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A Michelson interferometer divides light at a beam splitter, reflects the two arms at mirrors and recombines the returning fields. For vacuum arm lengths ${\ell }_1,{\ell }_2$, the delay magnitude is $|\tau |=2|{\ell }_2-{\ell }_1|/c$. At each coarse delay, a small mirror scan produces maxima and minima of the detected light intensity. A photodiode and its electrical readout resolve these changes; the scan should cover several carrier cycles while changing $|g^{(1)}|$ negligibly. MIT's laboratory guide uses a mirror actuator, photodiode and electrical drive to scan interference fringes and extract interferometric visibility.^[ref]

A verified motivating experiment is Shaoul Ezekiel's MIT demonstration of source-spectrum control. A polarizer selects one or several lines from a laser whose adjacent lines have orthogonal linear polarizations. A sampled beam goes to a scanning Fabry–Pérot interferometer to check the optical spectrum; the main beam enters a Michelson interferometer, and a lens projects its interference fringes onto a screen. With one selected line the fringes retain good contrast through the demonstrated delay range. With several lines their contrast falls and returns as the arm length changes. At a fixed delay with poor contrast, selecting one line restores it; at matched arms both selections give good contrast. This control isolates the role of spectral composition rather than simply the arm position. The presentation is a qualitative demonstration, not a calibrated linewidth measurement. ^[ref].

A deliberately simpler two-line model explains the return of contrast. Take two stationary, mutually uncorrelated monochromatic contributions of equal power at ${\nu }_0\pm \delta \nu /2$. Their normalized wave-field correlation isfor equal detected single-arm light intensities. At $\tau =1/(2\delta \nu )$ the two contributions' interference fringes cancel on averaging; at $\tau =1/\delta \nu$ they agree again. This is a deduction from the stipulated spectrum, not a fit to Ezekiel's two- or three-line source. Finite individual linewidths eventually damp these revivals.

For a quantitative version of the experiment, the visibility law suggests explicit controls. Measure each arm separately at every coarse delay, subtract the detector's dark signal and any measured background, then normalize by $2\sqrt{I_1{I}_2}/(I_1+I_2)$. Keep the returning beams aligned on the same detector region with parallel linear polarizations; keep detector gain fixed and avoid saturation. Resolve the small mirror scan electrically, rather than averaging over its moving interference fringes. Repeat scans to identify vibration or drift. These controls follow from the averaging and mode-matching assumptions in first-order optical coherence; loss of contrast alone cannot distinguish source decorrelation from apparatus motion, misalignment or insufficient spatial resolution. The finite travel range gives access only to a finite range of delays.

The practical gain is that wave interference measures field correlations even when the detector cannot follow the optical oscillation itself. A long optical coherence length allows unequal-arm measurements; a short one localizes high contrast near matched delays. The beautiful feature is that both fading and reappearance of interference fringes are sums of the same rotating complex amplitudes.

A nonpolarizing beam splitter is designed for approximately equal power splitting ratios for the two orthogonal linear polarizations, at specified wavelengths and angle of incidence. It need not be 50:50, lossless, or have equal transmission and reflection wave phases for both linear polarizations. A scalar beam splitter scattering matrix therefore still requires a specified electromagnetic polarization and operating range. Polarizing beam splitters instead aim to separate those linear polarizations into different paths.

Optical activity is the ability to rotate the direction of linear polarization of transmitted light. In an achiral solvent, the two members of an enantiomer pair have equal and opposite optical rotations at the same wavelength, temperature and concentration. The measured angle concerns a bulk sample; it is not a geometric rotation of the molecules.^[ref]

The rotating object in this description is the direction along which the electric field oscillates. At each position the transmitted light can still have linear polarization: its electric field moves back and forth along a line, while the orientation of that line changes along the sample. Nothing requires the molecules to turn bodily through the observed angle. This makes polarimetry a bridge from molecular chirality to a visible bulk measurement, and, with a suitable calibration, a way to follow concentration or the balance of enantiomers.

In a transparent, isotropic chemical solution, natural optical rotation can be understood as circular birefringence: the two opposite circular polarizations accumulate different wave phases, and their equal-amplitude sum remains linearly polarized along a rotated direction. Circular dichroism instead means unequal absorption of those components and can produce elliptical polarization. These are related responses, not interchangeable measurements. General birefringence need not produce this simple rotation of linear polarization.^[ref]

Optical nonreciprocity means that exchanging source and detector paths need not restore the corresponding transmission when external conditions, including the magnetic field, are held fixed. The Faraday effect gives a linear polarization rotation with the same laboratory sense on both traversals; see Faraday rotation on a round trip.

An optical path length measures how much propagation wave phase a path accumulates, expressed as the distance that would accumulate the same wave phase in vacuum. A centimetre in a uniform medium with refractive index $3/2$ has optical path length $1.5\mathrm{cm}$, although its geometric length remains $1\mathrm{cm}$. This makes optical path length useful whenever two contributions must arrive with matching wave phases, from an optical lens to a Michelson interferometer. ^[ref]

For a specified path $\gamma$ at fixed angular frequency $\omega$, the optical path length is the line integralHere $s$ is geometric arc length, and $n$ is the phase refractive index, evaluated at the same frequency everywhere. The propagation interpretation assumes a stationary linear medium with isotropy, a real positive scalar refractive index and negligible loss, on a valid geometrical optics branch. The path can curve; at sharp interfaces apply the appropriate matching conditions to each segment. A prescribed curve has a mathematical line integral, but it predicts a propagation wave phase only when it represents the physical branch. ^[ref] Optical fibers can require a mode wavenumber rather than the bulk refractive index along a guessed light ray. ^[ref]

Fix the vacuum wavelength ${\lambda }_0=c/f=2\pi c/\omega$, with frequency $f=\omega /(2\pi )$ and vacuum wavenumber $k_0=2\pi /{\lambda }_0$. In a locally uniform medium the wavelength is $\lambda ={\lambda }_0/n$, while the frequency stays unchanged at a stationary interface. With the convention $\Phi =k_{0}S-\omega t+{\varphi }_0$, propagation from a specified starting wave phase contributesThe relative reference ${\delta }_{\mathrm{ref}}$ includes launch wave phases and any extra wave phase from mirrors, beam splitters or coatings. Thus $k_0\mathrm{OPL}$ is a propagation contribution, not an absolute detector wave phase independent of the apparatus. If the reference contributions stay fixed, a change in optical path difference $h$ changes the phase difference by $2\pi h/{\lambda }_0$. Using the material wavelength again in this formula would count the refractive index twice. ^[ref]

A motivating measurement turns invisible optical oscillations into a slowly changing electrical signal. MIT's teaching experiment uses a laser, a beam splitter, two returning mirrors, an amplified photodiode, an oscilloscope and a calibrated mirror actuator driven by a function generator. The returning beams overlap on the photodiode; one channel records detector voltage and another the actuator drive. The measured observable is the changing light intensity, including its maxima and minima, rather than direct timing of an optical cycle. A calibrated displacement per drive voltage and the spacing of the extrema relate geometric motion to the phase difference. The guide infers the laser wavelength from known actuator calibration; conversely, with independently known vacuum wavelength and ambient refractive index, the same relation calibrates mirror motion. These two unknowns cannot both be inferred from fringe spacing alone. ^[ref]

For a proposed change-of-medium measurement, keep the mirrors fixed and put a transparent cell of internal length $L$ in one arm. Compare its contents at refractive indices $n_a$ and $n_b$, leaving its windows in place. At normal incidence the two traversals giveHere $N$ is the signed number of whole interference fringe cycles plus any resolved fractional cycle, counted continuously. A mirror displacement $x$ instead gives $\Delta \mathrm{OPL}=2n_{\mathrm{amb}}x$; distinguish this double-pass factor from a single pass in a Mach-Zehnder interferometer. OpenStax's gas-cell worked example provides the same double-pass reasoning; it is a teaching calculation rather than a published measurement dataset.^[ref]

An exact toy calculation makes the sensitivity visible. Stipulate ${\lambda }_0=600\mathrm{nm}$, $L=10\mathrm{mm}$ and $n_b-n_a=3/100000=0.00003$. Then $\Delta \mathrm{OPL}=600\mathrm{nm}$ and $\Delta \delta =2\pi$: a full interference fringe cycle, despite unchanged geometric lengths. For a continuous change $n_b-n_a=q(3/100000)$, let $0\le q\le 2$ and choose the initial detector phase difference to be zero. Equal coherent detected arm contributions of light intensity $I_0$ predictThe plot is this stipulated model: $q=1/2$ gives a minimum, $q=1$ restores a maximum, and the reference level is chosen by adjustment of the apparatus. It contains no measured values or assigned properties of a real gas. The equivalent mirror travel in vacuum is $300\mathrm{nm}$ per cycle. A single endpoint light intensity cannot distinguish $q=0$ from $q=1$; the continuous fringe count carries the missing information.

For an interpretable cell measurement, establish the initial phase difference with a small calibrated mirror scan, then track slow changes through successive interference fringes. Measure the dark signal and each arm separately, keep photodiode gain within its linear range, align both beams on the same region and match their linear polarizations. Monitor interferometric visibility and keep arm delays within the useful optical coherence time. Record cell length, contents, pressure and temperature; return to the starting condition to check drift. Changes in window deformation, cell length, external air or mirror position also change optical path length, so a reference cell or an independently monitored mechanical path is needed to isolate the contents' refractive index. These controls are deductions from the model, not a claim that the illustrated apparatus was operated. Loss of fringe count leaves a whole-cycle ambiguity; at an intensity extremum the first derivative with respect to phase difference vanishes, so a scan around phase quadrature is more sensitive to small changes. Broadband illumination, unstable optical coherence, vibration and strong loss can invalidate a simple one-frequency inference.

Use the electronic weak-field Zeeman splitting model, neglecting hyperfine structure. The Landé g-factor is $g_g=2$ for $3S_{1/2}$ and $g_e=2/3$ for $3P_{1/2}$. The shift of an absorption frequency is thereforeHere the magnetic field is weak relative to spin-orbit coupling, so $J$ remains a useful quantum number, but the spectral line widths are small enough to resolve its magnetic substates.

For linear polarization parallel to $z$, the optical electric field has only the field-axis $\pi$ component. The electric-dipole selection rule is $\Delta m=0$. Its two allowed transitions are $m_g=m_e=\pm 1/2$, givingFor propagation perpendicular to $z$ with linear polarization also perpendicular to $z$, the optical electric field is an equal-amplitude combination of the two field-axis circular components. It drives ${\sigma }^{\pm }$ transitions with $\Delta m=\pm 1$. These connect $-1/2\to +1/2$ and $+1/2\to -1/2$, givingThus each orientation selects two lines, with twice the separation in the perpendicular-polarization case. Neither pair has an unshifted central component. This is polarization-selective Zeeman splitting of an alkali D1 line, an anomalous Zeeman effect rather than a normal orbital triplet.

Use $Z|H⟩=|H⟩$, $Z|V⟩=-|V⟩$. Choose the eigenstate phasesThe inverse relations are $|H⟩=(|H^{\prime }⟩+|V^{\prime }⟩)/\sqrt{2}=(|R⟩+|L⟩)/\sqrt{2}$ and $|V⟩=(|H^{\prime }⟩-|V^{\prime }⟩)/\sqrt{2}=-i(|R⟩-|L⟩)/\sqrt{2}$. The X states are linearly polarized at $+45^{\circ }$ and $-45^{\circ }$ to horizontal, while the Y states have equal orthogonal amplitudes in quadrature and hence opposite circular polarizations. We define R as the $+Y$ state; descriptions as clockwise or anticlockwise must specify the observation direction and time convention.

A polarizing beam splitter and two single-photon detectors directly measure Z, routing H and V to separate detectors. For X, place a half-wave plate with its axis at $22.5^{\circ }$ before that splitter; it implements a Hadamard transformation up to a global phase. For Y, first apply a quarter-wave retardance $Q=\mathrm{diag}(1,-i)$ in the H/V axes and then the same half-wave plate. The total analyzer transformation is $HQ$, with $HQ|R⟩=|H⟩$ and $HQ|L⟩=|V⟩$. The physical fast/slow-axis assignment is selected to realize this stated retardance.

A full fringe scan is unnecessary. Use the same components to make three fixed projective measurements, on separate batches of identically prepared photons. Direct path detection after the polarizing beam splitter measures the $H/V$ populations. Recombine after matching the arm photon polarizations, once with $\theta =0$ and once with $\theta =\pi /2$, to measure the two quadratures of the quantum coherence.

In the $H/V$ basis, write the density matrix in terms of its Bloch vector as $\rho =(I+r_{x}X+r_{y}Y+r_{z}Z)/2$. The previous formula becomesThus three measured relative frequencies giveand determineThe two interferometric settings analyze the diagonal linear polarizations $(|H⟩\pm |V⟩)/\sqrt{2}$ and the circular polarizations $(|H⟩\pm i|V⟩)/\sqrt{2}$, with detector labels fixed by the calibrated quantum phase convention. This is informationally complete three-observable qubit tomography: direct measurements in three bases replace fitting an entire phase-dependent fringe. It also reconstructs a mixed input, for which $|r|\le 1$, although the source here is assumed pure.

Simply measuring transmission through one fixed polarizing beam splitter would determine only $r_z$. Rotating a linear analyzer through different angles can additionally determine $r_x$, but cannot distinguish the two signs of $r_y$ without a relative-phase measurement. For arbitrary, possibly elliptical photon polarization, all three settings above are needed; no extra analyzer element beyond the described path-phase control and photon polarization rotator is required.

The three contributions to the Frank elastic energy penalize distinct slow distortions of the nematic director. Nematic splay measures local spreading or convergence and is governed by $K_1(\nabla \cdot \mathbf{n}{)}^2/2$. Nematic twist measures rotation around an axis perpendicular to the director, with cost $K_2(\mathbf{n}\cdot \nabla \times \mathbf{n}{)}^2/2$. Nematic bend measures curvature of director integral lines; for a unit vector, $(\mathbf{n}\cdot \nabla )\mathbf{n}=-\mathbf{n}\times (\nabla \times \mathbf{n})$, giving cost $K_3|\mathbf{n}\times \nabla \times \mathbf{n}{|}^2/2$. These terms respect the head-tail symmetry of a nematic liquid crystal. Positive elastic constants oppose these distortions.

For the one-dimensional reduction assume lateral uniformity, so $\theta$ and $\varphi$ depend only on $z$. A prime below means $d/dz$. Directly differentiating the nematic director givesThe squared curl is ${\sin }^2\theta {\theta }^{\prime 2}+{\cos }^2\theta {\varphi }^{\prime 2}$. Subtracting the squared twist invariant givesConsequently the one-dimensional twisted nematic energy has elastic coefficientsA normal electric field has $\mathbf{E}\cdot \mathbf{n}=E_0\sin \theta$. There is a genuine factor-of-two inconsistency in the source: its initial functional places the electric term inside the overall factor $1/2$, whereas its subsequent density and threshold use the full electric term. For the requested density and threshold, use the usual orientational coupling $-ϵE_0^2{\sin }^2\theta /(8\pi )$. Here $ϵ$ is the dielectric anisotropy of a nematic, in the printed Gaussian-unit convention. If the first functional is instead read literally, this coupling is halved and the resulting threshold voltage is multiplied by $\sqrt{2}$.

Put $c=ϵE_0^2/(8\pi )$ and $\mathcal{L}=f{\theta }^{\prime 2}/2+g{\varphi }^{\prime 2}/2-c{\sin }^2\theta$. Apply the Euler-Lagrange equations for two fields. Variation in $\varphi$, which does not appear explicitly, givesVariation in $\theta$ gives $(f{\theta }^{\prime }{)}^{\prime }-f_{\theta }{\theta }^{\prime 2}/2-g_{\theta }{\varphi }^{\prime 2}/2+2c\sin \theta \cos \theta =0$, orAt $\theta =0$, $f=K_1$, $g=K_2$, and $f_{\theta }=g_{\theta }=0$. With zero field, ${\theta }^{\prime }=0$ and ${\varphi }^{\prime }=\pi /(2L)$ satisfy both equations and the imposed azimuthal endpoints. Thus the quarter-turn planar texture is a stationary solution. Stationarity alone is not a claim of stability for arbitrary elastic constants.

To find the Fréedericksz transition, impose the additional strong planar anchoring $\theta (0)=\theta (L)=0$. Write $\alpha =\pi /(2L)$. The small-angle expansions areThe conserved $g{\varphi }^{\prime }$ and fixed total twist imply ${\varphi }^{\prime }=\alpha +O({\theta }^2)$ for a pure tilt perturbation. Linearizing the tilt Euler-Lagrange equation therefore givesSubstitute $\theta ={\theta }_0\sin (\pi z/L)$. A nonzero neutral amplitude requiresSince $V_c=E_{c}L$, the quarter-turn nematic instability threshold isThis is the first zero-endpoint mode: higher sine modes replace $K_1$ by $n^2{K}_1$ and cost more elastic energy. The quadratic energy change is $\frac{1}{2}\int [K_1{\theta }^{\prime 2}+\{(K_3-2K_2){\alpha }^2-ϵE_0^2/(4\pi )\}{\theta }^2]dz$, so the first mode has negative stiffness above this voltage. The result assumes $ϵ>0$ and $K_1+(K_3-2K_2)/4>0$. If the latter is negative the planar state is already unstable at zero field; for negative dielectric anisotropy of a nematic, a normal field opposes this tilt. The literal initial electric prefactor would give $V_{c,\mathrm{literal}}=\sqrt{2}{V}_c$.

For reflective operation of the twisted nematic field effect, orient each polarizer's transmission axis along the adjacent planar nematic director. Light entering through the top polarizer initially has $y$ linear polarization. At low voltage, adiabatic optical following in a twisted nematic rotates it through a quarter turn to $x$ as it travels downward, so it passes the bottom polarizer and reaches the mirror. On returning, it follows the reverse twist back to $y$ and exits the top polarizer: the pixel appears bright. At sufficiently high voltage, positive dielectric anisotropy of a nematic aligns most of the director with the normal field. The bulk no longer produces the quarter-turn optical rotation, so the crossed bottom polarizer blocks the incident light and the pixel appears dark. Elastic relaxation restores the bright twisted state after voltage is removed. The adiabatic requirement is more precisely $2\pi |\Delta n|/\lambda \gg |{\varphi }^{\prime }|$, involving birefringence, not thickness alone. Voltage switches director orientation and hence the reflected brightness; it does not require the liquid crystal to emit light.

The phase has wavevector $\mathbf{k}=k{\mathbf{e}}_z$. The electric field has fixed direction $({\mathbf{e}}_x+{\mathbf{e}}_y)/\sqrt{2}$, so it has linear polarization at $45^{\circ }$ to the $x$ and $y$ axes, perpendicular to the wavevector. Substitution into the electromagnetic wave equation givesFor the usual positive-frequency convention, $\omega =c|k|$ and propagation is in the direction $\mathrm{sgn}(k){\mathbf{e}}_z$. If $k>0$, this reduces to $\omega =ck$. The following wave formulas assume the nonconstant case $\omega \ne 0$.

A simple polariscope contains a fixed polarizer and a rotatable optical analyzer, with a space for a specimen between them. A historically suitable implementation uses two Nicol prisms. Each selects one linearly polarized component from the incident light through the crystal's optical action. A lamp and a collimating arrangement illuminate the first prism, and the observer examines the transmitted field through the second.

With no specimen and ideal elements, Malus's law predictswhere $I_0$ is the intensity after the polarizer and $\theta$ is the relative transmission-axis angle. Parallel settings are brightest; crossed settings give extinction. Rotating the analyser supplies an immediate check on the first element's polarization.

Insert a transparent birefringent plate between crossed elements with its principal directions oblique to their axes. The two components can acquire different phases, so they need not recombine into the original linear polarization; light can then pass the crossed analyser. With white light, wavelength-dependent retardation gives colours. Rotation of the plate, and a check without it, distinguish this response from simple sample opacity or a poor extinction adjustment. An optically active sugar solution can instead rotate the polarization plane, as in Biot's saccharimeter. The instrument therefore reveals changes of polarization through an analysed intensity, rather than making polarization directly visible to the eye.

An optically active glucose solution rotates the direction of linear polarization. With monochromatic illumination, rotating the polarimeter analyzer again produces a maximum and an extinction, but at angles shifted from the water-filled reference tube. In an ideal transparent sample, Malus's law givesThe specific rotation $[\alpha {]}_{\lambda }^T$ is defined for a stated temperature, wavelength, solvent and concentration convention. With $\ell$ in decimetres and $c$ in grams per millilitre, the conventional numerical value of $[\alpha ]$ must use the matching units. A broadband source generally gives coloured appearances because the rotation varies with wavelength; one analyser angle need not extinguish every colour.

Biot's saccharimeter is an angular polarimeter: a polarizing element, a tube of accurately known optical path length holding the sample between plane end windows, and a rotatable analysing element with an angular scale. The original arrangement used a black-glass polarizing mirror and a calcite analyser; later two-prism designs implement the same angular measurement. It should be distinguished from a later quartz-wedge compensating saccharimeter, which cancels the sample's rotation instead of directly reading the analyser rotation.^[ref]

For a historical urine estimate, clarify a sample without changing its sugar inventory, fill the tube without bubbles, make a solvent blank at the same settings, and measure the signed rotation relative to that blank. With a known dilution factor $D$ and an appropriate equilibrium glucose calibration,Repeat the reading and check a known glucose solution. Mutarotation makes the time since preparation relevant. The inference is selective only to the extent that other optically active solutes have been excluded or accounted for; a net rotation is not proof that every contribution is glucose. This is a historical analytical application, not a replacement for present clinical assessment of diabetes.

The longitudinal Faraday effect produces circular birefringence: right- and left-circular components have different phase velocities. Linear input contains equal amplitudes of both circular eigenpolarizations. A relative phase ${\delta }_F$ rotates its linear polarization through ${\delta }_F/2$, preserving linearity in an ideal transparent sample. Conventionally the signed rotation is ${\vartheta }_F=V{B}_{\parallel }{d}_F$, with Verdet constant $V$.

A pure Faraday rotator cannot serve as the required switchable quarter-wave plate. It rotates a line into another line rather than producing the required relative phase between fixed orthogonal linear components. Applying a field that gives a nominal quarter-cycle circular-eigenmode retardation simply rotates linear polarization by $45^{\circ }$. Circular dichroism or a separate linear retarder would change the device, and is not the stated pure Faraday mechanism.

Sunlight is approximately unpolarized before its interaction with the environment. For Rayleigh scattering by small isotropic scatterers, the induced electric dipoles radiate only their components transverse to the viewing direction. For unpolarized incident light scattered through angle $\chi$, the perpendicular and parallel intensities are proportional to 1 and ${\cos }^2\chi$. The degree of linear polarization is thereforeAt $90^{\circ }$ single scattering is fully linearly polarized perpendicular to the scattering plane. This explains the polarized sky away from the Sun; multiple scattering and aerosols reduce the ideal contrast.

Reflection at a dielectric surface treats the two linear polarizations differently. At Brewster incidence, $\tan {\theta }_B=n_2/n_1$, the reflected p component vanishes for ideal transparent nonmagnetic media. The remaining reflected glare is s polarized, perpendicular to the incidence plane, approximately horizontal for a horizontal water surface. Away from this angle the reflection is generally partially polarized. The transmitted beam also has unequal s/p transmission.

Refraction and internal reflection in water droplets likewise have polarization-dependent Fresnel coefficients, producing strong linear polarization in rainbow light. These are generic scattering and boundary-condition mechanisms; no biological detection mechanism is required.

For propagation along z, write $\mathbf{E}=\mathrm{Re}\{(E_x\hat{\mathbf{x}}+E_y\hat{\mathbf{y}})e^{i(kz-\omega t)}\}$. A Jones vector contains the complex transverse components. Equal phase modulo $\pi$ gives linear polarization; equal magnitudes with phase difference $\pm \pi /2$ give circular polarization. A common overall phase has no effect on the polarization ellipse. A deterministic optical element acts by a Jones matrix, with successive matrices ordered from right to left along the light path.

A circular polarizer uses a linear polarizer transmitting x and a quarter-wave plate whose principal axes are at $\pm 45^{\circ }$. Up to an overall phase, convenient matrices in fixed physical x/y axes areThe plate is lossless, $Q{Q}^{†}=I_2$, and $Q(1,0{)}^T=(1,i{)}^T/\sqrt{2}$ is circular in the supplied convention. Put P on side A and Q on side B: forward passage is $J_{A\to B}=QP$.

Unpolarized light has no single Jones vector. Represent its intensity by the coherency matrix $C_{\mathrm{in}}=(I_0/2)I_2$ and propagate it as $C_{\mathrm{out}}=J{C}_{\mathrm{in}}{J}^{†}$. Forward transmission givesThus the transmitted light is circular with intensity $I_0/2$. The initial polarizer removes one incoherent linear component; the plate changes phase without restoring the lost intensity.

Reverse passage encounters Q before P and has $J_{B\to A}=PQ$ in the same fixed transverse axes. It givesThe output is linearly polarized along x and again has intensity $I_0/2$. A retarder alone leaves an unpolarized coherency matrix unchanged. Consequently an unpolarized scene appears darker from either side, although the transmitted polarization differs. No polarization-dependent response of the observer is needed to explain this reduction.

For the mirror calculations below, use an ideal near-normal isotropic reflector, $M=-I_2$ up to a common phase. In fixed physical transverse axes reflection leaves the polarization ellipse unchanged; it reverses circular helicity because the propagation direction reverses. A reciprocal plate has the same relative phase delay on the return traversal in this fixed-axis convention. In particular $Q^2=i\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$: two passes exchange x and y linear states.

Let ${\mathbf{u}}_{\theta }=(\cos \theta ,\sin \theta {)}^{\mathsf{T}}$ be the transmission direction. An ideal linear polarizer projects the incident electric field onto this direction, so its Jones matrix isSince ${\mathbf{u}}_{\beta }^{\mathsf{T}}{\mathbf{u}}_{\alpha }=\cos (\alpha -\beta )$, the amplitude transmitted by a sequence of polarizers is the product of the projections between successive axes. Here the four directions are $0,\omega t,-2\omega t,0$. After the first polarizer the unpolarized light input has intensity $I/2$ and is in linear polarization along $x$. The transmission through a sequence of linear polarizers therefore follows from successive applications of Malus's lawThe relative angle between the rotating polarizers is $3\omega t$, not $\omega t$. The first factor $1/2$ is an ensemble average over incident polarization: unpolarized light cannot be represented by a single deterministic Jones vector. Equivalently, its coherency matrix is $(I/2)I_2$; the total matrix $P(0)P(-2\omega t)P(\omega t)P(0)$ gives the same output intensity by the trace rule. The quasistatic optical-element approximation assumes the mechanical rotation is slow compared with the optical oscillation.

The Faraday effect rotates the plane of linear polarization when light propagates through a material with a magnetic field component along the ray. In the weak-field transparent regime,with a stated propagation and transverse-basis convention; $V$ is the Verdet constant. For a uniform longitudinal field and sample length $L$, the magnitude is $|VBL|$. A scalar intensity change without a corresponding polarization analysis would not establish this effect.

The microscopic field breaks the equivalence of the two circular polarization modes: the Lorentz force and the Zeeman shifts of optical transitions produce different susceptibilities for their opposite senses of rotation. A linearly polarized wave is an equal-amplitude superposition of these two modes. If their real refractive refractive indices are $n_{+},n_{-}$, their relative wave phase after the sample is $\Delta \varphi =(\omega L/c)(n_{+}-n_{-})$. Recombination rotates the line byEqual transmitted circular amplitudes preserve linear polarization. Unequal absorption produces circular dichroism and generally elliptical polarization as well, so pure rotation is an approximation appropriate away from strong absorption.

A modern motivating experiment sends polarized laser light through a glass rod inside a solenoid, then through a rotatable analyzer onto a photodiode. Scan the analyzer first to calibrate its angular response. Driving the solenoid sinusoidally and recording the detector synchronously separates a small rotation signal from unsynchronized noise; the conversion to angle still requires the analyzer slope and the magnetic-field integral through the rod. A first-party William & Mary laboratory manual specifies this apparatus and measurement procedure.^[ref]

Additional controls are equal-magnitude field reversals, zero-field readings before and after each pair, and a blocked-beam pickup test. The odd-in-field rotation is $[{\theta }_{\mathrm{obs}}(B)-{\theta }_{\mathrm{obs}}(-B)]/2$; a fixed analyzer offset cancels. Heating, stressed-glass birefringence, source drift and field-dependent transmission need separate checks. A single reading near crossed extinction loses the sign because its leading response is quadratic; an analyzer near $45^{\circ }$ gives a linear small-rotation response.

Faraday rotation is nonreciprocal. With the magnetic bias held fixed, retracing the path adds rotation in the same laboratory sense rather than undoing it, unlike reciprocal natural optical activity. This permits an optical isolator with a $45^{\circ }$ Faraday rotator between appropriately oriented polarizers. One must compare the returned polarization in a consistently defined laboratory basis; merely reversing the sign of $\mathbf{B}\cdot \hat{\mathbf{k}}$ while retaining the old ray-basis convention gives an incorrect cancellation.

Both Thomson scattering and Rayleigh scattering can be understood as radiation from a dipole driven by the incident electric field. Their different frequency dependences come from the response of a free charge compared with a bound charge.

For a free particle of charge $q$ and mass $m$, a harmonic field gives $m\ddot{\mathbf{x}}=q\mathbf{E}$, hence $\mathbf{p}=q\mathbf{x}=-q^2\mathbf{E}/(m{\omega }^2)$. The radiated field contains two time derivatives of $\mathbf{p}$, cancelling the response's ${\omega }^{-2}$. The Thomson cross section is consequently independent of frequency in the classical low-recoil regime:For incident linear polarization with unit polarization $\mathbf{ϵ}$, summing over outgoing polarizations gives $d\sigma /d\Omega =r_q^2[1-|\hat{\mathbf{n}}\cdot \mathbf{ϵ}{|}^2]$. Radiation vanishes along the acceleration and is strongest perpendicular to it. Averaging an incident unpolarized light beam giveswhere $\Theta$ is the scattering angle. Scattering through $90^{\circ }$ is fully linearly polarized perpendicular to the scattering plane for this ideal isotropic dipole model. The electron case dominates over equally charged heavier particles because the cross section scales as $m^{-2}$. recoil corrections become necessary when photon energy is no longer small compared with $m{c}^2$.

For a bound charge with natural frequency ${\omega }_0$ and damping rate ${\gamma }_d$,Far below resonance, ${\alpha }_{\mathrm{e}}$ is approximately constant, soThis is Rayleigh scattering. It requires a scatterer small compared with the wavelength and, for the simple expression, an isotropic linear dipole response. Atoms, molecules and small dielectric particles provide examples. The stronger scattering of shorter visible wavelengths explains blue skylight; absorption and the nonuniform solar spectrum prevent treating that simple scaling as a complete color calculation.^[ref]

For isotropic electric scatterers the normalized angular and polarization patterns are the same as in Thomson scattering. The frequency scaling distinguishes their limits: a free charge has a displacement proportional to ${\omega }^{-2}$, whereas a tightly bound charge has an approximately frequency-independent displacement. Near resonance, far outside the small-particle limit, or with anisotropic polarizability, the simple Rayleigh formula must be modified. In particular, the electric-only pattern must not silently be applied to a perfectly conducting sphere whose magnetic dipole response is also retained.

A scattering measurement should subtract stray-light and detector backgrounds, control scatterer density and optical thickness, and calibrate detector response and incident power when comparing wavelengths. A polarization analyzer at $90^{\circ }$ tests the dipole pattern; multiple scattering or unresolved nonisotropic contributions can reduce its polarization.

Polarization under a boost. Lorentz invariance of normalized plane-wave polarization preserves the polarization ellipse's shape, handedness and degree of polarization when its transverse basis is carried to the aberrated ray. The propagation direction changes; equality of a fixed laboratory Cartesian pair of field components is not the appropriate comparison.

Rotate the $yz$ axes so that a general incident vacuum plane wave has unit direction $\hat{\mathbf{n}}=(\cos \vartheta ,\sin \vartheta ,0)$. Choose the transverse basis ${\mathbf{e}}_p=(-\sin \vartheta ,\cos \vartheta ,0)$, ${\mathbf{e}}_s=(0,0,1)$, and arbitrary complex amplitudes $E_p,E_s$. ThenPutting $\beta =v/c$, the given transformation givesThe four-wavevector transforms with Doppler factor $\mathcal{D}=\gamma (1-\beta \cos \vartheta )>0$, givingIn the corresponding basis ${\mathbf{e}}_p^{\prime }=(-\sin {\vartheta }^{\prime },\cos {\vartheta }^{\prime },0)$, ${\mathbf{e}}_s^{\prime }={\mathbf{e}}_s$, the transformed field isBoth components of the Jones vector acquire the same positive real multiplier. Their amplitude ratio and relative wave phase are unchanged, proving preservation of arbitrary linear, circular or elliptical polarization. All Stokes parameters scale by ${\mathcal{D}}^2$, so their normalized values and the degree of polarization are unchanged. An unpolarized light ensemble remains unpolarized under this single-ray boost. A mixture of different ray directions requires separate transformations and is outside this statement.

Reflection from the moving slab. Take the upper interface as $z=0$, the material below it, and the incident ray along $-\hat{\mathbf{z}}$. This fixes the sign left implicit by “along the $z$-axis”; choosing the opposite surface convention changes no result below. Treat $n$ as the real refractive index in the slab's rest frame, with unit relative magnetic permeability. In the laboratory,Transform to the slab rest frame:The incidence is now oblique, in the $xz$ plane, at angle $i$ to the surface normal:The incoming ensemble remains unpolarized light. In this frame the ordinary stationary-interface Fresnel equations give the amplitude coefficients:For generic oblique incidence, $R_s=|r_s{|}^2$ and $R_p=|r_p{|}^2$ differ. Equal incident $s$ and $p$ intensities therefore produce unequal reflected intensities: the reflection is partially polarized. Its degree of linear polarization isThe inverse boost preserves this degree, so the same conclusion holds in the laboratory even though the laboratory incident ray is normal to the surface. The moving material cannot be treated as an isotropic stationary dielectric in that frame.

The reflected rest-frame wavevector is $(\omega /c)(\gamma ,-\gamma \beta ,0,+1)$. Boosting back gives $(\omega /c)(1,0,0,+1)$: the reflected laboratory ray is normal to the surface and has the incident frequency. There is no contradiction with the oblique rest-frame geometry. Translation tangential to a uniform interface does not itself introduce a moving normal mirror.

Complete linear polarization. For Brewster reflection from a tangentially moving dielectric, at Brewster's angle, $r_p=0$ and $\tan i_B=n$. Combining this with the aberration relation givesOnly the $s$ component is reflected. The plane of incidence in the slab frame is $xz$, so the reflected electric field is along $y$, perpendicular to the slab velocity. Boosting this $s$ field back keeps its electric direction along $y$. For example, the reflected rest-frame $s$ field has $B_z^{\prime }=-\beta E_y^{\prime }/c$; hence $E_y=\gamma (E_y^{\prime }+v{B}_z^{\prime })=E_y^{\prime }/\gamma$ and $E_x=E_z=0$.

This exact extinction assumes a transparent nonmagnetic dielectric with $n\ne 1$. At $v=0$ the reflected intensities are equal and the reflected beam is unpolarized. If $n=1$, both reflection coefficients vanish and there is no reflected beam whose polarization can be assigned. With dispersion, evaluate $n$ at the shifted frequency ${\omega }^{\prime }=\gamma \omega$; the Brewster-speed equation is then implicit. absorption generally prevents a real incidence angle from making the complex $r_p$ exactly zero.

Birefringence is polarization-dependent propagation in an optically anisotropic medium. Its permittivity tensor relates $\mathbf{D}$ to $\mathbf{E}$. In a lossless uniaxial dielectric, take the optic axis along $z$ and the principal relative permittivities as $(n_o^2,n_o^2,n_e^2)$. For a wave vector making angle $\vartheta$ with that axis, the ordinary mode has index $n_o$, while the extraordinary phase-index angular dependence isThe ordinary electric field is perpendicular to the plane containing the wave vector and optic axis. The extraordinary electric displacement field lies in that plane and is perpendicular to the wave vector, but its electric field need not be. Its energy-flow direction can therefore differ from the wave-normal direction: birefringent walk-off. Along the optic axis the two indices coincide. Refraction generally splits an incident beam into two differently polarized rays, producing double images, as in calcite.

A waveplate instead uses the relative phase of two eigenpolarizations. With the axes chosen for the propagation direction, the retardance is $\Delta =2\pi L(n_2-n_1)/{\lambda }_0$. A quarter-wave plate gives $|\Delta |=\pi /2$ modulo $2\pi$ and converts equal-amplitude orthogonal components between linear and circular polarization. A half-wave plate gives $|\Delta |=\pi$ and rotates a linear polarization. This is a phase effect; unequal absorption is a separate property.^[ref]