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An optical isolator transmits light preferentially in one direction. An ideal Faraday effect device combines a $45^{\circ }$ rotator with polarizers at $0^{\circ }$ and $45^{\circ }$: forward light passes, while backward light acquires another $45^{\circ }$ rotation and reaches the entrance polarizer at $90^{\circ }$. See Faraday rotation on a round trip.

The closest components above differ by $\Delta {\nu }_{\min }={\mu }_{B}B/(2h)\simeq 7.0\mathrm{GHz}{\mathrm{T}}^{-1}B$. At $B=0.1\mathrm{T}$, their separation is about $0.70\mathrm{GHz}$. A visible spectral line near $500\mathrm{nm}$ would consequently require spectral resolving power of order $\nu /\Delta {\nu }_{\min }\simeq 9\times 10^5$. An ordinary low-resolution spectrometer would merge the components.

A suitable arrangement places an excited low-pressure strontium discharge or a transversely observed atomic beam between wide electromagnet pole pieces. Collect the emission perpendicular to the magnetic field, select the desired spectral line with a prefilter or monochromator, and send it through a high-finesse scanning Fabry–Pérot interferometer to a detector. A rotatable linear polarizer separates the two $\pi$ components from the four $\sigma$ components. Calibrate both the interferometer's free spectral range and the local magnetic field; take zero-field and several field-strength scans, and fit component positions against $B$.

The magnetic field should be uniform across the viewed volume, with its associated Zeeman splitting larger than the combined natural linewidth, Doppler broadening, pressure broadening and instrumental width, but much smaller than the separations of the relevant atomic fine-structure levels. A collimated transverse atomic beam, low collision rate and restricted collection angle reduce unwanted spectral line broadening. Avoid appreciable power broadening, resolve or model isotope and hyperfine structure, and check for self-absorption. A zero-field scan and the predicted linear scaling help distinguish actual Zeeman splitting components from neighbouring spectral lines.

For quantum key distribution with Bell pairs, a source distributes one photon to Alice and its partner to Bob. One implementation is the Ekert protocol, using the spin singlet state $|{\psi }^{-}⟩=(|01⟩-|10⟩)/\sqrt{2}$. Each party independently chooses a polarization measurement setting after distribution. Equal settings give opposite outcomes; Bob reverses his outcome-to-bit assignment so these rounds give matching random bits. Other setting pairs test a Bell inequality, with ideal spin singlet state correlation $E(\mathbf{a},\mathbf{b})=-\mathbf{a}\cdot \mathbf{b}$ for the corresponding Bloch sphere axes. The optical polarizer angles correspond to twice those angles on the appropriate Bloch sphere great circle.

Over an authenticated public classical channel, Alice and Bob announce settings and detection times, select coincident detection rounds, and perform key sifting. They reveal a test sample to estimate the quantum bit error rate and the chosen Bell inequality signal, discarding all revealed bits. Excess disturbance causes an abort. Otherwise information reconciliation makes the retained strings agree and privacy amplification shortens them to remove an adversary's residual information, including information disclosed during reconciliation. The tested quantum entanglement supplies correlations unavailable to an independently prepared classical mixture; a security claim also requires stated detector and sampling assumptions.^[ref]

Each photon polarization analyzer uses a setting-dependent waveplate followed by a polarizing beam splitter. A single-photon detector on each output converts the routed photon into an electrical click, recording the outcome and its time. For example, a single-photon avalanche diode is a reverse-biased photodiode operated above breakdown: absorption creates a charge carrier, impact ionization triggers an avalanche, and the resulting pulse is registered. A quenching circuit lowers the bias to stop the avalanche and then resets the device. The detector dead time during recovery limits the count rate; dark counts produce false clicks, and imperfect detection loses rounds. The polarization information comes from the preceding analyzer, while the detector registers which output fired. Coincidence timing associates Alice's and Bob's records with the same emitted pair. ^[ref]

Q-switching stores excitation in a pumped laser gain medium while preventing oscillation, then rapidly changes the optical cavity from high loss, or low quality factor, to low loss, or high quality factor. The initially suppressed stimulated field can then grow rapidly from a seed and deplete a large population inversion in a short pulse. The resulting peak power is much higher than for continuous operation at the same average pumping rate. This is energy storage and release, rather than the phase locking of many longitudinal modes used in mode locking.

An active implementation uses a Pockels cell and a polarizer. An applied voltage makes the cell change the intracavity polarization, causing the polarizer to introduce sufficient loss to prevent lasing while pumping builds the inversion. Rapidly changing the voltage removes that loss and opens the high-Q state. The switching time should be short relative to the pulse build-up and depletion times. A saturable absorber provides an alternative passive Q switch: its absorption decreases as the intracavity intensity rises.

The normal to the section is $[010]$, so its in-plane principal vibration directions are $[100]$ and $[001]$. Their refractive-index difference isThe slow direction is $[001]$. The optical path retardation isAt an angle $\alpha$ between a principal direction and the polarizer, ideal crossed-polar transmission has the formwhere $I_P$ is the intensity after the first polarizer. The two components recombine parallel to the incident polarization when $\Gamma =m\lambda$, and the crossed analyser removes them. ConsequentlyThe next order is260nm, outside ordinary visible illumination. At exact principal-axis alignment, $\sin 2\alpha =0$ and every wavelength is extinguished, explaining the black appearance rather than a special single-wavelength cancellation.

The quartz compensator is a birefringent optical compensator: a retarder of known principal directions. With its principal axes at45degrees to the polarizer and analyser, the previously dark isotropic background acquires the compensator's own interference colour. The crystals are now viewed through two retarders.

When a specimen's slow axis is parallel to the compensator's slow axis, the retardations add. A90degree relative principal-axis rotation pairs the specimen's slow axis with the compensator's fast axis, and the signed retardations subtract:Thus different crystallites can shift to higher or lower interference orders even if their original hue was similar. In the subtracting orientation, equal magnitudes can compensate and yield extinction. These shifts identify fast and slow specimen directions.

At a general relative angle the full polarization transformation is required; simple scalar addition is only valid for aligned principal axes. The source does not specify the compensator retardation, so no unique list of resulting named colours can be assigned. The robust predictions are a changed background, orientation-dependent addition or subtraction, and possible compensation.

On rotating a fixed crystallite around the microscope axis, the angle $\alpha$ to the fixed polarizer changes. The factor ${\sin }^2(2\alpha )$ repeats every90degrees. Each birefringent region is therefore extinguished four times in a complete revolution and reaches maximum brightness45degrees from its own extinction positions.

Different regions need not extinguish together if their principal in-plane directions differ. For a fixed propagation direction through a uniform region, azimuthal rotation changes brightness more directly than its retardation colour: the underlying index difference and thickness remain the same. The isotropic unstressed matrix remains dark throughout rotation.

An optically isotropic film changes the phase of both transverse polarization components equally, so crossed polarizers still extinguish the transmitted light. The etched rod-and-gap composite has form birefringence: its effective permittivity depends on the direction of the electric field, even when the constituent material is intrinsically isotropic. In a simple subwavelength lamellar model,The precise effective-medium formula changes with rod geometry, but the anisotropy remains. Resolving incident light along the two principal directions produces a phase retardation. An ideal lossless specimen between crossed polarizers transmitswhere $I_P$ is the intensity after the first polarizer. Extinction occurs when a principal direction follows either polarizer; transmission is strongest near $45^{\circ }$.

The local slow direction follows the radial rods. Between crossed polarizers, the angular factor is ${\sin }^2(2\phi )$, so radial directions parallel to either polarizer are dark. The pattern is a Maltese cross in polarized-light microscopy, with four bright sectors separated by dark arms parallel to the polarizer directions. Its colours depend on the local optical path difference; tapering rods or changing fill fraction can make this vary with radius. The cross belongs to the local radial-axis pattern and does not imply a conoscopic interference figure.

Use time dependence $e^{-i\omega t}$, positive wavenumber $k$ and aperture normal $\mathbf{n}={\mathbf{e}}_x$. Write $F(y^{\prime },z^{\prime })=E_{\mathrm{inc},y}(0,y^{\prime },z^{\prime })$ on the aperture and extend $F$ by zero outside it. DefineThe outgoing branch has $\kappa \ge 0$ for propagating components and $\mathrm{Im}\kappa >0$ for evanescent components.

Fix the Green function normalization by $G_0(\mathbf{r})=e^{ik|\mathbf{r}|}/(4\pi |\mathbf{r}|)$, so $(\Delta +k^2)G_0=-\delta$. Its transverse Fourier transform satisfies $({\partial }_x^2+{\kappa }^2){\hat{G}}_0=-\delta (x)$. An outgoing solution on both sides is proportional to $e^{i\kappa |x|}$, and its derivative jump fixes ${\hat{G}}_0=i{e}^{i\kappa |x|}/(2\kappa )$. Thus the Weyl plane-wave representation for observation at $x>0$ isSince $\mathbf{n}\times {\mathbf{E}}_{\mathrm{inc}}=F{\mathbf{e}}_z$, the given aperture integral becomes a superposition of curls of ${\mathbf{e}}_z{e}^{i\mathbf{q}\cdot \mathbf{r}}$. Each curl supplies $i\mathbf{q}\times {\mathbf{e}}_z=i(\nu ,-\kappa ,0)$. Therefore, with the unit prefactor in that integral,Every vector amplitude is transverse to $\mathbf{q}$, as Maxwell's equations require. The polarizer specifies the incident aperture field; diffraction can regenerate an $x$ component outside the aperture. In contrast, its $z$ component remains zero in this representation. Evanescent plane waves are needed for the complete near field, not just the propagating disk ${\nu }^2+{\eta }^2\le k^2$.

The factor $1/2$ follows explicitly from the standard outgoing Green function for the three-dimensional Helmholtz equation and the printed unit curl prefactor. If the intended kernel is the doubled half-space kernel $2G_0$, or the aperture formula carries the usual compensating factor $2$, delete that $1/2$. The latter normalization makes the tangential output trace equal to $F$. Keeping the convention explicit avoids silently changing the supplied representation; the angular dependence and transversality of an electromagnetic plane wave are the same in either normalization. Reversing the specified normal also reverses the overall sign.

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.

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.

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.

For two-element polarization identification, assume ideal elements and use an intensity detector. For each source, test both element orders while rotating their relative axes. The circular source followed by the quarter-wave plate and then the polarizer is the combination that gives full intensity modulation. The plate converts circular light to linear light, and the downstream polarizer then obeys Malus's law.

If the polarizer comes first, it transmits half the circular source's intensity and the downstream lossless plate cannot change that intensity. Unpolarized light also gives constant half-intensity in either order: a wave plate leaves it unpolarized, while a polarizer followed by a wave plate has no final analyzer. Thus the modulating source is circular, its first element is the quarter-wave plate, and its second is the polarizer; the remaining source is unpolarized. Testing both orders avoids needing to know either source's original brightness.

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.

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.

Choose the first polarizer's pass axis as ${\mathbf{e}}_1$ and write the second as $\mathbf{a}=({\mathbf{e}}_1+{\mathbf{e}}_2)/\sqrt{2}$. The positive signed rotation takes ${\mathbf{e}}_1$ toward $\mathbf{a}$ by the short $\pi /4$ angle. Let $B_{\parallel }$ be signed according to $\theta ={\rm Y}L{B}_{\parallel }$. The Jones matrices of ideal polarizers are their orthogonal projection matrices,Removing the unimportant common propagation phase, the lossless glass block has the rotation matrixThe elements act from right to left, soWith the common optical phase retained, multiply this by $e^{i\bar{\varphi }}$. Extra scalar attenuation would also multiply the result, but none is supplied. A single Jones vector does not represent unpolarized light; the device's Jones matrix is nevertheless well defined for each incident component and can be applied to their statistical mixture.

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.