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The Faraday effect is magnetic field-induced circular birefringence: opposite circular polarizations acquire different wave phases, rotating transmitted linear polarization. For a transparent medium with isotropy before applying the magnetic field, in the linear magnetic-response regime at fixed vacuum wavelength ${\lambda }_0$ and temperature $T$,Here $V$ is the Verdet constant, $L$ is the material path length, and $B_{\parallel }$ is the applied magnetic flux density component along the ray. This $\mathbf{B}$ is the external bias, distinct from the oscillating magnetic field of the light itself. Michael Faraday discovered the effect in 1845.^[ref]

Fix the sign before using the formula. On the forward path choose a right-handed Cartesian coordinate system, propagation along $+z$, and measure $\theta$ from $+x$ toward $+y$ for the electric field's oscillation line. Positive $\theta$ is counterclockwise in the $x$-right, $y$-up end view seen from $z>0$, looking toward the source. Then $B_{\parallel }=B_z$ and positive $V$ means positive rotation for positive $B_z$. This convention differs from the clockwise-positive convention in optical rotation. A linear polarization line is unchanged by $180^{\circ }$; a measured orientation determines accumulated rotation only modulo $\pi$, unless continuity selects a branch.

For uniform $B_{\parallel }$, the formula reduces to $\theta =VBL$. Its appeal is that a weak change of refractive index becomes a visible change of orientation. Magnetic reversal measurement of Faraday rotation turns this into an inference of $V$ or of the magnetic field path integral. Faraday rotation on a round trip explains the directional behavior used by an optical isolator. In an astronomical cold plasma, the high-frequency limit instead leads to the rotation measure law $\theta =\mathrm{RM}{\lambda }_0^2$; that wavelength dependence is not a universal law for transparent solids.

A William & Mary teaching experiment places a $10\mathrm{cm}$ glass rod inside a solenoid. Align a polarized helium-neon laser through the rod's centre onto a photodiode after a rotatable polarimeter analyzer. First scan the analyzer from maximum to minimum transmission in $5^{\circ }$ steps and record detector voltage, establishing the angular response. A function generator and electronic amplifier then drive sinusoidal electric current through the solenoid; a multimeter measures the electric current and an oscilloscope reads the photodiode. Trigger the oscilloscope from the generator, remove the large constant detector component by AC coupling, and average repeated traces. Convert the small synchronized voltage modulation to rotation using the calibrated angular slope. The manual fits rotation against magnetic flux density, using its stated coil calibration and converting measured root mean square current to peak current. This is a specified laboratory procedure, rather than measured results reproduced here.^[ref]

To test the inference, add matched zero-electric current readings before and after each reversal, hold the analyzer fixed during paired readings, and compare equal positive and negative electric currents. A blocked beam tests electrical pickup in the detector channel. These are proposed controls. Joule heating scales as current squared, so reversing electric current at equal magnitude helps separate an odd rotation from even heating effects; it does not remove drift between readings. Record temperature and allow repeatable settling. Changes in transmission, mechanical motion, stressed-glass birefringence and detector pickup can imitate a signal. Synchronized averaging suppresses unsynchronized noise, but cannot reject an artifact synchronized to the drive.

Quantitative extraction of the Verdet constant also needs the integral of magnetic flux density through the rod: a finite solenoid can have a nonuniform axial magnetic field. Use its calibrated path integral, or justify a uniform-field approximation; do not silently multiply a central field by the whole rod length. A signed reversal or phase-sensitive readout tests the part odd in applied magnetic field. Its agreement with the simple rotation law supports the model under these controls, rather than proving that every change of transmitted light intensity is Faraday rotation.

Consider an original model at fixed vacuum wavelength $600\mathrm{nm}$ and temperature $20^{\circ }\mathrm{C}$. Stipulate a transparent sample with Verdet constant $V=20\mathrm{rad}{\mathrm{T}}^{-1}{\mathrm{m}}^{-1}$, length $L=0.050\mathrm{m}$, uniform adjustable $B_z$, and a field-independent analyzer-zero offset ${\theta }_0=0.020\mathrm{rad}$. These are chosen values, not measured glass properties. On a continuously tracked small-angle branch,Taking half the difference gives ${\theta }_{\mathrm{odd}}=0.100\mathrm{rad}$; half the sum recovers ${\theta }_0=0.020\mathrm{rad}$. Hence $V={\theta }_{\mathrm{odd}}/(LB)=20\mathrm{rad}{\mathrm{T}}^{-1}{\mathrm{m}}^{-1}$. A field-reversal difference of $0.200\mathrm{rad}$ is twice the single-field rotation, not that rotation itself. More generally this subtraction removes every contribution that is even under $B\mapsto -B$, provided the sample and apparatus remain otherwise matched.

An ideal polarimeter analyzer at laboratory angle $a$ reads normalized light intensity $I(a,B)/I_0={\cos }^2[a-{\theta }_{\mathrm{obs}}(B)]$. Here $I_0$ is the light intensity immediately before the analyzer, separately calibrated for each condition. A scan locates the maximum at $a={\theta }_{\mathrm{obs}}$ modulo $\pi$. At $a=\pi /4$, expansion about zero gives $I/I_0=1/2+{\theta }_{\mathrm{obs}}+O({\theta }_{\mathrm{obs}}^3)$, with angles in radians. At crossed extinction, the response instead begins as ${\theta }_{\mathrm{obs}}^2$ and loses the sign near zero. The favorable $45^{\circ }$ slope follows directly by differentiating Malus's law; its sign reverses if the angular convention is reversed.

General pure state. Use time dependence $e^{i(kz-\omega t)}$. The most general nonzero Jones vector can be writtenThere are four real parameters in the full complex field. Removing overall intensity $A^2$ and the unobservable common phase $\chi$ leaves two real degrees of freedom for pure polarization, $\alpha$ and $\delta$. Their endpoint degeneracies simply reflect the fact that the phase of a vanishing component is irrelevant. A single Jones vector cannot represent partially polarized light.

Three linear polarizers. Define the transmission axis ${\mathbf{a}}_{\theta }=(\cos \theta ,\sin \theta {)}^T$. An ideal linear polarizer has Jones matrix $P_{\theta }={\mathbf{a}}_{\theta }{\mathbf{a}}_{\theta }^T$. For definiteness label $(1,-i{)}^T/\sqrt{2}$ as right-circular in the chosen convention; reversing the handedness convention changes no intensity below. The first polarizer produces $(1,0{)}^T/\sqrt{2}$. The intermediate polarizer and final $y$ analyzer then giveso Malus's law givesThe maximum is $1/8$ at $\theta =\pi /4+m\pi /2$, and the transmission vanishes at $\theta =m\pi /2$. The first polarizer accounts for an initial factor $1/2$; retaining that factor is essential when quoting transmission relative to the incident circularly polarized beam.

Waveplate and quarter-wave matrix. A lossless waveplate has linear birefringence: its orthogonal fast and slow axes have indices $n_f<n_s$. A thickness $L$ produces slow-minus-fast phase retardationResolving the incident Jones vector into the plate axes, advancing the two phases and transforming back gives, up to a common phase,A quarter-wave plate has $\delta =\pi /2$ modulo $2\pi$. With $c_{\theta }=\cos \theta$ and $s_{\theta }=\sin \theta$, its Jones matrix is thereforeIn particular $Q(0)=\mathrm{diag}(1,i)$. Equal incident components along the two axes acquire a relative quarter-cycle phase, converting $45^{\circ }$ linear polarization to circular polarization; an incident component along either principal axis stays linearly polarized.

Half-wave replacement. A half-wave plate has $\delta =\pi$ and henceIt converts the $x$-polarized output of the first polarizer to a linear state at angle $2\theta$. The final $y$ analyzer givesThe maximum is $1/2$, because the ideal half-wave plate loses no light after the first polarizer.

Quarter-wave replacement. The transmitted $y$ amplitude is $(1-i)c_{\theta }{s}_{\theta }/\sqrt{2}$, givingMore generally, the retarder gives $I_{\mathrm{out}}/I_{\mathrm{in}}=\frac{1}{2}{\sin }^2(\delta /2){\sin }^2(2\theta )$. All three requested curves have the same zeros and angular period, but different heights:

Preparing an arbitrary ellipse. Yes: two rotatable quarter-wave plates already suffice. Let the input left-circular state be ${\mathbf{J}}_L=(1,i{)}^T/\sqrt{2}$. Up to an overall phase, any target pure polarization can be parameterized by its polarization ellipse:Here $\beta$ is the major-axis orientation, and $\eta$ sets the signed minor-to-major semiaxis ratio $\tan \eta$. First note directly from $Q(\theta )$ thatChoose the two fast-axis angles asThe first quarter-wave plate produces linear light at $\beta +\eta$. In the axes of the second, its components are $(\cos \eta ,\sin \eta )$; the second gives precisely ${\mathbf{J}}_{\mathrm{target}}$. This proves pure-polarization preparation with two quarter-wave plates, including the linear and circular limiting cases. A half-wave plate can also rotate the axes, but is not required. The phrase about left-handed input is understood as left-circular light in this construction; the explicit vector fixes the convention.

Switchable retardation. The electro-optic Kerr effect induces linear birefringence quadratic in a static transverse electric field. Defining the coefficient $K$ by $\Delta n={\lambda }_{0}K{E}_{\mathrm{dc}}^2$, its phase retardation isThe two eigenpolarizations are linear, parallel and perpendicular to the applied field. Thus an appropriate field produces a switchable quarter-wave plate:The sign of $K$ determines which axis is fast. Switching the field off removes the induced retardation, assuming no residual birefringence.

The Faraday effect instead gives circular birefringence for a longitudinal magnetic field. The two circular eigenmodes acquire different phases, rotating a linear polarization throughwith sign set by the handedness convention and the Verdet constant $V$. Its Jones matrix in the linear basis is a real rotation, up to a common phase. It preserves the ellipticity of a pure state, so a linear input remains linear. A $\pi /2$ phase difference between circular eigenmodes rotates a linear state by $\pi /4$; it does not make it circular. The Kerr device supplies the switchable quarter-wave plate; a Faraday rotator alone does not.

Flint glass is approximately optically isotropic without an applied magnetic field. A longitudinal magnetic field changes the motion of its bound Electrons: the magnetic part of the Lorentz force acts with opposite sense on the two circularly rotating components of a driven electronic oscillation. Their optical response, and hence their refractive indices, split. Equivalently, the transverse permittivity tensor acquires a field-odd antisymmetric contribution whose eigenvectors are opposite circular polarizations.

The resulting circular birefringence gives Faraday rotation. For a weak longitudinal bias it is linear in the field:Here ${\rm Y}$ is the Verdet constant and $L$ the path length. The magnetic field breaks time-reversal symmetry, which is why this rotation is nonreciprocal. Reversing the bias reverses the rotation; reversing propagation while keeping the bias fixed does not undo it when physical polarization directions are compared in fixed laboratory axes. A transverse magnetic bias can have different, higher-order effects; the setup here is the longitudinal Faraday effect.

Let $I_0$ be the incident light intensity. The first ideal polarizer transmits $I_0/2$ and produces linear polarization along ${\mathbf{e}}_1$. Faraday rotation turns that direction through $\theta ={\rm Y}L{B}_{\parallel }$, and Malus's law at the second polarizer givesEquivalently, use the input coherency matrix $C_{\mathrm{in}}=(I_0/2)I_2$ and $I_f=\mathrm{tr}(J{C}_{\mathrm{in}}{J}^{†})$. This produces the same one-half factor required by unpolarized light.

Maxima satisfy $\theta =\pi /4+m\pi$, with $m\in \mathbb{Z}$. For $L=1\mathrm{m}$ and ${\rm Y}=9\mathrm{rad}{\mathrm{T}}^{-1}{\mathrm{m}}^{-1}$, the smallest field magnitude isIts direction is the axial direction that rotates the polarization passed by $P_1$ toward the pass axis of $P_2$ through the shorter $45^{\circ }$. In the right-handed ray-basis convention where $\rho =+{\rm Y}{B}_{\parallel }$ and the depicted second axis is $+45^{\circ }$, this is $B_{\parallel }>0$, along $S\to O$. Reversing that field at the same magnitude gives $\theta =-\pi /4$ and zero transmission. A signed Verdet constant is meaningful only together with its rotation convention, so the polarization-alignment description fixes the physical direction without relying on ambiguous clockwise labels in a perspective diagram.

The rotatory power is the rotation of transmitted linear polarization per unit path length, in $\mathrm{rad}{\mathrm{m}}^{-1}$. With equally transmitted circular eigenmodes $({\mathbf{e}}_1\mp i{\mathbf{e}}_2)/\sqrt{2}$ and time convention $e^{-i\omega t}$, their accumulated phase difference gives $\theta =({\varphi }_{+}-{\varphi }_{-})/2$. For uniform circular birefringence, $\rho =\pi (n_{+}-n_{-})/{\lambda }_0$. Generic linear birefringence need not admit one rotation angle common to all incident polarization directions. In a weak longitudinal Faraday effect, $\rho =V{B}_{\parallel }$, with signed Verdet constant $V$.

The Verdet constant is the coefficient relating Faraday rotation to the magnetic flux density path integral. Its SI units are $\mathrm{rad}{\mathrm{T}}^{-1}{\mathrm{m}}^{-1}$. It depends on material, wavelength and temperature; a sign requires the stated viewing convention.^[ref]

The name commemorates Émile Verdet. For nonuniform material or temperature, use $\theta =\int V(\ell )B_{\parallel }(\ell )d\ell$ when the local scalar description remains valid. A single central magnetic field reading cannot generally replace this weighted integral. Strong additional linear birefringence or appreciable circular dichroism requires following the full polarization of an electromagnetic wave, rather than treating the output as merely a rotated line.