfill \\ \widehat{\mathbf{p}}={\widehat{\mathbf{k}}}_{In}\times \widehat{\mathbf{s}}=\left(0,\frac{\sqrt{3}}{2},-\frac{1}{2}\right),\hfill \\ {\widehat{\mathbf{s}}}^{\prime }=\widehat{\mathbf{s}}=(1,0,0),\hfill \\ {\widehat{\mathbf{p}}}^{\prime }={\widehat{\mathbf{k}}}_R\times {\widehat{\mathbf{s}}}^{\prime }=\left(0,-\frac{\sqrt{3}}{2},-\frac{1}{2}\right).\hfill \end{array}$$

Thus, the orthogonal transformation matrices are

$$\begin{array}{l}{\mathbf{O}}_{in}^{-1}=\left(\begin{array}{lll}{\widehat{s}}_x& {\widehat{s}}_y& {\widehat{s}}_z\\ {\widehat{p}}_x& {\widehat{p}}_y& {\widehat{p}}_z\\ {\widehat{k}}_{x,In}& {\widehat{k}}_{y,In}& {\widehat{k}}_{z,In}\end{array}\right)=\left(\begin{array}{lll}1& 0& 0\\ 0& \frac{\sqrt{3}}{2}& \frac{-1}{2}\\ 0& \frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right),\\ {\mathbf{O}}_{out}=\left(\begin{array}{lll}{\widehat{s}}_x& {{\widehat{p}}^{\prime }}_x& {\widehat{k}}_{x,R}\\ {\widehat{s}}_y& {{\widehat{p}}^{\prime }}_y& {\widehat{k}}_{y,R}\\ {\widehat{s}}_z& {{\widehat{p}}^{\prime }}_z& {\widehat{k}}_{z,R}\end{array}\right)=\left(\begin{array}{lll}1& 0& 0\\ 0& \frac{-\sqrt{3}}{2}& \frac{1}{2}\\ 0& \frac{-1}{2}& \frac{-\sqrt{3}}{2}\end{array}\right).\end{array}$$

The angle of incidence is π/6; hence, the Fresnel reflection coefficients are rs = − 0.2404 and rp = 0.1589. Thus, the Jones matrix for this reflection is

$$\mathbf{J}=\left(\begin{array}{ll}{r}_s& 0\\ 0& r_p\end{array}\right)=\left(\begin{array}{ll}-0.2404& 0\\ 0& 0.1589\end{array}\right).$$

Using Equation 9.15, the P matrix is

$$\begin{array}{l}\mathbf{P}={\mathbf{O}}_{out}{\mathbf{J}}_r{\mathbf{O}}_{in}^{-1}=\left(\begin{array}{lll}{\widehat{s}}_x& {{\widehat{p}}^{\prime }}_x& {\widehat{k}}_{x,R}\\ {\widehat{s}}_y& {{\widehat{p}}^{\prime }}_y& {\widehat{k}}_{y,R}\\ {\widehat{s}}_z& {{\widehat{p}}^{\prime }}_z& {\widehat{k}}_{z,R}\end{array}\right)\left(\begin{array}{lll}{r}_s& 0& 0\\ 0& r_p& 0\\ 0& 0& 1\end{array}\right)\left(\begin{array}{lll}{\widehat{s}}_x& {\widehat{s}}_y& {\widehat{s}}_z\\ {\widehat{p}}_x& {\widehat{p}}_y& {\widehat{p}}_z\\ {\widehat{k}}_{x,In}& {\widehat{k}}_{y,In}& {\widehat{k}}_{z,In}\end{array}\right)\\ \phantom{P}=\left(\begin{array}{lll}1& 0& 0\\ 0& \frac{-\sqrt{3}}{2}& \frac{1}{2}\\ 0& \frac{-1}{2}& \frac{-\sqrt{3}}{2}\end{array}\right)\left(\begin{array}{lll}-0.240408& 0& 0\\ 0& 0.1589& 0\\ 0& 0& 1\end{array}\right)\left(\begin{array}{lll}1& 0& 0\\ 0& \frac{\sqrt{3}}{2}& \frac{-1}{2}\\ 0& \frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right)\\ \phantom{P}=\left(\begin{array}{lll}-0.240408& 0& 0\\ 0& 0.130825& 0.501818\\ 0& -0.501818& -0.710275\end{array}\right).\end{array}$$

Example 9.5The P Matrix for a Half Wave Retarder

Consider a ray with $\widehat{\mathbf{k}}=(0,\text{sin}10°,\text{cos}10°)$ that passes through a half waveplate that has its fast axis oriented α from the local x-axis, surface normal $\widehat{\mathbf{\eta }}=(0,\text{sin}10°,\text{cos}10°)$, and Jones matrix $\left(\begin{array}{cc}\text{cos}2\alpha & \text{sin}2\alpha \\ \text{sin}2\alpha & -\text{cos}2\alpha \end{array}\right)$, as shown in Figure 9.4. The finite thickness of the half waveplate is ignored in this example.

Figure 9.4(Left) A half waveplate has its fast axis (pink line) oriented at α from the xlocal axis. (Right) The plate is tilted 10° from the z-axis. A normal incident ray (brown arrow) passes through the tilted waveplate in the global xyz coordinate systems.

For normal incidence, s- and p-polarizations are degenerate. $\widehat{\mathbf{s}}$ is chosen to be (1, 0, 0). Then, $\widehat{\mathbf{p}}=(0,\text{cos}10°,-\text{sin}10°)$, calculated by Equation 9.6, since the ray is undeviated. For this waveplate example, $\widehat{\mathbf{s}}={\widehat{\mathbf{s}}}^{\prime }$, $\widehat{\mathbf{p}}={\widehat{\mathbf{p}}}^{\prime }$, and $\widehat{\mathbf{k}}=\widehat{{\mathbf{k}}^{\prime }}$. By Equation 9.15, the P matrix for a ray normally incident on the tilted waveplate is

$$\begin{array}{l}\mathbf{P}=\left(\begin{array}{lll}1& 0& 0\\ 0& \text{cos}10°& \text{sin}10°\\ 0& -\text{sin}10°& \text{cos}10°\end{array}\right)\left(\begin{array}{lll}\text{cos}2\alpha & \text{sin}2\alpha & 0\\ \text{sin}2\alpha & -\text{cos}2\alpha & 0\\ 0& 0& 1\end{array}\right){\left(\begin{array}{lll}1& 0& 0\\ 0& \text{cos}10°& \text{sin}10°\\ 0& -\text{sin}10°& \text{cos}10°\end{array}\right)}^{-1}\\ \phantom{P}=\left(\begin{array}{lll}\text{cos}2\alpha & \text{cos}10°\text{sin}2\alpha & -\text{sin}10°\text{sin}2\alpha \\ \text{cos}10°\text{sin}2\alpha & -{\cos }^{2}10°\text{cos}2\alpha +{\text{sin}}^{2}10°& {\text{cos}}^2\alpha \text{sin}20°\\ -\text{sin}10°\text{sin}2\alpha & {\text{cos}}^2\alpha \text{sin}20°& {\text{cos}}^{2}10°-\text{cos}2\alpha {\text{sin}}^{2}10°\end{array}\right).\end{array}$$

The incident p-polarization $\left(\begin{array}{c}0\\ \text{cos}10°\\ -\text{sin}10°\end{array}\right)$, originally oriented at π/2 − α from the fast axis, is rotated to $\mathbf{P}\cdot \left(\begin{array}{c}0\\ \text{cos}10°\\ -\text{sin}10°\end{array}\right)=\left(\begin{array}{c}\text{sin}2\alpha \\ -\text{cos}10°\text{cos}2\alpha \\ \text{sin}10°\text{cos}2\alpha \end{array}\right)$, cos−1 (sin 2α) from the fast axis. The incident s-polarization, originally oriented at α from the fast axis, is rotated to $\mathbf{P}\cdot \left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)=\left(\begin{array}{c}\text{cos}2\alpha \\ \text{cos}10°\text{sin}2\alpha \\ -\text{sin}10°\text{sin}2\alpha \end{array}\right)$, 2α from the fast axis as shown in Figure 9.5.

Figure 9.5The incident electric field Es (pink) and Ep (blue) passes through a half waveplate shown in Figure 9.4 and rotates to ${{\mathbf{E}}^{\prime }}_{\text{s}}$ and ${{\mathbf{E}}^{\prime }}_p$.

9.4Diattenuation Calculation Using Singular Value Decomposition

One objective of a polarization ray trace is to understand how an optical system changes the polarization properties of incident light. Methods to analyze the polarization properties of Jones matrices are discussed in Chapters 5 and 14 and many texts.15–17 This section derives a related algorithm to define diattenuation of an arbitrary polarization ray tracing matrix, P. The algorithm presented assumes that the ray begins and ends in a medium with a refractive index of one, such as air or vacuum, but these results are readily generalized to other object and image space refractive indices.

Diattenuation, $\mathcal{D}$, characterizes the difference of the maximum, Tmax, and minimum, Tmin, intensity transmittances possible when considering all incident polarization states,

$$\mathcal{D}=\frac{T_{\text{max}}-T_{\text{min}}}{T_{\text{max}}+T_{\text{min}}}\text{ and }0\le \mathcal{D}\le 1.$$9.18

9.18

An ideal polarizer has diattenuation equal to one; one incident polarization state is fully transmitted, but another is fully attenuated. An element with a diattenuation of zero attenuates all polarization states equally.

The diattenuation calculation for general P matrices is complicated by the fact that the eigenvectors of P generally do not represent physical polarization states because, in most cases, rays enter and exit optical elements in different directions.18 Therefore, the singular value decomposition (SVD), not the eigenvectors, is appropriate to calculate the diattenuation of P.

Math Tip 9.2Singular Value Decomposition (SVD)

The SVD19,20 of a 3 × 3 square matrix B is the matrix decomposition

$$\mathbf{B}=\mathbf{U}\mathbf{D}{\mathbf{V}}^{†}=\left(\begin{array}{lll}{u}_{x,0}& u_{x,1}& u_{x,2}\\ u_{y,0}& u_{y,1}& u_{y,2}\\ u_{z,0}& u_{z,1}& u_{z,2}\end{array}\right)\left(\begin{array}{lll}{\Lambda }_0& 0& 0\\ 0& {\Lambda }_1& 0\\ 0& 0& {\Lambda }_2\end{array}\right)\left(\begin{array}{lll}{v}_{x,0}^{*}& v_{y,0}^{*}& v_{z,0}^{*}\\ v_{x,1}^{*}& v_{y,1}^{*}& v_{z,1}^{*}\\ v_{x,2}^{*}& v_{y,2}^{*}& v_{z,2}^{*}\end{array}\right),$$9.19

9.19

where U and V are unitary matrices and D is a diagonal matrix with non-negative real elements. † indicates Hermitian adjoint,

$${\mathbf{V}}^{†}={(\mathbf{V}\text{*})}^T.$$9.20

9.20

The columns of U are orthonormal eigenvectors of BB† and the columns of V are orthonormal eigenvectors of B†B.

The matrices BB† and B†B are Hermitian and share the same eigenvalues. The diagonal elements of D are named the singular values Λi21 and are the non-negative real square roots of the eigenvalues from BB† or B†B. Convention orders the singular values in descending order,

$${\Lambda }_0\ge {\Lambda }_1\ge {\Lambda }_2\ge 0.$$9.21

9.21

The order of singular values could be rearranged if the columns of U and V are also rearranged. Our preferred convention is placing the incident and exiting propagation vectors in the first columns of U and V. This can lead to a reordering of the singular values.

Since P was constructed such that $\mathbf{P}{\widehat{\mathbf{k}}}_{q-1}={\widehat{\mathbf{k}}}_q$, (1) one of P’s singular values is always one, (2) the associated column of V is ${\widehat{\mathbf{k}}}_0$, and (3) the associated column of U is ${\widehat{\mathbf{k}}}_Q$,

$$\mathbf{P}=\mathbf{U}\mathbf{D}{\mathbf{V}}^{†}=\left(\begin{array}{lll}{k}_{x,Q}& u_{x,1}& u_{x,2}\\ k_{y,Q}& u_{y,1}& u_{y,2}\\ k_{z,Q}& u_{z,1}& u_{z,2}\end{array}\right)\left(\begin{array}{lll}1& 0& 0\\ 0& {\Lambda }_1& 0\\ 0& 0& {\Lambda }_2\end{array}\right)\left(\begin{array}{lll}{k}_{x,0}^{*}& k_{y,0}^{*}& k_{z,0}^{*}\\ v_{x,1}^{*}& v_{y,1}^{*}& v_{z,1}^{*}\\ v_{x,2}^{*}& v_{y,2}^{*}& v_{z,2}^{*}\end{array}\right).$$9.22

9.22

The other two columns of V, v1 and v2, are the two special polarization states in the incident transverse plane that generate the maximum and minimum transmitted flux. v1 and v2 are always orthogonal. Similarly, the last two columns of U are the two corresponding orthogonal polarization statesu1 and u2 in the exiting transverse plane.

Relationships between P, its singular values, and these special polarization states, are

$$\mathbf{P}{\mathbf{v}}_1={\Lambda }_1{\mathbf{u}}_1,\mathbf{P}{\mathbf{v}}_2={\Lambda }_2{\mathbf{u}}_2,\text{and}\mathbf{P}{\widehat{\mathbf{k}}}_0={\widehat{\mathbf{k}}}_Q.$$9.23

9.23

When Λ1 ≠ Λ2, v1 and v2 are the only two orthogonal incident polarization states that remain orthogonal when they emerge from P as u1 and u2, respectively. Therefore, these two orthogonal sets of states, (v1, v2) and (u1, u2), form a canonical basis for incident and exiting polarization states. The SVD leads directly to these special polarization states, which are similar to eigenpolarizations.

An arbitrary normalized incident polarization state E can be expressed as a linear combination of v1 and v2 as

$$\mathbf{E}=\alpha {\mathbf{v}}_1+\beta {\mathbf{v}}_2,$$9.24

9.24

where α and β are in general complex, $\sqrt{\text{|}\alpha {\text{|}}^2+\text{|}\beta {\text{|}}^2}=1$. The transmitted electric field vector after P is PE. Therefore, the flux of the transmitted electric field is

$$I_{Trans}=|\mathbf{P}\mathbf{E}{|}^2={\mathbf{E}}^{†}{\mathbf{P}}^{†}\mathbf{P}\mathbf{E}.$$9.25

9.25

From Equations 9.23 and 9.25, the flux of the transmitted electric field is

$$I_{Trans}=\text{|}\alpha {\text{|}}^2{\Lambda }_1^2+\text{|}\beta {\text{|}}^2{\Lambda }_2^2=\text{|}\alpha {\text{|}}^2\left({\Lambda }_1^2-{\Lambda }_2^2\right)+{\Lambda }_2^2.$$9.26

9.26

Since both $\text{|}\alpha {\text{|}}^2\left({\Lambda }_1^2-{\Lambda }_2^2\right)\text{ and }{\Lambda }_2^2$ are positive and Λ1 ≥ Λ2 by construction, the maximum intensity transmittance occurs when the incident state is v1, and the minimum intensity occurs when the incident state is v2; that is,

$$I_{Trans}=\{\begin{array}{l}{I}_{\text{max}}={\Lambda }_1^2\text{if}\text{|}\alpha {\text{|}}^2=1\\ I_{\text{min}}={\Lambda }_2^2\text{if}\text{|}\alpha {\text{|}}^2=0\end{array},$$9.27

9.27

for any polarization ray tracing matrix P. Thus, the diattenuation of P is

$$\mathcal{D}=\frac{{\Lambda }_1^2-{\Lambda }_2^2}{{\Lambda }_1^2+{\Lambda }_2^2},$$9.28

9.28

and v1 = vmax and v2 = vmin are the incident polarization states for which P gives the maximum and minimum transmittance. u1 = Pv1 and u2 = Pv2 are the corresponding exiting polarization states. For an ideal polarizer, one incident polarization state is fully attenuated; thus, the singular value Λ2 = 0.

The derivation of an algorithm to define the retardance of P matrices is introduced in a separate chapter because of its complexity (see Chapter 17, Parallel Transport and the Calculation of Retardance).

9.5Example—Interferometer with a Polarizing Beam Splitter

In this section, an interferometer with a polarizing beam splitter (PBS) is used as an example for the polarization ray tracing calculus. Figure 9.6 shows the schematic of the interferometer. In the arms of the interferometer, combinations of quarter wave retarders and mirrors are used to route light through the PBSs such that light does not reflect back toward the laser, which is what would happen without the quarter wave retarders. This happens because circularly polarized light changes its handedness when it reflects from mirrors; that is, left reflects as right, and vice versa. Performing this calculation in three dimensions helps model these changes in a straightforward way.

Figure 9.6Interferometer with a polarizing beam splitter and quarter wave retarders in each arm. Circularly polarized light is incident on each mirror. Incident left circularly polarized light reflects from mirrors as right circularly polarized and vice versa. Thus, the s-polarized light that reflects downward from the polarizing beam splitter is converted to p-polarized light on its return path and transmits toward the screen. The two beams exiting upward from the PBS are in orthogonal polarization states and would not produce intensity interference fringes without the linear polarizer projecting them into a common 45° polarization state.

In this example, only the polarizing elements’ P matrices are calculated. For simplicity, all rays are propagating along the axes and normally incident at each mirror and retarder.

Consider a laser generating vertically polarized light. Before the PBS, all rays pass through a half wave linear retarder (HWLR) with fast axis at 22.5° from the horizontal axis. This produces a beam polarized along 45° such that an equal amount of s- and p-polarization is incident on the PBS. The Jones matrix for the HWLR at 22.5° is

$${\mathbf{J}}_{HWLR}=R\left[\frac{\pi }{8}\right]\cdot \left(\begin{array}{ll}1& 0\\ 0& -1\end{array}\right)\cdot R{\left[\frac{\pi }{8}\right]}^{-1}=\frac{1}{\sqrt{2}}\left(\begin{array}{ll}1& 1\\ 1& -1\end{array}\right).$$9.29

9.29

Since ${\widehat{\mathbf{\eta }}}_1={\widehat{\mathbf{k}}}_0=\widehat{\mathbf{z}}$, the ray does not bend at the interface; hence, Oin,1 = Oout,1. Therefore, the P matrix for the first ray intercept (HWLR) is

$$\begin{array}{l}{\mathbf{P}}_{HWLR}={\mathbf{O}}_{out,1}{\mathbf{J}}_{HWLR}{\mathbf{O}}_{in,1}^{-1}\\ \phantom{P_{HWLR}}=\left(\begin{array}{lll}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\cdot \frac{1}{\sqrt{2}}\left(\begin{array}{lll}1& 1& 0\\ 1& -1& 0\\ 0& 0& \sqrt{2}\end{array}\right)\cdot \left(\begin{array}{lll}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{lll}1& 1& 0\\ 1& -1& 0\\ 0& 0& \sqrt{2}\end{array}\right).\end{array}$$9.30

9.30

The next polarizing element is the PBS, which divides the beam into two paths: the reference path (which reflects and transmits) and the test path (which transmits and reflects). Since P is dependent on propagation direction, each path has unique P matrices.

9.5.1Ray Tracing the Reference Path

First consider the reference path. The light is split into s- and p-components at the hypotenuse of the PBS with a surface normal $\widehat{\mathbf{\eta }}=(1,0,1)/\sqrt{2}$. Thus,

$${\widehat{\mathbf{k}}}_{reference,2}=-\widehat{\mathbf{x}},{\widehat{\mathbf{s}}}_{reference,2}={{\widehat{\mathbf{s}}}^{\prime }}_{reference,2}=\widehat{\mathbf{y}},{\widehat{\mathbf{p}}}_{reference,2}=-\widehat{\mathbf{x}},{{\widehat{\mathbf{p}}}^{\prime }}_{reference,2}=-\widehat{\mathbf{z}}.$$9.31

9.31

The subscript reference indicates rays along the reference path. The P matrix for reflection from the PBS is

$$\begin{array}{l}{\mathbf{P}}_{PBS(R)}=\left(\begin{array}{lll}|& |& |\\ {{\widehat{\mathbf{s}}}^{\prime }}_{reference,2}& {{\widehat{\mathbf{p}}}^{\prime }}_{reference,2}& {\widehat{\mathbf{k}}}_{reference,2}\\ |& |& |\end{array}\right)\cdot \left(\begin{array}{lll}{r}_s& 0& 0\\ 0& r_p& 0\\ 0& 0& 1\end{array}\right)\cdot \left(\begin{array}{lll}-& {\widehat{\mathbf{s}}}_{reference,2}& -\\ -& {\widehat{\mathbf{p}}}_{reference,2}& -\\ -& {\widehat{\mathbf{k}}}_1& -\end{array}\right)\\ \phantom{P_{PBS(R)}}=\left(\begin{array}{lll}0& 0& -1\\ 1& 0& 0\\ 0& -1& 0\end{array}\right)\cdot \left(\begin{array}{lll}1& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right)\cdot \left(\begin{array}{lll}0& 1& 0\\ -1& 0& 0\\ 0& 0& 1\end{array}\right)=\left(\begin{array}{lll}0& 0& -1\\ 0& 1& 0\\ 0& 0& 0\end{array}\right).\end{array}$$9.32

9.32

Here, lines indicate that vectors occupy the corresponding matrix rows or columns.

Note that the middle matrix, a 3D Jones matrix, is in its local $(\widehat{\mathbf{s}},\widehat{\mathbf{p}},\widehat{\mathbf{k}})$ coordinates. PPBS(R) is in global coordinates and shows explicitly that the PBS reflects the propagation vector from $\widehat{\mathbf{z}}$ to $-\widehat{\mathbf{x}}$, reflects $\widehat{\mathbf{y}}$ polarization into $\widehat{\mathbf{y}}$ polarization, and removes the $\widehat{\mathbf{x}}$ polarized component of the incident ray.

The next element is a quarter wave linear retarder (QWLR) with a fast axis at 45° from the y toward the z axis, $\left(0,1/\sqrt{2},1/\sqrt{2}\right)$, with Jones matrix

$$\begin{array}{l}{\mathbf{J}}_{QWLR(45)}\hfill \\ \phantom{}=\mathbf{R}[-45°]\cdot \frac{1}{\sqrt{2}}\left(\begin{array}{ll}1-i& 0\\ 0& 1+i\end{array}\right)\cdot \mathbf{R}{[-45°]}^{-1}=\frac{1}{\sqrt{2}}\left(\begin{array}{ll}1-i& 0\\ 0& 1+i\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{ll}1& -i\\ -i& 1\end{array}\right).\hfill \end{array}$$9.33

9.33

The ray is normally incident on the QWLR; thus, we choose local coordinates following Equation 9.6; ${{\widehat{\mathbf{s}}}^{\prime }}_{reference,2}$ is used as ${\widehat{\mathbf{s}}}_{reference,3}$ and ${\widehat{\mathbf{p}}}_{reference,3}$ and ${{\widehat{\mathbf{p}}}^{\prime }}_{reference,3}$ are calculated;

$${\widehat{\mathbf{k}}}_{reference,3}=-\widehat{\mathbf{x}},{\widehat{\mathbf{s}}}_{reference,3}={{\widehat{\mathbf{s}}}^{\prime }}_{reference,3}={{\widehat{\mathbf{s}}}^{\prime }}_{reference,2}=\widehat{\mathbf{y}},{\widehat{\mathbf{p}}}_{reference,3}={{\widehat{\mathbf{p}}}^{\prime }}_{reference,3}=-\widehat{\mathbf{z}}.$$9.34

9.34

Thus, the P matrix for the QWLR is

$$\begin{array}{l}{\mathbf{P}}_{QWLR}={\mathbf{O}}_{out}\cdot {\mathbf{J}}_{QWLR(45)}\cdot {\mathbf{O}}_{in}^{-1}\hfill \\ =\left(\begin{array}{lll}0& 0& -1\\ 1& 0& 0\\ 0& -1& 0\end{array}\right)\cdot \frac{1}{\sqrt{2}}\left(\begin{array}{lll}1& -i& 0\\ -i& 1& 0\\ 0& 0& \sqrt{2}\end{array}\right)\cdot \left(\begin{array}{lll}0& 1& 0\\ 0& 0& -1\\ -1& 0& 0\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{lll}\sqrt{2}& 0& 0\\ 0& 1& i\\ 0& i& 1\end{array}\right).\hfill \end{array}$$9.35

9.35

The next element is an ideal mirror that retroreflects the ray. Since the ray is normally incident on the mirror,

$$\begin{array}{l}{\widehat{\mathbf{k}}}_{reference,4}=\widehat{\mathbf{x}},{\widehat{\mathbf{s}}}_{reference,4}={{\widehat{\mathbf{s}}}^{\prime }}_{reference,4}={{\widehat{\mathbf{s}}}^{\prime }}_{reference,3}=\widehat{\mathbf{y}},\hfill \\ {\widehat{\mathbf{p}}}_{reference,4}=-\widehat{\mathbf{z}},{{\widehat{\mathbf{p}}}^{\prime }}_{reference,4}=\widehat{\mathbf{z}}.\hfill \end{array}$$9.36

9.36

Using ${\mathbf{J}}_{mirror}=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$, the P matrix for the ideal mirror is

$$\begin{array}{l}{\mathbf{P}}_{M,1}={\mathbf{O}}_{out}\cdot {\mathbf{J}}_{mirror}\cdot {\mathbf{O}}_{in}^{-1}\hfill \\ =\left(\begin{array}{lll}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right)\cdot \left(\begin{array}{lll}1& 0& 0\\ 0& -1& 0\\ 0& 0& 1\end{array}\right)\cdot \left(\begin{array}{lll}0& 1& 0\\ 0& 0& -1\\ -1& 0& 0\end{array}\right)=\left(\begin{array}{lll}-1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right).\hfill \end{array}$$9.37

9.37

Then, the ray propagates back to QWLR in the opposite direction (along +x). The local coordinates are

$${\widehat{\mathbf{k}}}_{reference,5}=\widehat{\mathbf{x}},{\widehat{\mathbf{s}}}_{reference,5}={{\widehat{\mathbf{s}}}^{\prime }}_{reference,5}={{\widehat{\mathbf{s}}}^{\prime }}_{reference,4}=\widehat{\mathbf{y}},{\widehat{\mathbf{p}}}_{reference,5}={{\widehat{\mathbf{p}}}^{\prime }}_{reference,5}=\widehat{\mathbf{z}}.$$9.38

9.38

The fast axis of the QWLR is still at $\left(0,1/\sqrt{2},1/\sqrt{2}\right)$, but since the propagation direction is reversed, the fast axis orientation in Jones matrix local coordinates is at 135°. Hence,

$$\begin{array}{l}{\mathbf{P}}_{QWLR,2}={\mathbf{O}}_{out}\cdot {\mathbf{J}}_{QWLR(135)}\cdot {\mathbf{O}}_{in}^{-1}\hfill \\ =\left(\begin{array}{lll}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right)\cdot \frac{1}{\sqrt{2}}\left(\begin{array}{lll}1& i& 0\\ i& 1& 0\\ 0& 0& \sqrt{2}\end{array}\right)\cdot \left(\begin{array}{lll}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{lll}\sqrt{2}& 0& 0\\ 0& 1& i\\ 0& i& 1\end{array}\right).\hfill \end{array}$$9.39

9.39

Next, the ray transmits through the PBS with surface normal $\widehat{\mathbf{\eta }}=(1,0,1)/\sqrt{2}$,

$${\widehat{\mathbf{k}}}_{reference,6}=\widehat{\mathbf{x}},{\widehat{\mathbf{s}}}_{reference,6}={{\widehat{\mathbf{s}}}^{\prime }}_{reference,6}=-\widehat{\mathbf{y}},{\widehat{\mathbf{p}}}_{reference,5}={{\widehat{\mathbf{p}}}^{\prime }}_{reference,5}=-\widehat{\mathbf{z}}.$$9.40

9.40

The PBS reflects ${\widehat{\mathbf{s}}}_{reference,6}$ state out of the beam, transmits ${\widehat{\mathbf{p}}}_{reference,6}$, and transmits ${\widehat{\mathbf{k}}}_{reference,5}$ undeviated,

$${\mathbf{P}}_{PBS(T)}=\left(\begin{array}{lll}0& 0& 1\\ -1& 0& 0\\ 0& -1& 0\end{array}\right)\cdot \left(\begin{array}{lll}0& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\cdot \left(\begin{array}{lll}0& -1& 0\\ 0& 0& -1\\ 1& 0& 0\end{array}\right)=\left(\begin{array}{lll}1& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right).$$9.41

9.41

PPBS(T) transmits the $\widehat{\mathbf{z}}$ polarization, removes $\widehat{\mathbf{y}}$ polarization, and does not deviate the propagation vector of the incident ray.

9.5.2Ray Tracing through the Test Path

For the test path, the PBS transmits the p-polarization and the propagation vector does not change. The subscript test indicates rays along the test path. Using the surface normal of the PBS,

$${\widehat{\mathbf{k}}}_{test,2}=\widehat{\mathbf{z}},{\widehat{\mathbf{s}}}_{test,2}={{\widehat{\mathbf{s}}}^{\prime }}_{test,2}=\widehat{\mathbf{y}},{\widehat{\mathbf{p}}}_{test,2}={{\widehat{\mathbf{p}}}^{\prime }}_{test,2}=-\widehat{\mathbf{x}},$$9.42

9.42

$${\mathbf{P}}_{PBS(T)}=\left(\begin{array}{lll}0& -1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right)\cdot \left(\begin{array}{lll}0& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\cdot \left(\begin{array}{lll}0& 1& 0\\ -1& 0& 0\\ 0& 0& 1\end{array}\right)=\left(\begin{array}{lll}1& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right).$$9.43

9.43

PPBS(T) transmits the $\widehat{\mathbf{x}}$ polarization and blocks the $\widehat{\mathbf{y}}$ polarization. Then, the light propagates through a QWLR oriented at 45° between the x- and y-axis. The local coordinates and the corresponding P matrix are

$${\widehat{\mathbf{k}}}_{test,3}=\widehat{\mathbf{z}},{\widehat{\mathbf{s}}}_{test,3}={{\widehat{\mathbf{s}}}^{\prime }}_{test,3}={{\widehat{\mathbf{s}}}^{\prime }}_{test,2}=\widehat{\mathbf{y}},{\widehat{\mathbf{p}}}_{test,2}={{\widehat{\mathbf{p}}}^{\prime }}_{test,2}=-\widehat{\mathbf{x}},$$9.44

9.44

$$\begin{array}{l}{\mathbf{P}}_{QWLR}={\mathbf{O}}_{out}\cdot {\mathbf{J}}_{QWLR(45)}\cdot {\mathbf{O}}_{in}^{-1}\hfill \\ =\left(\begin{array}{lll}0& -1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right)\cdot \frac{1}{\sqrt{2}}\left(\begin{array}{lll}1& -i& 0\\ -i& 1& 0\\ 0& 0& \sqrt{2}\end{array}\right)\cdot \left(\begin{array}{lll}0& 1& 0\\ -1& 0& 0\\ 0& 0& 1\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{lll}1& i& 0\\ i& 1& 0\\ 0& 0& \sqrt{2}\end{array}\right).\hfill \end{array}$$9.45

9.45

Assuming the test mirror is also an ideal mirror, the light reflects, changing its propagation direction from z to −z.

$${\widehat{\mathbf{k}}}_{test,4}=\widehat{\mathbf{z}},{\widehat{\mathbf{s}}}_{test,4}={{\widehat{\mathbf{s}}}^{\prime }}_{test,4}={{\widehat{\mathbf{s}}}^{\prime }}_{test,3}=\widehat{\mathbf{y}},{\widehat{\mathbf{p}}}_{test,4}=-\widehat{\mathbf{x}},{{\widehat{\mathbf{p}}}^{\prime }}_{test,4}=\widehat{\mathbf{x}}.$$9.46

9.46

$$\begin{array}{l}{\mathbf{P}}_{M,2}={\mathbf{O}}_{out}\cdot {\mathbf{J}}_{mirror}\cdot {\mathbf{O}}_{in}^{-1}\hfill \\ =\left(\begin{array}{lll}0& 1& 0\\ 1& 0& 0\\ 0& 0& -1\end{array}\right)\cdot \left(\begin{array}{lll}1& 0& 0\\ 0& -1& 0\\ 0& 0& 1\end{array}\right)\cdot \left(\begin{array}{lll}0& 1& 0\\ -1& 0& 0\\ 0& 0& 1\end{array}\right)=\left(\begin{array}{lll}1& 0& 0\\ 0& 1& 0\\ 0& 0& -1\end{array}\right).\hfill \end{array}$$9.47

9.47

The light propagates back to QWLR in the opposite direction. Similar to the reference path case, the fast axis of the QWLR in local coordinates is now at 135°.

$${\widehat{\mathbf{k}}}_{test,5}=-\widehat{\mathbf{z}},{\widehat{\mathbf{s}}}_{test,5}={{\widehat{\mathbf{s}}}^{\prime }}_{test,5}={{\widehat{\mathbf{s}}}^{\prime }}_{test,4}=\widehat{\mathbf{y}},{\widehat{\mathbf{p}}}_{test,5}={{\widehat{\mathbf{p}}}^{\prime }}_{test,5}=\widehat{\mathbf{x}},$$9.48

9.48

$$\begin{array}{l}{\mathbf{P}}_{QWLR,2}={\mathbf{O}}_{out}\cdot {\mathbf{J}}_{QWLR(135)}\cdot {\mathbf{O}}_{in}^{-1}\hfill \\ =\left(\begin{array}{lll}0& 1& 0\\ 1& 0& 0\\ 0& 0& -1\end{array}\right)\cdot \frac{1}{\sqrt{2}}\left(\begin{array}{lll}1& i& 0\\ i& 1& 0\\ 0& 0& \sqrt{2}\end{array}\right)\cdot \left(\begin{array}{lll}0& 1& 0\\ 1& 0& 0\\ 0& 0& -1\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{lll}1& i& 0\\ i& 1& 0\\ 0& 0& \sqrt{2}\end{array}\right).\hfill \end{array}$$9.49

9.49

Finally, the ray reaches the PBS the second time. This time, the ray gets reflected and changes its propagation direction from −z into x-direction.

$${\widehat{\mathbf{k}}}_{test,6}=\widehat{\mathbf{x}},{\widehat{\mathbf{s}}}_{test,6}={{\widehat{\mathbf{s}}}^{\prime }}_{test,6}=\widehat{\mathbf{y}},{\widehat{\mathbf{p}}}_{test,6}=\widehat{\mathbf{x}},{{\widehat{\mathbf{p}}}^{\prime }}_{test,6}=\widehat{\mathbf{z}},$$9.50

9.50

$${\mathbf{P}}_{PBS(R)}=\left(\begin{array}{lll}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right)\cdot \left(\begin{array}{lll}1& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right)\cdot \left(\begin{array}{lll}0& 1& 0\\ 1& 0& 0\\ 0& 0& -1\end{array}\right)=\left(\begin{array}{lll}0& 0& -1\\ 0& 1& 0\\ 0& 0& 0\end{array}\right).$$9.51

9.51

PPBS(R) reflects the $\widehat{\mathbf{y}}$ polarization, while blocking the $\widehat{\mathbf{z}}$ polarization.

9.5.3Ray Tracing through the Analyzer

Exiting the PBS, both paths have rays propagating along the +x-axis and pass through a linear polarizer at 45° in the y–z plane. Using ${\mathbf{J}}_{LP[45]}=\left(\begin{array}{cc}0.5& 0.5\\ 0.5& 0.5\end{array}\right)$ and local coordinates, ${\widehat{\mathbf{k}}}_7=\widehat{\mathbf{x}},{\widehat{\mathbf{s}}}_7={{\widehat{\mathbf{s}}}^{\prime }}_7=\widehat{\mathbf{y}},{\widehat{\mathbf{p}}}_7={{\widehat{\mathbf{p}}}^{\prime }}_7=\widehat{\mathbf{z}},$ the polarizer PLP is

$$\begin{array}{l}{\mathbf{P}}_{LP}={\mathbf{O}}_{out}{\mathbf{J}}_{LP[45]}{\mathbf{O}}_{in}^{-1}\\ \phantom{P_{LP}}=\left(\begin{array}{lll}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right)\cdot \left(\begin{array}{lll}0.5& 0.5& 0\\ 0.5& 0.5& 0\\ 0& 0& 1\end{array}\right)\cdot \left(\begin{array}{lll}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right)=\left(\begin{array}{lll}1& 0& 0\\ 0& 0.5& 0.5\\ 0& 0.5& 0.5\end{array}\right).\end{array}$$9.52

9.52

9.5.4Cumulative P Matrix for Both Paths

A matrix sequence for the reference path (red path) is listed from right to left since the sequence of matrix multiplications is written from right to left:

$$\begin{array}{lllllll}\mathbf{L}\mathbf{P}[45]\leftarrow & \mathbf{P}\mathbf{B}\mathbf{S}(\text{T})\leftarrow & QWLR\leftarrow & Mirror\leftarrow & QWLR\leftarrow & PBS(\text{R})\leftarrow & HWLR\\ \left(\begin{array}{lll}1& 0& 0\\ 0& \frac{1}{2}& \frac{1}{2}\\ 0& \frac{1}{2}& \frac{1}{2}\end{array}\right)& \left(\begin{array}{lll}1& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right)& \left(\begin{array}{lll}1& 0& 0\\ 0& \frac{1}{\sqrt{2}}& \frac{i}{\sqrt{2}}\\ 0& \frac{i}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)& \left(\begin{array}{lll}-1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)& \left(\begin{array}{lll}1& 0& 0\\ 0& \frac{1}{\sqrt{2}}& \frac{i}{\sqrt{2}}\\ 0& \frac{i}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)& \left(\begin{array}{lll}0& 0& -1\\ 0& 1& 0\\ 0& 0& 0\end{array}\right)& \left(\begin{array}{lll}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}& 0\\ \frac{1}{\sqrt{2}}& \frac{-1}{\sqrt{2}}& 0\\ 0& 0& 1\end{array}\right)\end{array}.$$9.53

9.53

The cumulative P for the reference path is

$${\mathbf{P}}_{reference}=\left(\begin{array}{lll}0& 0& 1\\ \frac{i}{2\sqrt{2}}& \frac{-i}{2\sqrt{2}}& 0\\ \frac{i}{2\sqrt{2}}& \frac{-i}{2\sqrt{2}}& 0\end{array}\right),$$9.54

9.54

and the electric field from the reference path at the observation screen is

$${\mathbf{E}}_{reference}={\mathbf{P}}_{reference}\cdot \left(\begin{array}{l}{E}_x\\ E_y\\ 0\end{array}\right)=\left(\begin{array}{l}0\\ \frac{i}{2\sqrt{2}}(E_x-E_y)\\ \frac{i}{2\sqrt{2}}(E_x-E_y)\end{array}\right).$$9.55

9.55

A matrix sequence for the test path (blue path) is

$$\begin{array}{lllllll}LP[45]\leftarrow & PBS(\text{R})\leftarrow & QWLR\leftarrow & Mirror\leftarrow & QWLR\leftarrow & PBS(\text{T})\leftarrow & HWLR\\ \left(\begin{array}{lll}1& 0& 0\\ 0& \frac{1}{2}& \frac{1}{2}\\ 0& \frac{1}{2}& \frac{1}{2}\end{array}\right)& \left(\begin{array}{lll}0& 0& -1\\ 0& 1& 0\\ 0& 0& 0\end{array}\right)& \left(\begin{array}{lll}1& 0& 0\\ 0& \frac{1}{\sqrt{2}}& \frac{i}{\sqrt{2}}\\ 0& \frac{i}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)& \left(\begin{array}{lll}1& 0& 0\\ 0& 1& 0\\ 0& 0& -1\end{array}\right)& \left(\begin{array}{lll}1& 0& 0\\ 0& \frac{1}{\sqrt{2}}& \frac{i}{\sqrt{2}}\\ 0& \frac{i}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)& \left(\begin{array}{lll}1& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right)& \left(\begin{array}{lll}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}& 0\\ \frac{1}{\sqrt{2}}& \frac{-1}{\sqrt{2}}& 0\\ 0& 0& 1\end{array}\right).\end{array}$$9.56

9.56

The cumulative P for the test path is

$${\mathbf{P}}_{test}=\left(\begin{array}{lll}0& 0& 1\\ \frac{i}{2\sqrt{2}}& \frac{i}{2\sqrt{2}}& 0\\ \frac{i}{2\sqrt{2}}& \frac{i}{2\sqrt{2}}& 0\end{array}\right),$$9.57

9.57

and the electric field from the test path is

$${\mathbf{E}}_{test}={\mathbf{P}}_{test}\cdot \left(\begin{array}{l}{E}_x\\ E_y\\ 0\end{array}\right)=\left(\begin{array}{l}0\\ \frac{i}{2\sqrt{2}}(E_x+E_y)\\ \frac{i}{2\sqrt{2}}(E_x+E_y)\end{array}\right).$$9.58

9.58

For horizontally polarized incident light from the laser, a sequence of electric field vectors can be calculated for both paths by multiplying the laser light’s polarization vector E0 = (1,0,0) by each P matrix. Following the polarization state through the interferometer’s reference arm, the sequence of global polarization states is

$$\begin{array}{llllllll}LP[45]\leftarrow & PBS(\text{T})\leftarrow & QWLR\leftarrow & Mirror\leftarrow & QWLR\leftarrow & PBS(\text{R})\leftarrow & HWLR\leftarrow & E_{in}\\ \left(\begin{array}{l}0\\ \frac{i}{2\sqrt{2}}\\ \frac{i}{2\sqrt{2}}\end{array}\right)& \left(\begin{array}{l}0\\ 0\\ \frac{i}{\sqrt{2}}\end{array}\right)& \left(\begin{array}{l}0\\ 0\\ \frac{i}{\sqrt{2}}\end{array}\right)& \left(\begin{array}{l}0\\ \frac{1}{2}\\ \frac{i}{2}\end{array}\right)& \left(\begin{array}{l}0\\ \frac{1}{2}\\ \frac{i}{2}\end{array}\right)& \left(\begin{array}{l}0\\ \frac{1}{\sqrt{2}}\\ 0\end{array}\right)& \left(\begin{array}{l}\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\\ 0\end{array}\right)& \left(\begin{array}{l}1\\ 0\\ 0\end{array}\right)\end{array}.$$9.59

9.59

The corresponding sequence for the test path is

$$\begin{array}{llllllll}LP[45]\leftarrow \hfill & PBS(\text{R})\leftarrow \hfill & QWLR\leftarrow \hfill & Mirror\leftarrow \hfill & QWLR\leftarrow \hfill & PBS(\text{T})\leftarrow \hfill & HWLR\leftarrow \hfill & E_{in}\hfill \\ \left(\begin{array}{l}0\\ \frac{i}{2\sqrt{2}}\\ \frac{i}{2\sqrt{2}}\end{array}\right)\hfill & \left(\begin{array}{l}0\\ \frac{i}{\sqrt{2}}\\ 0\end{array}\right)\hfill & \left(\begin{array}{l}0\\ \frac{i}{\sqrt{2}}\\ 0\end{array}\right)\hfill & \left(\begin{array}{l}\frac{1}{2}\\ \frac{i}{2}\\ 0\end{array}\right)\hfill & \left(\begin{array}{l}\frac{1}{2}\\ \frac{i}{2}\\ 0\end{array}\right)\hfill & \left(\begin{array}{l}\frac{1}{\sqrt{2}}\\ 0\\ 0\end{array}\right)\hfill & \left(\begin{array}{l}\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\\ 0\end{array}\right)\hfill & \left(\begin{array}{l}1\\ 0\\ 0\end{array}\right)\hfill \end{array}.$$9.60

9.60

Since the P matrix is constructed so that the exiting propagation vector is returned when the incident propagation vector is multiplied to the P matrix, a sequence of propagation vectors for both paths can be calculated in a similar way. The sequence of propagation vectors for the reference path is

$$\begin{array}{llllllll}LP[45]\leftarrow & PBS(\text{T})\leftarrow & QWLR\leftarrow & Mirror\leftarrow & QWLR\leftarrow & PBS(\text{R})\leftarrow & HWLR\leftarrow & {\widehat{k}}_0\\ \left(\begin{array}{l}1\\ 0\\ 0\end{array}\right)& \left(\begin{array}{l}1\\ 0\\ 0\end{array}\right)& \left(\begin{array}{l}1\\ 0\\ 0\end{array}\right)& \left(\begin{array}{l}1\\ 0\\ 0\end{array}\right)& \left(\begin{array}{l}-1\\ 0\\ 0\end{array}\right)& \left(\begin{array}{l}-1\\ 0\\ 0\end{array}\right)& \left(\begin{array}{l}0\\ 0\\ 1\end{array}\right)& \left(\begin{array}{l}0\\ 0\\ 1\end{array}\right)\end{array},$$9.61

9.61

and the sequence of propagation vectors of the test path is

$$\begin{array}{llllllll}LP[45]\leftarrow & PBS(\text{R})\leftarrow & QWLR\leftarrow & Mirror\leftarrow & QWLR\leftarrow & PBS(\text{T})\leftarrow & HWLR\leftarrow & {\widehat{k}}_0\\ \left(\begin{array}{l}1\\ 0\\ 0\end{array}\right)& \left(\begin{array}{l}1\\ 0\\ 0\end{array}\right)& \left(\begin{array}{l}0\\ 0\\ -1\end{array}\right)& \left(\begin{array}{l}0\\ 0\\ -1\end{array}\right)& \left(\begin{array}{l}0\\ 0\\ 1\end{array}\right)& \left(\begin{array}{l}0\\ 0\\ 1\end{array}\right)& \left(\begin{array}{l}0\\ 0\\ 1\end{array}\right)& \left(\begin{array}{l}0\\ 0\\ 1\end{array}\right)\end{array}.$$9.62

9.62

If the test path has an unknown sample described by a Jones matrix instead of an ideal mirror, the matrix sequence becomes

$$\begin{array}{lllllll}LP[45]\leftarrow & PBS(\text{R})\leftarrow & QWLR\leftarrow & \text{Sample }\leftarrow & QWLR\leftarrow & PBS(\text{T})\leftarrow & HWLR\\ \left(\begin{array}{lll}1& 0& 0\\ 0& \frac{1}{2}& \frac{1}{2}\\ 0& \frac{1}{2}& \frac{1}{2}\end{array}\right)& \left(\begin{array}{lll}0& 0& -1\\ 0& 1& 0\\ 0& 0& 0\end{array}\right)& \left(\begin{array}{lll}1& 0& 0\\ 0& \frac{1}{\sqrt{2}}& \frac{i}{\sqrt{2}}\\ 0& \frac{i}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)& \left(\begin{array}{lll}-1& 0& 0\\ 0& j_{yy}& j_{yz}\\ 0& j_{zy}& j_{zz}\end{array}\right)& \left(\begin{array}{lll}1& 0& 0\\ 0& \frac{1}{\sqrt{2}}& \frac{i}{\sqrt{2}}\\ 0& \frac{i}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)& \left(\begin{array}{lll}1& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right)& \left(\begin{array}{lll}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}& 0\\ \frac{1}{\sqrt{2}}& \frac{-1}{\sqrt{2}}& 0\\ 0& 0& 1\end{array}\right)\end{array}.$$9.63

9.63

With a horizontally polarized incident light, the test path electric field is

$${\mathbf{E}}_{out}=\frac{(j_{yx}-j_{xy})+i(j_{xx}+j_{yy})}{4\sqrt{2}}\left(\begin{array}{l}0\\ 1\\ 1\end{array}\right).$$9.64

9.64

9.6The Addition Form of Polarization Ray Tracing Matrices

This section introduces a method for coherent combination (i.e., addition) of polarization ray tracing matrices. In Section 9.1, Equation 9.4 was chosen to uniquely define the P matrix for a given ray. However, Equation 9.4 is not the only choice. The transformation of all polarization states on a transverse plane perpendicular to the propagation vector ${\widehat{\mathbf{k}}}_{q-1}$ can be described as linear combinations of the transformations of any two linearly independent basis vectors, Ea and Eb,

$${{\mathbf{E}}^{\prime }}_a={\mathbf{P}}_q{\mathbf{E}}_a,{{\mathbf{E}}^{\prime }}_b={\mathbf{P}}_q{\mathbf{E}}_b.$$9.65

9.65

The relationship in Equation 9.65 yields six equations, one for each row, but Pq has nine elements. Thus, Equation 9.1 does not fully constrain Pq. To uniquely define Pq, an additional set of three constraints is applied by relating the incident and exiting propagation vectors,

$$P_q{\widehat{k}}_{q-1}=\gamma {\widehat{k}}_q.$$9.66

9.66

The choice of γ is arbitrary, but only two values, either 0 or 1, allow Pq to be repeatedly cascaded and maintain the value of γ. Both choices of γ describe the same polarization effects. With γ = 1, which was the choice made in Section 9.1, only ideal polarizers have singular matrices,

$${\mathbf{P}}_q{\widehat{\mathbf{k}}}_{q-1}={\widehat{\mathbf{k}}}_q.$$9.67

9.67

γ = 1 is the principal definition for P used for the majority of this book. With the inclusion of Equation 9.67, Pq is now uniquely defined for each ray.

With γ = 0, Pq is always singular and thus ${\mathbf{P}}_q^{-1}$ never exists. One of the singular values of Pq will always be zero, as will one of the eigenvalues. When the P matrix is defined with this convention, γ = 0, the addition form of the polarization ray tracing matrix is indicated with the overscript cup $\stackrel{⌣}{\mathbf{P}}$,

$${\stackrel{⌣}{P}}_q{\widehat{k}}_{q-1}=0.$$9.68

9.68

The definition Equation 9.68 simplifies the addition of beams such as with interferometers. Consider a ray entering a Mach–Zehnder interferometer generating two exiting rays with PA and PB. For an incident polarization state Ein, the exiting state is the sum of the exiting states for the individual beams

$${\mathbf{E}}_{out}=({\mathbf{P}}_A+P_B){\mathbf{E}}_{in}={{\mathbf{E}}^{\prime }}_A+{{\mathbf{E}}^{\prime }}_B.$$9.69

9.69

Eout can be written as

$${\mathbf{E}}_{out}=\mathbf{P}{\mathbf{E}}_{in},$$9.70

9.70

where P is the polarization ray tracing matrix of the interferometer. But applying Equation 9.4 to the propagation vector ${\widehat{\mathbf{k}}}_{in}$ yields

$$\mathbf{P}{\widehat{\mathbf{k}}}_{in}=({\mathbf{P}}_A+{\mathbf{P}}_B){\widehat{\mathbf{k}}}_{in}=2{\widehat{\mathbf{k}}}_{out},$$9.71

9.71

which is not what is desired, since $\mathbf{P}{\widehat{\mathbf{k}}}_{in}$ should be ${\widehat{\mathbf{k}}}_{out}$. For combining parallel beams during ray tracing through interferometers or birefringent filters, a change can be made in the polarization ray tracing matrix definition, to avoid the issue of doubling the k vector, by using the alternative definition, $\stackrel{⌣}{P}$, from Equation 9.68. The difference between P and $\stackrel{⌣}{P}$ is the dyad D formed by the outer product between ${\widehat{k}}_{in}$ and ${\widehat{k}}_{out}$,

$$\begin{array}{l}D=\left(\begin{array}{lll}{\widehat{k}}_{x,in}{\widehat{k}}_{x,out}& {\widehat{k}}_{y,in}{\widehat{k}}_{x,out}& {\widehat{k}}_{z,in}{\widehat{k}}_{x,out}\\ {\widehat{k}}_{x,in}{\widehat{k}}_{y,out}& {\widehat{k}}_{y,in}{\widehat{k}}_{y,out}& {\widehat{k}}_{z,in}{\widehat{k}}_{y,out}\\ {\widehat{k}}_{x,in}{\widehat{k}}_{z,out}& {\widehat{k}}_{y,in}{\widehat{k}}_{z,out}& {\widehat{k}}_{z,in}{\widehat{k}}_{z,out}\end{array}\right),\hfill \\ D{\widehat{k}}_{in}=\left(\begin{array}{lll}{\widehat{k}}_{x,in}{\widehat{k}}_{x,out}& {\widehat{k}}_{y,in}{\widehat{k}}_{x,out}& {\widehat{k}}_{z,in}{\widehat{k}}_{x,out}\\ {\widehat{k}}_{x,in}{\widehat{k}}_{y,out}& {\widehat{k}}_{y,in}{\widehat{k}}_{y,out}& {\widehat{k}}_{z,in}{\widehat{k}}_{y,out}\\ {\widehat{k}}_{x,in}{\widehat{k}}_{z,out}& {\widehat{k}}_{y,in}{\widehat{k}}_{z,out}& {\widehat{k}}_{z,in}{\widehat{k}}_{z,out}\end{array}\right)\left(\begin{array}{l}{\widehat{k}}_{x,in}\\ {\widehat{k}}_{y,in}\\ {\widehat{k}}_{z,in}\end{array}\right)=\left(\begin{array}{l}{\widehat{k}}_{x,out}\\ {\widehat{k}}_{y,out}\\ {\widehat{k}}_{z,out}\end{array}\right)={\widehat{k}}_{out}.\hfill \end{array}$$9.72

9.72

Thus, the equation

$$\stackrel{⌣}{P}=\mathbf{P}-\mathbf{D}$$9.73

9.73

allows for easy transformation between the polarization ray tracing matrix, which is convenient for multiplying P, and the form $\stackrel{⌣}{P}$, which is useful for addition.

9.6.1Combining P Matrices for the Interferometer Example

The P matrix of the interferometer system in Section 9.5 can be calculated using the $\stackrel{⌣}{P}$ for each path. Ptest and Pref are constructed using Equation 9.4. One beam enters the interferometer and splits into two beams, and both exit the system along the same direction. This system is a good candidate to add Ptest and Pref to calculate the combined P matrix using $\stackrel{⌣}{P}$. The dyad matrix for ${\widehat{\mathbf{k}}}_{in}=\widehat{\mathbf{z}}$ and ${\widehat{\mathbf{k}}}_{out}=\widehat{\mathbf{x}}$ is

$$\mathbf{D}=\left(\begin{array}{lll}0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right).$$9.74

9.74

Then, $\stackrel{⌣}{\mathbf{P}}$ can be calculated for both paths,

$${\stackrel{⌣}{\mathbf{P}}}_{test}={\mathbf{P}}_{test}-\mathbf{D}=\left(\begin{array}{lll}0& 0& 0\\ \frac{i}{2\sqrt{2}}& \frac{i}{2\sqrt{2}}& 0\\ \frac{i}{2\sqrt{2}}& \frac{i}{2\sqrt{2}}& 0\end{array}\right),{\stackrel{⌣}{\mathbf{P}}}_{ref}={\mathbf{P}}_{ref}-\mathbf{D}=\left(\begin{array}{lll}0& 0& 0\\ \frac{i}{2\sqrt{2}}& \frac{-i}{2\sqrt{2}}& 0\\ \frac{i}{2\sqrt{2}}& \frac{-i}{2\sqrt{2}}& 0\end{array}\right).$$9.75

9.75

The combined P matrix for the system Pcombined is

$${\mathbf{P}}_{combined}={\stackrel{⌣}{\mathbf{P}}}_{test}+{\stackrel{⌣}{\mathbf{P}}}_{ref}+\mathbf{D}=\left(\begin{array}{lll}0& 0& 1\\ \frac{i}{\sqrt{2}}& 0& 0\\ \frac{i}{\sqrt{2}}& 0& 0\end{array}\right).$$9.76

9.76

Using Pcombined, Eout can be calculated directly and provides the same result as adding Etest and Eref, which are calculated from Ptest and Pref, respectively,

$$\begin{array}{l}{\mathbf{E}}_{out}={\mathbf{P}}_{combined}\cdot {\mathbf{E}}_{in}={\mathbf{P}}_{combined}\cdot \left(\begin{array}{l}{E}_x\\ E_y\\ 0\end{array}\right)=\frac{i{E}_x}{\sqrt{2}}\left(\begin{array}{l}0\\ 1\\ 1\end{array}\right)\\ \phantom{E_{out}}={\mathbf{P}}_{test}\cdot {\mathbf{E}}_{test}+{\mathbf{P}}_{ref}\cdot {\mathbf{E}}_{ref}.\end{array}$$9.77

9.77

9.7Example—A Hollow Corner Cube

Corner cubes are commonly used as retroreflectors, and their polarization properties are well studied.22–24 A hollow aluminum-coated corner cube provides an example of an inhomogeneous polarization component, an element in which the diattenuation and retardance are not aligned. This example focuses on calculating the P matrix of the corner cube and diattenuation of the system.

Figure 9.7 shows the hollow corner cube consisting of three mutually perpendicular aluminum mirrors. The entrance face is intercept number 1 and the three mutually perpendicular reflecting surfaces are intercepts 2, 3, and 4. A refractive index of 0.77 + 6.06i is assumed for aluminum at 500 nm.

Figure 9.7A corner cube retroreflector (CCR) with surfaces labeled from 1 for the front surface to 4.

A collimated beam of incident light can take six different ray paths depending on its entry location.25Figure 9.8 shows one of the ray paths with propagation vectors marked with black arrows. The incident and the exiting propagation vectors are anti-parallel and aligned with the z-axis.

Figure 9.8Top view of a single ray path through a hollow aluminum-coated CCR. The three reflecting surfaces of the corner cube are mutually perpendicular.

Figure 9.9 shows the corner cube with the propagation vectors in black, s-local coordinate vectors in solid red, and p-local coordinate vectors in dashed blue in three different views. The left figure shows how the local coordinate bases (s, p) change as the ray propagates from the same viewpoint as in Figure 9.8. The middle and right figures are rotation about the y-axis.

Figure 9.9Three different views of the corner cube showing how local coordinate vectors (s-vectors in solid red, p-vectors in blue, and propagation vectors in black) change as the ray propagates through the corner cube.

Each reflecting surface of the corner cube is specified by its normal ${\widehat{\mathbf{\eta }}}_q$. ${\widehat{\mathbf{k}}}_{q-1}$ determines the local coordinates for each intercept q and Pq is uniquely defined. Table 9.2 summarizes this ray’s propagation vectors, local coordinates, and P matrix.

Table 9.2Propagation Vectors, Local Coordinate Basis Vectors, Surface Normal Vectors, and Polarization Ray Tracing Matrices Associated with a Ray Path through an Aluminum-Coated Hollow Corner Cube

The net polarization ray tracing matrix Pcc (cc for corner cube) for this ray path is calculated by cascading the three P matrices in Table 9.2,

$${\mathbf{P}}_{cc}={\mathbf{P}}_3{\mathbf{P}}_2{\mathbf{P}}_1=\left(\begin{array}{lll}0.39+0.78i& 0.01+0.02i& 0\\ -0.02i& 0.40+0.78i& 0\\ 0& 0& -1\end{array}\right).$$9.78

9.78

The SVD of Pcc gives

$$\begin{array}{l}{\mathbf{U}}_{cc}=\left(\begin{array}{lll}0& 0.63+0.15i& 0.74+0.17i\\ 0& 0.37-0.66i& -0.32+0.57i\\ 1& 0& 0\end{array}\right),\hfill \\ {\mathbf{D}}_{cc}=\left(\begin{array}{lll}1& 0& 0\\ 0& 0.88& 0\\ 0& 0& 0.87\end{array}\right),\hfill \\ {\mathbf{V}}_{cc}=\left(\begin{array}{lll}0& 0.43-0.49i& 0.47-0.6i\\ 0& -0.41-0.64i& 0.38+0.52i\\ -1& 0& 0\end{array}\right).\hfill \end{array}$$9.79

9.79

As shown in Equation 9.19, Vcc and Ucc have the incident and exiting propagation vectors as their first columns. Table 9.3 lists the maximum and minimum intensity transmittances assuming the incident electric field’s intensity is one and the diattenuation of the corner cube is calculated from the singular values of the Pcc matrix.

Table 9.3The Maximum Intensity of Output and Associated Incident Electric Field, the Minimum Intensity of Transmitted Electric Field and Associated Incident Electric Field, and the Diattenuation for the Ray through the Corner Cube System

Imax	v1	Imin	v2	$\mathcal{D}$
0.774	$\left(\begin{array}{l}0.65e^{-0.85i}\\ 0.76e^{-2.15i}\\ 0\end{array}\right)=e^{-2.15i}\left(\begin{array}{l}0.65e^{1.30i}\\ 0.76\\ 0\end{array}\right)$	0.757	$\left(\begin{array}{l}0.76e^{-0.91i}\\ 0.65e^{0.94i}\\ 0\end{array}\right)=e^{-0.91i}\left(\begin{array}{l}0.76\\ 0.65e^{1.85i}\\ 0\end{array}\right)$	0.014

The last two columns of Vcc and Ucc represent two incident polarization states (v1, v2) and exiting states (u1, u2) with the maximum and the minimum intensity transmittances. v1, v2 and u1, u2 are elliptically polarized; hence, this path through the corner cube acts as a weak elliptical diattenuator with a diattenuation of 0.014. v1 and v2 are the only pair of orthogonal incident polarization states that remain orthogonal upon exit. Figure 9.10 shows the polarization states associated with the maximum and the minimum intensity transmittances represented in local coordinate systems, $\left({\widehat{\mathbf{s}}}_0,{\widehat{\mathbf{p}}}_0,{\widehat{\mathbf{k}}}_0\right)$ and $\left({\widehat{\mathbf{s}}}_3,{\widehat{\mathbf{p}}}_3,{\widehat{\mathbf{k}}}_3\right)$, so that each ellipse’s propagation vector is coming out of the page. The red ellipses in Figure 9.10 (left) are the exiting (i.e., u1) polarization state (shown on the left), which has maximum transmittance, and the corresponding incident (i.e., v1) polarization state (shown on the right). Note that the handedness of the v1 and u1 are opposite to each other; after an odd number of reflections, the left-handed incident polarization state exits right handed since the handedness of the polarization state is defined in local coordinate systems. Similarly, the blue ellipses in Figure 9.10 (right) are the exiting (i.e., u2) polarization state, which has minimum transmittance, and the corresponding incident (i.e., v2) polarization state. The handedness of v2 and u2 is also opposite to each other.

Figure 9.10The polarization ellipses represented in local coordinate. (Left) The exiting state from the corner cube with the maximum intensity and corresponding incident state are shown in red. (Right) The exiting state with the minimum intensity and corresponding incident state are shown in blue.

The use of local coordinate systems to describe Jones vectors with opposite propagation directions would complicate the discussion of these polarization states and transformations. In global coordinates, as shown in Figure 9.11, the direction of rotation of the electric field is in the same direction for the corresponding incident and exiting states; v1 and u1 have the same handedness and v2 and u2 have the same handedness due to the anti-parallel propagation vectors, ${\widehat{\mathbf{k}}}_0\text{ and }{\widehat{\mathbf{k}}}_3$. All states in Figure 9.11 are represented in global coordinates (x–y plane) looking into the corner cube; since ${\widehat{\mathbf{k}}}_3=\widehat{\mathbf{z}}$, the exiting electric fields are coming out of the page while the incident electric fields are going into the page.

Figure 9.11The polarization ellipses looking into the corner cube. (Left) The exiting state from the corner cube with the maximum intensity and corresponding incident state are shown in red. (Right) The exiting state with the minimum intensity and corresponding incident state are shown in blue.

All states in Figure 9.11 are represented in an x–y plane looking into the corner cube; since ${\widehat{\mathbf{k}}}_3=\widehat{\mathbf{z}}$ and ${\widehat{\mathbf{k}}}_0$ and ${\widehat{\mathbf{k}}}_3$ are anti-parallel, the exiting polarization ellipses are coming out of the page while the incident polarization ellipses are going into the page.

9.8Conclusion

The three-by-three polarization ray tracing matrices enable ray tracing in global coordinates, which provide an easy basis to interpret polarization properties for optical systems where rays are constantly changing directions. The interferometer in Section 9.5 and the corner cube in Section 9.7 highlight how global coordinates provide a straightforward basis. Using local coordinates, different analysts may make different choices, complicating the interpretation of complex geometries. It remains straightforward to convert results from global coordinates into other interesting local coordinate bases.

A formalism for polarization ray tracing using three-by-three matrices has been developed and the relationship to the Jones calculus has been shown. Algorithms for reflection, refraction, and polarization elements are summarized with specific examples.

If the optical system includes anisotropic or birefringent media, the propagation portions take the form of retarder matrices for birefringent media and/or diattenuation matrices for dichroic media. Denoting the propagation effect from ray interface q to q+1 as Aq+1,q, the polarization ray tracing matrix for a ray through an optical system with anisotropic media (i.e., Equation 9.2) becomes

$${\mathbf{P}}_{Total}={\mathbf{P}}_N{\mathbf{A}}_{N,N-1}{\mathbf{P}}_{N-1}\cdots {\mathbf{A}}_{3,2}{\mathbf{P}}_2{\mathbf{A}}_{2,1}{\mathbf{P}}_1={\mathbf{P}}_N{\displaystyle \prod _{q=N-1,-1}^1{\mathbf{A}}_{q+1,q}{\mathbf{P}}_q}.$$9.80

9.80

This formulation works well for stress birefringent and weakly anisotropic materials. In strongly birefringent materials like calcite and rutile, birefraction between the two modes (ordinary and extraordinary) causes ray doubling. In this case, each of the separated rays refracting into a birefringent material needs a separate polarization ray tracing matrix. Since the modes in these birefringent media have a single polarization, the ray intercept Ps have the form of a polarizer. This polarizer matrix selects the incident state that couples into the specified mode. Further comments on ray tracing in anisotropic materials are beyond the scope of the present chapter and are covered in Chapter 19 (Birefringent Ray Trace).

The calculation of the diattenuation of a P matrix is achieved via the SVD. The maximum and minimum transmittances are related to singular values of the P matrix. The two unitary matrices of the SVD provide two canonical entering and exiting polarization states that are orthogonal to each other and related by the singular values. The incident propagation vector and the exiting propagation vector are related by P due to the constraint $\mathbf{P}{\widehat{\mathbf{k}}}_{q-1}={\widehat{\mathbf{k}}}_q$ applied to the definition of P. $\stackrel{⌣}{\mathbf{P}}$ and a propagation vector dyad are introduced to simplify the addition of P matrices for parallel rays.

A step-by-step example of performing ray tracing with the P matrix is illustrated using an interferometer system with a PBS. Finally, a numeric example of ray tracing through a hollow aluminum corner cube was presented.

9.9Problem Sets

1. Consider a unit vector V = (vx, vy, vz).

   1. Find a unit vector orthogonal to V. Is it unique?

   2. Find the matrix Dy that maps V into V, and maps all vectors orthogonal to V to zero vectors. Show that your matrix works properly. This type of projection matrix is called a dyad.

2. Given two orthonormal bases of three vectors, (A, B, C) and (E, F, G), define a 3 × 3 unit dyad, H, which maps unit vector A into E while mapping the two orthogonal components B and C to zero,

   $$\mathbf{H}\cdot \left(\begin{array}{lll}{A}_x& B_x& C_x\\ A_y& B_y& C_y\\ A_z& B_z& C_z\end{array}\right)=\left(\begin{array}{lll}{E}_x& 0& 0\\ E_y& 0& 0\\ E_z& 0& 0\end{array}\right).$$

   

   Find the following dyads:

   1. Maps (0, 1, 0) into (0, 1, 0).

   2. Maps (0, 1, 0) into (1, 0, 0).

   3. Maps (0, 0, 1) into (1, 0, 0).

   4. Maps (0, 0, 1) into (0, 0, −1).

   5. Maps (0, 0, 1) into (α, β, γ).

   6. Maps (α, β, γ) into (α, β, γ).

   7. Maps (α, β, γ) into (δ, ε, ζ).

   8. What are the eigenvalues of a unit dyad?

   9. What is the determinant of a dyad?

   10. Demonstrate that the rows of (g) are linearly dependent.

   11. Are dyad matrices unitary or Hermitian?

3. Light is propagating with the propagation vector ${\mathbf{k}}_1=(-2,-3,3)/\sqrt{2^2+3^2+3^2}.$

   1. Create the dyad matrix K1, for k1. Find two basis vectors for the transverse plane by the following procedure. Pick two simple but arbitrary vectors, v1 and v2, in different directions.

   2. Form the x-basis vector x1 by subtracting k1 · v1 from v1, to get an orthogonal vector, and normalize. Verify whether x1 is orthogonal to k1 and normalized.

   3. Form X1, the dyad matrix for x1.

   4. Draw pictures of the various vectors in the (k1, v1) plane and explain the mathematical operation.

   5. Form the y-basis vector y1 by subtracting K1 ·v2 and X1 · v2 from v2, to get an orthogonal vector, and normalizing. Verify whether y1 is orthogonal to k1. This is an example of the Gram–Schmidt orthonormalization algorithm in three dimensions.

   6. Is the orthonormal basis (k1, x1, y1) left handed or right handed?

   7. Write a polarization vector E1 for right circularly polarized light propagating along K1.

4. Orthogonal matrices

   1. Find the orthogonal matrix O1 to rotate x into u, y into v, z into w, where x = (1, 0, 0), y = (0, 1, 0), z = (0, 0, 1), $\mathbf{u}=\left(\frac{3}{5},\frac{4}{5},0\right),\mathbf{v}=(0,0,1),\mathbf{w}=\left(\frac{4}{5},-\frac{3}{5},0\right)$.

   2. Find the orthogonal matrix O2 to rotate r into x, s into y, t into z, where r = (0, 12, 5)/13, s = (1, 0, 0), t = (0, 5, −12)/13.

   3. Find the orthogonal matrix O3 to rotate r into u, s into v, t into w.

5. A flat glass plate with normal η = (sin(π/6), 0, cos(π/6)) has an incident beam propagating with kin = (0, sin(π/6), cos(π/6)). The refractive light has propagation vector ${\mathbf{k}}_1=\frac{1}{\sqrt{2}}\left(\frac{1}{4},\frac{1}{2},\frac{3\sqrt{3}}{4}\right)$.

   1. Find the s-basis vector.

   2. Find the p-basis vector.

   3. Find the orthogonal matrix to rotate the global (x, y, z) coordinates into the local (k, s, p).

   4. Find the s′ and p′ basis vectors after refraction.

   5. Find the orthogonal matrix to rotate the global (x, y, z) to local (k, s′, p′) coordinates.

   6. Find the angle of incidence and angle of refraction.

   7. If the incident refractive index is 1, find the index of refraction of the medium.

   Refraction at this interface has the Jones matrix $\mathbf{J}=\left(\begin{array}{cc}3/4& 0\\ 0& 6\sqrt{2}/11\end{array}\right)$.

   • h. Find the associated polarization ray tracing matrix P.

   • i. What is the associated singular value decomposition?

6. 1. What are the three singular values for the polarization ray tracing matrix of an ideal polarizer?

   2. What are the three singular values for the polarization ray tracing matrix of an ideal retarder?

7. Find the matrix inverse P−1 of the singular value decomposition in terms of U, D, and V where P = UDV†.

8. Given the following singular value decomposition,

   $$\left(\begin{array}{lll}0& 0& 1\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}& 0\\ \frac{i}{\sqrt{2}}& \frac{-i}{\sqrt{2}}& 0\end{array}\right)\left(\begin{array}{lll}\frac{3}{8}& 0& 0\\ 0& \frac{1}{8}& 0\\ 0& 0& 1\end{array}\right)\left(\begin{array}{lll}0& \frac{1}{\sqrt{2}}& \frac{-i}{\sqrt{2}}\\ 0& \frac{1}{\sqrt{2}}& \frac{i}{\sqrt{2}}\\ 1& 0& 0\end{array}\right).$$

   

   1. Find the diattenuation.

   2. Along which axes does the light enter and exit?

   3. Is this a linear, elliptical, or circular element?

9. In order to calculate reflected and/or refracted polarization state at an intercept, we often decompose the polarization state into s- and p-polarizations. When a ray enters the intercept at 0 angle of incidence, that is, normal incident to the surface, there is no difference between s- and p-polarizations. In this problem, we calculate Jones matrices and polarization ray tracing (P) matrices to describe reflection at normal incidence and compare the results.

   1. Consider a ray propagating in air (n1 = 1) along k1 = (0, 0, 1) incident on a glass surface with index n1 = 1.52 and a surface normal η = (0, 0, 1). Write a Jones matrix for reflection using Fresnel reflection coefficients, $\left(\begin{array}{ll}{r}_s\hfill & 0\hfill \\ 0\hfill & r_p\hfill \end{array}\right)$.

   2. Using the Jones matrix calculated in part (a), calculate the reflected Jones vector Eout for a right circularly polarized incident Jones vector ${\mathbf{E}}_{in}=(1,-i)\text{/}\sqrt{2}.$

   3. What polarization state is Eout in?

   4. Now, calculate the P matrix for the same ray.

   5. Using the P matrix in part (d), calculate the polarization vector E for right circularly polarized incident light, vector $(1,-i,0)\text{/}\sqrt{2}$, reflecting at normal incidence.

   6. Compare the results from parts (c) and (e). Which polarization state is the reflected electric field vector in? Are (c) and (e) in conflict? Explain.

10. For light propagating along the z-axis, a polarization element has the Jones matrix$\mathbf{J}=\left(\begin{array}{ll}{j}_{11}\hfill & j_{12}\hfill \\ j_{21}\hfill & j_{22}\hfill \end{array}\right)$. Now, the element is placed in a system where light is propagating along $\mathbf{k}=(1,1,1)\text{/}\sqrt{3}$ and the x-axis has been moved to ${\mathbf{x}}_1=(0,-1,1)\text{/}\sqrt{2}$.

    1. Find the polarization ray tracing matrix P as a function of the Jones matrix elements.

    2. If J is a quarter wave linear retarder with the fast axis at 45°, find P.

    3. Which incident polarization vector E1 will yield left circularly polarized output with flux 16?

    4. Find the singular value decomposition of P = VDW†.

11. Light propagating in the direction ${\mathbf{k}}_0=(1,0,2)\text{/}\sqrt{5}$ is incident on a wire grid polarizer with normal ${\mathbf{\eta }}_0=(1,\sqrt{4},1)\text{/}\sqrt{6}$. All of the p-polarized light is transmitted through the polarizer and exits in the same direction, k0.

    1. Find the polarization ray tracing matrix Pt for the polarizer in transmission.

    2. Find the reflected propagation vector kr.

    3. If all of the s-polarized light is reflected, find the matrix for reflection Pr.

    4. The wire grid polarizer is replaced with another wire grid polarizer, which reflects 0.8 of the s-polarized light amplitude and delays the phase by π/3. It reflects 0.1 of the p-polarized light and delays the phase by π/2. Find the polarization ray tracing matrix for reflection.

12. A reflection linear quarter wave retarder has a surface normal η1 = (0, 0, 1). It advances the phase of the horizontal light (fast axis is x) by π/4 and retards the orthogonal state by π/4. The incident ray has a propagation vector k0 = (0, sin 30°, cos 30°). Calculate the 3D polarization ray tracing matrix.

    1. Calculate (s1, p1, k0) and $\left({s^{\prime }}_1,{p^{\prime }}_1,{k^{\prime }}_0\right)$.

    2. Calculate P matrix using a Jones matrix for the horizontal fast axis linear quarter wave retarder, $\mathbf{J}=\left(\begin{array}{cc}{e}^{-i\pi /4}& 0\\ 0& e^{i\pi /4}\end{array}\right)$.

    3. When right circularly polarized light is incident, what is the orientation of the major axis in three dimensions of the reflected polarization state?

13. Light with a propagation vector $\mathbf{k}=(1,1,0)\text{/}\sqrt{2}$ is normally incident on the front face of a Wollaston prism. At the hypotenuse with normal η = (cos 55°, sin 55°, 0), the s-component of the light, which is polarized in the z-direction, is refracted and exits the cube propagating in the direction kα = (cos 48°, sin 48°, 0). The orthogonal component of the incident light, the p-component at the interface, is refracted in the opposite direction and exits the cube in the direction kβ = (cos 42°, sin 42°, 0). Assume both beams lose 9% of their incident flux, 4% entering the cube, 1% reflection loss at the hypotenuse, and 4% exiting into air. Our equations will analyze just the polarization ray tracing matrices between the incident and exiting beams in air. We will not write P matrices at the hypotenuse.

    1. Find the s-, p-, and k-components of the incident light, in air.

    2. Find the matrix for the amplitude transmissions of the Wollaston polarizer for beam α, including the polarizing effect of the interface and the end-to-end losses. The Fresnel coefficients for the beams at the interfaces are as (asα, apα, asβ, apβ).

    3. Find the orthogonal transformation for beam α from global to local coordinates ${\mathbf{O}}_{\alpha }^{-1}=\left(\begin{array}{ccc}{s}_x& s_y& s_z\\ p_x& p_y& p_z\\ k_x& k_y& k_z\end{array}\right)$.

    4. Find the orthogonal transformation for beam α from local to global coordinates ${\mathbf{O}}_{\alpha }=\left(\begin{array}{ccc}{s}_x& {p^{\prime }}_x& k{\alpha }_x\\ s_y& {p^{\prime }}_y& k{\alpha }_y\\ s_z& {p^{\prime }}_z& k{\alpha }_z\end{array}\right)$.

    5. Find the polarization ray tracing matrix for the α polarizer path.

    6. When left circularly polarized light is incident, with $EL=\left(\frac{1}{2}-\frac{i}{2},\frac{-1}{2}+\frac{i}{2},\frac{1+i}{\sqrt{2}}\right)$, find the E vector for the exiting light. What is the exiting flux and polarization state?

    7. Find the polarization ray tracing matrix for the β polarizer path.

14. Light propagating along the y-axis encounters a roof mirror with mirror normals η1 and η2. ${\mathbf{k}}_0=(0,1,0),{\mathbf{\eta }}_1=(1,1,0)\text{/}\sqrt{2},{\mathbf{\eta }}_2=(-1,1,0)\text{/}\sqrt{2}$. One half of the beam reflects from mirror 1 and mirror 2 and exits along k2 = (0, −1, 0).

    1. Assuming Fresnel reflection coefficients rs and rp for each mirror, find the polarization ray tracing matrices for the individual reflections, P1 and P2, and for the entire path P. Keep rs and rp as variables.

    2. The other half of the beam strikes mirror 2 before mirror 1 and exits in the same direction. Find the polarization ray tracing matrices for the individual reflections and for the entire path.

    3. Find the Jones vector, which will emerge as right circularly polarized light.

15. A linear polarizer with a transmission axis at 45° has a surface normal η1 = (0,0,1). The incident ray has a propagation vector k0 = (0, sin30°, cos30°). The polarizer does not deviate the propagation vector for the transmitted ray.

    1. Using η1 and k0, calculate (s1, p1, k0) and $\left({{\mathbf{s}}^{\prime }}_1,{{\mathbf{p}}^{\prime }}_1,{{\mathbf{k}}^{\prime }}_0\right)$.

    2. Generate the orthogonal matrices Oin and Oout.

    3. Calculate the P matrix using the Jones matrix for a linear polarizer at 45°.

16. An adjustable three-mirror reflector takes incident light along the z-axis, reflects it three times such that the light also exits along the z-axis. The mirrors are ideal reflectors, rs = 1, rp = 1. As the adjustment θ is performed, the propagation vectors after each mirror vary as follows:

    $$\begin{array}{l}{\mathbf{k}}_0=(0,1,0),\hfill \\ {\mathbf{k}}_1=(\text{sin}\theta ,0,\text{cos}\theta ),\hfill \\ {\mathbf{k}}_2=(0,\text{sin}\theta ,\text{cos}\theta ),\hfill \\ {\mathbf{k}}_3={\mathbf{k}}_0.\hfill \end{array}$$

    

    1. For the case of θ = 45°, find the first mirror’s reflection basis vectors s1, p1, ${{\mathbf{p}}^{\prime }}_1$. You can use the alternative expression for the s-basis vector ${\mathbf{s}}_q=\frac{{\mathbf{k}}_q\times {\mathbf{k}}_{q-1}}{|{\mathbf{k}}_q\times {\mathbf{k}}_{q-1}|}$.

    2. Create the orthogonal transformation matrices Oin,1 and Oout,1 for θ = 45°.

    3. Calculate the polarization ray tracing matrix P1.

    4. Let the incident light be x-polarized. Find the orientation of the exiting polarized light for θ = 45°.

    5. Create a program that calculates P1, P2, P3 and a total ray tracing matrix P when a value for θ is input.

    6. For x-polarized incident light, plot the orientation of the exiting polarized light as a function of θ.

    7. For what θ is the light approximately 45° polarized?

    8. For what angle is the light y-polarized?

17. The polarization ray tracing matrix for a ray path is defined by the three matrix equations

    $$\begin{array}{l}{\mathbf{E}}_{\text{a},1}=\mathbf{P}\cdot {\mathbf{E}}_{\text{a},0}\hfill \\ {\mathbf{E}}_{b,1}=\mathbf{P}\cdot {\mathbf{E}}_{b,0}\hfill \\ {\mathbf{k}}_1=\mathbf{P}\cdot {\mathbf{k}}_0,\hfill \end{array}$$

    

    where Ea,0 and Eb,0 are two of the columns of V, which are attenuated by Λ1 and Λ2, respectively, when propagating along the ray path. Now, light is sent the reverse direction along the ray path, and the two states defined by the columns of U are attenuated by Λ1 and Λ2, respectively. Find the PRT matrix for the opposite direction $\stackrel{⌣}{\mathbf{P}}$. Because the attenuation is the same in both directions, $\stackrel{⌣}{\mathbf{P}}$ is not quite the matrix inverse of P.

18. Find $\stackrel{⌣}{\mathbf{P}}$ for the following P. For each, find the corresponding incident and exiting propagation vectors, k0 and k1, and the diattenuation.

    a. ${\mathbf{P}}_1=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0.9& 0\\ 0& 0& 0.5\end{array}\right)$	b. ${\mathbf{P}}_2=\left(\begin{array}{ccc}0.5& 0& 0\\ 0& 0.9& 0\\ 0& 0& 1\end{array}\right)$		c. ${\mathbf{P}}_3=\left(\begin{array}{ccc}0& 0& 1\\ 0& 0.9& 0\\ 0.5& 0& 0\end{array}\right)$
    d. ${\mathbf{P}}_4=\left(\begin{array}{ccc}0& 0.9& 0\\ 0& 0& 1\\ 0.5& 0& 0\end{array}\right)$	e. ${\mathbf{P}}_5=\frac{1}{8}\left(\begin{array}{ccc}7& -1& 0\\ -1& 7& 0\\ 0& 0& 4\end{array}\right)$		f. ${\mathbf{P}}_6=\frac{1}{8}\left(\begin{array}{ccc}1& 7& 0\\ -7& -1& 0\\ 0& 0& 4\end{array}\right)$
    g. ${\mathbf{P}}_7=\left(\begin{array}{ccc}0.714& 0.01& -0.52\\ -0.102& 0.827& 0.296\\ 0.289& -0.106& 0.394\end{array}\right)$		h. ${\mathbf{P}}_8=\left(\begin{array}{ccc}-0.67& 0.183& 0.052\\ 0.01& 0.036& -0.822\\ 0.19& -0.719& -0.311\end{array}\right)$

19. Consider the phase-shifting Twyman–Green interferometer, as shown below, using a PBS in conjunction with two linear quarter wave retarders to get the light through the system with minimal loss.

    1. Once the beams are recombined, an analyzer is used to get the fringes. What are the P matrices for the two paths before the 45° linear polarizer? What is the P matrix for the entire system? Be sure to account for unmatched phase between the arms.

       

    2. Suppose one of the quarter waveplates is rotated (from 45°) by some small angle δ. What is the new P matrix for that arm? For the system? How will this affect the measured phase?

    3. Suppose the retardance of one of the waveplates is off by a small amount δ. What is the new P matrix for that arm? For the overall system? How will this affect the measured phase?

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1 Propagation vectors are normalized throughout this book.

2 For gratings and holographic elements, different diffraction orders would naturally have different Jq.