16Image Formation with Polarization Aberration

16.1Introduction

A fundamental metric for optical imaging systems is the point spread function(PSF), which describes the image of a point object. To calculate a PSF, the methods of geometrical optics and ray tracing are augmented by physical optics and Fourier optics calculations. For systems with polarization aberration, the PSF depends on the incident polarization state.1 In this chapter, the concept of a point spread function is generalized so that it describes imaging of arbitrary polarization states by introducing an amplitude response matrix(ARM) and Mueller point spread matrix (MPSM). It will be shown how the point spread function calculation of a conventional optical design program can be generalized through the use of Jones matrices and polarization ray tracing matrices. Next, the calculation of images for objects with spatially varying polarization states is considered. Polarization aberration affects the image quality of a system. Chapter 12 (Fresnel Aberrations) and Chapter 15 (Paraxial Polarization Aberrations) provided examples of polarization aberrations in optical systems and their effects on the PSF. Given a wavefront aberration function and Jones pupil function, the MPSM is calculated, which shows the polarization structure of the PSF and how the PSF varies with the incident polarization state. From the MPSM, an image of a point source with arbitrary polarization state can be calculated and one can vary incident polarization states and observe the resulting image polarization structure. Similarly, the optical transfer function (OTF) of conventional optics can be extended to optical transfer matrix (OTM) to show how the spatial filtering of an object during imaging depends on the incident polarization state.

The image of an extended object is calculated by replacing each point in the object with the corresponding response function. In lenses and other imaging systems, the response function varies across the field of view as the aberrations steadily change, moving from on-axis to other parts of the field. An isoplanatic patch is a region of the field over which the changes of the PSF are small. Then, simple linear system analysis methods (the ARM and MPSM) can be applied over space-invariant patches of the object plane, whose size depends on the rate of change of variation in wavefront aberration function and JonesPupil(x, y).2

The flowcharts of the algorithms for coherent and incoherent imaging of scalar waves and for the imaging of systems with polarization aberrations are compared in Figures 16.1 and 16.2. Each operand and operation will be described in detail in this chapter.

Figure 16.1Flowcharts for coherent and incoherent imaging calculations. The pupil function P represents the wavefront function at the exit pupil of an optical system. The spatial variable of the system is x; the wavelength of light is λ; the focal length of the imaging system is f. The coherent point spread function is cPSF (i.e., amplitude response function, ARF); the coherent transfer function is CTF; the incoherent point spread function is iPSF; the optical transfer function is OTF; the modulation transfer function is MTF; and the phase transfer function is PTF. $\Im$ is the Fourier transform operation. ★ is the auto-correlation operation.

Figure 16.2The imaging calculation for polarized light. The 2 × 2 amplitude response matrix is the ARM; the 4 × 4 Mueller point spread matrix is the MPSM; the 4 × 4 optical transfer matrix is the OTM; the 4 × 4 modulation transfer matrix is the MTM; and the 4 × 4 phase transfer matrix is the PTM.

16.2Discrete Fourier Transformation

The calculation of diffraction in image formation involves Fourier transformation of the exit pupil functions. Most textbooks on diffraction and image formation provide examples that involve simple Fourier transform pairs such as Fourier transforming a rectangle function into a sinc function or a Gaussian into a Gaussian function. In optical design, it is necessary, however, to Fourier transform arbitrary functions such as high-order aberrations and irregular pupil shapes to calculate PSFs. Further, the results from ray tracing are also sampled functions and not continuous functions. For such pupil functions in optical design, evaluating the Fourier transforms as continuous functions is difficult and impractical. The discrete Fourier transform (DFT)3–5 provides a straightforward algorithm to Fourier transform arbitrary functions on a regular sampled grid. The DFT is used in most optical design software diffraction calculations. All mathematical software packages such as MATLAB, Mathematica, and so on, provide built-in DFT algorithms; hence, only a brief summary is given in Math Tip 16.1.

Math Tip 16.11D Discrete Fourier Transformation

The one-dimensional (1D) discrete Fourier transformation of an array u = (u1, u2,…) is

$$U_s=\frac{1}{\sqrt{n}}{\displaystyle \sum _{r=1}^n{u}_r{e}^{2\pi i(r-1)(s-1)\text{/}n}},$$16.1

16.1

where r − 1 and s − 1 occur since the index for the arrays u and U are counted from one. Equation 16.1 can be written as a matrix multiplication,

$$\left(\begin{array}{l}{U}_1\\ U_2\\ U_3\\ \begin{array}{l}{U}_4\\ ⋮\end{array}\end{array}\right)=\frac{1}{\sqrt{n}}\left(\begin{array}{l}\begin{array}{llll}1& e^0& e^0& e^0\\ e^0& e^{\omega }& e^{2\omega }& e^{3\omega }\\ e^0& e^{2\omega }& e^{4\omega }& e^{6\omega }\\ e^0& e^{3\omega }& e^{6\omega }& e^{9\omega }\end{array}\cdots \\ \begin{array}{llll}& ⋮& & \ddots \end{array}\end{array}\right)\left(\begin{array}{l}{u}_1\\ u_2\\ u_3\\ \begin{array}{l}{u}_4\\ ⋮\end{array}\end{array}\right)$$16.2

16.2

where ω = i2π/n. $\frac{1}{\sqrt{n}}$ is the normalization of the DFT.

The inverse discrete Fourier transformation of array U = (U1, U2,…) is

$$u_r=\frac{1}{\sqrt{n}}{\displaystyle \sum _{s=1}^n{U}_s{e}^{-2\pi i(r-1)(s-1)\text{/}n}}.$$16.3

16.3

Example 16.11D DFT Examples

Calculate the 1D DFT of the following arrays.

1. u = ($\sqrt{2}$, 0)

   $$\begin{array}{l}{U}_1=\frac{1}{\sqrt{2}}{\displaystyle \sum _{r=1}^2{u}_r{e}^{i2\pi (r-1)(1-1)\text{/}2}}=\frac{1}{\sqrt{2}}(\sqrt{2}+0)=1.\\ U_2=\frac{1}{\sqrt{2}}{\displaystyle \sum _{r=1}^2{u}_r{e}^{i2\pi (r-1)(2-1)\text{/}2}}=\frac{1}{\sqrt{2}}(\sqrt{2}+0)=1.\end{array}$$

   

   Thus, U = (1, 1).

2. u = (0, 2, 0, 2)

   $$\left(\begin{array}{l}{U}_1\\ U_2\\ U_3\\ U_3\end{array}\right)=\frac{1}{\sqrt{4}}\left(\begin{array}{llll}1& e^0& e^0& e^0\\ e^0& e^{i2\pi /4}& e^{i2\times 2\pi /4}& e^{i3\times 2\pi \text{/}4}\\ e^0& e^{i2\times 2\pi /4}& e^{i4\times 2\pi /4}& e^{i6\times 2\pi \text{/}4}\\ e^0& e^{i3\times 2\pi /4}& e^{i6\times 2\pi /4}& e^{i9\times 2\pi \text{/}4}\end{array}\right)\left(\begin{array}{l}0\\ 2\\ 0\\ 2\end{array}\right)=\frac{1}{\sqrt{4}}\left(\begin{array}{llll}1& 1& 1& 1\\ 1& i& -1& -i\\ 1& -1& 1& -1\\ 1& -i& -1& i\end{array}\right)\left(\begin{array}{l}0\\ 2\\ 0\\ 2\end{array}\right)=\left(\begin{array}{l}2\\ 0\\ -2\\ 0\end{array}\right).$$

   

   Thus, U = (2, 0, −2, 0).

The input function u can be written in terms of its DFT coefficients as

$$u=\frac{1}{\sqrt{4}}\left(U_1{e}^{-i2\pi (r-1)0}+0+U_3{e}^{-i2\pi (r-1)(3-1)\text{/}4}\right),$$

which expresses the input array as the sum of complex exponentials.

Math Tip 16.22D Discrete Fourier Transformation

The two-dimensional (2D) DFT transformation of an array uq,r is

$$U_{s,t}=\frac{1}{n}{\displaystyle \sum _{q=1}^n{\displaystyle \sum _{r=1}^n{u}_{q,r}{e}^{2\pi i\left[(r-1)(s-1)+(q-1)(t-1)\right]\text{/}n}}}.$$16.4

16.4

This can be calculated by taking 1D DFTs of each row and then taking the DFTs of each column.

Math Tip 16.3Shifting Functions

The discrete Fourier transform of 2D functions places the origin of the DFT, the constant component, at the (1, 1) element. This constant component is frequently referred to as the DC component, and is the mean of the input array. Because the “center of the DFT” is located in the corner, the DFT of most functions is divided between its four corners as in the left side of Figure 16.3. This makes viewing of most 2D DFTs difficult. Viewing DFT functions, such as the PSF, is easier when the DFT origin is at the center following a shifting operation; thus, throughout the chapter, DFT origins have been shifted to the center.

Figure 16.3Fourier transform of a centered rectangular aperture (left) is concentrated in the four corners. After shifting the origin to the center, the transform has the expected appearance (right).

Math Tip 16.4Padding for Higher Resolution

In order to increase the resolution of the DFT, the input array is often padded with zeros. The Jones pupil, a grid of Jones matrices at the exit pupil, can be padded with zero matrices $\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$, as shown in Figure 16.4, to achieve a higher resolution in Fourier transform domain. The padded grid should be at least twice the size of the original grid in order to avoid aliasing. Increasing padding by more than two times improves the resolution by having better interpolation in the Fourier transform domain.

Figure 16.4An element of an array is padded around the outside with zeros to provide higher resolution of its Fourier transform. The red circle indicates the exit pupil of the system.

16.3Jones Exit Pupil and Jones Pupil Function

The polarization aberration of an imaging system can be described by a grid of Jones matrices at the exit pupil, that is, the Jones exit pupil described in Chapter 15 (Paraxial Polarization Aberrations). Figure 16.5 shows a Cassegrain telescope, which is used as this chapter’s image formation example.

Figure 16.5A side view of the Cassegrain telescope system with an on-axis grid of rays. The parabolic primary mirror is in blue and the hyperbolic secondary mirror is in pink.

A grid of rays is traced through the Cassegrain telescope and the Jones pupil for an on-axis object is calculated. Double pole local coordinates were used to describe the Jones matrices on the spherical reference surface in the exit pupil (described in Chapter 11 [The Jones Pupil and Local Coordinate Systems]). Figure 16.6 shows the amplitude and phase of the Jones pupil. The system has zero wavefront aberration but some polarization aberration from its aluminum mirrors.

Figure 16.6The amplitude (left) and phase (right) of the Cassegrain Jones pupil. A grid of on-axis rays was traced and each ray’s Jones matrix at the exit pupil is plotted.

The full description of the wavefront and polarization at the exit pupil is divided into a combination of four functions: the wavefront aberration function, amplitude function, aperture function, and polarization aberration function at the exit pupil. This combination is called the Jones pupil,

$$JonesPupil(x,y)=aperture(x,y)\cdot a(x,y)\cdot J(x,y)\cdot e^{-2\pi iW(x,y)}.$$16.5

16.5

aperture(x, y) is an aperture function, which is one inside the aperture and zero outside the aperture. a(x, y) is an amplitude function that describes the transmission along the ray path, also known as the apodization. W(x, y) is the wavefront aberration function that characterizes the optical path difference between each ray and the chief ray. J(x, y) is the Jones matrix along ray paths from the entrance to exit pupil. (x, y) is the pupil coordinate. The JonesPupil(x, y) is spatially varying and is the starting point for the polarization image formation calculations. Two example Jones pupil functions are discussed in Examples 16.2 and 16.3.

Example 16.2One Wave of Defocus with a Circular Aperture

Consider an optical system with a circular aperture with a radius of one, a wavefront aberration with one wave of defocus, uniform amplitude, and no polarization aberration. Since there is no polarization aberration, J(x, y) is an identity matrix and the Jones pupil is

$$\begin{array}{l}aperture(x,y)=\text{If }(x^2+y^2\le 1,1,0),\hfill \\ a(x,y)=1,J(x,y)=\left(\begin{array}{ll}1& 0\\ 0& 1\end{array}\right),\hfill \\ W(x,y)=W_{020}(x^2+y^2),W_{020}=1,\hfill \\ JonesPupil(x,y)=aperture(x,y)\cdot a(x,y)\cdot J(x,y)\cdot e^{-2\pi iW(x,y)}.\hfill \end{array}$$

The amplitude and phase of the (1, 1) element of the JonesPupil(x, y) are shown in Figure 16.7. The amplitude is one inside the aperture and a quadratic phase is seen on the right.

Figure 16.7The amplitude (left) and phase (right) of the (1, 1) element of the JonesPupil(x, y) for one wave of defocus wavefront aberration. The Jones pupil has been sampled onto a square array of points.

Example 16.3Quadratic Radially Oriented Retardance

One of the polarization aberrations described in Chapter 15 (Paraxial Polarization Aberrations) is retardance defocus with retardance magnitude that varies quadratically with the pupil radius δo (x2 + y2). Assume a retardance magnitude of δo = 1 wave at the edge of the pupil, that is, one wave of retardance defocus. Further assume a tangentially varying fast axis orientation, tan−1(y/x). Each ray path at the exit pupil has a retarder Jones matrix. Because of the retardance defocus aberration, each ray in the exit pupil experiences different magnitude and orientation of retardance as shown in Figure 16.8.

Figure 16.8Retardance defocus aberration is plotted where the length of the line indicates the magnitude of the retardance and the orientation of the line indicates the orientation of the retardance. Mirror systems, such as Cassegrain telescopes, have tangentially oriented retardance of this form, but the retardance magnitude is much larger in this example.

Assume there is no wavefront aberration and the amplitude function is one. The Jones pupil function is

$$\begin{array}{l}aperture(x,y)=\text{If }(x^2+y^2\le 1,1,0),\hfill \\ a(x,y)=1,\hfill \\ J(x,y)=\left(\begin{array}{ll}\frac{e^{-i{\delta }_o(x^2+y^2)\text{/}2}{x}^2+e^{i{\delta }_o(x^2+y^2)\text{/}2}{y}^2}{x^2+y^2}& \frac{-2ixy\sin \left({\delta }_o(x^2+y^2)\text{/}2\right)}{x^2+y^2}\\ \frac{-2ixy\sin \left({\delta }_o(x^2+y^2)\text{/}2\right)}{x^2+y^2}& \frac{e^{i{\delta }_o(x^2+y^2)\text{/}2}{x}^2+e^{-i{\delta }_o(x^2+y^2)\text{/}2}{y}^2}{x^2+y^2}\end{array}\right),\hfill \\ W(x,y)=0,\hfill \\ JonesPupil(x,y)=aperture(x,y)\cdot a(x,y)\cdot J(x,y)\cdot e^{-2\pi iW(x,y)}.\hfill \end{array}$$

Figure 16.9 shows the amplitude and phase of each of the four elements of the JonesPupil(x, y).

Figure 16.9The amplitude (left) and phase (right) of the JonesPupil(x,y) for retardance defocus.

The polarization aberration of the Cassegrain telescope is a combination of retardance defocus and diattenuation defocus, with a larger retardance contribution.

16.4Amplitude Response Matrix (ARM)

In conventional optics, the amplitude response function (ARF) at the image is the Fourier transform of the pupil function. To incorporate polarization aberration, the amplitude response function is generalized to the amplitude response matrix (ARM), which is the impulse response of the Jones pupil to coherent light. Using the Fraunhofer diffraction equation, the ARM describing the image of an optical system with a continuous Jones pupil can be calculated by 2D Fourier transformation of the Jones pupil function observed at z,

$$JonesARM(\xi ,\eta )={\displaystyle \underset{aperture}{\iint }JonesPupil(x,y)\exp \left[\frac{i2\pi }{\lambda z}(x\xi +y\eta )\right]dxdy}.$$16.6

16.6

The ARM of a Jones pupil that has been regularly sampled by polarization ray tracing can be calculated by the discrete Fourier transform $\left(\Im \right)$ of each of the 2 × 2 elements of JonesPupil(x, y), as stated earlier,

$$\begin{array}{l}ARM(r^{\prime })=\left(\begin{array}{ll}AR{M}_{xx}(r^{\prime })& AR{M}_{yx}(r^{\prime })\\ AR{M}_{xy}(r^{\prime })& AR{M}_{yy}(r^{\prime })\end{array}\right)\\ \phantom{ARM(r^{\prime })}=\left(\begin{array}{ll}\Im \left[JonesPupi{l}_{x,x}(x,y)\right]& \Im \left[JonesPupi{l}_{y,x}(x,y)\right]\\ \Im \left[JonesPupi{l}_{x,y}(x,y)\right]& \Im \left[JonesPupi{l}_{y,y}(x,y)\right]\end{array}\right)\end{array},$$16.7

16.7

where $\Im$(JonesPupill,m(x, y)) = $\Im$ (aperture(x,y) · a(x, y) · Jl,m(x, y) · e−2πiW(x,y)) and r′ = (x′, y′) is the image plane coordinate. From the grid of ray parameters, the ARM is calculated by discrete Fourier transform from the Jones pupil. Each component of the JonesPupil(x, y) is a 2D array of electric field values cumulated at the pupil, and each component of the ARM is a 2D array characterizing the impulse response at the image plane.

For a given point source with Jones vector $E_{object}=\left(\begin{array}{c}{E}_x\\ E_y\end{array}\right)$, the amplitude response matrix multiplied by Eobject yields an electric field distribution Eimage(x, y) for the image, which characterizes the polarization variation. For coherent imaging, an x-polarized incident illumination produces an image with an x-polarized $\Im$[JonesPupilx,x(x, y)] and a y-polarized $\Im$[JonesPupilx,y(x, y)] amplitude response functions. The corresponding intensity point spread function is ${\left|\Im \left[JonesPupi{l}_{x,x}(x,y)\right]\right|}^2+{\left|\Im \left[JonesPupi{l}_{x,y}(x,y)\right]\right|}^2$, which can also be calculated with the Mueller point spread matrix MPSM demonstrated in Section 16.5.

The four elements of the ARM describe the four ARFs seen when the imaging system is located between the four polarizer pairs (H&H, V&H, H&V, and V&V), where H stands for a horizontal polarizer and V stands for a vertical polarizer.

Example 16.4ARM of an Aberration-Free System

For an imaging system with a circular aperture and no wavefront, amplitude, or polarization aberration, the Jones exit pupil is an identity matrix with circular apertures as shown in Figure 16.10 left. Thus, the diagonal elements of the ARM are real-valued Somb (sombrero or Airy) functions,6 as shown in Figure 16.10 (right).

Thus, an image of a point object after this non-polarizing system has no polarization mixing since the ARM is diagonal; the polarization of the image is uniform and equal everywhere to the incident Jones vector.

Figure 16.10(Left) Jones pupil function and (right) amplitude response matrix of an aberration-free optical system. Both functions have had their origin shifted to the center of each array.

Figure 16.11 shows the amplitude of the Jones ARM for the example Cassegrain telescope of Figure 16.5 with aluminum mirrors (n = 0.958 + 6.69i) and marginal ray angles of incidence of 36° and 42° on the primary and secondary mirrors. The right figure zooms in to the center portion. This Cassegrain telescope ARM has off-diagonal components, which indicates a polarization mixing; for example, a horizontally polarized point source has a small, vertically polarized component in its image due to the four bumps in the off-diagonal components of the ARM. However, since the relative magnitudes of the diagonal elements are significantly larger than that of the off-diagonal elements (2.388: 0.101), the polarization mixing is small.

Figure 16.11The amplitude response matrix (ARM) for the Cassegrain telescope (left) plots the entire array and (right) zooms into the center of the pupil.

16.5Mueller Point Spread Matrix (MPSM)

For incoherent light, a point spread function (PSF) describes the response of the system to a point source object in intensity space. Similarly, a Mueller point spread matrix (MPSM) is calculated by transforming the Jones ARM into an MPSM, a Mueller matrix representation using Equation 6.106 or 6.107. The MPSM relates the Stokes parameters of a point object to the Stokes parameter distribution of its image.

Simulating imaging within an isoplanatic patch simplifies the imaging calculation into a convolution.6,7 The calculation of the images for coherent and incoherent objects involves the same JonesPupil(x, y) and discrete Fourier transform, but the calculation ends at different response functions, the amplitude response function for coherent objects or the point spread function for incoherent objects. Figure 16.12 illustrates this relationship between the object field and the diffracted field. The calculation of the scale of the ARM and MPSM, the distance between array points in the image, is explained in Section 16.6.

Figure 16.12The convolution (*) of an example object with three bars (left) with a point spread function (PSF) yields a calculated image (right). In this case, the image is a blurred version of the three bars.

Math Tip 16.5Convolution

A convolution of functions f and g is

$$(f\ast g)(t)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f(\tau )g(t-\tau )d\tau }.$$16.8

16.8

Example 16.5MPSM of an Aberration-Free System

An MPSM of the aberration-free system with the ARM of Example 16.4 has an Airy disk along the diagonal elements and zeros in the off-diagonal elements as shown in Figure 16.13, calculated by converting the Jones matrices of Figure 16.11 into Mueller matrices.

Figure 16.13The MPSM of the polarization aberration-free optical system with a circular aperture is a diagonal matrix of sampled Airy functions.

Figure 16.14 shows the MPSM of the Cassegrain telescope system in a three-dimensional (3D) plot (left) and a raster plot (right). Note the relative magnitude difference between the diagonal elements and the off-diagonal elements; the presence of off-diagonal elements indicates polarization mixing, almost universally undesired, on the image plane, but here the small magnitude of the off-diagonal elements indicates only a small amount of polarization mixing. Note that the scale of the contour plot saturates the diagonal elements for a better visualization of the much smaller off-diagonal elements.

Figure 16.14The MPSM of the Cassegrain telescope system in a 3D view (left) and the corresponding overexposed raster plot to show the dimmer feature (right).

Padding the Jones exit pupil array with zeros (as shown in Math Tip 16.3) increases the resolution of the MPSM without changing the description of the wavefront at the image plane. To obtain the ARM and MPSM shown above with good resolution of structures, it was necessary to pad the Jones pupil array with zero matrices into an array eight times larger than the pupil. Without the padding, only a few points are calculated within the image such that the intensity distribution is difficult to visualize.

16.6The Scale of the ARM and MPSM

The DFT of the Jones pupil provides an array of amplitude values for an image but does not provide the size of the image or the spacing between the elements of the ARM array. The Jones ARM and MPSM in this chapter are calculated using a discrete Fourier transformation of the JonesPupil(x, y). The scale of the ARM and MPSM is calculated by relating the Fraunhofer diffraction equation and DFT equation.

For simplicity, consider a 1D Fourier transformation. Then, Equation 16.6 becomes

$$JonesARM(\xi )={\displaystyle \int JonesPupil(x)\exp \left(\frac{i2\pi }{\lambda z}(x\xi )\right)dx},$$16.9

16.9

where z is the location of the image plane from the exit pupil. The DFT equation is

$$U_s=\frac{1}{\sqrt{n}}{\displaystyle \sum _{r=1}^n{u}_r{e}^{2\pi i(r-1)(s-1)\text{/}n}},$$16.10

16.10

where n is the length of the ur array. Comparing Equations 16.9 and 16.10, JonesPupil(x) = ur and the exponent in Fourier transformation is equivalent to the exponent in DFT. Since s and r are discrete, x and ξ can be written as x = Δxr and ξ = Δss, where Δx and Δs are the unit spacing in the spatial domain x and the spacing in the Fourier transform domain ξ. Thus,

$$\frac{2\pi i}{\lambda z}x\xi =\frac{2\pi i}{\lambda z}{∆}_{x}r{∆}_{s}s=\frac{2\pi i}{n}rs.$$16.11

16.11

Therefore, the grid spacing in the ARM is

$${∆}_s=\frac{\lambda z}{n{∆}_x}.$$16.12

16.12

The denominator nΔx is the size of the Jones pupil array including the zero padding.

Example 16.6Scale of the ARM and MPSM

A lens has a 20 mm exit pupil diameter (XPD = 20) and operates at a magnification of M = 1/3. The exit pupil is z = L = 80 mm from the image plane. A Jones pupil array has been calculated for λ = 1.064 μm by ray tracing a square grid with 41 rays across the exit pupil. The Jones pupil is then padded to a size of 200 × 200 Jones matrices (n = 200) with zero matrices, $\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$. The 2 × 2 × 200 × 200 amplitude response matrix is calculated by taking the 2D Fourier transforms of each 200 × 200 Jones pupil array. What is the spacing in millimeters between the elements of the ARM?

The spacing in the exit pupil is Δx = XPD/41 = 0.4878 mm. Thus, the grid spacing in the ARM is Δs = λL/(200 · Δx) = 0.00087 mm. The spacing between the elements of the MPSM array is also Δs.

The spacing Δs can be interpreted as shown in Figure 16.15. Consider a plane wave exiting the exit pupil perpendicular to the z-axis. This plane wave maps to the center pixel in the image plane in Fourier domain. Then, a plane wave tilted by Δθ (as shown in dark green) maps to one pixel away from the center pixel. The optical path varies between the two plane waves by λ. Hence,

Figure 16.15Consider a plane wave with one wave of tilt across the entire Jones grid (not just the exit pupil). This plane wave corresponds to the array spacing in the ARM.

$$∆\theta =\frac{\lambda }{n{∆}_x},$$16.13

16.13

and from the geometry,

$${∆}_s=z∆\theta =\frac{\lambda L}{n{∆}_x}.$$16.14

16.14

16.7Polarization Structure of Images

The polarization structure of images can be calculated using the ARM or MPSM. An example of the image of a linearly polarized point source, expressed as a Jones vector image, is shown in Figure 16.16. The Jones ARM multiplies an incident Jones vector, yielding a 3D plot of an image represented as a Jones vector. On the right side, a polarization state is plotted for each ray on the grid; the length of the line indicates the amplitude of the Jones vector and the orientation of the line indicates the orientation of the Jones vector. Since the off-diagonal elements of ARM are much smaller than the diagonal elements (as shown in Figure 16.11), the Jones vector image is mostly horizontally polarized with varying amplitudes.

Figure 16.16Jones vector image calculated from a Jones ARM when a horizontally polarized Jones vector point object is incident to the Cassegrain telescope.

A Stokes parameter image can be calculated from the MPSM with the object Stokes parameters; Figure 16.17 shows the Stokes parameter image when a horizontally polarized Stokes parameter is incident on the Cassegrain telescope. The Stokes image in this case is the sum of the first two columns of the MPSM. Note that the center of diagonal elements has been overexposed to reveal faint details. Because of the MPSM’s off-diagonal elements, the image of a horizontally polarized point object contains spatially varying S2 and S3 components.

Figure 16.17The MPSM operates on horizontally polarized Stokes parameters yielding a Stokes parameter image. The color scale is the same as in Figure 16.14.

16.8Optical Transfer Matrix (OTM)

The optical transfer function (OTF) of conventional (scalar) optics describes imaging as a spatial filtering operation. In general, an imaging system accurately images the lowest frequencies of the object, but the amplitude of higher spatial frequencies is reduced until, at and beyond the cutoff frequency, the highest spatial frequency components are completely attenuated. As illustrated in Figure 16.18, the Fourier transform of an image is a Fourier transformation of an object multiplied by the optical system’s OTF.8 The OTF is usually specified in units of line pairs per millimeter. When imaging within an isoplanatic patch, the imaging operation is linear and linear systems theory applies. Within this isoplanatic patch, the image of a cosinusoidal object is always cosinusoidal.6,7

Figure 16.18In the Fourier domain, the Fourier transform of the object of Figure 16.12 (left) times the OTF (center) is the Fourier transform of the image (right). The object of Figure 16.12 has a pair of strong frequency components about the central lobe, which become significantly reduced by the system’s optical transfer function, as is seen in the Fourier transform of the image on the right, which corresponds to the blurred image on the right of Figure 16.12.

The extension of the OTF to incoherent imaging systems with polarization aberration is performed with the Mueller optical transfer matrix (OTM), a complex valued matrix function. A matrix is necessary because each of the object’s Stokes parameters can couple into all four of the image’s Stokes parameters. The Mueller OTM characterizes the spatial filtering of an object for each of the object’s Stokes parameters. The Mueller OTM operates on the Fourier transform of the object’s Stokes parameter. The modulus of the complex valued OTM (i.e., a Mueller modulation transfer matrix [MTM]) describes the change in modulation of each spatial frequency. Figure 16.19 shows the Cassegrain telescope MTM as a 3D plot (left) and raster plot (right). Mueller matrices are real, but in general, the OTM is not real but complex, because the MPSM functions may have both even and odd components, and real odd functions have imaginary valued Fourier transforms. Physically, this means that the positions of the images of cosinusoidal objects are shifted from their geometrical images. Thus, the phase of the complex valued OTM (i.e., a Mueller phase transfer matrix [PTM]) describes the shift of each cosinusoidal spatial frequency from its geometrical image; a value of the PTM of π corresponds to a half period shift. Examples of the image degradation due to the phase transfer function are addressed in the work of Gaskill7 beginning on page 236. Figure 16.20 shows the Cassegrain telescope PTM in 3D plot (left) and raster plot (right).

Figure 16.19Amplitude of OTM of the Cassegrain telescope in 3D plot (left) and raster plot (right, overexposed).

Figure 16.20Phase of OTM of the Cassegrain telescope in 3D plot (left) and raster plot (right).

The raster plot of MTM is overexposed for the diagonal elements to show more details in the off-diagonal elements. The MTM is always normalized to one at the center of the pupil in the m00 element. This means that, for any incident Stokes parameters ${\left(\begin{array}{cccc}{S}_{0,obj}& S_{1,obj}& S_{2,obj}& S_{3,obj}\end{array}\right)}^T$, the Mueller MTM provides image Stokes parameters that have a maximum of S0,image = S0,obj at the center pixel of the array.

Example 16.7Scale of OTM

Continuing from Example 16.6, calculate the spacing in line pairs per millimeters between the elements of the OTM Δξ and calculate the cutoff frequency in the OTM.

The size of the MPSM array is nΔs. First, the lowest spatial frequency pimage of an MPSM is one period over nΔs, pimage = nΔs = 0.174496.

Since the system has a magnification of 1/3, the fundamental period at the object, that is, a spatial frequency in line pairs per millimeter at the object, is

$$p_{object}=p_{image}/M=\mathrm{0.523488.}$$

Thus, the spatial frequency interval Δξ in the OTM in line pairs per millimeter is

$$\Delta \xi =1/p_{object}=1.91026,$$

and the cutoff frequency ΔξNyquist is ΔξNyquist = Δξn/2 = 191.026 in line pairs per millimeter.

16.9Example—Polarized Pupil with Unpolarized Object

As an incoherent imaging example with large and easily seen polarization effects, consider the following diffraction-limited imaging system with a square aperture. Let the center 2/3 of the aperture (−2/3 ≤ x ≤2/3) be filled with a vertical polarizer. The remainder (−1 ≤ x −2/3 and 2/3 ≤ x ≤ 1) is filled with a horizontal polarizer as shown in Figure 16.21 (left). The JonesPupil(x, y) of this striped polarizer system is shown in Figure 16.21 (right), a square aperture with uniform amplitude and zero wavefront aberration. The non-zero part of the J11 component represents the horizontal polarizer at the edges of the pupil, while the J22 component represents the vertical polarizer over the center 2/3 of the pupil.

Figure 16.21(Left) A square aperture where the center 2/3 is filled with a vertical polarizer and the remainder filled with a horizontal polarizer is shown. (Right) JonesPupil(x, y) for a square aperture with no wavefront aberration and uniform amplitude is shown. The center 2/3 of the square is a vertical polarizer and the remainder is a horizontal polarizer.

The Jones ARM of this striped polarizer system is calculated using Equation 16.6 and is shown in Figure 16.22 and the J11 component (left) and the J22 component (right) are shown in Figure 16.23. Since each Jones pupil element is an even function, each element of the Jones ARM is real. The J11 component resembles an interference pattern since the horizontal polarizer portion of the JonesPupil(x, y) is equivalent to a double slit separated along the x-axis. The J22 component of the ARM is a sinc function with different widths in x and y since the vertically polarized aperture is equivalent to a single slit with wider dimension along the y-axis. Thus, the J22 component of the ARM is wider along the x-axis than along the y-axis.

Figure 16.22The Jones ARM of horizontal/vertical polarizer JonesPupil(x, y).

Figure 16.23(Left) The (1, 1) component of the ARM is an interference pattern formed from the horizontal polarizer double slit JonesPupil(x, y). The amplitude varies between −4.40 and 4.71. (Right) The (2, 2) component of the ARM Fraunhofer diffraction pattern results from the vertical polarizer as a rectangular single slit JonesPupil(x, y). The amplitude varies between −2.04 and 9.42.

As shown in Figure 16.1, the MPSM is calculated by converting the Jones ARM to a Mueller matrix. A scalar PSF of the system when a horizontally polarized object is viewed from a horizontal polarizer can be calculated by cascading the MPSM in between two horizontal polarizer Mueller matrices; this is shown in Figure 16.24 (left). This scalar PSF is called the horizontal–horizontal MPSM. Similarly, the scalar PSF of the system when a vertically polarized object is viewed through a vertical polarizer can be calculated and is shown in Figure 16.24 (right).

Figure 16.24(Left) The PSF of the striped polarizer system viewed between two horizontal polarizers is a two-slit interference pattern with maximum and minimum intensity 22.18 and 0. (Right) The PSF viewed between two vertical polarizers has maximum and minimum intensity 88.66 and 0.

The Mueller OTM is calculated and plotted in Figure 16.25. Figure 16.26 shows the OTM (0, 0) element from two different viewpoints. The transfer function along the x-direction and y-direction exhibits drastically different profiles.

Figure 16.25Mueller OTM of the example system.

Figure 16.26Two views of a scalar OTF when an unpolarized object enters the example system looking along the y-axis (left) and along the x-axis (right) showing different resolution. One band of spatial frequencies around 200 line pairs per millimeter (left) is completely removed where the OTM goes to zero. The resulting filtering is seen in Figure 16.29 (right).

Figure 16.27 shows the Mueller OTM viewed in between two horizontal polarizers (horizontal–horizontal OTM) on the left and the OTM viewed in between two vertical polarizers (vertical–vertical OTM) on the right in two different views (top and bottom). For a horizontally polarized light, the OTM has three peaks since the JonesPupil(x, y) is a double slit. For a vertically polarized light, the OTM follows a familiar triangular shape since the JonesPupil(x, y) is a rectangular aperture with uniform amplitude.

Figure 16.27(Left) Mueller OTM viewed between two horizontal polarizers and (right) Mueller OTM viewed between two vertical polarizers. H–H OTM has a center peak value of 0.325 and V–V OTM has a center peak value of 0.675.

Consider an unpolarized object consisting of three strips as shown in Figure 16.28. The Fourier transform of the image’s Stokes parameters are calculated by matrix multiplying the Mueller OTM of the optical system to the Fourier transform of the object,

Figure 16.28The s0 Stokes parameter of the unpolarized object is g(x, y) with three vertical strips is plotted.

$$\Im \left(\begin{array}{l}{s}_{image,0}\\ s_{image,1}\\ s_{image,2}\\ s_{image,3}\end{array}\right)=OTM\cdot \Im \left(\begin{array}{l}{s}_{object,0}\\ s_{object,1}\\ s_{object,2}\\ s_{object,3}\end{array}\right),$$16.15

16.15

where the unpolarized object is $S_{object}={\left(\begin{array}{cccc}g(x,y)& 0& 0& 0\end{array}\right)}^T$ and g(x, y) has the object’s intensity shown in Figure 16.28. The object g(x, y) has a spatial frequency that is half the spatial frequency of the horizontal–horizontal MPSM of the system shown in Figure 16.24 (left).

Then, the inverse Fourier transform of Equation 16.15 provides the image of the object. Since the Mueller OTM is non-zero in m00 and m10 components, the unpolarized object has two orthogonal polarization components, horizontal and vertical, as shown in Figure 16.29.

Figure 16.29(Left) Horizontally polarized component of the image and (right) vertically polarized component of the image drawn in the same scale.

The horizontal and vertical components of the image are completely different. For the vertical component, the image is a blurred version of the three strips object due to the loss in high frequency from a limited aperture size of the optical system. This is straightforward and can be observed from the scalar PSF/OTF calculations. The horizontal component of the image, however, requires more explanations. Since the object has a spatial frequency that is half of the horizontal–horizontal MPSM of the system, Fourier transform of the object has a frequency that is twice the horizontal–horizontal OTM of the system. This is why the horizontally polarized component of the image does not show any peaks (or spatial frequency of the object) but has an envelope that shows the blurred extent of the object. Thus, it is seen how a polarization-dependent image can be understood using the MPSM/OTM of the optical system.

16.10Example—Solid Corner Cube Retroreflector

An interesting example of polarization image formation occurs with solid corner cube retroreflectors (CCRs), which have a large polarization aberration due to the total internal reflection (TIR) at the three back surfaces. First, a typical CCR is analyzed with a mid-refractive index glass, n = 1.6. Higher indices are preferred for solid corner cubes because the field of view, which is limited by the TIR failure, is larger. Then, a cube with a low refractive index chosen to operate at the critical angle n = 1.227 is considered for its theoretical interest.

A CCR consists of three mutually perpendicular reflective surfaces. As shown in Figure 16.30, a corner cube (shown in red) is literally the corner of a cube. For a solid glass cube, rays refracting into the front face undergo three TIRs from the back surfaces and exit the front face with their propagation vector reversed. For a normally incident ray, the angle of incidence (AOI) at each of the three reflecting faces is arcos (1/$\sqrt{3}$) = 54.74°. For an example cube with n = 1.6, these rays experience three linear retardances of δ = 48.94°, but each retardance is at a different orientation; hence they cascade into an elliptical retarder.

Figure 16.30A corner cube retroreflector (red) has three mutually perpendicular reflecting surfaces. Rays refracting into the front face reflect three times and exit the front face with their propagation vector, thus retroreflecting the light.

Figure 16.31 shows how each surface of the CCR is numbered. There are six subapertures in the CCR; depending on the position of the entering ray in surface 1, six different ray paths exist. The retarder eigenstates are different for each of the six ray paths. Figure 16.32 lists the order of surface intercepts for the six subapertures viewed looking into surface 1. A Polaris-M polarization ray trace has been performed and Figure 16.33 follows the evolution of polarization for the six subapertures (each in a different column) for three different polarization states (different rows). The polarization ellipses are shown after each surface in 3D. For the linear states in the top two rows, the exiting light is always elliptical. Figure 16.34 shows polarization propagation for three more rays.

Figure 16.31The front surface of the CCR where light enters and exits is numbered 1, and the three reflecting surfaces are numbered 2, 3, and 4.

Figure 16.32Depending on where light enters the front face, it encounters a different order of surface reflections as labeled in the figure. The corner cube exit pupil contains six distinct subapertures.

Figure 16.33Polarization propagations through the CCR for horizontal (top row), vertical (middle), and left circularly polarized incident beams are shown in 3D following each surface. Each column follows one of the six subapertures.

Figure 16.34The polarization change along the three different ray paths is shown for three different incident polarization states.

Figure 16.35 plots the amplitude and phase of the Jones pupil calculated by the polarization ray trace. Figure 16.36 shows eigenvectors of the 3D polarization ray tracing (PRT) matrix for each subaperture. Since these are for a retroreflection PRT, the polarization ellipses are associated with vectors propagating in opposite directions to the incident propagation vector. The ellipticities are the same in each subaperture. The ellipses’ major axes rotate by 120° between subapertures.

Figure 16.35Amplitude (left) and phase (right) of the CCR’s Jones pupil.

Figure 16.36The two eigenpolarizations are plotted for each subaperture.

Assuming a hexagonal aperture with an amplitude transmittance a(x, y) = 1 and free of wavefront aberration W(x, y) = 0, the far field diffraction pattern, the Jones ARM calculated by discrete Fourier transformation of the JonesPupil(x, y), is plotted in Figure 16.37. The ellipse map on the right is the Jones vector image for a horizontally polarized point source. Unlike Figure 16.16, the image’s polarization has substantial spatial variation in orientation and ellipticity since the off-diagonal elements of the Jones ARM are comparable in magnitude to the diagonal elements. Interestingly, at the radius of the blue dashed circle, Figure 16.37 (right) shows a 720° rotation of linear polarization.

Figure 16.37The magnitude of the Jones ARMs for the n = 1.6 CCR (left) and a zoomed-in view of the center (middle) is shown. The Jones vector image for a horizontally polarized point source (right) is plotted with blue dashed lines indicating the orientation of the six subapertures. A 720° rotation of linear polarization can be observed following the blue dashed circle.

The MPSM is calculated by transforming the Jones ARM into a Mueller matrix image, which is plotted in Figure 16.38. The first column contains the Stokes image for an unpolarized point source. The m00 element has six islands surrounding the central peak. The m10 and m20 elements show that the light at the same blue dashed line’s radius has linear polarization that rotates (due to the six alternating lobes) by 540° around the center. Thus, even the PSF of unpolarized light has partially polarized regions.

Figure 16.38The Mueller point spread matrix of the n = 1.6 CCR shows large and complex amounts of polarization coupling between the incident and exiting Stokes parameters. The images are overexposed to show the structure of the dimmer regions.

The Mueller matrix optical transfer matrix (MOTM) is calculated by taking the discrete Fourier transformation of each element of the MPSM. The elements of the modulation transfer matrix and the phase transfer matrix are plotted in Figure 16.39. Because of this large polarization aberration, image formation is far from diffraction limited. The m00 element does not resemble a diffraction-limited transfer function for example.

Figure 16.39Amplitude (left) and phase (right) of the Mueller OTM.

16.11Example—Critical Angle Corner Cube Retroreflector

A corner cube without retardance can be constructed by adjusting the refractive index to operate at the critical angle at the three reflections for normally incident light. This critical angle corner cube retroreflector has a very interesting property of rotating linear polarization in all six subapertures. Since the angle of incidence is 54.73°, the index of this cube is ncritical = 1.2247. TIR is associated with a retardance that varies rapidly with the angle of incidence, as plotted in Figure 16.40 for ncritical. At the critical angle, the retardance is zero, and the reflection is ideal and non-polarizing with rs = 1 and rp = 1. Since an on-axis ray experiences zero retardance at the three ray intercepts, it might seem that this cube is non-polarizing. However, by the time the beam changes direction three times and is retroreflected, the polarization states become rotated. This change of polarization is called geometrical transformation or skew aberration. This geometrical transformation is calculated using the Q matrix method of Chapter 17 (Parallel Transport and the Calculation of Retardance) and is understood as skew aberration as discussed in Chapter 18 (A Skew Aberration). In Figure 16.40, note that, at the critical angle, the slope of the retardance is infinite; hence, actually operating at the critical angle would require an impossibly perfectly collimated and perfectly aligned beam, rendering it impractical. The theoretical properties of this cube, however, provide a fascinating polarization aberration example.

Figure 16.40Retardance as a function of angles of incidence for internal reflection ncritical = 1.2247.

Geometric transformation describes how the polarization changes as each ray bends or reflects at a non-polarizing surface (free of diattenuation and retardance). Geometric transformation does not change the ellipticity or the amplitude of the electric field. Since the retardance at each ray intercept is zero, the critical angle CCR has no polarization aberration, only geometric transformation. Figure 16.41 shows the Jones pupil for a grid of rays through the critical angle CCR.

Figure 16.41Jones exit pupil of the critical angle CCR1 is real valued everywhere, corresponding to three pairs of different circular retarders.

This Jones pupil contains only real values since the phase of the Jones pupil elements is either π or −π. Each triangular subaperture (as labeled in Figure 16.42) has a Jones matrix, which is a rotation matrix along the z-axis with a rotation amount of 0°, 120°, and 240°. The Jones matrix of opposite pairs of subapertures is identical,

Figure 16.42Labels for the subapertures of the CCR.

$$J_A,J_D=\left(\begin{array}{ll}-1& 0\\ 0& -1\end{array}\right),J_B,J_E=\left(\begin{array}{ll}1\text{/}2& -\sqrt{3}\text{/}2\\ \sqrt{3}\text{/}2& 1\text{/}2\end{array}\right),J_C,J_F=\left(\begin{array}{ll}1\text{/}2& \sqrt{3}\text{/}2\\ -\sqrt{3}\text{/}2& 1\text{/}2\end{array}\right).$$16.16

16.16

Thus, each subaperture of the Jones pupil has a circular retarder Jones matrix that rotates all polarization states by θ = 0°, 120°, and 240° shown in Figure 16.43, demonstrating the polarization rotation after three reflections. All three reflections become perfect total internal reflections at the critical angle. Therefore, the resultant circular retardance of the reflected light is solely induced by the geometry of the ray paths. The polarization rotations around the exit surface are 0°, 120°, and 240°.

Figure 16.43(a, b, and c) Six incident rays incident at different regions of the entrance surface follow different paths within the corner cube, and the corresponding exiting states have a different amount of polarization rotations for vertically, horizontally, and circularly polarized incident rays. (c) Circular retardance induced by the corner cube alters the absolute phase of circular incident light. The amount of circular retardance depends on the ray path inside the corner cube. (d) The rotation associated with the induced circular retardance are 0°, 120°, and 240° around the exit pupil.

These properties and symmetries of the critical angle CCR yield a fascinating amplitude response matrix shown in Figure 16.44 (left), one of the most interesting and symmetric the authors have ever seen. An analysis shows that the ARM has circular retarder Jones matrices everywhere. The circular retardance magnitude of the ARM is plotted in Figure 16.44 (right). The center of the ARM is completely dark because the three components of the light, 0°, 120°, and 240° for example, destructively interfere. Around a small circle, at about 0.25 from the center to the edge of the figure, an 8π retardance change is observed circling around the pupil; this will cause a steady 4π rotation of linear polarization around the pupil at this distance. The circular retardance is zero along the horizontal and vertical directions and has 2π/3 retardance in the +60° radial direction and −2π/3 retardance along the −60° direction. Along each of these lines, two polarization singularities are found at zeros of the ARM; on the x-axis, the singularities are found at about 0.5 and 0.8 from the center to edge of the figure. Moving around a circle between these two zeros (about 0.65 on the x-axis), a continuous 16π retardance change is observed.

Figure 16.44The amplitude of the amplitude response matrix (left) is purely real. A circular retardance map of the critical angle corner cube (right) is plotted in radians.

For a horizontally polarized incident plane wave, the Jones ARM yields a Jones vector image in the far field as shown in Figure 16.45. Because of the off-diagonal elements of the Jones ARM, vertically polarized light (lower) has been mixed into the horizontally polarized state and there is a spatial linear polarization variation on the image plane. The six subapertures that are visible in the circular retardance map in Figure 16.44 match the six islands in Figure 16.45. Different amounts of rotation of the horizontal polarization state are observed for each of the six islands.

Figure 16.45The diffraction pattern from the critical angle corner cube with horizontally polarized incident light shows a dark center and six inner islands of light at 0°, 120°, and −120° orientations. The x-amplitude (left top) and y-amplitude (left bottom) of the Jones ARM vector show a complex structure that, when represented as polarization ellipses (middle), clearly shows the six islands with varying polarization states. The 720° rotation of linear polarization can be observed by moving around the diffraction pattern at the radius of the six islands. (Right) The line lengths are magnified two times to show additional polarization structure further from the center; the lines in the center are suppressed. Since the ARM Jones matrices are pure circular retarders everywhere, this diffraction pattern is linearly polarized everywhere, free of elliptical polarization.

The critical angle CCR’s MPSM is shown in Figure 16.46. The three zero elements in the top row show that the intensity distribution is independent of the incident polarization state; there is no diattenuation. The first column shows that for unpolarized incident light, the image is unpolarized everywhere; there is no polarizance. The first and last columns show that a right circularly polarized point source object generates a right circularly polarized image point and a left circularly polarized object forms a left circularly polarized image. The middle four elements are particularly interesting, having the form of a spatially varying circular retarder. For a linearly polarized source, the image is linearly polarized everywhere, with a rotating polarization state. For a circularly polarized object, the MPSM gives circularly polarized image Stokes parameters with the same handedness as the object, the opposite of the n = 1.6 CCR. The MPSM is dark at the center for all polarization states; a right circularly polarized point source object will have a right circularly polarized image point, and an unpolarized point source object will have an unpolarized image point.

Figure 16.46An MPSM for the critical angle CCR (left) operating on a horizontal linearly polarized set of Stokes parameters forms a diffraction pattern with spatially varying Stokes parameters corresponding to a varying polarization orientation.

Figure 16.47 shows the Mueller OTM’s magnitude (left) and the phase (right). Again, the MPSM is a circular retarder for all spatial frequencies and orientations. The OTM contains only real values with 0 or π phase. Figure 16.48 shows a cross section of the Mueller MTM along the horizontal direction.

Figure 16.47The Mueller MTM (left) and PTM (right) for the critical angle CCR, overexposed to reveal faint structures.

Figure 16.48Cross section of the Mueller MTM along the horizontal direction.

The critical angle CCR provides a unique polarization imaging example. All it does is rotate polarization state between subapertures, which yields complex and highly symmetric PSFs and OTFs. This circular retardance-like change in polarization is present even though the critical angle CCR has no retardance or diattenuation. Such intrinsic geometric transformation is explored in more detail in Chapter 17 (Parallel Transport and the Calculation of Retardance) and Chapter 18 (A Skew Aberration).

16.12Discussion and Conclusion

The analysis of the imaging system has been generalized to include polarization aberration. The optical system is described by an aperture function, amplitude function, wavefront aberration, and polarization aberration, which are combined into the JonesPupil(x, y). The ARM, which describes image formation of a coherent optical system, can be calculated by applying the DFT to each of the four elements of the JonesPupil(x, y). For an incoherent imaging system, the MPSM can be calculated by converting the Jones ARM into a Mueller matrix. An additional DFT results in modulation transfer functions between all the pairs of incident and exiting Stokes parameters in the form of the Mueller OTM.

The Jones ARM, MPSM, and Mueller OTM provide a full description of the image quality including polarization aberration. Often, these matrices need to be reduced to a scalar version of the corresponding metric. The Jones ARM can be reduced to an ARF when an object Jones vector is applied to the Jones ARM and viewed through a linear polarizer. Similarly, the MPSM or OTM can be reduced to the PSF and OTF by applying an object Stokes parameters to the corresponding Mueller matrix and viewing the image through a polarizer.

Off-diagonal elements in ARM, MPSM, or OTM describe the coupling between polarization components in the image. The relative magnitude of the off-diagonal elements with respect to the diagonal elements tells the amount of polarization mixing that occurs in the image plane. For example, the Cassegrain telescope system has small coupling between polarization states; thus, the image polarization was almost the same as the object polarization state. The two corner cube retroreflectors had significant off-diagonal elements in their ARM, MPSM, and OTM; hence, substantial spatially varying polarization coupling was observed in their images.

16.13Problem Sets

1. Perform a 1D discrete Fourier transformation of the following arrays u. Identify the functions as sin, cos, constant, and so on.

   1. $u=\left(\sqrt{2},0\right)$

   2. u = (0, 2, 0, 2)

   3. u = (0, 2i, 0, −2i)

   4. u = (0, 0, 4, 0)

   5. $u=\left(0,0,2\sqrt{2},0,0,0,0,0\right)$

   6. u = (4, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0,)

2. Calculate the following 2D discrete Fourier transforms Us,t of the following sampled functions uq,r. Identify the functions as sin, cos, constant, and so on.

   • a. $U=\left(\begin{array}{cccc}4& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right),$ b. $U=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 2& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 2\end{array}\right),$ c. $U=\left(\begin{array}{cccc}0& 0& 0& 0\\ 2& 0& 0& 0\\ 0& 0& 0& 0\\ 2& 0& 0& 0\end{array}\right),$

   • d. $U=\left(\begin{array}{cccc}0& 2& 0& 2\\ 2& 0& 0& 0\\ 0& 0& 0& 0\\ 2& 0& 0& 0\end{array}\right),$ e. $U=\left(\begin{array}{cccc}0& 2& 0& 2\\ 2& 0& 2& 0\\ 0& 2& 0& 2\\ 2& 0& 2& 0\end{array}\right),$

   • f. $U=\left(\begin{array}{cccccccc}0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 4& 0& 0& 0& 4& 0\\ 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right),$ and g. $U=\left(\begin{array}{cccccccc}0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 8& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right).$

   What is the highest frequency wave that can be described by a four-element sampled function? An eight-element sampled function?

3. Calculate the PSM for retardance defocus following the procedure below.

   The point spread functions will be calculated for the polarization aberration “retardance defocus.” A uniformly illuminated circular aperture is assumed with no wavefront aberration. The magnitude of the retardance defocus, δmag, is 1 radian. The exit pupil diameter is 2 mm. In J(x, y), the Jones matrices for a retarder have been expanded into linear and quadratic Taylor series terms.

   $$\begin{array}{l}a(x,y)=1,\\ J(x,y)=\left(\begin{array}{ll}1-\frac{i{\delta }_{mag}(x^2-y^2)}{2}& -i{\delta }_{mag}xy\\ -i{\delta }_{mag}xy& 1-\frac{i{\delta }_{mag}(-x^2+y^2)}{2}\end{array}\right),\\ W(x,y=0).\end{array}$$

   

   Use a computer program (MATLAB or Mathematica) to create a pupil array with odd dimensions such as 129 × 129.

   1. Create an exit pupil array with 33 rays across the circular exit pupil and padded with zeros out to 129 × 129. Graph the real and imaginary part of the 2 × 2 × 129 × 129 Jones Pupil.

   2. Calculate the 2 × 2 × 129 × 129 amplitude response matrix (ARM), the Jones matrix form of the coherent point spread function. Each 129 × 129 element of the 2 × 2 Jones pupil is separately Fourier transformed. Graph the real and imaginary part of the 2 × 2 × 129 × 129 ARM.

   3. Calculate and graph the 4 × 4 × 129 × 129 MPSM.

   Based on the amplitude response matrix in parts a and b, answer the following questions:

   1. If the optic is illuminated with horizontal polarized light and viewed through a vertical polarizer, what form does the coherent point spread function have?

   2. What wavefront aberration is present in the Jxx element? How does it compare to the aberration in the Jyy element?

   3. Describe and explain the difference between the horizontal-in horizontal-analyzed coherent point spread function and the vertical-in vertical-analyzed coherent point spread function.

   Based on the MPSM in part c, answer the following questions:

   1. What is the polarization of the image of an unpolarized point source?

   2. How does the intensity distribution of the point spread function (the S0 component) depend on the incident polarization state?

4. Refer to Figures 16.35 and 16.37. When right circularly polarized light is incident, is the center of the image right circularly polarized or left circularly polarized? Why? Will the six islands in the first diffraction ring have the same polarization as the center?

5. Figure 16.46 showed exiting Stokes parameters for the horizontally polarized light after the CC MPSM. Match the incident polarization state to the following exiting Stokes parameters.

   • a. $S=\left(\begin{array}{c}1\\ -1\\ 0\\ 0\end{array}\right)$ b. $S=\left(\begin{array}{c}1\\ 0\\ 1\\ 0\end{array}\right)$ c. $S=\left(\begin{array}{c}1\\ 0\\ -1\\ 0\end{array}\right)$ d. $S=\left(\begin{array}{c}1\\ 0\\ 0\\ 1\end{array}\right)$ e. $S=\left(\begin{array}{c}1\\ 0\\ 0\\ -1\end{array}\right)$

     

6. What calculations are performed on the Jones pupil to calculate the ARM in regions near focus but not in focus?

7. Consider vortex retarders that are half wave linear retarders with their fast axes rotated m/2 times around the center. Their Jones matrix is

   $$J_m=\left(\begin{array}{ll}\cos (m\cdot \arctan (x,y))& \sin (m\cdot \arctan (x,y))\\ \sin (m\cdot \arctan (x,y))& -\cos (m\cdot \arctan (x,y))\end{array}\right).$$

   

   The retardance pupil map is shown below:

   

   Calculate PSM of Jones matrices for m = 1, 2, 3, and 4.

8. A collimated beam of 500 nm light is incident on a 1 mm × 1 mm square aperture over a Wollaston prism with the following polarization aberration function and pupil functions:

   $$\begin{array}{l}a(x,y)=1,\\ J(x,y)=\left(\begin{array}{ll}\exp (-i\pi x\text{/}1)& 0\\ 0& \exp (i\pi x\text{/}1)\end{array}\right),\\ W(x,y=0).\end{array}$$

   

   The prism is followed by a non-polarizing lens. The exit pupil is 100 mm from the image. Choose nine pixels across the aperture in a Jones pupil sized 2 × 2 × 199 × 199.

   1. Calculate the 2 × 2 × 199 × 199 Jones pupil. Graph the magnitude and phase of a region just larger than the pupil.

   2. Calculate the 2 × 2 × 199 × 199 amplitude response matrix. Graph the magnitude and phase of the center of the ARM.

   3. Graph the magnitude and phase of the amplitude point spread function for horizontally polarized incident light, both Ex and Ey. Only graph the center where the majority of the light is located.

   4. Graph the magnitude and phase of the center of the amplitude point spread function for vertically polarized incident light.

   5. Graph the amplitude point spread function for 45° polarized incident light.

   6. Calculate and graph the optical transfer matrix for horizontally polarized incident light.

   7. Determine and graph the optical transfer matrix for 45° polarized incident light.

   Based on the Jones pupil in part (a):

   1. What aberrations are present?

   Based on the amplitude point spread function for horizontal polarized incident light in part (c):

   1. What is the polarization state of the amplitude point spread function for horizontally polarized incident light?

   Based on the amplitude point spread function for vertical polarized incident light in part (d):

   1. What is the polarization state of the amplitude point spread function for vertically polarized incident light?

   Based on the amplitude point spread functions in parts (c) and (d):

   1. How are horizontal point spread functions located relative to the vertical point spread functions? Explain.

   Based on the amplitude point spread function for 45° incident light in part (e):

   1. Describe the distribution of polarization states for 45° polarized point spread functions.

   Based on the optical transfer matrix for horizontally polarized incident light in part (f):

   1. How much is the image degraded by the polarization aberration? Can it be compared to a diffraction-limited optical transfer function?

   Based on the optical transfer matrix for 45° polarized incident light in part (g):

   1. Which spatial frequencies are filtered out (completely removed) by the system?

   2. What is the separation between pixels in the ARM in millimeters?

References

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2J. P. McGuire and R. A. Chipman, Diffraction image formation in optical systems with polarization aberrations. I: Formulation and example, J. Opt. Soc. Am. A 7(9) (1990).

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1 The Jones pupil is calculated using local coordinates defined by the incident and exiting propagation vectors for the CCR.