Waveplate

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.

A detected lower-path photon with vertical polarization evolves under an angle-dependent Jones matrix. For fast-axis angle $\eta$, its horizontal amplitude is $(1-i)\sin \eta \cos \eta$ and its vertical amplitude is ${\sin }^2\eta +i{\cos }^2\eta$. A later vertical-path probability therefore depends on the plate orientation and on whether the earlier lower-path event has been conditioned on.

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 $90^{\circ }$.