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
Brief solution
1. Method. For a linear analyser at angle \(\theta\), the transmitted intensity is
2. Key step.
Reading |
Optics before the detector |
Information supplied |
|---|---|---|
\(I_H\) |
Linear polarizer at \(0^\circ\) |
\(I_H=(S_0+S_1)/2\) |
\(I_V\) |
Linear polarizer at \(90^\circ\) |
\(I_V=(S_0-S_1)/2\) |
\(I_D\) |
Linear polarizer at \(45^\circ\) |
\(I_D=(S_0+S_2)/2\) |
\(I_A\) |
Linear polarizer at \(135^\circ\) |
\(I_A=(S_0-S_2)/2\) |
\(I_R\) |
Quarter-wave plate and polarizer set as a right-circular analyser |
\(I_R=(S_0+S_3)/2\) |
\(I_L\) |
Quarter-wave plate and polarizer set as a left-circular analyser |
\(I_L=(S_0-S_3)/2\) |
Ideally, \(I_H+I_V=I_D+I_A=I_R+I_L\). In a real experiment, use the average of these three sums for a lower-noise estimate of \(S_0\), after correcting the channels for detector gain and optical throughput.
3. Answer.
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 \(I_H\), \(I_V\), \(I_D\), and \(I_R\). Then use \(S_0=I_H+I_V\), \(S_1=I_H-I_V\), \(S_2=2I_D-S_0\), and \(S_3=2I_R-S_0\). The six-reading method is normally preferable because each component is formed from a balanced difference.
Show detailed steps
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 \(I_H\), \(I_V\), \(I_D\), and \(I_A\) the intensities passed by linear analysers at \(0^\circ\), \(90^\circ\), \(45^\circ\), and \(135^\circ\), respectively. Denote by \(I_R\) and \(I_L\) 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 \(+45^\circ\) or \(-45^\circ\) relative to the polarizer axis. Which setting is called right-handed depends on the viewing and time convention; label it so that \(S_3=I_R-I_L\), as in this chapter.
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 \(\theta\), the transmitted intensity is
The two circular-analyser readings are
A rotating linear polarizer alone can therefore determine \(S_0\), \(S_1\), and \(S_2\), but not \(S_3\): 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.
Reading |
Optics before the detector |
Information supplied |
|---|---|---|
\(I_H\) |
Linear polarizer at \(0^\circ\) |
\(I_H=(S_0+S_1)/2\) |
\(I_V\) |
Linear polarizer at \(90^\circ\) |
\(I_V=(S_0-S_1)/2\) |
\(I_D\) |
Linear polarizer at \(45^\circ\) |
\(I_D=(S_0+S_2)/2\) |
\(I_A\) |
Linear polarizer at \(135^\circ\) |
\(I_A=(S_0-S_2)/2\) |
\(I_R\) |
Quarter-wave plate and polarizer set as a right-circular analyser |
\(I_R=(S_0+S_3)/2\) |
\(I_L\) |
Quarter-wave plate and polarizer set as a left-circular analyser |
\(I_L=(S_0-S_3)/2\) |
Subtracting each orthogonal pair isolates one signed Stokes component, while adding either member pair gives the total intensity:
Ideally, \(I_H+I_V=I_D+I_A=I_R+I_L\). In a real experiment, use the average of these three sums for a lower-noise estimate of \(S_0\), after correcting the channels for detector gain and optical throughput.
Step 4 — State the numbered result. The requested method returns the complete Stokes vector
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 \(I_H\), \(I_V\), \(I_D\), and \(I_R\). Then use \(S_0=I_H+I_V\), \(S_1=I_H-I_V\), \(S_2=2I_D-S_0\), and \(S_3=2I_R-S_0\). 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
The degree of polarization is \(P=\sqrt{S_1^2+S_2^2+S_3^2}/S_0\), so \(0\leq P\leq1\). Useful calibration states are horizontal linear light, \((S_0,S_0,0,0)\), \(45^\circ\) linear light, \((S_0,0,S_0,0)\), and right-circular light, \((S_0,0,0,S_0)\) 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
Brief solution
1. Method. The working uses algebraic rearrangement and dimensional checks.
2. Reasoning and answer.
Show detailed steps
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.
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 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.
Detailed step 1. \(\operatorname{diag}(1,j)^2=\operatorname{diag}(1,-1)\),
Detailed step 2. a half-wave plate.
Detailed step 3. Orthogonal fast axes give \(\operatorname{diag}(1,j)\operatorname{diag}(j,1)=jI\),
Detailed step 4. so polarization is unchanged apart from global phase.
Step 4 — State the numbered result. The principal result obtained in the working is
Step 5 — Check. Equation (6) 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
Brief solution
1. Method. The working uses matrix multiplication and eigenvalue rules, trigonometric and small-angle identities, and algebraic rearrangement and dimensional checks.
2. Key step.
\(T(\theta)=R(-\theta)\operatorname{diag}(1,0)R(\theta)\) evaluates to \(\boxed{\begin{bmatrix}\cos^2\theta&\sin\theta\cos\theta\\ \sin\theta\cos\theta&\sin^2\theta\end{bmatrix}}\).
3. Answer.
Show detailed steps
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.
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 matrix multiplication and eigenvalue rules, trigonometric and small-angle identities, and 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.
\(T(\theta)=R(-\theta)\operatorname{diag}(1,0)R(\theta)\) evaluates to \(\boxed{\begin{bmatrix}\cos^2\theta&\sin\theta\cos\theta\\ \sin\theta\cos\theta&\sin^2\theta\end{bmatrix}}\).
Step 4 — State the numbered result. The principal result obtained in the working is
Step 5 — Check. Equation (7) 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
Brief solution
1. Method. The working uses exponential, logarithmic, and phasor identities.
Show detailed steps
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.
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 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.
Detailed step 1. The polarizer eigenvectors are its pass/block linear axes,
Detailed step 2. eigenvalues 1,0; the retarder eigenvectors are its fast/slow linear axes,
Detailed step 3. eigenvalues \(1,e^{-j\Gamma}\); the rotator eigenvectors are RCP/LCP,
Detailed step 4. eigenvalues \(e^{\mp j\theta}\).
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
Brief solution
1. Method. The working uses trigonometric and small-angle identities and algebraic rearrangement and dimensional checks.
2. Reasoning and answer.
Show detailed steps
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.
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 trigonometric and small-angle identities and 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.
Detailed step 1. \(\theta_B=\tan^{-1}(1.5)=\boxed{56.31^\circ}\) from the normal.
Detailed step 2. The internal angle is \(33.69^\circ\),
Detailed step 3. which is the reverse-interface Brewster angle,
Detailed step 4. so TM reflection vanishes at both parallel faces.
Step 4 — State the numbered result. The principal result obtained in the working is
Step 5 — Check. Equation (8) 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
Brief solution
1. Method. The working uses algebraic rearrangement and dimensional checks.
2. Reasoning and answer.
Show detailed steps
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.
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 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.
Detailed step 1. As \(\sigma\to\infty\),
Detailed step 2. impedance tends to zero and \(R\to1\).
Detailed step 3. The Hagen–Rubens result \(R\simeq1-2\sqrt{2\epsilon_0\omega/\sigma}\) gives copper reflectances \(\boxed{0.9534}\) at 1.06 micrometres and \(\boxed{0.9853}\) at 10.6 micrometres.
Detailed step 4. In the lossless sub-plasma-frequency Drude region the index is imaginary,
Detailed step 5. so no net transmitted power exists and \(R=1\).
Step 4 — State the numbered result. The principal result obtained in the working is
Step 5 — Check. Equation (9) 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
Brief solution
1. Method. The working uses algebraic rearrangement and dimensional checks.
Show detailed steps
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.
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 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.
Detailed step 1. Circular eigenindices satisfy \(n_\pm\simeq n\pm G/(2n)\) for \(G\ll n\).
Detailed step 2. Linear polarization is their equal superposition,
Detailed step 3. so its rotation per length is half their phase difference: \(\boxed{\rho=(k_0/2)(n_+-n_-)\simeq k_0G/(2n)}\).
End-of-chapter problems
Step 4 — State the numbered result. The principal result obtained in the working is
Step 5 — Check. Equation (10) 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
Brief solution
1. Method. The working uses matrix multiplication and eigenvalue rules and the Stokes-to-ellipse relations \(2\psi=\operatorname{atan2}(S_2,S_1)\) and \(\sin 2\chi=S_3/S_0\).
2. Key step.
Consequently, \(\operatorname{atan2}(-S_2,-S_1)=\operatorname{atan2}(S_2,S_1)+\pi\). Halving this angle gives \(\psi_2=\psi_1+\pi/2\) modulo \(\pi\), so the major axes are perpendicular. Also \(\sin 2\chi_2=-\sin 2\chi_1\), hence \(\chi_2=-\chi_1\): the two fields rotate in opposite senses.
3. Answer.
Show detailed steps
Definitions and setup. Remove the physically irrelevant common phase and write one normalized, fully polarized state as \(\mathbf J_1=(a,b e^{j\delta})^T\), where \(a,b\geq0\) and \(a^2+b^2=1\). Jones states are orthogonal when their Hermitian inner product is zero. For a noncircular ellipse, let \(\psi\) denote the major-axis angle and \(\chi\) its ellipticity angle; the sign of \(\chi\) specifies handedness.
Mathematical formulas used. The working uses matrix multiplication and eigenvalue rules and the Stokes-to-ellipse relations \(2\psi=\operatorname{atan2}(S_2,S_1)\) and \(\sin 2\chi=S_3/S_0\).
Worked derivation. A normalized vector orthogonal to \(\mathbf J_1\) is
For \(\mathbf J_1\), the three polarization-dependent Stokes components are \(S_1=a^2-b^2\), \(S_2=2ab\cos\delta\), and (with the chapter’s handedness convention) \(S_3=-2ab\sin\delta\). Substitution of \(\mathbf J_2\) changes the sign of every one of them but leaves \(S_0=1\) unchanged:
Consequently, \(\operatorname{atan2}(-S_2,-S_1)=\operatorname{atan2}(S_2,S_1)+\pi\). Halving this angle gives \(\psi_2=\psi_1+\pi/2\) modulo \(\pi\), so the major axes are perpendicular. Also \(\sin 2\chi_2=-\sin 2\chi_1\), hence \(\chi_2=-\chi_1\): the two fields rotate in opposite senses.
Numbered result. Orthogonal polarization states occupy antipodal points on the Poincaré sphere, which gives
Check. For \(\delta=0\), both states are linear and their Jones vectors point along perpendicular lines. For \(a=b\) and \(|\delta|=90^\circ\), 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
Brief solution
1. Method. The working uses matrix multiplication and eigenvalue rules and \(R(a)R(b)=R(a+b)\).
2. Key step.
The cancellation holds for every \(\alpha\); no direction in the transverse plane is preferred by an ideal optical rotator.
3. Answer.
Show detailed steps
Definitions and setup. The chapter’s passive coordinate-rotation matrix and the Jones matrix of a physical rotator through \(\theta\) are
Rotate the coordinate axes through an arbitrary angle \(\alpha\); this changes the matrix representation to \(T'=R(\alpha)TR(-\alpha)\).
Mathematical formulas used. The working uses matrix multiplication and eigenvalue rules and \(R(a)R(b)=R(a+b)\).
Worked derivation. Two-dimensional rotations commute, so
The cancellation holds for every \(\alpha\); no direction in the transverse plane is preferred by an ideal optical rotator.
Numbered result.
Check. Direct multiplication gives the same four matrix elements. Setting \(\alpha=90^\circ\) also leaves \(T\) unchanged, whereas the matrix of a fixed-axis retarder generally changes under that operation.
Problem 6.1-7 — Half-wave plate
Brief solution
1. Method. The working uses trigonometric and small-angle identities and the rotated-device rule \(H(\beta)=R(-\beta)H_0R(\beta)\).
2. Key step.
A true rotator would instead produce \(\mathbf J_{\rm lin}(\theta+\gamma)\) with one fixed \(\gamma\) for every input angle. The half-wave plate’s change, \(2(\beta-\theta)\), depends on the input orientation. It also reverses the handedness of circular or elliptical light, which an ideal rotator does not.
3. Answer.
for a fast axis along \(x\).
Show detailed steps
Definitions and setup. Take the fast axis as \(x\). Apart from a common phase, an ideal half-wave retarder has \(H_0=\operatorname{diag}(1,-1)\). The input is linearly polarized at \(\theta\) to that axis, so \(\mathbf J_{\rm in}=(\cos\theta,\sin\theta)^T\).
Mathematical formulas used. The working uses trigonometric and small-angle identities and the rotated-device rule \(H(\beta)=R(-\beta)H_0R(\beta)\).
Worked derivation. With the plate axes unrotated,
Thus the output polarization lies at \(-\theta\); relative to the input it has turned through \(-2\theta\) (a magnitude \(2\theta\), toward and through the fast axis). More generally, if the fast axis is at \(\beta\), application of the rotated matrix gives
A true rotator would instead produce \(\mathbf J_{\rm lin}(\theta+\gamma)\) with one fixed \(\gamma\) for every input angle. The half-wave plate’s change, \(2(\beta-\theta)\), depends on the input orientation. It also reverses the handedness of circular or elliptical light, which an ideal rotator does not.
Numbered result.
for a fast axis along \(x\).
Check. Input along either plate axis remains on that axis. For \(\theta=45^\circ\), the output is at \(-45^\circ\), a \(90^\circ\) turn in magnitude, as expected for a half-wave plate.
Problem 6.1-8 — Three retarders
Brief solution
1. Method. The working uses matrix multiplication and eigenvalue rules and the coordinate-rotation matrix defined in Problem 6.1-6.
2. Key step.
Changing the sign convention for retarder phase can multiply intermediate matrices by common phases, but it does not change either observable rotation.
3. Answer.
Show detailed steps
Definitions and setup. A retarder with fast axis \(x\) and phase delay \(\Gamma\) has \(W_x(\Gamma)=\operatorname{diag}(1,e^{-j\Gamma})\). The light first meets (a), then (b), then (c), so the system matrix is \(T_{abc}=T_cT_bT_a\); the first element encountered stands at the right.
Mathematical formulas used. The working uses matrix multiplication and eigenvalue rules and the coordinate-rotation matrix defined in Problem 6.1-6.
Worked derivation. The three requested Jones matrices are
Multiplying in propagation order gives
It sends \(x\) polarization to \(y\) and \(y\) polarization to \(-x\), precisely a \(+90^\circ\) rotation. If the physical order is reversed, the matrix order is also reversed:
Changing the sign convention for retarder phase can multiply intermediate matrices by common phases, but it does not change either observable rotation.
Numbered result.
Check. Both products are unitary with determinant \(+1\). 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
Brief solution
1. Method. The working uses the Jones vectors \(\mathbf e_R=(1,j)^T/\sqrt2\) and \(\mathbf e_L=(1,-j)^T/\sqrt2\) 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.
2. Key step.
where \(\doteq\) means equality apart from a physically irrelevant common phase.
3. Answer.
Show detailed steps
Definitions and setup. Consider normal reflection from an ideal isotropic mirror. Let the incident propagation basis \((\hat{\mathbf x}_i,\hat{\mathbf y}_i,\hat{\mathbf z}_i)\) be right-handed, with \(\hat{\mathbf z}_i\) along the incident wavevector. After reflection \(\hat{\mathbf z}_r=-\hat{\mathbf z}_i\). A convenient right-handed reflected basis is therefore \(\hat{\mathbf x}_r=\hat{\mathbf x}_i\) and \(\hat{\mathbf y}_r=-\hat{\mathbf y}_i\).
Mathematical formulas used. The working uses the Jones vectors \(\mathbf e_R=(1,j)^T/\sqrt2\) and \(\mathbf e_L=(1,-j)^T/\sqrt2\) 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 \(-1\), the laboratory components become \((-A_x,-A_y)\). Expressing the reflected field in its local basis flips the sign of its \(y\) coordinate once more, hence
Apply this map to the two circular states:
where \(\doteq\) means equality apart from a physically irrelevant common phase.
Numbered result.
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
Brief solution
1. Method. A \(45^\circ\) 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.
2. Key step.
The first polarizer converts arbitrary background light to linear \(x\) polarization.
The quarter-wave plate resolves that field equally along its fast and slow axes and adds a \(90^\circ\) relative phase, producing circular light.
Reflection at the glass reverses its handedness.
A second passage through the same retarder adds the complementary quarter-wave delay. The returning field is therefore linear along \(y\), rather than \(x\).
The polarizer rejects this returned \(y\) 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:
3. Answer.
Show detailed steps
Definitions and setup. Place, from the observer toward the window, a linear polarizer followed by a quarter-wave retarder. Let the polarizer pass \(x\) polarization and set either principal axis of the retarder at \(45^\circ\) to \(x\). 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 \(45^\circ\) 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.
The first polarizer converts arbitrary background light to linear \(x\) polarization.
The quarter-wave plate resolves that field equally along its fast and slow axes and adds a \(90^\circ\) relative phase, producing circular light.
Reflection at the glass reverses its handedness.
A second passage through the same retarder adds the complementary quarter-wave delay. The returning field is therefore linear along \(y\), rather than \(x\).
The polarizer rejects this returned \(y\) 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:
Numbered result.
Check. Without the quarter-wave plate, the reflected \(x\) 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
Brief solution
1. Method. Apply continuity of tangential \(\mathbf E\) and \(\mathbf H\), Snell’s law, and \(\eta_i=\eta_0/n_i\) for nonmagnetic lossless dielectrics. The algebra uses trigonometric and small-angle identities.
2. Key step.
This is the requested reflection relation, Eq. (6.2-6). For nonmagnetic dielectrics, substitute \(\eta_i=\eta_0/n_i\) and cancel the common factors to obtain the TE Fresnel coefficient.
3. Answer.
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.
Show detailed steps
Definitions and setup. A TE-polarized plane wave is incident from medium 1 onto medium 2. The media have characteristic impedances \(\eta_1,\eta_2\), refractive indices \(n_1,n_2\), and propagation angles \(\theta_1,\theta_2\) measured from the interface normal. Normalize the incident electric amplitude to one and write the reflected and transmitted amplitudes as \(r_s\) and \(t_s\).
Mathematical formulas used. Apply continuity of tangential \(\mathbf E\) and \(\mathbf H\), Snell’s law, and \(\eta_i=\eta_0/n_i\) for nonmagnetic lossless dielectrics. The algebra uses trigonometric and small-angle identities.
Worked derivation. For TE polarization the electric field is wholly tangential. The magnetic field has tangential magnitude \(E\cos\theta/\eta\); its sign reverses for the reflected wave. The two boundary equations are therefore
Insert \(t_s=1+r_s\) into the magnetic-field equation and collect the terms in \(r_s\):
This is the requested reflection relation, Eq. (6.2-6). For nonmagnetic dielectrics, substitute \(\eta_i=\eta_0/n_i\) and cancel the common factors to obtain the TE Fresnel coefficient.
Numbered result.
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 \(r_s=(n_1-n_2)/(n_1+n_2)\). If \(n_1=n_2\), Snell’s law gives \(\theta_1=\theta_2\) and the coefficient vanishes, as it must when there is no optical discontinuity.
Problem 6.2-4 — Glass at 45 degrees
Brief solution
1. Method. Use Snell’s law and the TE/TM Fresnel coefficients. For the TM magnitude it is convenient to use \(r_p=(n_2\cos\theta_1-n_1\cos\theta_2)/ (n_2\cos\theta_1+n_1\cos\theta_2)\); a different reflected-axis convention may reverse its sign but not its power reflectance.
2. Key step.
Squaring gives \(R_s=0.092013\) and \(R_p=0.0084665\). Unpolarized light carries equal mean power in the two orthogonal modes, so its reflectance is their arithmetic mean.
3. Answer.
Show detailed steps
Definitions and setup. Light travels from air, \(n_1=1\), into glass, \(n_2=1.5\), at \(\theta_1=45^\circ\). For lossless media the power reflectance is the squared magnitude of the electric-field reflection coefficient: \(R_s=|r_s|^2\) and \(R_p=|r_p|^2\).
Mathematical formulas used. Use Snell’s law and the TE/TM Fresnel coefficients. For the TM magnitude it is convenient to use \(r_p=(n_2\cos\theta_1-n_1\cos\theta_2)/ (n_2\cos\theta_1+n_1\cos\theta_2)\); a different reflected-axis convention may reverse its sign but not its power reflectance.
Worked derivation. First find the transmitted angle:
Now substitute \(\cos45^\circ=0.7071068\) and \(\cos28.1255^\circ=0.8819171\):
Squaring gives \(R_s=0.092013\) and \(R_p=0.0084665\). Unpolarized light carries equal mean power in the two orthogonal modes, so its reflectance is their arithmetic mean.
Numbered result.
Check. The TM reflectance is small because \(45^\circ\) is fairly close to the air–glass Brewster angle \(\tan^{-1}(1.5)=56.31^\circ\). Both values lie between zero and one, and the unpolarized result lies exactly between them.
Problem 6.2-5 — Brewster geometry
Brief solution
1. Method. The working uses Snell’s law and trigonometric and small-angle identities.
2. Key step.
Since the reflected angle equals \(\theta_B\), the reflected and refracted rays differ by \(180^\circ-(\theta_B+\theta_2)=90^\circ\).
3. Answer.
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.
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Definitions and setup. At the Brewster angle the TM reflection coefficient vanishes. Its numerator gives the supplied condition \(n_1\sec\theta_1=n_2\sec\theta_2\); phase matching also requires \(n_1\sin\theta_1=n_2\sin\theta_2\).
Mathematical formulas used. The working uses Snell’s law and trigonometric and small-angle identities.
Worked derivation. Rewrite the zero-reflection condition as \(n_1\cos\theta_2=n_2\cos\theta_1\). Together with Snell’s law this gives
Therefore \(\sin\theta_1\cos\theta_1 =\sin\theta_2\cos\theta_2\), or \(\sin2\theta_1=\sin2\theta_2\). For refraction into a different medium, the nontrivial physical solution is \(2\theta_1=\pi-2\theta_2\), hence \(\theta_1+\theta_2=90^\circ\). Put \(\sin\theta_2=\cos\theta_1\) into Snell’s law:
Since the reflected angle equals \(\theta_B\), the reflected and refracted rays differ by \(180^\circ-(\theta_B+\theta_2)=90^\circ\).
Numbered result.
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 \(56.31^\circ\) and Snell’s law gives \(33.69^\circ\); their sum is \(90^\circ\) and direct substitution makes the TM Fresnel numerator zero.
Problem 6.2-6 — TIR retardance
Brief solution
1. Method. Above the critical angle, define \(q=\sqrt{\sin^2\theta-m^2}\). The unit-magnitude Fresnel coefficients have phases \(\phi_s=-2\tan^{-1}(q/\cos\theta)\) and \(\phi_p=-2\tan^{-1}[q/(m^2\cos\theta)]\) in this phase convention.
2. Key step.
The negative sign says that, under the chosen phasor convention, the TM phase lags the TE phase by \(44.98^\circ\). Interchanging the definition of retardance or the time-harmonic convention reverses the sign but not the physical phase difference.
3. Answer.
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Definitions and setup. Reflection is from glass \(n_1=1.5\) into air \(n_2=1\). Define \(m=n_2/n_1=2/3\) and measure phase as in \(r_{s,p}=e^{j\phi_{s,p}}\). The requested retardance of TM relative to TE is \(\Gamma=\phi_p-\phi_s\).
Mathematical formulas used. Above the critical angle, define \(q=\sqrt{\sin^2\theta-m^2}\). The unit-magnitude Fresnel coefficients have phases \(\phi_s=-2\tan^{-1}(q/\cos\theta)\) and \(\phi_p=-2\tan^{-1}[q/(m^2\cos\theta)]\) in this phase convention.
Worked derivation. The critical and specified angles are
At this angle, \(q=\sqrt{\sin^2(50.1724^\circ)-(2/3)^2}\). Substitution into the two phase expressions gives
The negative sign says that, under the chosen phasor convention, the TM phase lags the TE phase by \(44.98^\circ\). Interchanging the definition of retardance or the time-harmonic convention reverses the sign but not the physical phase difference.
Numbered result.
Check. Both reflection magnitudes equal one, so TIR changes phase but not power. At \(\theta=\theta_c\), \(q=0\) 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
Brief solution
1. Method. With \(m=n_2/n_1=\sin\theta_c\) and \(q=\sqrt{\sin^2\theta-m^2}\), use the TE phase from Problem 6.2-6, \(\phi_s=-2\tan^{-1}(q/\cos\theta)\). The transverse wavevector is \(k_x=k\sin\theta\), so \(dk_x=k\cos\theta\,d\theta\).
2. Key step.
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.
3. Answer.
Reversing the positive \(x\) direction or phase convention reverses the reported sign; the measurable magnitude is unchanged.
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Definitions and setup. Two equal-amplitude TE plane waves in the denser medium have incidence angles \(\theta\) and \(\theta+d\theta\). Let \(x\) run along the interface and let \(k\) be their common wavenumber. Under TIR the reflection coefficient is \(r_s=e^{j\phi_s(\theta)}\); write \(d\phi_s=\dot\phi_s\,d\theta\).
Mathematical formulas used. With \(m=n_2/n_1=\sin\theta_c\) and \(q=\sqrt{\sin^2\theta-m^2}\), use the TE phase from Problem 6.2-6, \(\phi_s=-2\tan^{-1}(q/\cos\theta)\). The transverse wavevector is \(k_x=k\sin\theta\), so \(dk_x=k\cos\theta\,d\theta\).
Worked derivation. Differentiate the TIR phase. Since \(dq/d\theta=\sin\theta\cos\theta/q\), the chain rule simplifies to
At the interface, the two incident waves form fringes proportional to \(\cos^2(dk_x x/2)\). Reflection adds \(d\phi_s\) to the phase difference, so the reflected fringes are proportional to \(\cos^2[(dk_xx+d\phi_s)/2]\). A reflected maximum therefore occurs where the corresponding incident maximum would occur after a translation \(\Delta\) satisfying
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,
Reversing the positive \(x\) direction or phase convention reverses the reported sign; the measurable magnitude is unchanged.
Check. The result has dimensions \(1/k\), 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
Brief solution
1. Method. Apply Maxwell’s normal-incidence relation \(H=E/\eta\) and continuity of tangential \(E\) and \(H\) at the boundary.
2. Key step.
The problem defines the reflected Jones component along the reflected wave’s local transverse axis, which is opposite to this fixed-axis amplitude. Thus \(r=-r_E\), and insertion of \(\widetilde n\) gives the requested form.
3. Answer.
The convention-independent power reflectance is
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Definitions and setup. Light is normally incident from free space onto a nonmagnetic absorbing medium of real refractive index \(n\) and intensity absorption coefficient \(\alpha\). Define
The factor two occurs because intensity decays as \(e^{-\alpha z}\) while field amplitude decays as \(e^{-\alpha z/2}\).
Mathematical formulas used. Apply Maxwell’s normal-incidence relation \(H=E/\eta\) and continuity of tangential \(E\) and \(H\) at the boundary.
Worked derivation. In one common fixed-laboratory-axis convention, normalizing the incident field gives \(1+r_E=t\) and \(1-r_E=\widetilde n t\). Solving,
The problem defines the reflected Jones component along the reflected wave’s local transverse axis, which is opposite to this fixed-axis amplitude. Thus \(r=-r_E\), and insertion of \(\widetilde n\) gives the requested form.
Numbered result.
The convention-independent power reflectance is
Check. When \(\alpha=0\), this reduces in magnitude to the ordinary normal-incidence Fresnel result. When \(n=1\) and \(\alpha=0\), the interface disappears and \(r=0\); as absorption becomes very large, \(R\rightarrow1\).
Problem 6.3-1 — Quartz retardation
Brief solution
1. Method. The relative phase accumulated through thickness \(d\) is \(\Gamma=k_0(n_e-n_o)d=2\pi(n_e-n_o)d/\lambda_0\).
2. Key step.
Solving gives all possible thicknesses. Values with even \(q\) have retardance \(+\pi/2\) modulo \(2\pi\); odd \(q\) give the complementary \(-\pi/2\) action.
3. Answer.
The thinnest plate is therefore \(17.58\ \mathrm{\mu m}\); the next is \(52.75\ \mathrm{\mu m}\).
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Definitions and setup. Quartz is positive uniaxial with \(n_e=1.553\), \(n_o=1.544\), and vacuum wavelength \(\lambda_0=633\ \mathrm{nm}\). Retardation is largest when the wavevector is perpendicular to the optic axis, because the two eigenmodes then sample \(n_e\) and \(n_o\) directly.
Mathematical formulas used. The relative phase accumulated through thickness \(d\) is \(\Gamma=k_0(n_e-n_o)d=2\pi(n_e-n_o)d/\lambda_0\).
Worked derivation. The maximum birefringence is \(\Delta n=n_e-n_o=0.009\). For \(d=1\ \mathrm{mm}\),
Thus one millimetre produces 14.218 full retardation cycles. A quarter-wave retarder needs an odd multiple of \(\pi/2\):
Solving gives all possible thicknesses. Values with even \(q\) have retardance \(+\pi/2\) modulo \(2\pi\); odd \(q\) give the complementary \(-\pi/2\) action.
Numbered result.
The thinnest plate is therefore \(17.58\ \mathrm{\mu m}\); the next is \(52.75\ \mathrm{\mu m}\).
Check. Substituting the first thickness gives \(\Delta n d/\lambda_0=1/4\), hence \(\Gamma=\pi/2\). The expression has units of length and every increase by \(\lambda_0/(2\Delta n)\) adds exactly \(\pi\) of retardance.
Problem 6.3-2 — Maximum extraordinary walk-off
Brief solution
1. Method. The normal to the extraordinary \(k\)-surface gives
2. Key step.
The two tangents are reciprocals, so \(\phi_{\max}=90^\circ-\theta_{\max}\). Numerically, \(\theta_{\max}=45.1665^\circ\), \(\phi_{\max}=44.8335^\circ\), and their difference is \(0.3330^\circ\).
3. Answer.
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Definitions and setup. Let \(\theta\) be the angle from the optic axis to the extraordinary wavevector \(\mathbf k\), and \(\phi\) the angle from that axis to the Poynting vector \(\mathbf S\). Their difference \(\rho=\theta-\phi\) is the extraordinary ray walk-off. For quartz, \(n_e=1.553\) and \(n_o=1.544\).
Mathematical formulas used. The normal to the extraordinary \(k\)-surface gives
Maximize \(\rho\) using the stationary-value condition.
Worked derivation. Differentiate \(\rho(\theta)=\theta-\tan^{-1}(r\tan\theta)\):
Let \(u=\tan\theta\). Rearrangement gives \(1+r^2u^2=r(1+u^2)\), and because \(r\ne1\), \(u^2=1/r\). In the first quadrant,
The two tangents are reciprocals, so \(\phi_{\max}=90^\circ-\theta_{\max}\). Numerically, \(\theta_{\max}=45.1665^\circ\), \(\phi_{\max}=44.8335^\circ\), and their difference is \(0.3330^\circ\).
Numbered result.
Check. At \(\theta=0^\circ\) or \(90^\circ\), symmetry forces \(\mathbf S\parallel\mathbf k\), so the walk-off vanishes. If \(n_e=n_o\), then \(r=1\) and it vanishes at every angle, as required for an isotropic medium.
Problem 6.3-3 — Double refraction in quartz
Brief solution
1. Method. Tangential phase matching requires \(n\sin\theta=\sin\theta_i\). For the extraordinary wave, whose wavevector is at \(\gamma=\theta_e+60^\circ\) to the optic axis, use
2. Key step.
Iteration gives \(\theta_e=18.78566^\circ\), \(\gamma=78.78566^\circ\), and \(n(\gamma)=1.552657\). The normal to the extraordinary \(k\)-surface then gives \(\phi=78.65792^\circ\) from the optic axis. Since the axis is at \(-60^\circ\), the extraordinary ray is at \(78.65792^\circ-60^\circ=18.65792^\circ\) from the surface normal.
3. Answer.
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Definitions and setup. An unpolarized wave in air strikes quartz at \(\theta_i=30^\circ\). Quartz has \(n_o=1.544\) and \(n_e=1.553\). 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 \(-60^\circ\) from the inward surface normal. Angles \(\theta_o\) and \(\theta_e\) below are wavevector angles from that normal.
Mathematical formulas used. Tangential phase matching requires \(n\sin\theta=\sin\theta_i\). For the extraordinary wave, whose wavevector is at \(\gamma=\theta_e+60^\circ\) to the optic axis, use
The extraordinary ray direction \(\phi\) measured from the optic axis satisfies \(\tan\phi=(n_o/n_e)^2\tan\gamma\).
Worked derivation. The ordinary refractive index is direction independent, hence
Its spherical \(k\)-surface makes the ordinary Poynting vector parallel to its wavevector, so the ordinary wave and ray have this same direction.
For the extraordinary component, \(n\) depends on the unknown direction; solve the single scalar equation
Iteration gives \(\theta_e=18.78566^\circ\), \(\gamma=78.78566^\circ\), and \(n(\gamma)=1.552657\). The normal to the extraordinary \(k\)-surface then gives \(\phi=78.65792^\circ\) from the optic axis. Since the axis is at \(-60^\circ\), the extraordinary ray is at \(78.65792^\circ-60^\circ=18.65792^\circ\) from the surface normal.
Numbered result.
Check. Substitution gives \(1.544\sin18.89496^\circ=0.5\) and \(1.552657\sin18.78566^\circ=0.5\), confirming tangential phase matching. The extraordinary ray differs from its wavevector by \(0.12774^\circ\), whereas the ordinary ray has zero walk-off.
Problem 6.3-4 — Geometry for largest separation
Brief solution
1. Method. For a positive uniaxial crystal, the extraordinary ray and wave-normal angles obey
2. Key step.
The ordinary ray travels straight through the plate. The extraordinary ray tilts by \(\rho_{\max}\) toward the optic axis, so the two spots on the second face are separated by \(L\tan\rho_{\max}\). 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.
3. Answer.
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Definitions and setup. Let a parallel-sided plate of a positive uniaxial crystal have thickness \(L\), ordinary index \(n_o\), and principal extraordinary index \(n_e>n_o\). 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 \(\theta\) be the angle from that axis to the common wavevector. The ordinary ray is parallel to its wavevector. The extraordinary ray makes an angle \(\phi\) with the optic axis and therefore walks away from the ordinary ray by \(\rho=\theta-\phi\).
Mathematical formulas used. For a positive uniaxial crystal, the extraordinary ray and wave-normal angles obey
Problem 6.3-2 showed that the stationary (and maximum) walk-off occurs when \(\tan\theta=n_e/n_o\). A ray crossing a distance \(L\) at angle \(\rho\) to the face normal acquires lateral displacement \(\Delta x=L\tan\rho\).
Worked derivation. At the optimum orientation,
Consequently, the largest ray–wavevector angle is
The ordinary ray travels straight through the plate. The extraordinary ray tilts by \(\rho_{\max}\) toward the optic axis, so the two spots on the second face are separated by \(L\tan\rho_{\max}\). 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
Check. If \(n_e=n_o\), the material becomes isotropic; \(\theta_*=45^\circ\), \(\rho_{\max}=0\), and the separation vanishes. The result also scales linearly with \(L\), 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
Brief solution
1. Method. For a wavevector at angle \(\theta\) to the optic axis, the extraordinary phase index and ray angle are
2. Key step.
The integer 689 represents complete cycles. The observable residual is \(0.53097\) cycle, or \(191.15^\circ\), although the unwrapped phase is \(\Gamma=4332.45\ \mathrm{rad}\).
3. Answer.
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Definitions and setup. The He–Ne wavelength is \(\lambda_0=633\ \mathrm{nm}\). The LiNbO3 plate has \(n_o=2.20\), \(n_e=2.29\), and thickness \(L=1.00\ \mathrm{cm}\). Its optic axis is at \(\theta=45^\circ\) 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 \(\theta\) to the optic axis, the extraordinary phase index and ray angle are
Thus the walk-off is \(\rho=\theta-\phi\), the lateral separation at the exit face is \(\Delta x=L\tan\rho\), and the relative phase accumulated between the extraordinary and ordinary waves is
Worked derivation. First calculate the phase index without prematurely rounding it:
The extraordinary ray direction and walk-off are then
It follows that
For the retardation,
The integer 689 represents complete cycles. The observable residual is \(0.53097\) cycle, or \(191.15^\circ\), although the unwrapped phase is \(\Gamma=4332.45\ \mathrm{rad}\).
Numbered result. At the output face,
Check. The calculated index satisfies \(n_o<n(45^\circ)<n_e\), as it must. Because this is a positive uniaxial crystal, the extraordinary ray bends toward the optic axis, giving the positive shift above. Finally, \(10\ \mathrm{mm}\times\tan(2.3^\circ)\approx0.40\ \mathrm{mm}\), an independent magnitude check.
Problem 6.3-6 — Conical refraction
Brief solution
1. Method. At fixed frequency the group velocity is normal to the constant-frequency \(k\) surface,
2. Key step.
After propagating through the plate, the conical rays intersect the second face on a circle of radius
All of these modes nevertheless share the axial wavevector at the conical contact point. Hence its tangential component at the parallel output face is zero for every azimuth. Tangential phase matching makes every ray refract normally into air. The emerging rays are parallel, but originate around the circle, forming a hollow cylindrical bundle; a transverse screen records a bright ring.
3. Answer.
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Definitions and setup. Consider a parallel-sided biaxial-crystal plate of thickness \(L\). One optic axis is normal to both faces, and a narrow ray in air is normally incident along that axis. Let \(\beta\) denote the semi-angle of the cone of energy-flow directions inside the crystal. Wavevector \(\mathbf k\) specifies phase propagation, whereas the ray direction is the group velocity (or Poynting-vector direction); in an anisotropic crystal these directions need not coincide.
Mathematical formulas used. At fixed frequency the group velocity is normal to the constant-frequency \(k\) surface,
The radius reached by a ray that crosses axial thickness \(L\) at cone angle \(\beta\) is simply \(R=L\tan\beta\).
Worked derivation. Away from an optic axis, a specified wavevector intersects two smooth sheets of the biaxial \(k\) surface and therefore has two normal modes. On an optic axis those sheets touch. Near the contact point their common surface is locally conical, so it has not one unique normal but a continuous family of normals indexed by azimuth. The normally incident field can therefore excite a continuum of energy-flow directions. Those normals all make the same semi-angle \(\beta\) with the optic axis: the refracted rays fill a cone rather than separating into only two rays.
After propagating through the plate, the conical rays intersect the second face on a circle of radius
All of these modes nevertheless share the axial wavevector at the conical contact point. Hence its tangential component at the parallel output face is zero for every azimuth. Tangential phase matching makes every ray refract normally into air. The emerging rays are parallel, but originate around the circle, forming a hollow cylindrical bundle; a transverse screen records a bright ring.
Numbered result. The geometrical outcome is
Check. Rotational symmetry around the chosen optic axis requires a circle rather than a preferred transverse direction. In the isotropic or uniaxial limiting case the conical contact disappears, \(\beta\rightarrow0\), and the ring collapses to the ordinary on-axis spot. Doubling \(L\) doubles the ring radius, consistent with straight ray propagation inside the plate.
Problem 6.6-1 — Circular dichroic selector
Brief solution
1. Method. If \(C=[\mathbf e_R\;\mathbf e_L]\) changes circular-basis components into linear-basis components, then \(T=CT_cC^\dagger\). Equivalently, an ideal selector is the outer-product projector \(T=\mathbf e_R\mathbf e_R^\dagger\).
2. Key step.
Thus every nonzero transmitted field is proportional to \(\mathbf e_R\) and is right circularly polarized. The qualification is important: a passive selector cannot produce light from a pure LCP input; that input is extinguished rather than converted with nonzero efficiency.
3. Answer.
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Definitions and setup. Use the book’s circular-polarization convention
An ideal right-circular dichroic selector transmits the RCP component with unit amplitude and absorbs the LCP component completely. In the circular basis its Jones matrix is therefore \(T_c=\operatorname{diag}(1,0)\).
Mathematical formulas used. If \(C=[\mathbf e_R\;\mathbf e_L]\) changes circular-basis components into linear-basis components, then \(T=CT_cC^\dagger\). Equivalently, an ideal selector is the outer-product projector \(T=\mathbf e_R\mathbf e_R^\dagger\).
Worked derivation. Carrying out the outer product gives
For an arbitrary incident Jones vector \(\mathbf J=[A_x\;A_y]^T\),
Thus every nonzero transmitted field is proportional to \(\mathbf e_R\) and is right circularly polarized. The qualification is important: a passive selector cannot produce light from a pure LCP input; that input is extinguished rather than converted with nonzero efficiency.
Numbered result. In the linear \(x,y\) basis, the required Jones matrix is
Check. Direct multiplication gives \(T_R\mathbf e_R=\mathbf e_R\), \(T_R\mathbf e_L=\mathbf0\), and \(T_R^2=T_R\). These are exactly the transmission, absorption, and projector properties required of the ideal circular dichroic device.
Problem 6.6-2 — Many weakly rotated polarizers
Brief solution
1. Method. An ideal polarizer is a projector, \(P_m=\mathbf u_m\mathbf u_m^T\). Adjacent axes differ by \(\theta\), and hence
2. Key step.
Exponentiating shows that both the amplitude factor and the power transmittance tend to unity even though the polarization turns through \(90^\circ\).
3. Answer.
Show detailed steps
Definitions and setup. The incident field is linearly polarized along \(x\), with normalized Jones vector \(\mathbf u_0=[1\;0]^T\). There are \(N\) ideal linear polarizers. The \(m`th transmission axis is at :math:\)alpha_m=mtheta`, where \(\theta=\pi/(2N)\) and \(m=1,\ldots,N\). Define a unit vector along that axis by
The final axis is \(\alpha_N=N\theta=\pi/2\), so it points along \(y\).
Mathematical formulas used. An ideal polarizer is a projector, \(P_m=\mathbf u_m\mathbf u_m^T\). Adjacent axes differ by \(\theta\), and hence
Each projection therefore multiplies the field amplitude by \(\cos\theta\) and aligns it with the new axis.
Worked derivation. After the first polarizer, \(\mathbf J_1=P_1\mathbf u_0=\cos\theta\,\mathbf u_1\). If after \(m-1\) polarizers \(\mathbf J_{m-1}=\cos^{m-1}\theta\,\mathbf u_{m-1}\), then
Induction to \(m=N\) proves that the output points along \(y\) and has amplitude factor \(\cos^N[\pi/(2N)]\). The power transmittance is the square of that factor. To evaluate the large-\(N\) limit, use \(\ln\cos x=-x^2/2+O(x^4)\):
Exponentiating shows that both the amplitude factor and the power transmittance tend to unity even though the polarization turns through \(90^\circ\).
Numbered result. The transmitted field and power are
Check. For \(N=1\), the single polarizer is crossed with the input, and the formula gives zero transmission. For \(N=2\), the axes are at \(45^\circ\) and \(90^\circ\); the amplitude is \((1/\sqrt2)^2=1/2\) and the power is \(1/4\), agreeing with two successive applications of Malus’s law.