Chapter 6: Polarization Optics ============================== Source: Saleh and Teich, *Fundamentals of Photonics*, second edition, Chapter 6. Global Jones phases are physically immaterial. In-text exercises ----------------- Exercise 6.1-1 — Measuring Stokes parameters ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Step 1 — Definitions and setup.** Send the light to a calibrated photodetector through * a rotatable ideal linear polarizer, and * a removable quarter-wave plate whose fast axis can also be rotated. The detector readings below are background-subtracted intensities. Denote by :math:`I_H`, :math:`I_V`, :math:`I_D`, and :math:`I_A` the intensities passed by linear analysers at :math:`0^\circ`, :math:`90^\circ`, :math:`45^\circ`, and :math:`135^\circ`, respectively. Denote by :math:`I_R` and :math:`I_L` the readings of right- and left-circular analysers. The latter are made by placing the quarter-wave plate **before** the linear polarizer and setting its fast axis at :math:`+45^\circ` or :math:`-45^\circ` relative to the polarizer axis. Which setting is called right-handed depends on the viewing and time convention; label it so that :math:`S_3=I_R-I_L`, as in this chapter. .. _fop-exercise-6-1-1-illustration: .. figure:: /_static/knowledge_base/worked_exercises/fundamentals_of_photonics/exercise_illustrations/exercise_06_01_01.svg :alt: Illustrated calculation map for Exercise 6.1-1, Measuring Stokes parameters :align: center :width: 95% **Figure 48 — Exercise 6.1-1: Measuring Stokes parameters.** The diagram identifies the input quantities, physical operation, requested result, variable meanings, and an independent verification route. Every symbol in the variable strip is labeled on the model itself. **Step 2 — Mathematical formulas used.** For a linear analyser at angle :math:`\theta`, the transmitted intensity is .. math:: :label: fop-exercise-6-1-1-linear-analyser I_{\mathrm{lin}}(\theta) =\frac{1}{2}\left(S_0+S_1\cos 2\theta+S_2\sin 2\theta\right). The two circular-analyser readings are .. math:: :label: fop-exercise-6-1-1-circular-analyser I_R=\frac{1}{2}(S_0+S_3), \qquad I_L=\frac{1}{2}(S_0-S_3). A rotating linear polarizer alone can therefore determine :math:`S_0`, :math:`S_1`, and :math:`S_2`, but not :math:`S_3`: the quarter-wave plate is what converts circular polarization (the quadrature component) into a linear intensity difference. **Step 3 — Worked derivation.** Keep the incident beam power constant while recording the following six settings. A six-state measurement is slightly redundant, but the redundancy exposes source drift and analyser errors. .. list-table:: :header-rows: 1 :widths: 18 34 48 * - Reading - Optics before the detector - Information supplied * - :math:`I_H` - Linear polarizer at :math:`0^\circ` - :math:`I_H=(S_0+S_1)/2` * - :math:`I_V` - Linear polarizer at :math:`90^\circ` - :math:`I_V=(S_0-S_1)/2` * - :math:`I_D` - Linear polarizer at :math:`45^\circ` - :math:`I_D=(S_0+S_2)/2` * - :math:`I_A` - Linear polarizer at :math:`135^\circ` - :math:`I_A=(S_0-S_2)/2` * - :math:`I_R` - Quarter-wave plate and polarizer set as a right-circular analyser - :math:`I_R=(S_0+S_3)/2` * - :math:`I_L` - Quarter-wave plate and polarizer set as a left-circular analyser - :math:`I_L=(S_0-S_3)/2` Subtracting each orthogonal pair isolates one signed Stokes component, while adding either member pair gives the total intensity: .. math:: :label: fop-exercise-6-1-1-six-state-reconstruction \begin{aligned} S_0 &= I_H+I_V,\\ S_1 &= I_H-I_V,\\ S_2 &= I_D-I_A,\\ S_3 &= I_R-I_L. \end{aligned} Ideally, :math:`I_H+I_V=I_D+I_A=I_R+I_L`. In a real experiment, use the average of these three sums for a lower-noise estimate of :math:`S_0`, after correcting the channels for detector gain and optical throughput. **Step 4 — State the numbered result.** The requested method returns the complete Stokes vector .. math:: :label: fop-exercise-6-1-1-result \boxed{ \mathbf S= \begin{bmatrix}S_0\\S_1\\S_2\\S_3\end{bmatrix} = \begin{bmatrix} I_H+I_V\\ I_H-I_V\\ I_D-I_A\\ I_R-I_L \end{bmatrix}} This works for fully, partially, or unpolarized stationary light; it does not require the light to possess a Jones vector. If only the minimum number of readings is desired, measure :math:`I_H`, :math:`I_V`, :math:`I_D`, and :math:`I_R`. Then use :math:`S_0=I_H+I_V`, :math:`S_1=I_H-I_V`, :math:`S_2=2I_D-S_0`, and :math:`S_3=2I_R-S_0`. The six-reading method is normally preferable because each component is formed from a balanced difference. **Step 5 — Check.** A physical Stokes vector must satisfy .. math:: :label: fop-exercise-6-1-1-physicality S_0\geq 0, \qquad S_1^2+S_2^2+S_3^2\leq S_0^2. The degree of polarization is :math:`P=\sqrt{S_1^2+S_2^2+S_3^2}/S_0`, so :math:`0\leq P\leq1`. Useful calibration states are horizontal linear light, :math:`(S_0,S_0,0,0)`, :math:`45^\circ` linear light, :math:`(S_0,0,S_0,0)`, and right-circular light, :math:`(S_0,0,0,S_0)` in the adopted handedness convention. If the source fluctuates appreciably during sequential readings, split the beam into six simultaneous analyser channels, or monitor its power with a reference detector and normalize every reading before taking the differences. Exercise 6.1-2 — Cascaded quarter-wave plates ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Step 1 — Definitions and setup.** Symbols are local to this item and follow the chapter convention. Each physical quantity and supplied numerical value is introduced at its first use below; angles are in radians unless a degree symbol is shown, and units are retained through numerical substitution. .. _fop-exercise-6-1-2-illustration: .. figure:: /_static/knowledge_base/worked_exercises/fundamentals_of_photonics/exercise_illustrations/exercise_06_01_02.svg :alt: Illustrated calculation map for Exercise 6.1-2, Cascaded quarter-wave plates :align: center :width: 95% **Figure 49 — Exercise 6.1-2: Cascaded quarter-wave plates.** The diagram identifies the input quantities, physical operation, requested result, variable meanings, and an independent verification route. Every symbol in the variable strip is labeled on the model itself. **Step 2 — Mathematical formulas used.** The working uses :ref:`algebraic rearrangement and dimensional checks `. **Step 3 — Worked derivation.** The calculation is kept in symbolic form until the governing relation has been rearranged for the requested quantity. :math:`\operatorname{diag}(1,j)^2=\operatorname{diag}(1,-1)`, a half-wave plate. Orthogonal fast axes give :math:`\operatorname{diag}(1,j)\operatorname{diag}(j,1)=jI`, so polarization is unchanged apart from global phase. **Step 4 — State the numbered result.** The principal result obtained in the working is .. math:: :label: fop-exercise-6-1-2-result \operatorname{diag}(1,j)\operatorname{diag}(j,1)=jI **Step 5 — Check.** Equation :eq:`fop-exercise-6-1-2-result` can be checked by substituting it back into the preceding governing relation and reversing the algebraic steps. Check that dimensions agree term by term, then test the simplest symmetry or limiting case for the expected sign and scale. Exercise 6.1-3 — Rotated polarizer ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Step 1 — Definitions and setup.** Symbols are local to this item and follow the chapter convention. Each physical quantity and supplied numerical value is introduced at its first use below; angles are in radians unless a degree symbol is shown, and units are retained through numerical substitution. .. _fop-exercise-6-1-3-illustration: .. figure:: /_static/knowledge_base/worked_exercises/fundamentals_of_photonics/exercise_illustrations/exercise_06_01_03.svg :alt: Illustrated calculation map for Exercise 6.1-3, Rotated polarizer :align: center :width: 95% **Figure 50 — Exercise 6.1-3: Rotated polarizer.** The diagram identifies the input quantities, physical operation, requested result, variable meanings, and an independent verification route. Every symbol in the variable strip is labeled on the model itself. **Step 2 — Mathematical formulas used.** The working uses :ref:`matrix multiplication and eigenvalue rules `, :ref:`trigonometric and small-angle identities `, and :ref:`algebraic rearrangement and dimensional checks `. **Step 3 — Worked derivation.** The calculation is kept in symbolic form until the governing relation has been rearranged for the requested quantity. :math:`T(\theta)=R(-\theta)\operatorname{diag}(1,0)R(\theta)` evaluates to :math:`\boxed{\begin{bmatrix}\cos^2\theta&\sin\theta\cos\theta\\ \sin\theta\cos\theta&\sin^2\theta\end{bmatrix}}`. **Step 4 — State the numbered result.** The principal result obtained in the working is .. math:: :label: fop-exercise-6-1-3-result \boxed{\begin{bmatrix}\cos^2\theta&\sin\theta\cos\theta\\ \sin\theta\cos\theta&\sin^2\theta\end{bmatrix}} **Step 5 — Check.** Equation :eq:`fop-exercise-6-1-3-result` can be checked by substituting it back into the preceding governing relation and reversing the algebraic steps. Multiply the matrices independently in the stated input-to-output order and verify that every product has compatible dimensions. Exercise 6.1-4 — Normal polarization modes ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Step 1 — Definitions and setup.** Symbols are local to this item and follow the chapter convention. Each physical quantity and supplied numerical value is introduced at its first use below; angles are in radians unless a degree symbol is shown, and units are retained through numerical substitution. .. _fop-exercise-6-1-4-illustration: .. figure:: /_static/knowledge_base/worked_exercises/fundamentals_of_photonics/exercise_illustrations/exercise_06_01_04.svg :alt: Illustrated calculation map for Exercise 6.1-4, Normal polarization modes :align: center :width: 95% **Figure 51 — Exercise 6.1-4: Normal polarization modes.** The diagram identifies the input quantities, physical operation, requested result, variable meanings, and an independent verification route. Every symbol in the variable strip is labeled on the model itself. **Step 2 — Mathematical formulas used.** The working uses :ref:`exponential, logarithmic, and phasor identities `. **Step 3 — Worked derivation.** The calculation is kept in symbolic form until the governing relation has been rearranged for the requested quantity. The polarizer eigenvectors are its pass/block linear axes, eigenvalues 1,0; the retarder eigenvectors are its fast/slow linear axes, eigenvalues :math:`1,e^{-j\Gamma}`; the rotator eigenvectors are RCP/LCP, eigenvalues :math:`e^{\mp j\theta}`. **Step 4 — Interpret the result.** The final relation or conclusion in Step 3 is the requested result. Read its sign, scale, or physical classification using the conventions fixed in Step 1. **Step 5 — Check.** Check that dimensions agree term by term, then test the simplest symmetry or limiting case for the expected sign and scale. Exercise 6.2-1 — Brewster window ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Step 1 — Definitions and setup.** Symbols are local to this item and follow the chapter convention. Each physical quantity and supplied numerical value is introduced at its first use below; angles are in radians unless a degree symbol is shown, and units are retained through numerical substitution. .. _fop-exercise-6-2-1-illustration: .. figure:: /_static/knowledge_base/worked_exercises/fundamentals_of_photonics/exercise_illustrations/exercise_06_02_01.svg :alt: Illustrated calculation map for Exercise 6.2-1, Brewster window :align: center :width: 95% **Figure 52 — Exercise 6.2-1: Brewster window.** The diagram identifies the input quantities, physical operation, requested result, variable meanings, and an independent verification route. Every symbol in the variable strip is labeled on the model itself. **Step 2 — Mathematical formulas used.** The working uses :ref:`trigonometric and small-angle identities ` and :ref:`algebraic rearrangement and dimensional checks `. **Step 3 — Worked derivation.** The calculation is kept in symbolic form until the governing relation has been rearranged for the requested quantity. :math:`\theta_B=\tan^{-1}(1.5)=\boxed{56.31^\circ}` from the normal. The internal angle is :math:`33.69^\circ`, which is the reverse-interface Brewster angle, so TM reflection vanishes at both parallel faces. **Step 4 — State the numbered result.** The principal result obtained in the working is .. math:: :label: fop-exercise-6-2-1-result \theta_B=\tan^{-1}(1.5)=\boxed{56.31^\circ} **Step 5 — Check.** Equation :eq:`fop-exercise-6-2-1-result` can be checked by substituting it back into the preceding governing relation and reversing the algebraic steps. The zero-angle or paraxial limit supplies an independent sign and magnitude check whenever that limit is part of the model. Repeat the substitution with unrounded intermediate values and retain the displayed units; the final unit must have the requested dimension. Exercise 6.2-2 — Conductive reflector ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Step 1 — Definitions and setup.** Symbols are local to this item and follow the chapter convention. Each physical quantity and supplied numerical value is introduced at its first use below; angles are in radians unless a degree symbol is shown, and units are retained through numerical substitution. .. _fop-exercise-6-2-2-illustration: .. figure:: /_static/knowledge_base/worked_exercises/fundamentals_of_photonics/exercise_illustrations/exercise_06_02_02.svg :alt: Illustrated calculation map for Exercise 6.2-2, Conductive reflector :align: center :width: 95% **Figure 53 — Exercise 6.2-2: Conductive reflector.** The diagram identifies the input quantities, physical operation, requested result, variable meanings, and an independent verification route. Every symbol in the variable strip is labeled on the model itself. **Step 2 — Mathematical formulas used.** The working uses :ref:`algebraic rearrangement and dimensional checks `. **Step 3 — Worked derivation.** The calculation is kept in symbolic form until the governing relation has been rearranged for the requested quantity. As :math:`\sigma\to\infty`, impedance tends to zero and :math:`R\to1`. The Hagen--Rubens result :math:`R\simeq1-2\sqrt{2\epsilon_0\omega/\sigma}` gives copper reflectances :math:`\boxed{0.9534}` at 1.06 micrometres and :math:`\boxed{0.9853}` at 10.6 micrometres. In the lossless sub-plasma-frequency Drude region the index is imaginary, so no net transmitted power exists and :math:`R=1`. **Step 4 — State the numbered result.** The principal result obtained in the working is .. math:: :label: fop-exercise-6-2-2-result \boxed{0.9853} **Step 5 — Check.** Equation :eq:`fop-exercise-6-2-2-result` can be checked by substituting it back into the preceding governing relation and reversing the algebraic steps. Check that dimensions agree term by term, then test the simplest symmetry or limiting case for the expected sign and scale. Exercise 6.4-1 — Optical rotatory power ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Step 1 — Definitions and setup.** Symbols are local to this item and follow the chapter convention. Each physical quantity and supplied numerical value is introduced at its first use below; angles are in radians unless a degree symbol is shown, and units are retained through numerical substitution. .. _fop-exercise-6-4-1-illustration: .. figure:: /_static/knowledge_base/worked_exercises/fundamentals_of_photonics/exercise_illustrations/exercise_06_04_01.svg :alt: Illustrated calculation map for Exercise 6.4-1, Optical rotatory power :align: center :width: 95% **Figure 54 — Exercise 6.4-1: Optical rotatory power.** The diagram identifies the input quantities, physical operation, requested result, variable meanings, and an independent verification route. Every symbol in the variable strip is labeled on the model itself. **Step 2 — Mathematical formulas used.** The working uses :ref:`algebraic rearrangement and dimensional checks `. **Step 3 — Worked derivation.** The calculation is kept in symbolic form until the governing relation has been rearranged for the requested quantity. Circular eigenindices satisfy :math:`n_\pm\simeq n\pm G/(2n)` for :math:`G\ll n`. Linear polarization is their equal superposition, so its rotation per length is half their phase difference: :math:`\boxed{\rho=(k_0/2)(n_+-n_-)\simeq k_0G/(2n)}`. End-of-chapter problems ----------------------- **Step 4 — State the numbered result.** The principal result obtained in the working is .. math:: :label: fop-exercise-6-4-1-result \boxed{\rho=(k_0/2)(n_+-n_-)\simeq k_0G/(2n)} **Step 5 — Check.** Equation :eq:`fop-exercise-6-4-1-result` can be checked by substituting it back into the preceding governing relation and reversing the algebraic steps. Check that dimensions agree term by term, then test the simplest symmetry or limiting case for the expected sign and scale. Problem 6.1-5 — Orthogonal ellipses ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** Remove the physically irrelevant common phase and write one normalized, fully polarized state as :math:`\mathbf J_1=(a,b e^{j\delta})^T`, where :math:`a,b\geq0` and :math:`a^2+b^2=1`. Jones states are orthogonal when their Hermitian inner product is zero. For a noncircular ellipse, let :math:`\psi` denote the major-axis angle and :math:`\chi` its ellipticity angle; the sign of :math:`\chi` specifies handedness. **Mathematical formulas used.** The working uses :ref:`matrix multiplication and eigenvalue rules ` and the Stokes-to-ellipse relations :math:`2\psi=\operatorname{atan2}(S_2,S_1)` and :math:`\sin 2\chi=S_3/S_0`. **Worked derivation.** A normalized vector orthogonal to :math:`\mathbf J_1` is .. math:: :label: fop-problem-6-1-5-orthogonal-vector \mathbf J_2= \begin{bmatrix}b\\-a e^{j\delta}\end{bmatrix}, \qquad \mathbf J_1^\dagger\mathbf J_2 =ab-ab=0. For :math:`\mathbf J_1`, the three polarization-dependent Stokes components are :math:`S_1=a^2-b^2`, :math:`S_2=2ab\cos\delta`, and (with the chapter's handedness convention) :math:`S_3=-2ab\sin\delta`. Substitution of :math:`\mathbf J_2` changes the sign of every one of them but leaves :math:`S_0=1` unchanged: .. math:: :label: fop-problem-6-1-5-antipodal-stokes (S_1,S_2,S_3)_2=-(S_1,S_2,S_3)_1. Consequently, :math:`\operatorname{atan2}(-S_2,-S_1)=\operatorname{atan2}(S_2,S_1)+\pi`. Halving this angle gives :math:`\psi_2=\psi_1+\pi/2` modulo :math:`\pi`, so the major axes are perpendicular. Also :math:`\sin 2\chi_2=-\sin 2\chi_1`, hence :math:`\chi_2=-\chi_1`: the two fields rotate in opposite senses. **Numbered result.** Orthogonal polarization states occupy antipodal points on the Poincaré sphere, which gives .. math:: :label: fop-problem-6-1-5-result \boxed{\psi_2=\psi_1+90^\circ\pmod{180^\circ}, \qquad \chi_2=-\chi_1.} **Check.** For :math:`\delta=0`, both states are linear and their Jones vectors point along perpendicular lines. For :math:`a=b` and :math:`|\delta|=90^\circ`, the pair is right- and left-circular; handedness is still opposite, although a circle has no unique major-axis direction. Problem 6.1-6 — Rotator under coordinate rotation ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** The chapter's passive coordinate-rotation matrix and the Jones matrix of a physical rotator through :math:`\theta` are .. math:: :label: fop-problem-6-1-6-matrices R(\alpha)= \begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}, \qquad T(\theta)= \begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta& \cos\theta\end{bmatrix}=R(-\theta). Rotate the coordinate axes through an arbitrary angle :math:`\alpha`; this changes the matrix representation to :math:`T'=R(\alpha)TR(-\alpha)`. **Mathematical formulas used.** The working uses :ref:`matrix multiplication and eigenvalue rules ` and :math:`R(a)R(b)=R(a+b)`. **Worked derivation.** Two-dimensional rotations commute, so .. math:: :label: fop-problem-6-1-6-transformation \begin{aligned} T'&=R(\alpha)R(-\theta)R(-\alpha)\\ &=R(\alpha-\theta-\alpha)\\ &=R(-\theta)=T. \end{aligned} The cancellation holds for every :math:`\alpha`; no direction in the transverse plane is preferred by an ideal optical rotator. **Numbered result.** .. math:: :label: fop-problem-6-1-6-result \boxed{R(\alpha)T(\theta)R(-\alpha)=T(\theta).} **Check.** Direct multiplication gives the same four matrix elements. Setting :math:`\alpha=90^\circ` also leaves :math:`T` unchanged, whereas the matrix of a fixed-axis retarder generally changes under that operation. Problem 6.1-7 — Half-wave plate ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** Take the fast axis as :math:`x`. Apart from a common phase, an ideal half-wave retarder has :math:`H_0=\operatorname{diag}(1,-1)`. The input is linearly polarized at :math:`\theta` to that axis, so :math:`\mathbf J_{\rm in}=(\cos\theta,\sin\theta)^T`. **Mathematical formulas used.** The working uses :ref:`trigonometric and small-angle identities ` and the rotated-device rule :math:`H(\beta)=R(-\beta)H_0R(\beta)`. **Worked derivation.** With the plate axes unrotated, .. math:: :label: fop-problem-6-1-7-axis-aligned \mathbf J_{\rm out}=H_0\mathbf J_{\rm in} =\begin{bmatrix}\cos\theta\\-\sin\theta\end{bmatrix} =\mathbf J_{\rm lin}(-\theta). Thus the output polarization lies at :math:`-\theta`; relative to the input it has turned through :math:`-2\theta` (a magnitude :math:`2\theta`, toward and through the fast axis). More generally, if the fast axis is at :math:`\beta`, application of the rotated matrix gives .. math:: :label: fop-problem-6-1-7-rotated-plate H(\beta)\mathbf J_{\rm lin}(\theta) =\mathbf J_{\rm lin}(2\beta-\theta). A true rotator would instead produce :math:`\mathbf J_{\rm lin}(\theta+\gamma)` with one fixed :math:`\gamma` for every input angle. The half-wave plate's change, :math:`2(\beta-\theta)`, depends on the input orientation. It also reverses the handedness of circular or elliptical light, which an ideal rotator does not. **Numbered result.** .. math:: :label: fop-problem-6-1-7-result \boxed{\theta_{\rm out}=-\theta, \qquad \Delta\theta=-2\theta} for a fast axis along :math:`x`. **Check.** Input along either plate axis remains on that axis. For :math:`\theta=45^\circ`, the output is at :math:`-45^\circ`, a :math:`90^\circ` turn in magnitude, as expected for a half-wave plate. Problem 6.1-8 — Three retarders ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** A retarder with fast axis :math:`x` and phase delay :math:`\Gamma` has :math:`W_x(\Gamma)=\operatorname{diag}(1,e^{-j\Gamma})`. The light first meets (a), then (b), then (c), so the system matrix is :math:`T_{abc}=T_cT_bT_a`; the first element encountered stands at the right. **Mathematical formulas used.** The working uses :ref:`matrix multiplication and eigenvalue rules ` and the coordinate-rotation matrix defined in Problem 6.1-6. **Worked derivation.** The three requested Jones matrices are .. math:: :label: fop-problem-6-1-8-element-matrices Q_x=\begin{bmatrix}1&0\\0&-j\end{bmatrix},\qquad H_{45}=R(-45^\circ) \begin{bmatrix}1&0\\0&-1\end{bmatrix}R(45^\circ) =\begin{bmatrix}0&1\\1&0\end{bmatrix},\qquad Q_y=\begin{bmatrix}-j&0\\0&1\end{bmatrix}. Multiplying in propagation order gives .. math:: :label: fop-problem-6-1-8-forward-product T_{abc}=Q_yH_{45}Q_x =\begin{bmatrix}0&-1\\1&0\end{bmatrix} =T(+90^\circ). It sends :math:`x` polarization to :math:`y` and :math:`y` polarization to :math:`-x`, precisely a :math:`+90^\circ` rotation. If the physical order is reversed, the matrix order is also reversed: .. math:: :label: fop-problem-6-1-8-reverse-product T_{cba}=Q_xH_{45}Q_y =\begin{bmatrix}0&1\\-1&0\end{bmatrix} =T(-90^\circ). Changing the sign convention for retarder phase can multiply intermediate matrices by common phases, but it does not change either observable rotation. **Numbered result.** .. math:: :label: fop-problem-6-1-8-result \boxed{T_{abc}=T(+90^\circ),\qquad T_{cba}=T(-90^\circ).} **Check.** Both products are unitary with determinant :math:`+1`. Their product is the identity, confirming that reversing this noncommuting sequence reverses the rotation rather than reproducing it. Problem 6.1-9 — Circular polarization at reflection ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** Consider normal reflection from an ideal isotropic mirror. Let the incident propagation basis :math:`(\hat{\mathbf x}_i,\hat{\mathbf y}_i,\hat{\mathbf z}_i)` be right-handed, with :math:`\hat{\mathbf z}_i` along the incident wavevector. After reflection :math:`\hat{\mathbf z}_r=-\hat{\mathbf z}_i`. A convenient right-handed reflected basis is therefore :math:`\hat{\mathbf x}_r=\hat{\mathbf x}_i` and :math:`\hat{\mathbf y}_r=-\hat{\mathbf y}_i`. **Mathematical formulas used.** The working uses the Jones vectors :math:`\mathbf e_R=(1,j)^T/\sqrt2` and :math:`\mathbf e_L=(1,-j)^T/\sqrt2` in each local right-handed propagation basis. Reversing the viewing direction is essential: handedness is defined relative to the direction of propagation, not to fixed laboratory axes. **Worked derivation.** At an ideal mirror both tangential laboratory-field components acquire the same reflection phase. Taking that phase as :math:`-1`, the laboratory components become :math:`(-A_x,-A_y)`. Expressing the reflected field in its local basis flips the sign of its :math:`y` coordinate once more, hence .. math:: :label: fop-problem-6-1-9-reflection-map \begin{bmatrix}A_x\\A_y\end{bmatrix}_{i} \longmapsto \begin{bmatrix}-A_x\\A_y\end{bmatrix}_{r}. Apply this map to the two circular states: .. math:: :label: fop-problem-6-1-9-circular-map \mathbf e_R\longmapsto \frac{1}{\sqrt2}\begin{bmatrix}-1\\j\end{bmatrix} \doteq\mathbf e_L, \qquad \mathbf e_L\longmapsto \frac{1}{\sqrt2}\begin{bmatrix}-1\\-j\end{bmatrix} \doteq\mathbf e_R, where :math:`\doteq` means equality apart from a physically irrelevant common phase. **Numbered result.** .. math:: :label: fop-problem-6-1-9-result \boxed{R\ \overset{\text{mirror reflection}}{\longleftrightarrow}\ L.} **Check.** Two successive mirror reflections restore the original propagation direction and handedness. For a real mirror at oblique incidence, unequal TE and TM phases can additionally make the state elliptical; the pure handedness swap above is the ideal normal-incidence result asked for here. Problem 6.1-10 — Anti-glare screen ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** Place, from the observer toward the window, a linear polarizer followed by a quarter-wave retarder. Let the polarizer pass :math:`x` polarization and set either principal axis of the retarder at :math:`45^\circ` to :math:`x`. Background light travels from the observer's side toward the window, reflects, and returns through the same elements. Light from the self-luminous object makes only the final one-way passage toward the observer. **Mathematical formulas used.** A :math:`45^\circ` quarter-wave plate converts linear light into circular light on the outward pass. Problem 6.1-9 shows that mirror reflection reverses circular handedness. On the return pass, the same reciprocal plate converts that reversed circular state into the linear state orthogonal to the original one. **Worked derivation.** 1. The first polarizer converts arbitrary background light to linear :math:`x` polarization. 2. The quarter-wave plate resolves that field equally along its fast and slow axes and adds a :math:`90^\circ` relative phase, producing circular light. 3. Reflection at the glass reverses its handedness. 4. A second passage through the same retarder adds the complementary quarter-wave delay. The returning field is therefore linear along :math:`y`, rather than :math:`x`. 5. The polarizer rejects this returned :math:`y` component, removing the specular background glare. Object light approaching from behind the window is not first prepared by the polarizer; one component survives its single pass and remains visible, although attenuated. Equivalently, the double passage through the retarder, together with the handedness reversal, acts as a half-wave transformation between the outward and return passes: .. math:: :label: fop-problem-6-1-10-round-trip \mathbf e_x\ \longrightarrow\ \mathbf e_R \ \xrightarrow{\rm reflection}\ \mathbf e_L \ \longrightarrow\ \mathbf e_y, \qquad P_x\mathbf e_y=0. **Numbered result.** .. math:: :label: fop-problem-6-1-10-result \boxed{\text{linear polarizer} \;\rightarrow\;45^\circ\text{ quarter-wave plate} \;\rightarrow\;\text{window}.} **Check.** Without the quarter-wave plate, the reflected :math:`x` state would pass back through the polarizer, so the retarder's role is essential. The screen is **not** an optical isolator: all of its elements are reciprocal, it suppresses only the prepared reflection path, and it attenuates desired unpolarized object light. A true isolator requires a nonreciprocal element such as a Faraday rotator. Problem 6.2-3 — Fresnel TE coefficient ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** A TE-polarized plane wave is incident from medium 1 onto medium 2. The media have characteristic impedances :math:`\eta_1,\eta_2`, refractive indices :math:`n_1,n_2`, and propagation angles :math:`\theta_1,\theta_2` measured from the interface normal. Normalize the incident electric amplitude to one and write the reflected and transmitted amplitudes as :math:`r_s` and :math:`t_s`. **Mathematical formulas used.** Apply continuity of tangential :math:`\mathbf E` and :math:`\mathbf H`, Snell's law, and :math:`\eta_i=\eta_0/n_i` for nonmagnetic lossless dielectrics. The algebra uses :ref:`trigonometric and small-angle identities `. **Worked derivation.** For TE polarization the electric field is wholly tangential. The magnetic field has tangential magnitude :math:`E\cos\theta/\eta`; its sign reverses for the reflected wave. The two boundary equations are therefore .. math:: :label: fop-problem-6-2-3-boundary-conditions 1+r_s=t_s, \qquad \frac{\cos\theta_1}{\eta_1}(1-r_s) =\frac{\cos\theta_2}{\eta_2}t_s. Insert :math:`t_s=1+r_s` into the magnetic-field equation and collect the terms in :math:`r_s`: .. math:: :label: fop-problem-6-2-3-impedance-result r_s= \frac{\eta_2\sec\theta_2-\eta_1\sec\theta_1} {\eta_2\sec\theta_2+\eta_1\sec\theta_1}, \qquad t_s=1+r_s. This is the requested reflection relation, Eq. (6.2-6). For nonmagnetic dielectrics, substitute :math:`\eta_i=\eta_0/n_i` and cancel the common factors to obtain the TE Fresnel coefficient. **Numbered result.** .. math:: :label: fop-problem-6-2-3-result \boxed{r_s= \frac{n_1\cos\theta_1-n_2\cos\theta_2} {n_1\cos\theta_1+n_2\cos\theta_2}.} For a finite beam, first Fourier-decompose its transverse field into an angular spectrum of plane waves. For each transverse wavevector, define its own plane of incidence, resolve the field into TE and TM components, multiply by the corresponding Fresnel coefficient, include the reflected propagation phase, and inverse-transform the spectrum. Replacing the whole beam by the coefficient at its central angle is only a narrow-angle approximation and misses effects such as the Goos--Hänchen shift. **Check.** At normal incidence this becomes :math:`r_s=(n_1-n_2)/(n_1+n_2)`. If :math:`n_1=n_2`, Snell's law gives :math:`\theta_1=\theta_2` and the coefficient vanishes, as it must when there is no optical discontinuity. Problem 6.2-4 — Glass at 45 degrees ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** Light travels from air, :math:`n_1=1`, into glass, :math:`n_2=1.5`, at :math:`\theta_1=45^\circ`. For lossless media the power reflectance is the squared magnitude of the electric-field reflection coefficient: :math:`R_s=|r_s|^2` and :math:`R_p=|r_p|^2`. **Mathematical formulas used.** Use Snell's law and the TE/TM Fresnel coefficients. For the TM magnitude it is convenient to use :math:`r_p=(n_2\cos\theta_1-n_1\cos\theta_2)/ (n_2\cos\theta_1+n_1\cos\theta_2)`; a different reflected-axis convention may reverse its sign but not its power reflectance. **Worked derivation.** First find the transmitted angle: .. math:: :label: fop-problem-6-2-4-snell \theta_2=\sin^{-1}\!\left(\frac{n_1}{n_2}\sin45^\circ\right) =\sin^{-1}(0.4714045)=28.1255^\circ. Now substitute :math:`\cos45^\circ=0.7071068` and :math:`\cos28.1255^\circ=0.8819171`: .. math:: :label: fop-problem-6-2-4-amplitudes \begin{aligned} r_s&=\frac{1(0.7071068)-1.5(0.8819171)} {1(0.7071068)+1.5(0.8819171)}=-0.303337,\\ r_p&=\frac{1.5(0.7071068)-1(0.8819171)} {1.5(0.7071068)+1(0.8819171)}=0.0920134. \end{aligned} Squaring gives :math:`R_s=0.092013` and :math:`R_p=0.0084665`. Unpolarized light carries equal mean power in the two orthogonal modes, so its reflectance is their arithmetic mean. **Numbered result.** .. math:: :label: fop-problem-6-2-4-result \boxed{R_{\rm TE}=9.201\%,\qquad R_{\rm TM}=0.8467\%,\qquad R_{\rm unpol}=\frac{R_{\rm TE}+R_{\rm TM}}{2}=5.024\%.} **Check.** The TM reflectance is small because :math:`45^\circ` is fairly close to the air--glass Brewster angle :math:`\tan^{-1}(1.5)=56.31^\circ`. Both values lie between zero and one, and the unpolarized result lies exactly between them. Problem 6.2-5 — Brewster geometry ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** At the Brewster angle the TM reflection coefficient vanishes. Its numerator gives the supplied condition :math:`n_1\sec\theta_1=n_2\sec\theta_2`; phase matching also requires :math:`n_1\sin\theta_1=n_2\sin\theta_2`. **Mathematical formulas used.** The working uses Snell's law and :ref:`trigonometric and small-angle identities `. **Worked derivation.** Rewrite the zero-reflection condition as :math:`n_1\cos\theta_2=n_2\cos\theta_1`. Together with Snell's law this gives .. math:: :label: fop-problem-6-2-5-ratio \frac{\sin\theta_1}{\sin\theta_2} =\frac{n_2}{n_1} =\frac{\cos\theta_2}{\cos\theta_1}. Therefore :math:`\sin\theta_1\cos\theta_1 =\sin\theta_2\cos\theta_2`, or :math:`\sin2\theta_1=\sin2\theta_2`. For refraction into a different medium, the nontrivial physical solution is :math:`2\theta_1=\pi-2\theta_2`, hence :math:`\theta_1+\theta_2=90^\circ`. Put :math:`\sin\theta_2=\cos\theta_1` into Snell's law: .. math:: :label: fop-problem-6-2-5-brewster n_1\sin\theta_B=n_2\cos\theta_B \quad\Longrightarrow\quad \tan\theta_B=\frac{n_2}{n_1}. Since the reflected angle equals :math:`\theta_B`, the reflected and refracted rays differ by :math:`180^\circ-(\theta_B+\theta_2)=90^\circ`. **Numbered result.** .. math:: :label: fop-problem-6-2-5-result \boxed{\theta_B=\tan^{-1}\!\left(\frac{n_2}{n_1}\right), \qquad \theta_B+\theta_2=90^\circ.} The transmitted TM electric field lies in the plane of incidence and is perpendicular to its propagation direction. It is consequently parallel to the reflected-ray direction. In the dipole-scattering picture, a dipole cannot radiate along its own oscillation axis, explaining the missing TM reflection. **Check.** For air to glass, the formula gives :math:`56.31^\circ` and Snell's law gives :math:`33.69^\circ`; their sum is :math:`90^\circ` and direct substitution makes the TM Fresnel numerator zero. Problem 6.2-6 — TIR retardance ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** Reflection is from glass :math:`n_1=1.5` into air :math:`n_2=1`. Define :math:`m=n_2/n_1=2/3` and measure phase as in :math:`r_{s,p}=e^{j\phi_{s,p}}`. The requested retardance of TM relative to TE is :math:`\Gamma=\phi_p-\phi_s`. **Mathematical formulas used.** Above the critical angle, define :math:`q=\sqrt{\sin^2\theta-m^2}`. The unit-magnitude Fresnel coefficients have phases :math:`\phi_s=-2\tan^{-1}(q/\cos\theta)` and :math:`\phi_p=-2\tan^{-1}[q/(m^2\cos\theta)]` in this phase convention. **Worked derivation.** The critical and specified angles are .. math:: :label: fop-problem-6-2-6-angles \theta_c=\sin^{-1}\!\left(\frac{1}{1.5}\right)=41.8103^\circ, \qquad \theta=1.2\theta_c=50.1724^\circ. At this angle, :math:`q=\sqrt{\sin^2(50.1724^\circ)-(2/3)^2}`. Substitution into the two phase expressions gives .. math:: :label: fop-problem-6-2-6-phases \phi_s=-61.52^\circ, \qquad \phi_p=-106.50^\circ, \qquad \Gamma=\phi_p-\phi_s=-44.98^\circ. The negative sign says that, under the chosen phasor convention, the TM phase lags the TE phase by :math:`44.98^\circ`. Interchanging the definition of retardance or the time-harmonic convention reverses the sign but not the physical phase difference. **Numbered result.** .. math:: :label: fop-problem-6-2-6-result \boxed{|\Gamma|=44.98^\circ\ \text{per reflection}.} **Check.** Both reflection magnitudes equal one, so TIR changes phase but not power. At :math:`\theta=\theta_c`, :math:`q=0` and both phases vanish; at grazing incidence both approach the same limiting phase, so their difference again tends to zero. Problem 6.2-7 — Goos--Hänchen shift ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** Two equal-amplitude TE plane waves in the denser medium have incidence angles :math:`\theta` and :math:`\theta+d\theta`. Let :math:`x` run along the interface and let :math:`k` be their common wavenumber. Under TIR the reflection coefficient is :math:`r_s=e^{j\phi_s(\theta)}`; write :math:`d\phi_s=\dot\phi_s\,d\theta`. **Mathematical formulas used.** With :math:`m=n_2/n_1=\sin\theta_c` and :math:`q=\sqrt{\sin^2\theta-m^2}`, use the TE phase from Problem 6.2-6, :math:`\phi_s=-2\tan^{-1}(q/\cos\theta)`. The transverse wavevector is :math:`k_x=k\sin\theta`, so :math:`dk_x=k\cos\theta\,d\theta`. **Worked derivation.** Differentiate the TIR phase. Since :math:`dq/d\theta=\sin\theta\cos\theta/q`, the chain rule simplifies to .. math:: :label: fop-problem-6-2-7-phase-slope \dot\phi_s=\frac{d\phi_s}{d\theta} =-\frac{2\sin\theta} {\sqrt{\sin^2\theta-\sin^2\theta_c}}. At the interface, the two incident waves form fringes proportional to :math:`\cos^2(dk_x x/2)`. Reflection adds :math:`d\phi_s` to the phase difference, so the reflected fringes are proportional to :math:`\cos^2[(dk_xx+d\phi_s)/2]`. A reflected maximum therefore occurs where the corresponding incident maximum would occur after a translation :math:`\Delta` satisfying .. math:: :label: fop-problem-6-2-7-fringe-translation dk_x\Delta+d\phi_s=0 \quad\Longrightarrow\quad \Delta=-\frac{1}{k\cos\theta}\frac{d\phi_s}{d\theta}. A finite beam is a continuous superposition of just such neighboring angular components. The same spectral phase slope therefore displaces its reflected envelope—the Goos--Hänchen effect. **Numbered result.** In the present sign convention, .. math:: :label: fop-problem-6-2-7-result \boxed{\Delta= \frac{2\tan\theta} {k\sqrt{\sin^2\theta-\sin^2\theta_c}}.} Reversing the positive :math:`x` direction or phase convention reverses the reported sign; the measurable magnitude is unchanged. **Check.** The result has dimensions :math:`1/k`, hence length. The two-plane-wave expression grows near the critical angle because the reflection phase changes rapidly there; a real beam's finite angular width regularizes that idealized divergence. Problem 6.2-8 — Absorbing-medium reflection ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** Light is normally incident from free space onto a nonmagnetic absorbing medium of real refractive index :math:`n` and intensity absorption coefficient :math:`\alpha`. Define .. math:: :label: fop-problem-6-2-8-complex-index \widetilde n=n-j\kappa, \qquad \kappa=\frac{\alpha c_0}{2\omega}, \qquad \widetilde\eta=\frac{\eta_0}{\widetilde n}. The factor two occurs because intensity decays as :math:`e^{-\alpha z}` while field amplitude decays as :math:`e^{-\alpha z/2}`. **Mathematical formulas used.** Apply Maxwell's normal-incidence relation :math:`H=E/\eta` and continuity of tangential :math:`E` and :math:`H` at the boundary. **Worked derivation.** In one common fixed-laboratory-axis convention, normalizing the incident field gives :math:`1+r_E=t` and :math:`1-r_E=\widetilde n t`. Solving, .. math:: :label: fop-problem-6-2-8-laboratory-coefficient r_E=\frac{1-\widetilde n}{1+\widetilde n}. The problem defines the reflected Jones component along the reflected wave's local transverse axis, which is opposite to this fixed-axis amplitude. Thus :math:`r=-r_E`, and insertion of :math:`\widetilde n` gives the requested form. **Numbered result.** .. math:: :label: fop-problem-6-2-8-result \boxed{r= \frac{(n-j\alpha c_0/2\omega)-1} {(n-j\alpha c_0/2\omega)+1}.} The convention-independent power reflectance is .. math:: :label: fop-problem-6-2-8-power R=|r|^2= \frac{(n-1)^2+\kappa^2}{(n+1)^2+\kappa^2}. **Check.** When :math:`\alpha=0`, this reduces in magnitude to the ordinary normal-incidence Fresnel result. When :math:`n=1` and :math:`\alpha=0`, the interface disappears and :math:`r=0`; as absorption becomes very large, :math:`R\rightarrow1`. Problem 6.3-1 — Quartz retardation ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** Quartz is positive uniaxial with :math:`n_e=1.553`, :math:`n_o=1.544`, and vacuum wavelength :math:`\lambda_0=633\ \mathrm{nm}`. Retardation is largest when the wavevector is perpendicular to the optic axis, because the two eigenmodes then sample :math:`n_e` and :math:`n_o` directly. **Mathematical formulas used.** The relative phase accumulated through thickness :math:`d` is :math:`\Gamma=k_0(n_e-n_o)d=2\pi(n_e-n_o)d/\lambda_0`. **Worked derivation.** The maximum birefringence is :math:`\Delta n=n_e-n_o=0.009`. For :math:`d=1\ \mathrm{mm}`, .. math:: :label: fop-problem-6-3-1-per-millimetre \Gamma_{1\,\mathrm{mm}} =2\pi\frac{0.009(1.000\times10^{-3})}{633\times10^{-9}} =89.33\ \mathrm{rad} =2\pi(14.218). Thus one millimetre produces 14.218 full retardation cycles. A quarter-wave retarder needs an odd multiple of :math:`\pi/2`: .. math:: :label: fop-problem-6-3-1-quarter-wave-condition \frac{2\pi\Delta n d}{\lambda_0} =\frac{(2q+1)\pi}{2}, \qquad q=0,1,2,\ldots Solving gives all possible thicknesses. Values with even :math:`q` have retardance :math:`+\pi/2` modulo :math:`2\pi`; odd :math:`q` give the complementary :math:`-\pi/2` action. **Numbered result.** .. math:: :label: fop-problem-6-3-1-result \boxed{d_q=\frac{(2q+1)\lambda_0}{4\Delta n} =(2q+1)(17.58\ \mathrm{\mu m}), \qquad q=0,1,2,\ldots} The thinnest plate is therefore :math:`17.58\ \mathrm{\mu m}`; the next is :math:`52.75\ \mathrm{\mu m}`. **Check.** Substituting the first thickness gives :math:`\Delta n d/\lambda_0=1/4`, hence :math:`\Gamma=\pi/2`. The expression has units of length and every increase by :math:`\lambda_0/(2\Delta n)` adds exactly :math:`\pi` of retardance. Problem 6.3-2 — Maximum extraordinary walk-off ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** Let :math:`\theta` be the angle from the optic axis to the extraordinary wavevector :math:`\mathbf k`, and :math:`\phi` the angle from that axis to the Poynting vector :math:`\mathbf S`. Their difference :math:`\rho=\theta-\phi` is the extraordinary ray walk-off. For quartz, :math:`n_e=1.553` and :math:`n_o=1.544`. **Mathematical formulas used.** The normal to the extraordinary :math:`k`-surface gives .. math:: :label: fop-problem-6-3-2-ray-wave-relation \tan\phi=r\tan\theta, \qquad r=\left(\frac{n_o}{n_e}\right)^2. Maximize :math:`\rho` using the :ref:`stationary-value condition `. **Worked derivation.** Differentiate :math:`\rho(\theta)=\theta-\tan^{-1}(r\tan\theta)`: .. math:: :label: fop-problem-6-3-2-stationary-condition \frac{d\rho}{d\theta} =1-\frac{r\sec^2\theta}{1+r^2\tan^2\theta}=0. Let :math:`u=\tan\theta`. Rearrangement gives :math:`1+r^2u^2=r(1+u^2)`, and because :math:`r\ne1`, :math:`u^2=1/r`. In the first quadrant, .. math:: :label: fop-problem-6-3-2-optimum-angle \tan\theta_{\max}=\frac{1}{\sqrt r}=\frac{n_e}{n_o}, \qquad \tan\phi_{\max}=\sqrt r=\frac{n_o}{n_e}. The two tangents are reciprocals, so :math:`\phi_{\max}=90^\circ-\theta_{\max}`. Numerically, :math:`\theta_{\max}=45.1665^\circ`, :math:`\phi_{\max}=44.8335^\circ`, and their difference is :math:`0.3330^\circ`. **Numbered result.** .. math:: :label: fop-problem-6-3-2-result \boxed{\theta_{\max}=45.1665^\circ\ \text{from the optic axis}, \qquad \rho_{\max}=0.3330^\circ.} **Check.** At :math:`\theta=0^\circ` or :math:`90^\circ`, symmetry forces :math:`\mathbf S\parallel\mathbf k`, so the walk-off vanishes. If :math:`n_e=n_o`, then :math:`r=1` and it vanishes at every angle, as required for an isotropic medium. Problem 6.3-3 — Double refraction in quartz ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** An unpolarized wave in air strikes quartz at :math:`\theta_i=30^\circ`. Quartz has :math:`n_o=1.544` and :math:`n_e=1.553`. The optic axis lies in the incidence plane and is perpendicular to the incident wavevector; because an optic axis is an unoriented line, take it at :math:`-60^\circ` from the inward surface normal. Angles :math:`\theta_o` and :math:`\theta_e` below are wavevector angles from that normal. **Mathematical formulas used.** Tangential phase matching requires :math:`n\sin\theta=\sin\theta_i`. For the extraordinary wave, whose wavevector is at :math:`\gamma=\theta_e+60^\circ` to the optic axis, use .. math:: :label: fop-problem-6-3-3-extraordinary-index \frac{1}{n^2(\gamma)} =\frac{\cos^2\gamma}{n_o^2} +\frac{\sin^2\gamma}{n_e^2}. The extraordinary ray direction :math:`\phi` measured from the optic axis satisfies :math:`\tan\phi=(n_o/n_e)^2\tan\gamma`. **Worked derivation.** The ordinary refractive index is direction independent, hence .. math:: :label: fop-problem-6-3-3-ordinary-angle \theta_o=\sin^{-1}\!\left(\frac{\sin30^\circ}{1.544}\right) =18.89496^\circ. Its spherical :math:`k`-surface makes the ordinary Poynting vector parallel to its wavevector, so the ordinary wave and ray have this same direction. For the extraordinary component, :math:`n` depends on the unknown direction; solve the single scalar equation .. math:: :label: fop-problem-6-3-3-extraordinary-phase-match n(\theta_e+60^\circ)\sin\theta_e=sin30^\circ. Iteration gives :math:`\theta_e=18.78566^\circ`, :math:`\gamma=78.78566^\circ`, and :math:`n(\gamma)=1.552657`. The normal to the extraordinary :math:`k`-surface then gives :math:`\phi=78.65792^\circ` from the optic axis. Since the axis is at :math:`-60^\circ`, the extraordinary ray is at :math:`78.65792^\circ-60^\circ=18.65792^\circ` from the surface normal. **Numbered result.** .. math:: :label: fop-problem-6-3-3-result \boxed{\begin{array}{c|cc} &\text{wavevector from normal}&\text{ray from normal}\\ \hline \text{ordinary}&18.89496^\circ&18.89496^\circ\\ \text{extraordinary}&18.78566^\circ&18.65792^\circ \end{array}} **Check.** Substitution gives :math:`1.544\sin18.89496^\circ=0.5` and :math:`1.552657\sin18.78566^\circ=0.5`, confirming tangential phase matching. The extraordinary ray differs from its wavevector by :math:`0.12774^\circ`, whereas the ordinary ray has zero walk-off. Problem 6.3-4 — Geometry for largest separation ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** Let a parallel-sided plate of a positive uniaxial crystal have thickness :math:`L`, ordinary index :math:`n_o`, and principal extraordinary index :math:`n_e>n_o`. Send the light normally into the plate, so both transmitted wavevectors are normal to its faces. Put the optic axis in the plane in which the beam separation is wanted, and let :math:`\theta` be the angle from that axis to the common wavevector. The ordinary ray is parallel to its wavevector. The extraordinary ray makes an angle :math:`\phi` with the optic axis and therefore walks away from the ordinary ray by :math:`\rho=\theta-\phi`. **Mathematical formulas used.** For a positive uniaxial crystal, the extraordinary ray and wave-normal angles obey .. math:: :label: fop-problem-6-3-4-ray-angle \tan\phi=\left(\frac{n_o}{n_e}\right)^2\tan\theta . Problem 6.3-2 showed that the stationary (and maximum) walk-off occurs when :math:`\tan\theta=n_e/n_o`. A ray crossing a distance :math:`L` at angle :math:`\rho` to the face normal acquires lateral displacement :math:`\Delta x=L\tan\rho`. **Worked derivation.** At the optimum orientation, .. math:: :label: fop-problem-6-3-4-optimum-angles \theta_* = \tan^{-1}\!\left(\frac{n_e}{n_o}\right), \qquad \phi_* = \tan^{-1}\!\left(\frac{n_o}{n_e}\right) =90^\circ-\theta_* . Consequently, the largest ray--wavevector angle is .. math:: :label: fop-problem-6-3-4-max-walkoff \rho_{\max}=\theta_*-\phi_* =2\tan^{-1}\!\left(\frac{n_e}{n_o}\right)-90^\circ . The ordinary ray travels straight through the plate. The extraordinary ray tilts by :math:`\rho_{\max}` *toward the optic axis*, so the two spots on the second face are separated by :math:`L\tan\rho_{\max}`. At that parallel exit face both wavevectors have zero tangential component; after refraction into air the two output beams are again normal to the faces and parallel to one another. The separation produced inside the plate remains. **Numbered result.** The required cut and maximum lateral separation are .. math:: :label: fop-problem-6-3-4-result \boxed{\begin{gathered} \text{optic-axis angle to the plate normal:}\quad \theta_*=\tan^{-1}(n_e/n_o),\\ \Delta x_{\max}=L\tan\!\left[ 2\tan^{-1}(n_e/n_o)-90^\circ\right]. \end{gathered}} **Check.** If :math:`n_e=n_o`, the material becomes isotropic; :math:`\theta_*=45^\circ`, :math:`\rho_{\max}=0`, and the separation vanishes. The result also scales linearly with :math:`L`, as a geometrical displacement must. Choosing the optic axis either parallel or perpendicular to the wavevector gives zero walk-off, confirming that the optimum lies between those orientations. Problem 6.3-5 — One-centimetre LiNbO3 plate ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ **Definitions and setup.** The He--Ne wavelength is :math:`\lambda_0=633\ \mathrm{nm}`. The LiNbO\ :sub:`3` plate has :math:`n_o=2.20`, :math:`n_e=2.29`, and thickness :math:`L=1.00\ \mathrm{cm}`. Its optic axis is at :math:`\theta=45^\circ` to the plate normal. Normal incidence makes the ordinary and extraordinary *wavevectors* normal to the faces, but the extraordinary energy ray walks toward the optic axis. Unpolarized input is resolved into incoherent ordinary and extraordinary components; the same geometry and phase difference apply to a coherent input containing both components. **Mathematical formulas used.** For a wavevector at angle :math:`\theta` to the optic axis, the extraordinary phase index and ray angle are .. math:: :label: fop-problem-6-3-5-formulas \frac{1}{n^2(\theta)}= \frac{\cos^2\theta}{n_o^2}+\frac{\sin^2\theta}{n_e^2}, \qquad \tan\phi=\left(\frac{n_o}{n_e}\right)^2\tan\theta . Thus the walk-off is :math:`\rho=\theta-\phi`, the lateral separation at the exit face is :math:`\Delta x=L\tan\rho`, and the relative phase accumulated between the extraordinary and ordinary waves is .. math:: :label: fop-problem-6-3-5-retardance-formula \Gamma=\frac{2\pi L}{\lambda_0}\,[n(\theta)-n_o]. **Worked derivation.** First calculate the phase index without prematurely rounding it: .. math:: :label: fop-problem-6-3-5-index n(45^\circ)= \left[\frac{1/2}{(2.20)^2}+\frac{1/2}{(2.29)^2}\right]^{-1/2} =2.2436473. The extraordinary ray direction and walk-off are then .. math:: :label: fop-problem-6-3-5-walkoff \phi=\tan^{-1}\!\left[\left(\frac{2.20}{2.29}\right)^2\right] =42.7052^\circ, \qquad \rho=45^\circ-\phi=2.29479^\circ. It follows that .. math:: :label: fop-problem-6-3-5-shift \Delta x=(10.0\ \mathrm{mm})\tan(2.29479^\circ) =0.40073\ \mathrm{mm}. For the retardation, .. math:: :label: fop-problem-6-3-5-retardance \frac{\Gamma}{2\pi} =\frac{(2.2436473-2.20)(0.0100\ \mathrm{m})} {633\times10^{-9}\ \mathrm{m}} =689.53097. The integer 689 represents complete cycles. The observable residual is :math:`0.53097` cycle, or :math:`191.15^\circ`, although the unwrapped phase is :math:`\Gamma=4332.45\ \mathrm{rad}`. **Numbered result.** At the output face, .. math:: :label: fop-problem-6-3-5-result \boxed{\Delta x=0.4007\ \mathrm{mm},\qquad \Gamma=2\pi(689.531)=4332.45\ \mathrm{rad} \equiv191.15^\circ\pmod{360^\circ}.} **Check.** The calculated index satisfies :math:`n_o