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A submerged object looks shallower because its emerging light rays bend away from the surface normal vector. Their backward extensions give an approximate virtual image nearer the surface. For an object at perpendicular depth $d$ in a medium of refractive index $n_{\mathrm{obj}}$, viewed through a flat boundary from a medium of refractive index $n_{\mathrm{obs}}$, the near-normal paraxial approximation gives $d_{\mathrm{par}}=(n_{\mathrm{obs}}/n_{\mathrm{obj}})d$. Thus the model water depth $12\mathrm{cm}$ appears approximately $9\mathrm{cm}$ from air. This is a bundle-of-light rays statement, not a deflection of the axial light ray.^[ref]

For general transmitted light rays, specify what “depth” means. Put the object at $(0,-d)$ below a horizontal boundary $z=0$. A light ray reaches the boundary at $(x,0)$, with object-side angle ${\theta }_{\mathrm{obj}}$ and observer-side angle ${\theta }_{\mathrm{obs}}$, both measured from the upward normal vector. For $x>0$, its backward extension meets the object's normal line $x=0$ at $z=-d_{\mathrm{axis}}$, whereThis axis-intercept depth is exact for that one transmitted light ray in the stated model. It tends to $d_{\mathrm{par}}$ as the angles tend to zero; at exactly zero the quotient is interpreted as a limit. These elementary tangent, sine and cosine relations show precisely where the small-angle approximation enters.

The original calculations below use $d=12\mathrm{cm}$, $n_{\mathrm{obj}}=4/3$ and $n_{\mathrm{obs}}=1$:

At appreciable obliquity these depths differ, so the full outgoing bundle has no single exact virtual image on the object's normal line. An oblique eye samples a narrow bundle; its local apparent position also depends on that bundle and cannot be inferred by declaring $d_{\mathrm{axis}}$ to be a universal image depth. A plane refracting boundary gives approximate image formation near normal incidence, with departures at larger angles. The construction assumes a flat surface, uniform transparent media with isotropy, and a propagating transmitted light ray; it does not extend to total internal reflection.

In the sharp planar, nonmagnetic ionosphere model with frequency above the plasma frequency, the refractive index is $n=\sqrt{1-{\omega }_p^2/{\omega }^2}<1$. From air, the Brewster angle satisfies $\tan {\theta }_B=n$. At that angle the reflected p-polarized wave vanishes, while s polarization retains a nonzero reflected component. A ground-to-ground single reflection from height $h$ spans $D=2h\tan {\theta }_B=2hn$. This is below the critical angle for total internal reflection for total internal reflection, which reflects both polarizations in this lossless isotropic model. The distinction diagnoses ordinary Brewster reflection rather than total internal reflection.

A photon emitted at angle ${\theta }_C$ to a normally incident particle can cross a planar boundary from index $n_1$ to $n_2$ only if Snell's law permits a real transmitted angle. Combining $\cos {\theta }_C=1/(n_1\beta )$ with that condition gives $n_1^2-{\beta }^{-2}\le n_2^2$. Otherwise the photon undergoes total internal reflection. For a relativistic particle in glass of index $1.55$ followed by air, the condition fails despite emission inside the glass.

The specular reflection of a wave at a planar dielectric interface with amplitudes described by the Fresnel equations. Below the critical angle, the reflected fraction is generally nonzero even when transmission is allowed; above the critical threshold, ideal lossless total internal reflection has unit reflected power. The reflected electric-field sign depends on polarization and the chosen basis conventions.

Frustrated total internal reflection occurs when a nearby second high-index dielectric couples to the evanescent field in total internal reflection, allowing power to cross a finite low-index gap. The extra boundary changes the dielectric-interface boundary conditions; the gap field need not be a single decaying evanescent wave. Consequently, the zero normal Poynting vector calculation for two infinite half-spaces does not forbid transmission in this three-medium geometry.

Harvard’s actual demonstration uses two right-angle glass prisms, a helium-neon laser, a small clamp and two perpendicular screens as detectors. The contact faces are cleaned and placed opposite each other, with the clamp protecting the corners. Increasing pressure narrows the intervening air film; the observed beam routing changes gradually from total internal reflection toward onward transmission. The broad-gap reflecting configuration supplies the comparison for the narrow-gap transmitting configuration. This is a qualitative readout: the page explicitly says the gap is not measured.^[ref]

The inference is that the lower-index region can mediate electromagnetic energy transfer when another boundary is nearby. The demonstration does not determine a numerical evanescent penetration depth, resolve gap uniformity, or establish a three-layer transmission coefficient. For the constructed $50^{\circ }$ example, $\delta \approx 178\mathrm{nm}$ identifies the scale over which the isolated-interface field remains appreciable; substituting $e^{-2g/\delta }$ for a gap transmission fraction would omit the changed boundary fields and their interference. A quantitative gap calculation must solve the full Maxwell equations with both boundaries.

Geometrical optics describes propagation by rays and Snell's law, neglecting diffraction on scales much larger than the wavelength.

A light ray is a path in an approximation, not a physical thread of zero width. In a transparent, homogeneous, isotropic medium it is straight and perpendicular to a wavefront; boundaries redirect it by the law of reflection or Snell's law. The approximation is useful when the wavelength is small compared with relevant apertures, surface curvature scales, and distances over which the refractive index changes. For an illustrative wavelength of $500\mathrm{nm}$ and a $10\mathrm{mm}$ aperture, their ratio is $5\times 10^{-5}$. That small ratio supports ray tracing for image location, but does not eliminate diffraction when calculating the fine structure of a focused spot. At a wave caustic, geometrical optics can fail despite large optical elements. ^[ref].

The mechanism of refraction is a change in phase velocity. For transparent isotropic media at the chosen frequency, the refractive index iswhere $c$ is the vacuum speed of light and the angles are measured from the interface normal vector. At a stationary planar interface the frequency and tangential wavevector match across the boundary. Since the wavenumber is $k=n\omega /c$, that matching gives Snell's law. The transmitted wavelength changes while the frequency stays fixed. Here $n$ uses phase velocity, which need not equal group velocity. ^[ref]. ^[ref].

Choose model air with $n_1=1$ and model glass with $n_2=1.5$, without claiming these are exact values for every sample or wavelength. At ${\theta }_1=45^{\circ }$, Snell's law givesThe transmitted light ray bends toward the normal vector; its phase velocity is $c/1.5$, and its wavelength is two thirds of the air value. By reversibility of an optical ray, incidence from glass at $28.1255^{\circ }$ returns the outgoing air angle $45^{\circ }$.

Incidence from glass at $45^{\circ }$ is a different experiment. It would require $\sin {\theta }_2=1.5\sin 45^{\circ }\approx 1.06066$, so there is no real transmitted angle. The critical angle for total internal reflection isAbove this angle, an ideal lossless glass-air boundary produces total internal reflection. The physical optics description still permits an evanescent wave in the air, although no propagating transmitted light ray crosses the boundary. ^[ref]. ^[ref].

The usefulness of geometrical optics is that complicated instruments can be assembled from these local rules. Applying Snell's law at successive curved surfaces changes a spreading bundle into a converging one. The Fermat principle unifies the construction through stationary optical path length; stationarity does not require a global minimum. ^[ref]. Fang's MIT notes explicitly connect this principle to specular reflection, refraction, and thin lens imaging, providing a route from individual paths to complete instruments. ^[ref].

Efficiency has several denominators. Specify a steady electric current $I>0$, positive terminal electric potential difference $V_f$, and a simplified narrow emission spectrum with photon energy $E_{\gamma }$. Assume at most one radiative photon per injected electron and electron hole pair. Let ${\eta }_{\mathrm{inj}}$ be the fraction of the electrical pair-supply rate $I/q$ reaching the active layer, ${\eta }_{\mathrm{rad}}$ the fraction of those pairs recombining radiatively, and ${\eta }_{\mathrm{ext}}$ the fraction of internally emitted photons that escape. Then counting the successive events givesDefine internal quantum efficiency here conditionally on pairs reaching the active layer: ${\eta }_{\mathrm{IQE}}={\eta }_{\mathrm{rad}}$. The internal photon yield per electrical pair is instead ${\eta }_{\mathrm{inj}}{\eta }_{\mathrm{IQE}}$. Stating this convention matters when another definition folds injection into internal quantum efficiency. The external quantum efficiency and wall-plug efficiency areThese are photon-count and energy-flow identities for the declared model. For a broad spectrum, replace $E_{\gamma }$ by the mean escaping photon energy, weighted by photon number, when computing optical power.

A stipulated example chooses $I=20\mathrm{mA}$, $V_f=3.20\mathrm{V}$, $\lambda =450\mathrm{nm}$, ${\eta }_{\mathrm{inj}}=0.90$, ${\eta }_{\mathrm{IQE}}=0.80$ and ${\eta }_{\mathrm{ext}}=0.50$. For every 100 electrical pairs, 90 reach the active layer, 72 emit internally, and 36 photons escape. Thus ${\eta }_{\mathrm{EQE}}=0.36$, not $0.80$. Using $q=1.602176634\times 10^{-19}\mathrm{C}$ givesThe dimensional analysis is transparent: $A/C={\mathrm{s}}^{-1}$, $J/s=\mathrm{W}$, and $\mathrm{CV}=\mathrm{J}$. The optical and electrical efficiencies differ because $E_{\gamma }/q=2.75520\mathrm{V}$ differs from the chosen $V_f$.

The factors suggest different improvements. Better confinement can reduce injection leakage; suppressing nonradiative recombination raises the radiative fraction; improving light extraction efficiency helps already-emitted photons escape. A photon trapped by total internal reflection has already passed the radiative step and can still be lost through absorption. Improving extraction therefore need not change the internal radiative fraction.

Shuji Nakamura's lecture cautions that why indium gallium nitride emits efficiently despite many defects was not completely settled in that account.^[ref]

As an independent consistency check on the 1994 reported values, treating its broad spectrum as monochromatic at the $450\mathrm{nm}$ peak givesThis agrees at the quoted precision with the reported $2.7\%$. Its optical/electrical power ratio is instead $1.5\mathrm{mW}/(20\mathrm{mA}\times 3.6\mathrm{V})=2.0833\%$. The first estimate uses a peak in place of a photon-number-weighted mean, so agreement is a consistency check rather than an exact reanalysis of the measured spectrum.

Use a slit light source, a rectangular transparent block on plain paper, pencil, ruler and protractor. Trace the block outline, keep it fixed, and mark widely separated points on incident and emergent light rays. Remove the block and join the entry and exit points to reconstruct its internal light ray. Draw the surface normal vector at entry; measure $i$ outside and $r$ inside from that normal line. Choose a path exiting the opposite parallel face rather than a side edge. Repeat for a range of $i$ with the same block and source colour; record $i,r,\sin i,\sin r$ and repeat-reading spread. Plot $\sin i$ vertically against $\sin r$ horizontally: Snell's law predicts slope $n_{\mathrm{block}}/n_{\mathrm{air}}$, approximately the block's refractive index. The emergent light ray should be parallel to the incident one. Beam width, pencil marks, protractor resolution, block movement and colour spread limit precision; a narrow-colour source reduces spread from optical dispersion. These are measured angular readings, not a direct timing of phase velocity.^[ref]

A normal incidence trial checks the drawn normal and the absence of bending, but the ratio $\sin i/\sin r$ is then $0/0$. Small nonzero angles also make that ratio sensitive to reading errors. Compare several trials through the slope rather than relying on one nearly axial pair. This rectangular-block measurement complements the existing total internal reflection experiment, which probes a different boundary condition.

For a light ray launched on the centreline in the plane of a circular optical fiber bend, let $R$ be centreline radius, $r$ core radius and $\beta$ the internal launch angle from the tangent. The straight segment's distance from the bend centre is $R\cos \beta$. Thus its outer-wall angle of incidence satisfies $\sin i=R\cos \beta /(R+r)$. Equating this to $n_2/n_1$ gives the limiting total internal reflection threshold in the title, provided $n_1\cos \beta >n_2$. Radially displaced or nonplanar launches need different geometry; this is not a universal wave-optical loss cutoff.

For the lower-face-first ray path through an equilateral optical prism, take external entry angle $\theta$, internal refraction angle $r$, base incidence $\alpha -r$ and one total internal reflection at the horizontal base. At the other sloped face the internal incidence equals $r$, so Snell's law gives exit angle $\theta$ again. Its counterclockwise, oriented direction change is the displayed expression with $\alpha =60^{\circ }$. It is independent of refractive index for this route, but whether the base reflection is total is not. Under normal dispersion, a setting at the green critical threshold sends shorter wavelengths along the total-reflection route while longer wavelengths can leak through the base.

An optical fiber is a thin, flexible electromagnetic waveguide that carries light along its length. In a conventional step-index optical fiber, an optical fiber core with refractive index $n_1$ is surrounded by an optical fiber cladding with lower refractive index $n_2$. Suitable light rays remain confined by total internal reflection at this interface. The cladding is an optical part of the guide, distinct from its protective coating.^[ref]

The useful surprise is that a transparent strand can deliver light around a route that a free beam would leave. This makes optical fibers useful for communications and for carrying collected light to a fiber-fed spectrograph. Guidance, successful launch, and surviving optical power are separate questions: the optical fiber acceptance angle constrains launch geometry, while optical fiber attenuation describes loss along the route.

The optical fiber acceptance angle is the largest launch half-angle admitted by the specified guide model. For a flat entrance face perpendicular to the optical axis, external refractive index $n_0$, and a straight step-index optical fiber, Snell's law at the entrance givesAt the core–cladding interface, the critical angle for total internal reflection $i_c$ satisfiesFor a meridional ray, strict total internal reflection requires $i>i_c$. At the limiting geometry $i=i_c$, the refracted light ray is tangent to the interface, soThis identifies the numerical aperture with an entrance cone under these assumptions. With $n_0=1$ for air, the conventional inclusive boundary condition isfor a nonnegative launch angle $\theta$; a practical launch stays inside the boundary. If $\mathrm{NA}>n_0$, this ideal model imposes no further restriction within the forward hemisphere, and $\arcsin (\mathrm{NA}/n_0)$ must not be used. The light ray must also intersect the optical fiber core: a suitable angle alone does not correct a misplaced beam.

For an original example, take $n_1=1.50$, $n_2=1.48$ and air at the entrance. ThenThe internal limiting angle is only ${\varphi }_{\max }\simeq 9.37^{\circ }$, while the critical angle for total internal reflection is $i_c\simeq 80.63^{\circ }$. A $10^{\circ }$ launch has $\varphi \simeq 6.65^{\circ }$ and $i\simeq 83.35^{\circ }$, so it satisfies total internal reflection. A $20^{\circ }$ launch has $\varphi \simeq 13.18^{\circ }$ and $i\simeq 76.82^{\circ }$, so it fails this test.

This cone is a geometrical optics result for meridional rays. A single-mode optical fiber requires the spatial overlap of the incident electromagnetic wave with its guided normal mode; its numerical aperture does not impose an exact angular cutoff on arbitrary beams. Diffraction, finite cladding, and bends require a physical optics description.^[ref]

The optical material surrounding the optical fiber core. In a conventional step-index optical fiber, its lower refractive index permits total internal reflection; it is distinct from the protective coating.

Guiding light and retaining useful optical power are different achievements. Charles Kuen Kao and George Alfred Hockham argued that impurities need not impose the measured losses of existing glasses on every future optical fiber. Their 1966 paper treats wave absorption, scattering of light and guidance separately; reducing impurity absorption leaves intrinsic material losses and geometric losses to address. This was a route toward practical communication, building on existing optical fibers, rather than the invention of every use of guided light.^[ref]

The launch experiment used clad glass optical fibers fixed in capillary tubes with epoxy for end polishing. Laser illumination gave transverse normal mode patterns; input-spot displacement and polarization of an electromagnetic wave controlled excitation. These observations tested guidance, not kilometre-scale optical fiber attenuation.^[ref]

Charles Kuen Kao's lecture describes the complementary material experiment: short, polished glass samples, including plasma-deposited fused silicon dioxide, were examined with purpose-built single-beam and later double-beam spectrophotometers. A separate ellipsometry measurement assessed end-surface specular reflection, a check against confusing interface loss with bulk wave absorption. The readout was loss versus wavelength. Comparing glass samples supported impurity reduction as a route to lower loss. Surface contamination and the tiny bulk loss of short samples limited inference; the result did not itself demonstrate a long drawn optical fiber. The lecture also credits subsequent manufacturing work at Corning.^[ref]

An original scale comparison makes the material problem tangible. In the uniform optical fiber attenuation model, a stipulated $20dB/km$ leaves $10^{-2}=0.01$ of launched optical power after $1\mathrm{km}$; a stipulated $0.2dB/km$ leaves $10^{-0.02}\simeq 0.95499$. Both guides could satisfy the same total internal reflection condition. A successful spot pattern at the output therefore cannot by itself establish a useful communication distance.

Snell's law is $n_a\sin {\theta }_a=n_b\sin {\theta }_b$, with both angles measured from the interface normal vector. Let $\psi$ be the meridional ray's angle to the optical axis inside the core and let $n_0$ be the incident medium's refractive index. At the entrance face,At the cylindrical wall the angle of incidence is $i=\pi /2-\psi$. For $n_{\mathrm{inner}}>n_{\mathrm{outer}}$, the critical angle for total internal reflection satisfies $\sin i_c=n_{\mathrm{outer}}/n_{\mathrm{inner}}$. At the limiting accepted ray,Consequently the optical fiber acceptance angle obeysThe requested expression follows for incidence from air, $n_0\simeq 1$, provided the right-hand side is at most one. Strict total internal reflection occurs inside this limiting cone. The derivation uses the standard meridional ray model; skew launches need additional geometry.

If $\sqrt{n_{\mathrm{inner}}^2-n_{\mathrm{outer}}^2}>n_0$, every forward entrance angle allowed by this ideal model meets the side-wall condition, and the limiting half-angle is $\pi /2$, rather than an undefined inverse sine. For example, $n_{\mathrm{inner}}=1.6$, $n_{\mathrm{outer}}=1.1$, $n_0=1$ gives $\sqrt{1.35}>1$. Thus the unsaturated printed formula is not valid for every possible pair of refractive indices.

A real transmitted normal wavenumber exists only ifso the critical angle for total internal reflection isFor ${\theta }_I>{\theta }_I^{\mathrm{cr}}$, putFor incident amplitude $E_0\hat{\mathbf{y}}$, the transverse-electric Fresnel reflection coefficient and transmission coefficient areA boundary-matched solution iswith each magnetic component obtained from ${\mathbf{B}}_X={\mathbf{k}}_X\times {\mathbf{E}}_X/\omega$, now using the complex transmitted wavevector $(i\kappa ,0,k_z)$. Since $|r|=1$, the wave undergoes total internal reflection; the transmitted field is an evanescent wave that propagates parallel to the interface and decays exponentially into $x>0$.

At $p=600\mathrm{MeV}/c$, the velocities are $\beta =p/\sqrt{p^2+m^2{c}^2}$, giving approximately $1.0000$, $0.98485$ and $0.53874$ for electrons, muons and protons. The glass threshold is $\beta >1/1.55=0.64516$. Thus the first two produce Cherenkov radiation, with internal angleswhereas a primary proton at this momentum is below threshold.

If an optical arrangement transports those cone angles without changing them, a plane $L=2\mathrm{m}$ downstream would intersect rings of radiusThese are conditional geometric estimates, not the ray trace for a flat slab in air. The $0.1\mathrm{m}$ emission depth alone spreads the radius by order $0.12\mathrm{m}$, already larger than the approximately $0.062\mathrm{m}$ separation. The slab is one radiation length: electrons lose appreciable energy by bremsstrahlung and begin an electromagnetic shower, producing a broader and less clean pattern. Muons mainly ionize and retain a comparatively clean cone, with some Multiple Coulomb scattering. A below-threshold primary proton gives no direct cone, although energetic delta electrons and other secondary interactions can produce light.

There is an essential optical qualification. For a normal beam through a parallel-faced glass slab followed by air, the internal critical angle is ${\theta }_{\mathrm{crit}}=\arcsin (1/1.55)\simeq 40.18^{\circ }$. Both emission angles exceed it. Snell's law would require $\sin {\theta }_{\mathrm{air}}=1.55\sin {\theta }_C>1$, so the ideal photons undergo total internal reflection at the downstream face and do not form those direct downstream rings. Optical coupling, a prism or different surface geometry must be specified to extract them. The stated material parameters alone do not define a unique detected pattern; this flat-slab geometry is a concrete counterexample to silently assuming straight transmitted rays.

Follow the branch of a light ray that undergoes one internal specular reflection in the spherical drop. At entry, refraction turns its direction by $\alpha -\beta$. The internal specular reflection turns it by $\pi -2\beta$, because the radius is the normal vector and the incidence and reflection angles are both $\beta$. At exit the second refraction contributes another $\alpha -\beta$. Thus the total deviation angle isSnell's law at entry gives $\sin \alpha =n\sin \beta$, taking the surrounding air's refractive index as one. The same internal angle at successive encounters follows from the isosceles triangles formed by each chord and the two radii. Internal specular reflection here need not be total internal reflection; the transmitted branch at that encounter does not belong to the one-reflection trajectory.

Huygens' principle treats every point of an advancing wavefront as a source of secondary wavelets; their common envelope forms the next wavefront. Consider two points of an incident plane wave reaching the interface at different times. During the same interval $\Delta t$, the remaining incident portion travels $v_1\Delta t$ while the first transmitted wavelet expands a distance $v_2\Delta t$. The two wavefront right triangles share their interface segment, soThis is Snell's law, with both angles measured from the surface normal. The speeds are phase velocities at the common frequency in stationary isotropic media. If the equation requires $|\sin {\theta }_2|>1$, there is no propagating refracted ray and total internal reflection applies instead.

For $n=1.5$, the critical angle is $i_c=\arcsin (1/1.5)\simeq 41.8103^{\circ }$. Here $0\le r<i_c<\alpha =60^{\circ }$, so the condition at B is $\alpha -r>i_c$. At the threshold,Thus the requested positive-angle interval for strict total internal reflection isAt the upper equality the transmitted ray is grazing; normal incidence $\theta =0$ also gives total internal reflection at B if the endpoint is included. The distinction follows the question's positive-angle convention.