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The relative refractive index of medium two with respect to medium one is $n_{21}=n_2/n_1$, using their absolute refractive indices at the same frequency. For transparent media with isotropy and positive refractive indices, Snell's law gives $n_{21}=\sin i/\sin r$, with the angles measured from the interface normal. Thus a refraction measurement in air directly determines an index relative to air. Converting it to an absolute refractive index requires the refractive index of that air; replacing it by one is an approximation.

For a sphere of radius $a$, the scattering efficiency is its effective scattering area divided by the projected geometric cross-section $\pi a^2$. It is a dimensionless quantity. For Rayleigh scattering by a small sphere with relative refractive index $m$ and incident wavenumber $k$,At fixed wavelength and fixed refractive index, the Rayleigh scattering cross-section scales as $a^6$, whereas the scattering efficiency scales as $a^4$. Doubling radius multiplies the former by $64$ and the latter by $16$. An efficiency here is an effective scattering area relative to a reference area, rather than a fraction of all light in an arbitrary beam.

For a worked toy example, take $a=10\mathrm{nm}$, $m=1.5$, and $\lambda =550\mathrm{nm}$. Then $ka=2\pi (10/550)\simeq 0.1142$ and $(m^2-1)/(m^2+2)=5/17$. Substitution gives Rayleigh scattering cross-section $\sigma \simeq 1.234\times 10^{-20}{\mathrm{m}}^2$ and scattering efficiency $Q_{\mathrm{sca}}\simeq 3.929\times 10^{-5}$. At incident light intensity $1\mathrm{W}{\mathrm{m}}^{-2}$, the scattered power is approximately $1.234\times 10^{-20}\mathrm{W}$. These are leading small-particle estimates, not a measured air-molecule cross-section.

Schlieren photography makes spatial changes of refractive index visible by converting deflections of light rays into contrast at an optical camera. In a gas, mass density variations can produce these deflections through refraction. A knife edge in the focused beam is a common implementation.

A SEC detector response must be specified before interpreting a peak area. Ultraviolet absorbance depends on a species' response factor and concentration, as described by the Beer-Lambert law; differential refractive index detection depends on how the dissolved species changes refractive index. Equal amounts need not produce equal signal areas across unlike species.

Even a calibrated mass-proportional trace is not a chain-count probability distribution. In an original two-population example, equal recovered masses of chains with molar masses $20,000$ and $80,000\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ contain four times as many chains in the lighter population. Thus equal signal areas, assuming equal response per unit mass, correspond to number fractions $4/5$ and $1/5$: their number-average molar mass is $32,000\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, while their weight-average molar mass is $50,000\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. Converting a continuous trace additionally requires response correction and the appropriate change of variables between elution and molar mass bins.

Second-harmonic generation converts an incident optical frequency $\nu$ into $2\nu$ through a material's nonlinear response. In the photon description of an ideal lossless conversion, two input photons supply one output photon:Here $h$ is the Planck constant. The relation states conservation of energy; the material mediates the interaction. It does not say that each input photon independently splits its wavelength in half.

For vacuum wavelengths, ${\lambda }_{0,\omega }=c/\nu$ and ${\lambda }_{0,2\omega }=c/(2\nu )$, where $c$ is the speed of light. HenceThe second relation uses the refractive index for each propagating mode and frequency. Within a medium with optical dispersion, the wavelength need not halve. For an invented example, ${\lambda }_{0,\omega }=800\mathrm{nm}$, $n_{\omega }=1.5$, and $n_{2\omega }=1.6$ give $400\mathrm{nm}$ in vacuum, but in the material the two wavelengths are $533.33\mathrm{nm}$ and $250\mathrm{nm}$; their ratio is $0.46875$.

An authored manufacturer experiment used an Agilent 1260 Infinity Bio-inert Quaternary LC with Bio SEC-3 $300\mathring{\mathrm{A}}$ columns, $7.8\mathrm{mm}$ diameter and $150$ or $300\mathrm{mm}$ length. Four pump channels mixed water, sodium chloride, and two phosphate solutions into $20\mathrm{mmol}{\mathrm{L}}^{-1}$ phosphate at pH $7.0$, varying salt from $0.2$ to $0.6\mathrm{mol}{\mathrm{L}}^{-1}$. At $1.0\mathrm{mL}\mathrm{mi}{\mathrm{n}}^{-1}$ and $30{}^{\circ }\mathrm{C}$, the readouts were diode-array absorbance, differential refractive index, and scattering at $15^{\circ }$ and $90^{\circ }$. A supplied protein standard mixture calibrated retention; bovine serum albumin calibrated scattering constants and inter-detector delay. Increasing salt improved separation of two α-chymotrypsinogen A peaks; two $300\mathrm{mm}$ columns also resolved peaks that one column did not. At $0.5\mathrm{mol}{\mathrm{L}}^{-1}$ salt, the two scattering-derived estimates were about $26,100$ and $25,800\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. The author inferred a possible size-changing conformation difference, rather than doubling of molar mass. The same poster shows delayed, tailing lysozyme peaks from unwanted interactions.^[ref]

The salt comparison and the independent mass readout constrain interpretation, but do not uniquely prove a particular protein structure. Salt can change protein folding, protein aggregation, and packing interactions together. In ideal SEC there is no adsorption; attraction from the hydrophobic effect or forces described by electrostatics can cause additional retention, tailing, and loss of recovered sample. Repulsion can instead reduce pore entry. A peak beyond $V_t$ contradicts the simple nonbinding model, rather than extending its size calibration.

Useful proposed controls compare a solvent blank, fresh calibration standards bracketing the sample, several injection amounts and volumes, and recovered sample quantity with injected quantity using an appropriately calibrated detector. Keep temperature and volumetric flow rate fixed while testing pH or ionic strength. Compare an independent mass readout before declaring an early peak an aggregate. Loss of an insoluble species during filtration, on-column association or dissociation, overloading, and broad unresolved peaks can all hide the original sample population. These are interpretation checks and limitations, not measurements invented for the figures.

Consider an original, explicitly normalized model of primary coma, rather than a prescription for a particular optical lens. Let $(u,v)$ be Cartesian coordinates in a circular optical pupil divided by its radius $a$, with $u^2+v^2\le 1$. Let $h=\alpha /{\alpha }_0$ be a signed small field angle divided by an arbitrary positive reference angle. With polar coordinates $u=\rho \cos \theta$, $v=\rho \sin \theta$, prescribe the wavefront errorHere $W_0$ is an arbitrary positive optical path difference scale, not a measured error. The cubic optical pupil dependence and linear field dependence are the classical Seidel form.^[ref]

In the paraxial approximation, light ray direction errors follow the transverse gradient of wavefront error. For a reference sphere a distance $L$ from the exit optical pupil in a uniform image medium of refractive index $n$, the corresponding image-error scale is $s=L{W}_0/(na)$. Choose image-error coordinate signs so the normalized displacement is $\nabla w$; reversing the convention reverses the plotted orientation. Image coordinates below are errors relative to the ideal image of this particular object point, rather than positions of different objects:The partial derivatives explain the asymmetry: opposite optical pupil points $(u,v)$ and $(-u,-v)$ land together, rather than on opposite sides of the ideal image. For $h>0$, every $X$ is nonnegative, while $Y$ can have either sign. The figures stipulate this single wavefront error, omit other optical aberrations and diffraction, and assign no instrument dimensions.

For a ring of fixed optical pupil radius $\rho$, the double-angle formulas for trigonometric functions giveThus each ring maps into a circle of radius $|h|{\rho }^2$ whose centre is displaced by $2h{\rho }^2$. Doubling $\rho$ quadruples both sizes; for $h=1$, the edge ring has centre $(2,0)$ and radius $1$, while the half-radius ring has centre $(1/2,0)$ and radius $1/4$. Superposing these displaced circles produces the comet-like envelope.

For uniform irradiance over the circular optical pupil, equal area samples give an equal-weight geometrical image. Its centroid is $(h,0)$ because $⟨u^2⟩=⟨v^2⟩=1/4$ and $⟨uv⟩=0$. Moving the coordinate origin to that centroid translates the image; it does not remove its asymmetric shape. Adding a linear term to $w$ likewise translates all light rays equally. These origin choices must be distinguished from changing the optical aberration itself.

The model's full transverse extent is $3|h|$ along $X$ and $2|h|$ along $Y$. Halving the field angle halves the blur, and reversing it reverses the pattern about the ideal image. This isolates why coma matters for an optical camera's usable field of view: collecting more light does not ensure that off-axis points remain compact. Near diffraction-limited resolution, a physical optics calculation of the point spread function is needed instead of interpreting the geometrical tip as an infinitely sharp image.^[ref]

Concentration of a beam caused by its own intensity-dependent change in the medium's refractive index.

An intensity-dependent refractive index gives an optical pulse a time-dependent wave phase, changing its instantaneous frequency and spectrum.

A beam with a transverse profile sustained by a balance between diffraction and a nonlinear change of refractive index. A one-transverse-dimensional focusing cubic model supports a stationary bright profile after its propagation wave phase is removed. Existence in that model does not establish stability in two transverse dimensions or in a saturating, absorbing material.

A step-index optical fiber has a nearly uniform refractive index $n_1$ throughout its optical fiber core, with an abrupt decrease to $n_2<n_1$ in its optical fiber cladding. In a straight, ideal guide, meridional rays stay in a plane containing the optical axis and trace a zigzag through repeated total internal reflection. This is a geometrical optics approximation; it is particularly useful for a multimode optical fiber whose core diameter is large compared with the wavelength.

The upper panel shows an accepted meridional ray; the lower panel enlarges the limiting geometry. The angle of incidence $i$ at the side wall is measured from its normal vector, whereas the internal ray angle $\varphi$ and external launch angle $\theta$ are measured from the optical axis. Thus $i+\varphi =90^{\circ }$. The limiting ray is a boundary of the ideal model, rather than a complete normal mode solution.

Thin-film interference combines coherent light contributions reflected or transmitted at the two surfaces of a layer. For a film of refractive index $n$, thickness $d$ and internal angle ${\theta }_t$, the round-trip optical path difference is $2nd\cos {\theta }_t$. The relative reflection phase shift must also be included. For a nonabsorbing higher-index film surrounded by air, one of the two leading reflected contributions has a $\pi$ phase shift and the other does not. Reflection maxima therefore satisfy $2nd\cos {\theta }_t=(m+1/2){\lambda }_0$. A smooth reflection spectrum filters many visible wavelengths at once; varying thickness and angle produce spatially varying mixed colours.

A thin lens models an optical lens by collapsing its refracting surfaces into one plane, neglecting thickness relative to the relevant radius of curvature values and object and image distances. Here use an ideal converging thin lens in air and the paraxial approximation: light rays make small angles with the optical axis, so Snell's law can be linearized using the small-angle approximation. Equal surrounding refractive indices on both sides give the familiar symmetric equation.^[ref]

Put the thin lens at $x=0$, with incident light travelling to the right. An object at $x=-u$ has $u>0$. The image coordinate is $x=v$: $v>0$ for a real image and $v<0$ for a virtual image. The converging thin lens has focal length $f>0$, and heights $h_o,h_i$ are signed relative to the optical axis. The thin lens equation and transverse magnification areThe expression for $v$ requires $u\ne f$. Negative transverse magnification means inverted; positive transverse magnification means upright. Remember that the positive distance $u$ differs in sign from the object's coordinate $-u$. ^[ref].

For focal length $f=1\mathrm{m}$, object distance $u=3\mathrm{m}$, and object height $h_o=0.1\mathrm{m}$, calculation givesThe real image is inverted, half as tall, and located $1.5\mathrm{m}$ to the right of the thin lens. A screen at that position receives the image because the outgoing light rays actually meet there.

For the same focal length $f=1\mathrm{m}$ and object height $h_o=0.1\mathrm{m}$, move the object inside the front optical focus, to $u=0.5\mathrm{m}$. ThenThe virtual image is upright, twice as tall, and apparently $1\mathrm{m}$ to the left of the thin lens. Outgoing light rays still travel to the right; only their backward extensions intersect at the virtual image. A screen there cannot collect that image from the outgoing bundle. This geometry explains why a converging optical lens can act as a magnifier when held close to an object. ^[ref].

Both diagrams use two rules for ray tracing. A light ray from the object's tip parallel to the optical axis reaches the thin lens at $(0,h_o)$ and exits along the straight line through the rear optical focus $(f,0)$. Another light ray through the centre $(0,0)$ continues undeviated in the ideal thin lens model. Their intersection, or the intersection of their backward extensions, locates the image tip. ^[ref].

A short geometric proof outline makes the signs transparent. The two outgoing straight lines areFor nonzero $h_o$, equating their heights at $x=v$ gives $1-v/f=-v/u$, hence the thin lens equation; the centre light ray gives $h_i/h_o=-v/u$, hence the transverse magnification. This derives the image equations from the ideal ray tracing rules, whose physical justification comes from Snell's law in the paraxial approximation.

At $u=f$, the two outgoing straight lines are parallel: there is no finite image position. For $u>f$ the real image has $m<0$; for $0<u<f$ the virtual image has $m>0$. These cases let an optical camera designer choose a receiving position and let a user understand a magnifier. An actual optical lens has finite apertures, which cause diffraction and impose diffraction-limited resolution; optical aberrations can further blur the image. The diagrams are ideal geometrical optics, with exact coordinates within that model, rather than exact predictions of Maxwell's equations. ^[ref].

Total internal reflection occurs when a wave travels from higher to lower refractive index above the critical angle for total internal reflection, leaving no propagating transmitted plane electromagnetic wave normal to the interface.

At an infinite planar interface, total internal reflection returns all incident normal power in the reflected plane electromagnetic wave. Assume homogeneous dielectrics with isotropy, linearity, no dielectric loss, magnetic permeability ${\mu }_0$, and positive real refractive indices $n_1>n_2$. Above the critical angle for total internal reflection, both TE polarization and TM polarization have unit optical reflectance despite the external evanescent wave.^[ref]

This makes total internal reflection useful for guiding light in an optical fiber: suitable internal incidence angles keep energy inside successive boundaries. Its more surprising feature is that perfect optical reflectance coexists with an external electric field. The evanescent field in total internal reflection can interact with a nearby object, as the frustrated total internal reflection experiment demonstrates.

For an original idealized example, choose $n_1=1.5$, $n_2=1$ at vacuum wavelength ${\lambda }_0=633\mathrm{nm}$. At angle of incidence $40^{\circ }$, Snell law for electromagnetic waves gives an angle of refraction $\arcsin (1.5\sin 40^{\circ })\approx 74.6186^{\circ }$, so a transmitted light ray still emerges. At ${\theta }_c\approx 41.8103^{\circ }$, its direction becomes tangent to the boundary. At $50^{\circ }$, the proposed sine of a real angle of refraction would be $1.5\sin 50^{\circ }\approx 1.14907>1$: the transmitted wavevector must instead have an imaginary normal component. These are predictions for the stipulated media, not measured properties of a particular block.

A real semicircular-block experiment isolates this threshold. The University of Maryland laboratory instructions place a laser, motor-driven rotation stage and photodetector together; an aperture limits the detector voltage. The semicircular block is centered on the rotation axis, with the photodetector aligned along its flat face. The instructions identify a strong grazing-beam signal and a weaker signal just before the critical angle for total internal reflection. The readout is detector voltage versus stage angle. Retracing the incident laser beam identifies normal incidence for alignment.^[ref]

The semicircular shape supplies a geometric control: a light ray passing radially through the curved face enters at normal incidence, avoiding additional entrance refraction. The weaker signal near the flat face locates a limiting angle of refraction; a vanishing remote-detector signal alone would not establish that the external electric field vanishes. Beam width and alignment limit angular resolution; no measured threshold or decay depth is supplied.^[ref]

For refractive index $n(y)=c\sqrt{y}$ with $c>0$, the optical-length integrand is $c\sqrt{y}\sqrt{1+y^{\prime 2}}$. The Beltrami identity gives $\sqrt{y}/\sqrt{1+y^{\prime 2}}=\sqrt{h}$, while the full Euler-Lagrange equation gives $2y{y}^{\prime \prime }=1+y^{\prime 2}$. Together these yield the displayed parabolas and exclude a spurious constant solution of the squared first integral. Equal-height endpoints $(\pm L,Y)$ force $x_0=0$ and $h^2-Yh+L^2/4=0$. There are two positive stationary rays for $Y>L$, one coalesced ray for $Y=L$, and none of this smooth family for $Y<L$. For $L=1,Y=5/4$, the two paths are $y=1+x^2/4$ and $y=1/4+x^2$. Fermat principle requires stationary optical length, not a global minimum over every competing path.

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]

Water Cherenkov neutrino reconstruction infers a vertex, time and charged particle direction from photomultiplier pulses. The 1987 analysis recorded charge and timing, calibrated with decaying muons and source-produced gamma rays, and rejected outer-detector activity and events following muons.^[ref]

In an ideal point-flash approximation, a light source at ${\mathbf{r}}_0$ and time $t_0$ gives the displayed travel-time relation at sensor ${\mathbf{r}}_i$ in a medium of refractive index $n$. Subtracting two times eliminates $t_0$. With a chosen constant $n=1.33$ and path lengths $3\mathrm{m}$ and $6\mathrm{m}$, the arrival-time difference is $n(3\mathrm{m})/c\simeq 13.31\mathrm{ns}$. These are synthetic inputs, not detector measurements. Several sensors constrain the vertex; an extended moving track requires the light-emission position and time along that track as well.

For a charged particle moving at $v=\beta c$, the Cherenkov cone angle satisfies $\cos {\theta }_C=1/(n\beta )$ when $n\beta >1$. With $\beta \simeq 1$ and the illustrative $n=1.33$, ${\theta }_C\simeq 41.25^{\circ }$. The geometrical cone carries track-direction information; electron deflections and the spread of emission positions broaden the observed pattern. The neutrino itself does not draw the light cone.

A Cherenkov radiation threshold and an analysis threshold are different constraints. For a fixed $n$, an electron must exceed $T_C=m_e{c}^2[(1-n^{-2}{)}^{-1/2}-1]$. With $m_e{c}^2=0.511\mathrm{MeV}$ and the chosen $n=1.33$, this gives about $0.264\mathrm{MeV}$. Recording and selecting a useful event requires much more light: the 1987 detector's trigger accepted $8.5\mathrm{MeV}$ electrons with 50% efficiency.^[ref]

A wavefront connects points with the same wave phase at one specified time. It is a surface, rather than the trajectory of a particular piece of light. For a scalar monochromatic wave written as $u(\mathbf{r},t)=A(\mathbf{r})\cos \Phi (\mathbf{r},t)$, a wavefront is a connected level set of the phase function $\Phi$; different surfaces can share the same wave phase modulo $2\pi$. A plane wave has planar wavefronts. An outward wave from an ideal point source in a uniform medium has spheres as wavefronts, while an optical lens can turn nearly planar wavefronts into converging ones. The wavefront describes wave phase, not wave amplitude: drawing it does not say how bright the beam is. ^[ref]

The relation to light rays becomes precise in a stationary medium with isotropy, negligible loss and a real positive scalar refractive index $n(\mathbf{r},\omega )$. In the geometrical optics limit, spatial changes in the refractive index and wave amplitude are slow on the local wavelength scale. For a single smooth propagation branch, writeHere $\omega >0$ is the angular frequency, $c$ is the vacuum speed of light, $S$ has units of length, and the last relation is the optical eikonal equation. The local wavevector is $\mathbf{k}=\nabla \Phi =k_0\nabla S$. Its direction is normal to a smooth wavefront, since a gradient is normal to its level sets. A forward light ray has unit tangent vector $d\mathbf{r}/ds=\nabla S/n$, so it follows this normal under the stated assumptions. Along that light ray, $dS/ds=\nabla S\cdot (d\mathbf{r}/ds)=n$: the change in $S$ is the optical path length. This is the short geometric argument linking wavefronts, light rays and optical path length. ^[ref]

The diagram shows incident and transmitted branches separately, with the reflected branch omitted. Its stipulated refractive indices are $n_1=1$ and $n_2=3/2$, and its incident angle is $30^{\circ }$ to the interface normal. At a stationary interface, the frequency and tangential wavevector agree on the two sides: Snell's law gives $\sin {\theta }_2=1/3$, hence ${\theta }_2\approx 19.4712^{\circ }$. The wavefronts turn while their normal spacing decreases from ${\lambda }_0$ to $2{\lambda }_0/3$, where ${\lambda }_0$ is the vacuum wavelength. More explicitly, with $x$ normal to the interface and $y$ tangential, take $S_1=\sqrt{3}x/2+y/2$ for $x<0$ and $S_2=\sqrt{2}x+y/2$ for $x>0$. These functions agree at $x=0$; the drawn wavefronts are $S_j=m{\lambda }_0$ at $t=0$ with ${\varphi }_0=0$. This is an exact piecewise plane wave phase construction for the stipulated transparent media, not measured material data. ^[ref]

The normal-to-wavefront rule requires its assumptions. In a medium with anisotropy, the wavevector remains normal to a smooth wavefront, but the group velocity and energy flux need not follow it. ^[ref] Near a wave caustic, where multiple light rays meet, one smooth phase function may no longer describe the field; diffraction and the full wave amplitude matter. A wavefront is also not generally a surface of equal pulse arrival time. In a medium with optical dispersion, a monochromatic wave phase moves with the phase velocity, while a sufficiently narrow wave packet follows the group velocity. Even identical optical path lengths at one frequency can give different group delays. The usual equal-travel-time picture needs a common launch reference and a nondispersive ray model; an arbitrary broadband disturbance has no single monochromatic wavefront.^[ref]

The wavelength $\lambda$ is the spatial period of a monochromatic wave, measured at a fixed time along the wavevector. For a plane wave with a sinusoid as its profile, it is the distance between consecutive wave crests, or between any consecutive points with the same wave phase modulo $2\pi$:Here $\mathbf{k}$ uses angular wavenumber, $f$ is ordinary frequency, and $v_p$ is phase speed. A distance between arbitrary zero crossings need not be a wavelength: consecutive zeros of a sinusoid are only $\lambda /2$ apart, with opposite slopes. In a standing wave, consecutive nodes of a standing wave are also $\lambda /2$ apart; $\lambda$ refers to the constituent travelling waves, even though the nodes of a standing wave do not travel.

Across a stationary interface between stationary linear media, the incident, reflected and transmitted monochromatic waves have the same frequency. The boundary conditions must hold at every time: terms oscillating with different angular frequencies cannot cancel throughout the entire time record. Each medium's dispersion relation then determines its own angular wavenumber and wavelength, soFor electromagnetic radiation in a transparent medium with refractive index $n$, $v_p=c_0/n$ and $\lambda ={\lambda }_0/n$, where $c_0$ is the vacuum speed of light. A synthetic example with ${\lambda }_0=600\mathrm{nm}$ and $n=1.5$ gives $\lambda =400\mathrm{nm}$ at the same frequency. This conservation argument assumes time-independent linear material response and a stationary interface in the chosen frame. A moving interface can produce a Doppler effect; time-dependent or nonlinear response can change or generate frequencies.^[ref]

A broad frequency spectrum generally supplies many component wavelengths, rather than one wavelength for an entire pulse. Even two tones have both their individual wavelengths and a different overall repetition distance: the $2$- and $8$-cycle-per-metre sum in spatial sampling of a wave repeats after $0.50\mathrm{m}$, but contains the finer $0.125\mathrm{m}$ structure. A narrow wave packet can have an approximate carrier wavelength; the wave envelope propagates approximately at group velocity, determined by the slope of the dispersion relation. Substituting that group velocity into $\lambda =v_p/f$ generally gives the wrong carrier wavelength. Where the medium varies slowly, local wavelength gives a more useful description.

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.