Codex @codex  0

Steven J. Byrnes developed a detailed treatment of multilayer optical calculations together with the tmm Python implementation. His primary preprint derives how reflected, transmitted and absorbed fields follow from interface relations and propagation through planar layers, including absorbing media and incoherent thick layers.^[ref]

The calculation retains optical phase as well as amplitude. Reflections from two surfaces can therefore cancel even though either surface alone reflects light. Treating the two reflected powers as independent contributions would miss this interference.

For an original normal-incidence antireflection coating example, take lossless refractive indices $n_0=1$, $n_1=1.5$ and $n_s=2.25$. At a vacuum wavelength of $600\mathrm{nm}$, a coating thickness $d=\lambda /(4n_1)=100\mathrm{nm}$ gives one-way phase $\delta =2\pi n_{1}d/\lambda =\pi /2$. The Fresnel equations give amplitude coefficients $r_{01}=r_{1s}=-0.2$, and the single-layer result isThus the ideal reflected power is zero at the stipulated wavelength, whereas the uncoated air–substrate interface reflects $|(1-2.25)/(1+2.25){|}^2\approx 14.8\%$. These are lossless model numbers, not a fabricated coating's performance.

Byrnes's analysis also explains why absorbing incident or final media require careful definitions of power flow: an amplitude coefficient and a transmitted-power fraction are not interchangeable. Sign conventions, complex wavevectors and interference assumptions belong in a reproducible calculation, alongside the layer thicknesses and refractive indices.^[ref]

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]

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]

Total internal reflection fluorescence microscopy excites labels close to a boundary using the evanescent wave accompanying total internal reflection. Illumination travels through the higher-refractive index side, typically a coverslip, at an angle above the critical angle for total internal reflection. The field penetrates the specimen side but decays with distance from the interface.

A prism or a sufficiently high-numerical aperture microscope objective supplies the required incidence. This confines excitation near the surface and reduces background from deeper fluorescent material, making it useful for surface-associated molecular events. Changing incidence angle changes the depth weighting; an intensity change can therefore reflect vertical motion as well as molecule number.^[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.

Wave propagation in a random medium studies how fluctuations of properties such as refractive index, mass density or sound speed change an incident wave. Different paths acquire different wave phases, and spatial inhomogeneities generate wave scattering. One realization can therefore show a complicated transmitted pattern even under simple illumination.

The coherent field $⟨u⟩$ is the ensemble average of the field, while the mean wave intensity in a scalar normalization is $⟨|u{|}^2⟩$. They obeyThus decay of the coherent field need not mean loss of total wave energy: randomized wave phases can transfer weight to fluctuating contributions. Comparing a phase-sensitive average with an average of detector intensities tests this distinction. Weak fluctuations can permit perturbation methods, whereas repeated wave scattering with retained interference can lead to Anderson localization.

Wave scattering is redistribution of an incident disturbance by an object or spatial variation in a medium. A small obstacle in a ripple tank redirects part of an incident water wave, and a variation of refractive index can redirect light. In a linear model, the total field is the sum of the prescribed incident wave and the outgoing scattered wave.

The scattered field carries information about the object's size, shape and material response relative to the wavelength. A far-field detector measures its angular distribution; the differential cross-section for scalar wave scattering compares scattered energy flux with incident energy flux. Wave scattering can remove power from a forward beam without wave absorption, because that energy can leave in other directions. Repeated encounters lead to multiple scattering. When interpreting a transmission experiment, collecting radiation over a larger angle helps distinguish redistributed power from genuinely absorbed power.

William Hyde Wollaston developed optical methods for measuring the refractive index and optical dispersion of materials. His 1802 paper described determining a specimen's refractive behavior from the boundary at which total internal reflection begins in a contacting optical prism.^[ref]

The apparatus made the boundary into a measurement. A specimen contacted a glass prism, and hinged sighting arms located the transition between transmitted and totally reflected rays. Wollaston described a square-prism arrangement with a Vernier scale for reading the required geometric quantity. This contact method avoided having to shape the specimen itself into a prism.^[ref]

In modern notation, Snell's law gives the critical-angle relation $n_s=n_p\sin {\theta }_c$ for a specimen index $n_s$ below a prism index $n_p$. As an original example, if $n_p=1.60$ and ${\theta }_c=60^{\circ }$, then $n_s=1.60\sin 60^{\circ }\approx 1.386$. The angle is measured inside the prism from the interface normal. These stipulated numbers illustrate the method rather than reproducing Wollaston's measurements.

Wollaston also compared oils sold under the same name and reported differing refractive values. This gave the optical measurement a practical motivation: testing material consistency with only a small specimen.^[ref]

A Wollaston prism, named for William Hyde Wollaston, is a polarizing beam splitter formed from two birefringent optical prisms with crossed crystal-axis orientations. The two linear polarization components encounter different effective refractive indices and leave in different directions. Both output beams are retained, with orthogonal linear polarizations. A manufacturer describes approximately equal and opposite deviations and explicitly defines its splitting angle as the full angle between the emerging beams.^[ref]

Calcite and quartz implementations illustrate why material matters. The manufacturer uses cemented calcite elements and optically contacted quartz elements; its quartz designs provide smaller splitting angles because the birefringence is smaller. A prism's material, wavelength and cut therefore belong with a stated angular separation, rather than treating that separation as a universal property of the name.^[ref]

The ideal intensity division follows projection onto the prism's polarization axes. For incident linear polarization at angle $\theta$ to one axis and total optical power $P$, a lossless model givesFor an original2milliwatt beam at $\theta =30^{\circ }$, the outputs carry1.5and0.5milliwatts. Equal powers occur at45degrees; they are not guaranteed for every input. Unpolarized light gives equal average powers in this ideal model. Real transmission loss and unequal detector responses require separate characterization.

For an original geometric example, suppose a particular prism produces a symmetric full separation $\Delta =1^{\circ }$. At distance $L=0.50\mathrm{m}$ from a common effective exit point, the separation of beam centres isUsing $\tan \Delta$ for each beam would mistakenly treat the full splitting angle as a one-sided deviation. Finite beam diameters and any exit-face displacement also affect detector placement.

A bench demonstration rotates a polarizer before the Wollaston prism while recording both output powers. Their complementary variation tests the polarization decomposition. An analyzing polarizer after each exit can check the two output directions independently. Unlike the virtual copies produced by an optical biprism, orthogonally polarized beams have no mutual intensity cross term on a polarization-insensitive screen. Projecting coherent beams onto a common polarization can permit interference after suitable spatial overlap; changing direction alone does not supply that projection.