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.

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 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.

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 $90^{\circ }$.