Birefringence ($\Delta n=n_e-n_o$)

Birefringence is the existence of different optical propagation constants for different eigenpolarizations in an anisotropic material. It provides the propagation-mode separation needed for adiabatic optical following in a twisted nematic.

Two different refractive indices can produce two displaced images, as in the calcite double-image experiment, or a phase retardation between overlapping components, as in a waveplate. Spatial separation concerns where energy flux travels; phase retardation concerns the relative wave phase. A specimen need not show two visibly separated images to have birefringence.

Photoelasticity is the dependence of a material’s optical response on stress, used to reveal otherwise invisible mechanical patterns. A transparent material that is optically isotropic before loading can become birefringent under load: orthogonal polarizations acquire different refractive indices and therefore a relative phase retardation.^[ref]

For a thin, initially isotropic specimen in a suitable linear regime, the stress-optic law relates induced birefringence to the difference of in-plane principal stresses:Here $C$ is a material coefficient with units $\mathrm{P}{\mathrm{a}}^{-1}$, $t$ is the optical thickness, $\lambda$ the vacuum wavelength and $\delta$ the phase delay. This simple form assumes the relevant stress directions and difference do not change along the light path. A general three-dimensional specimen needs a more careful optical and mechanical model.

With ideal crossed polarizers, intensity normalized to the light after the first polarizer iswhere $\theta$ is the angle between the first polarization direction and a principal optical axis. A dark region can therefore arise from axis alignment or from an integer multiple of $2\pi$ retardation. Darkness alone does not prove that stress is zero.

For illustrative $C=3\times 10^{-12}\mathrm{P}{\mathrm{a}}^{-1}$, $t=5\mathrm{mm}$ and $\lambda =500\mathrm{nm}$, one full phase cycle corresponds toThese chosen values illustrate conversion from optical delay to a stress difference, not a calibration for every plastic. The method does not directly recover a hydrostatic stress shared equally by the two in-plane directions.

A polariscope makes the measurement visible. The Illinois teaching apparatus places a specimen between a polarizer and analyzer, applies load with a manual leadscrew, measures it with a load cell and records the pattern with a camera. Monochromatic illumination and adjustable polarization optics separate information about stress directions and differences.^[ref] Comparing patterns as load changes is a motivating experiment in connecting an optical observable to a mechanical model.

The stress-optic law relates stress-induced birefringence to a mechanical stress difference. For an initially optically isotropic material in a suitable linear regime,where ${\sigma }_1,{\sigma }_2$ are the principal stresses normal to the light path and $C(\lambda )$ is the stress-optic coefficient. Its units are $\mathrm{P}{\mathrm{a}}^{-1}$, and its wavelength dependence matters when comparing optical measurements.^[ref]

For uniform thickness $t$ and uniform relevant stress directions along the path, the measured phase retardation isThe measurement determines a difference, not both stresses separately. States $(20,5)\mathrm{MPa}$ and $(40,25)\mathrm{MPa}$ have the same difference and therefore the same retardation in this model. Extra mechanical information is needed to separate their absolute stress components.

A uniaxial calibration specimen gives a practical coefficient measurement. With axial force $F$, known sectional area $A$ and negligible transverse stress, ${\sigma }_1-{\sigma }_2=F/A$. If retardation plotted against force has slope $s=d\delta /dF$, thenFor illustrative $s=0.003\mathrm{rad}{\mathrm{N}}^{-1}$, $\lambda =532\mathrm{nm}$, $A=20\mathrm{m}{\mathrm{m}}^2$ and $t=2\mathrm{mm}$, this gives $C\approx 2.54\times 10^{-12}\mathrm{P}{\mathrm{a}}^{-1}$. A $200\mathrm{N}$ load would give $10\mathrm{MPa}$ stress difference and $0.60$ radians retardation. These are chosen calibration numbers, not a universal coefficient for a plastic.

A polariscope with controlled loading, a force readout and polarization analysis supplies the relevant apparatus.^[ref] Background birefringence, thickness, temperature and wavelength should be accounted for. A measured intensity alone can have multiple compatible phase delays, and a changing optical-axis direction through a three-dimensional specimen can invalidate the simple scalar-path formula. The law is most useful when its optical and mechanical assumptions are checked together.

The stress-optic coefficient $C$ measures the birefringence induced per unit principal-stress difference in the linear stress-optic law:Here $\Delta n$ is a difference of refractive indices, so $C$ has SI units ${\mathrm{Pa}}^{-1}$. The wavelength $\lambda$ matters: a coefficient determined with one illumination wavelength should not simply be reused with another. The signs of $C$ and $\Delta n$ also require a consistent ordering of the optical axes and stresses.^[ref]

For a uniform specimen of thickness $t$ viewed along a principal optical direction, the measured phase retardation isA known uniaxial load supplies $\Delta \sigma =F/A$ away from grips and other stress concentrations. Fitting the slope $s=d\delta /dF$ givesSubtracting the unloaded retardation removes a constant background, but a load-dependent background must be investigated rather than absorbed into the coefficient.

For an original illustrative measurement, let $\lambda =500\mathrm{nm}$, $t=2\mathrm{mm}$ and $\Delta \sigma =20\mathrm{MPa}$ produce $\delta =0.80\mathrm{rad}$. ThenDoubling the stress difference doubles the retardation only within the same linear regime. This numerical value is illustrative, not a material specification.

An equivalent magnitude convention uses the photoelastic fringe value $f_{\sigma }=\lambda /|C|$. One additional full retardation cycle corresponds to $|\Delta \sigma |=f_{\sigma }/t$ for the uniform model. A fringe magnitude alone does not determine the sign of $C$ or the separate principal stresses.

The Illinois primary apparatus description makes the measurement chain concrete: a leadscrew loads the specimen, a load cell measures force, monochromatic light passes through the specimen between polarizing elements, and a camera records fringes. Establishing $C$ connects those optical observations to a mechanical stress scale. Wavelength, thickness, unloaded birefringence and loading conditions belong with the reported coefficient.^[ref]

The photoelastic fringe value $f_{\sigma }$ is the material-and-wavelength scale connecting photoelastic fringe order to principal-stress difference times specimen thickness in the linear stress-optic law. With vacuum wavelength $\lambda$ and stress-optic coefficient $C$, use the magnitude conventionHere $N=|\delta |/(2\pi )$ counts full relative-phase cycles and $t$ is specimen thickness. Since $C$ has units $\mathrm{P}{\mathrm{a}}^{-1}$, $f_{\sigma }$ has units $\mathrm{Pa}\mathrm{m}=N/m$. A value expressed in $\mathrm{MPa}\mathrm{mm}$ equals the same numerical value in $kN/m$. The wavelength dependence of $C$ must be retained.^[ref]

For an original example, take $\lambda =600\mathrm{nm}$ and $|C|=50\times 10^{-12}\mathrm{P}{\mathrm{a}}^{-1}$. Then $f_{\sigma }=12000N/m$. A3mm specimen at fringe order2.5 has stress-difference magnitude $2.5(12000)/0.003=10\mathrm{MPa}$. One additional complete phase cycle changes the inferred magnitude by4MPa for that thickness.

A uniaxial calibration supplies a known stress difference away from grips and other stress concentrations. For a rectangular strip of width10mm and thickness3mm, load300N gives stress10MPa. If the corresponding referenced fringe order is2.5, the measured fringe value is $10(3)/2.5=12\mathrm{MPa}\mathrm{mm}$. Repeating at several loads tests the assumed linear regime; counting from a referenced unloaded state helps separate the applied-load change from pre-existing retardation.

A primary instructional apparatus uses a manual leadscrew, load-cell readout, monochromatic illumination, polarizing elements and a camera. Those components connect a known mechanical load to the observed optical pattern.^[ref] A single intensity value does not uniquely identify an absolute phase cycle, so fringe order requires continuity, reference information or another phase-resolving method.

The magnitude convention loses the sign of $C$ and does not separately determine the two principal stresses. It also assumes the stated through-thickness optical/stress model. A varying stress field along the light path requires the appropriate retardation model rather than inserting one arbitrary local stress into the uniform-specimen formula.

The photoelastic fringe order $N$ expresses the magnitude of the full relative phase retardation between two principal optical components in cycles:For a uniform specimen obeying the linear stress-optic law, $N=t|{\sigma }_1-{\sigma }_2|/f_{\sigma }$, where $t$ is thickness and $f_{\sigma }$ is the photoelastic fringe value. The order is dimensionless and need not be an integer. One unit means one additional complete relative-phase cycle, not a separate stress component.

The connection to a visible fringe depends on the optical arrangement. For ideal crossed linear polarizers and a uniform retarder with optical axes at angle $\alpha$ to the incident polarization, the transmitted intensity satisfiesHere $I_0$ is the intensity entering the specimen after the first polarizer, and $\delta$ denotes the full relative phase. At $\alpha =45^{\circ }$, integer orders are dark and half-integer orders are bright. At axis alignment $\alpha =0$, the image is dark for every order. The crossed-polarizer derivation therefore separates retardation-related extinction from axis-alignment extinction.^[ref]

For an original ambiguity example at45°, orders0.25,0.75 and1.25 all give $I/I_0=1/2$. With $f_{\sigma }=12000N/m$ and $t=3\mathrm{mm}$, they correspond to stress-difference magnitudes1,3 and5MPa. Brightness alone does not choose among them. Referencing an unloaded state and tracking changes continuously during loading can supply information for photoelastic phase unwrapping.

The primary Illinois apparatus makes these controls concrete: a leadscrew loads the specimen, a load cell records force, monochromatic light and polarizing elements reveal the pattern, and a camera records it. Its magnitude-observation setting uses quarter-wave plates to remove the direction-related fringes.^[ref] That changes the observation geometry but does not remove the need to establish absolute order.

An order map under a known photoelastic fringe value gives principal-stress difference in the stated specimen model. It does not by itself give the two stresses individually or the sign of the stress-optic coefficient. Wavelength, initial retardation and the through-thickness stress model remain part of the interpretation.

Photoelastic phase unwrapping assigns consistent cycle counts to retardation observations that determine phase only modulo a full cycle. If a phase-resolved method reports ${\varphi }_k\in (-\pi ,\pi ]$, the full relative phase retardation has the formThe integers carry the fringe-order information that the wrapped values omit. A known starting order and sufficiently small changes can make the branch choices recoverable.

For an original one-dimensional example, take wrapped phases $2.80,3.05,-2.95,-2.70$ radians. Choosing the first value as the known starting branch, the nearby increasing continuation isor approximately $2.80,3.05,3.3332,3.5832$ radians. The apparent large negative jump is a change of representation at $\pi$, not a demonstrated physical reversal. The final photoelastic fringe order on this branch is about0.5703.

A nearest-branch rule is justified in a noiseless sequence when successive true phase changes have magnitude strictly below $\pi$: among increments differing by $2\pi$, exactly one lies in that interval. With phase errors bounded by $ϵ$, the stronger condition $|{\delta }_{k+1}-{\delta }_k|+2ϵ<\pi$ is sufficient for that local choice. Larger steps, discontinuities or inadequate signal can cause a photoelastic branch-slip error. One wrong cycle changes inferred stress-difference magnitude by $f_{\sigma }/t$ in the uniform-specimen model.

The algorithm assumes phase or equivalent branch information is available. A single crossed-polarizer intensity $I/I_0={\sin }^2(\pi N)$ does not itself supply a unique signed wrapped phase. Its ambiguity includes reflection within a cycle as well as the integer count. Additional polarization measurements, a known load path or other reference information must justify phase recovery before applying the simple continuation rule. The primary stress-analysis derivation explicitly shows how intensity depends on retardation and axis orientation.^[ref]

The Illinois apparatus supplies controlled loading, a load-cell record and camera observations, making a load-indexed sequence physically meaningful.^[ref] Continuity alone still leaves a common $2\pi$ offset unless the absolute starting order is known. An unloaded image can reveal residual retardation rather than automatically establish zero stress or zero order.

A photoelastic branch-slip error is an incorrect integer-cycle choice during photoelastic phase unwrapping. If the measured wrapped phase retardation is ${\varphi }_k$ and the true unwrapped value is ${\delta }_k={\varphi }_k+2\pi m_k$, replacing the correct integer $m_k$ by $m_k+j$ changes the recovered phase by $2\pi j$. A small error in one branch decision can therefore become a full-cycle offset in many later observations.

For an original noiseless counterexample, suppose a known starting phase is $2.2\pi$ and the next true phase is $3.4\pi$. Their wrapped values in $(-\pi ,\pi ]$ are $0.2\pi$ and $-0.6\pi$. The nearest-increment rule chooses $-0.8\pi$ instead of the true $1.2\pi$, producing $1.4\pi$ as the second recovered phase. The recovered photoelastic fringe order is $0.7$ rather than $1.7$. Every later small increment can be locally consistent while retaining that wrong cycle.

In the uniform-thickness stress-optical model, $N=t|{\sigma }_1-{\sigma }_2|/f_{\sigma }$, where $f_{\sigma }$ is the material fringe value. While staying on a known nonnegative stress-difference branch, one cycle changes the inferred stress difference by $f_{\sigma }/t$. Stipulated $f_{\sigma }=10\mathrm{N}\mathrm{m}{\mathrm{m}}^{-1}$ and thickness $t=5\mathrm{mm}$ make the example's one-cycle error $2\mathrm{MPa}$. The material coefficient depends on the optical conditions, including wavelength; the primary photoelastic derivation explains that dependence and the analyzer's periodic intensity response.^[ref]

A sufficient local noise margin iswhen each phase error is bounded in magnitude by $ϵ$. The triangle inequality bounds the error in the phase increment by $2ϵ$, leaving it inside the nearest-branch interval. Equality gives no strict guarantee. Reducing the loading increment helps only if the assumed phase recovery and error bound are valid.

The Illinois polariscope provides controlled loading, a load readout and camera observation, which support a proposed check using smaller repeated load increments.^[ref] Agreement between a coarse sequence and a finer sequence tests branch stability over those measurements. It cannot establish an unknown common starting order. A lone crossed-polarizer intensity also does not supply a unique signed phase: the additional phase-recovery information assumed by the algorithm must be measured or independently justified.

A known retarder inserted in a polarization microscope to compare phase delay with an unknown specimen. Parallel slow directions add signed retardations; a slow direction paired with a fast one subtracts them. General relative angles require composing polarization transformations rather than scalar addition.

A quartz retarder used to identify specimen fast and slow directions through changes of interference colour or compensation. Its own retardation and orientation must be specified before unique final colours can be predicted.

For a lossless uniaxial dielectric, $n_{\mathrm{ex}}^{-2}={\cos }^2\vartheta /n_o^2+{\sin }^2\vartheta /n_e^2$, where $\vartheta$ is the angle between the wave vector and optic axis. Along the axis $n_{\mathrm{ex}}=n_o$, while perpendicular propagation gives $n_{\mathrm{ex}}=n_e$.

Colour produced when wavelength-dependent interference selectively transmits or reflects part of white light. Between crossed polarizers, a uniform birefringent plate gives a factor ${\sin }^2(\pi \Gamma /\lambda )$, where $\Gamma$ is its optical path difference. The observed colour also depends on illumination and the spectral response of the observer.

A birefringent plate with retardation $\Gamma =t\Delta n$ gives $I/I_P={\sin }^2(2\alpha ){\sin }^2(\pi \Gamma /\lambda )$. At non-extinction azimuths, wavelengths satisfying $\Gamma =m\lambda$ are removed by a crossed analyser. At a principal-axis azimuth every wavelength is extinguished.

Birefringence produced by subwavelength structure rather than solely by intrinsic crystalline anisotropy. For ideal dielectric lamellae, continuity of tangential electric field gives ${ϵ}_{\parallel }=f{ϵ}_1+(1-f){ϵ}_2$, while continuity of normal electric displacement field gives ${ϵ}_{\perp }^{-1}=f/{ϵ}_1+(1-f)/{ϵ}_2$. The unequal effective permittivities give different principal refractive indices.

Linear birefringence means that two orthogonal linear polarizations have different refractive indices. Their different propagation phases produce phase retardation. In a lossless waveplate, with fast-axis angle $\theta$ and slow-minus-fast retardation $\delta$, the Jones matrix, up to a common phase, isusing time dependence $e^{-i\omega t}$. Circular birefringence instead has circular eigenpolarizations.

Place a clear calcite specimen on a transparency bearing printed words on an overhead projector. The projected readout is two copies of the lettering. Put a rotatable linear polarizer after the specimen: suitable orientations suppress one copy or the other. These are the actual apparatus and observations in the University of Iowa demonstration.^[ref]

For a useful control, remove the specimen while keeping the transparency, projector and screen fixed: the doubling should disappear. Replace it, hold it stationary and rotate only the polarizer; complementary brightness changes distinguish two linear polarizations from an accidental duplicate transparency. These are suggested controls, rather than additional reported measurements. Ideally the two copies obey complementary Malus's law curves, $I_o(\alpha )=I_{o,\max }{\cos }^2\alpha$ and $I_e(\alpha )=I_{e,\max }{\sin }^2\alpha$, with angle $\alpha$ measured from the exiting ordinary wave's linear polarization. Unequal maxima do not invalidate the test. Finite polarizer extinction, poor focus and specimen defects limit visibility; image separation alone does not determine either refractive index without the specimen geometry.

A transparent, nonmagnetic linear anisotropic dielectric is uniaxial when two principal permittivities agree. With the optic axis along $z$, its permittivity tensor has the displayed diagonal matrix form. Here $n_o$ and $n_e$ are principal refractive indices at a specified frequency; $n_e$ is not the extraordinary wave's index in every direction.^[ref]

The symmetry direction of a uniaxial optical medium. For a wavevector along it, the two refractive indices coincide: there is no birefringence for that propagation direction, even though $n_e\ne n_o$.

The ordinary wave's electric field is perpendicular to the plane containing the wavevector and optic axis. Its refractive index $n_o$ is direction independent at fixed frequency.

The extraordinary wave's electric displacement field lies in that plane and is perpendicular to the wavevector; its electric field need not be. For angle $\theta$ between the wavevector and optic axis,Thus $n_{\mathrm{ex}}(0)=n_o$ and $n_{\mathrm{ex}}(\pi /2)=n_e$.^[ref]

A short derivation starts from the existing plane wave in an anisotropic dielectric equation. Put $\mathbf{k}=(k_x,0,k_z)$ and $k_0=\omega /c$, with angular frequency $\omega$ and vacuum speed of light $c$. The $y$ component gives the ordinary wave relation $k_x^2+k_z^2=n_o^2{k}_0^2$. The determinant of the $x,z$ system givesSubstituting $k_x=n_{\mathrm{ex}}{k}_0\sin \theta$ and $k_z=n_{\mathrm{ex}}{k}_0\cos \theta$ yields the displayed directional refractive index.

The extraordinary wave's Poynting vector can differ in direction from its wavevector. The latter specifies wave phase propagation, while the former specifies energy flux. This angular separation is walk-off.^[ref]

In the lossless, nondispersive model above, take the gradient of $\omega =c\sqrt{k_x^2/n_e^2+k_z^2/n_o^2}$ to obtain the group velocity, parallel to the Poynting vector. If its angle to the optic axis is $\varphi$, this gives $\tan \varphi =(n_o^2/n_e^2)\tan \theta$. Consequently the extraordinary wave generally carries energy away from its wavevector direction. An internal wavevector normal to a slab face therefore does not guarantee a normal energy path. Birefringent walk-off vanishes at $\theta =0$ and $\theta =\pi /2$; the latter direction still has unequal refractive indices. This is a useful waveplate geometry: overlapping components can acquire phase retardation without spatial splitting. A thin specimen, insufficient imaging resolution, or excitation of only one linear polarization can also hide a double image, for different reasons.

For a lossless nonmagnetic uniaxial optical medium, the electric displacement field of a plane electromagnetic wave is perpendicular to its wavevector. The magnetic field is perpendicular to the plane containing the electric field and wavevector, while the Poynting vector is perpendicular to the electric field in that same plane. Thus the angle between $\mathbf{E}$ and $\mathbf{D}$ equals the walk-off angle between energy flow and phase propagation. At normal incidence on a parallel slab of thickness $L$, a lateral separation $\Delta$ of the ordinary and extraordinary exiting rays gives ${\alpha }_e=\arctan (|\Delta |/L)$, while ${\alpha }_o=0$. The output rays can be parallel even though their internal energy paths differ.

A waveplate uses birefringence to change polarization of an electromagnetic wave through phase retardation. Its fast axis has the smaller refractive index, hence the larger phase velocity; its slow axis has the larger index. Consider normal incidence with the optic axis in the plate face, so both components follow the same straight path. A linear polarization at $45^{\circ }$ to these axes excites equal wave amplitudes.

A complementary experiment uses two sheet polarizers with a cellophane sheet between them. First cross the polarizers and remove the sheet: the field is dark. Insert and rotate the sheet about the beam: light returns at intermediate orientations, but extinction recurs at two perpendicular orientations. The sheet has changed the polarization of an electromagnetic wave, without requiring a resolved double image. With white light, different wavelengths can produce different transmitted colors.^[ref]. Brightness alone does not establish circular polarization: an ideal rotating linear polarizer also gives orientation-independent light intensity for unpolarized light.

Suppose an earlier path selection prepares a photon in path $L$ with vertical polarization $|V⟩$. Conditional on that selection, the normalized Jones vector is $(0,1{)}^{\mathsf{T}}$. For time convention $e^{-i\omega t}$, a quarter-wave plate with fast-axis angle $\eta$ acts asConsequently,A subsequent horizontal/vertical measurement of linear photon polarization in that same path givesIf the earlier path event has probability $p_L$, the joint probabilities are $P(L,H)=p_{L}P(H\mid L)$ and $P(L,V)=p_{L}P(V\mid L)$. Normalizing after a selection changes a joint event into a conditional one; it does not remove the selection's success cost. If paths are coherently recombined instead of selected, their relative amplitudes and phases must be retained in the joint quantum state.

The relative phase accumulated by orthogonal eigenpolarizations is $\Delta =2\pi L(n_2-n_1)/{\lambda }_0$. A quarter-wave plate has relative phase $\pi /2$ modulo $2\pi$, while a half-wave plate has $\pi$ modulo $2\pi$.

For the overlapping monochromatic wave components, let $d$ be the thickness and ${\lambda }_0$ the vacuum wavelength. Define $\Delta n=n_s-n_f>0$ using the slow and fast refractive indices. With complex amplitude convention $e^{i(kz-\omega t)}$, the accumulated slow-minus-fast phase difference isThis is the optical path difference $\Delta nd$ expressed as a wave phase. It is a signed accumulation convention; with the opposite time convention the sign reverses. The outgoing polarization of an electromagnetic wave depends on $\delta$ modulo $2\pi$. Here the positive slow-minus-fast $\Delta n$ differs from the signed principal-index $n_e-n_o$ used in the parent birefringence entry when $n_e<n_o$.

For an original worked example, stipulate $\Delta n=0.0100$ at ${\lambda }_0=600\mathrm{nm}$ and a uniform thickness $d=15.0\mu \mathrm{m}$. These are design assumptions, not measured material data. The optical path difference is $150\mathrm{nm}$, soIt is a quarter-wave plate. At the same wavelength and refractive indices, doubling $d$ to $30.0\mu \mathrm{m}$ gives $\delta =\pi$, a half-wave plate. A $1.0\mu \mathrm{m}$ thickness increase changes $\delta$ by $\pi /30$, or $6^{\circ }$; retaining the same birefringence at $500\mathrm{nm}$ would instead give $\delta =3\pi /5$ for the $15.0\mu \mathrm{m}$ plate. Real refractive indices also vary with wavelength, so this last calculation explicitly neglects optical dispersion.

For these equal transmitted wave amplitudes, orthogonal projection onto a polarizer crossed with the initial $45^{\circ }$ linear polarization gives a simple independent readout. After removing a common wave phase, the component complex amplitudes are proportional to $(1,e^{i\delta })/\sqrt{2}$; orthogonal projection onto $(1,-1)/\sqrt{2}$ gives $(1-e^{i\delta })/2$. Squaring its modulus yields $I_{\perp }/I_{\mathrm{in}}={\sin }^2(\delta /2)$: zero at $d=0$, one half at $15\mu \mathrm{m}$, and one at $30\mu \mathrm{m}$. This assumes ideal polarizers, negligible specular reflection and wave absorption, overlapping components and stable relative wave phase. It measures a periodic function of phase retardation, so one brightness measurement cannot determine the thickness uniquely.

A zero-order quarter-wave plate satisfies ${\omega }_{0}d\Delta n({\omega }_0)/c=\pi /2$. Its retardance error at nearby angular frequency is $ϵ=\omega d\Delta n(\omega )/c-\pi /2$. Taylor expansion gives the title expression, including material dispersion. Neglecting that dispersion yields $ϵ=(\pi /2)\Delta \omega /{\omega }_0$. A half-wave plate made from twice the same material thickness has exactly twice this phase error. Common propagation phase can be removed from its Jones matrix without changing polarization.

A quarter-wave plate gives a quarter-cycle phase retardation. Equal transmitted wave amplitudes initially in linear polarization at $45^{\circ }$ become circular polarization; unequal wave amplitudes generally give elliptical polarization. Reversing the sign reverses the traversal of the same circle.

Take a half-wave plate with fast axis at $15^{\circ }$ to a linearly polarized input, followed by a quarter-wave plate at a further $60^{\circ }$, so the second axis is at $75^{\circ }$ to the input. With shared quarter-wave retardance error $ϵ$ and half-wave error $2ϵ$, use $Q=\mathrm{diag}(1,i{e}^{iϵ})$ and $H=\mathrm{diag}(1,-e^{2iϵ})$. Rotating each matrix into the input axes and multiplying gives $\mathbf{J}(0)=e^{i75^{\circ }}(1,-i{)}^{\mathsf{T}}/\sqrt{2}$ and the title expansion. Its first-order change is common phase only, so the output remains circular to first order. This input-specific local compensation does not prove a frequency-independent quarter-wave operator for arbitrary input across a finite spectral band.

Suppose the unknown ideal elements are a linear polarizer and a quarter-wave plate, and the two unknown sources are unpolarized and circularly polarized. Test both element orders while rotating their relative axes and monitoring output intensity. Only circular input followed by the quarter-wave plate and then the polarizer gives complete modulation: the plate makes linear light and the analyzer obeys Malus's law. Polarizer-first output is constant half-intensity because the following lossless plate changes polarization but not intensity. Unpolarized input is constant in either order. This identifies both elements and both sources without comparing source brightnesses.

Two rotatable ideal quarter-wave plates can transform a fixed circular polarization into any pure polarization represented by a Jones vector. Use time dependence $e^{-i\omega t}$, $Q(\theta )=R(-\theta )\mathrm{diag}(1,i)R(\theta )$ and input $J_L=(1,i{)}^T/\sqrt{2}$. Every target polarization ellipse can be written, up to a common phase, as $R(-\beta )(\cos \eta ,i\sin \eta {)}^T$, with signed ellipticity angle $-\pi /4\le \eta \le \pi /4$. The first quarter-wave plate at ${\theta }_1=\beta +\eta +\pi /4$ produces linear polarization at $\beta +\eta$. The second, at ${\theta }_2=\beta$, therefore produces the required target. The two angles supply the two independent parameters of pure polarization.

For a reciprocal ideal quarter-wave plate followed by a mirror at normal incidence, use the same fixed transverse laboratory axes for the forward and returning Jones vectors. If the axes coincide with the plate's fast and slow axes, each pass has Jones matrix $J=\mathrm{diag}(1,i)$ up to a common phase factor. A polarization-independent mirror has scalar reflection coefficient $r_m$, so the returned Jones vector is $r_m{J}^2\mathbf{e}=r_m\mathrm{diag}(1,-1)\mathbf{e}$. The plate acts twice as a half-wave plate, rather than undoing its phase retardation.

For input linear polarization at angle $\alpha$ to the fast axis, the returned line is at $-\alpha$ modulo $\pi$. At $\alpha =45^{\circ }$ it is perpendicular to the original line. An axis-aligned input instead returns along the same line. General elliptical polarization requires its full input Jones vector; no unique returned state follows from the word “polarized” alone. The propagation direction reverses, so comparing descriptions of circular polarization must also specify the viewing convention.

Take a linear polarizer $P=\mathrm{diag}(1,0)$ and a reciprocal quarter-wave plate $Q=2^{-1/2}\left(\begin{array}{cc}1& i\\ i& 1\end{array}\right)$ in fixed physical transverse axes. A near-normal isotropic mirror has $M=-I_2$ up to global phase. If the source/detector side encounters P before Q, the round-trip matrix is $PQMQP=-P{Q}^{2}P=0$. Reversing the filter gives $QPMPQ=-QPQ$, which transmits half an initially unpolarized intensity. The result is double-pass rejection, not optical nonreciprocity; changing direction without changing the input polarization does not preserve the same physical input state.

A half-wave plate reverses the relative sign of the two components. For initial linear polarization at angle $\alpha$ to the fast axis, the outgoing line is at $-\alpha$ modulo $\pi$. The $45^{\circ }$ input therefore rotates by $90^{\circ }$.

Circular birefringence means different refractive indices for opposite circular polarizations. In a transparent isotropic medium, their relative wave phase changes rotate the direction of transmitted linear polarization; optical activity gives the solution measurement and convention.

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 $V$.