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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.

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.

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.

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 $x>0$.

Two interface effects. For $n_2<n_1$, Snell's law gives $\sin {\theta }_t=(n_1/n_2)\sin {\theta }_i$. If $\sin {\theta }_i>n_2/n_1$, no propagating transmitted ray exists: there is total internal reflection, with an evanescent wave on the far side. An optical fibre uses this to guide light.

For ordinary isotropic nonmagnetic media, $\tan {\theta }_i=n_2/n_1$ is the Brewster angle. The reflected p-polarized component, whose electric field lies in the incidence plane, vanishes; the reflected s-polarized component generally remains. Reflection polarizers and polarized glare reduction exploit this selectivity.

The plasma response. Neglect collisions, thermal-pressure corrections and background magnetic field. Let the wave vary as $e^{-i\omega t}$. An electron displacement obeysIts dipole moment is $-e{\mathbf{x}}_e$, so the electron electric polarization is ${\mathbf{P}}_e=-n{e}^2\mathbf{E}/(m_e{\omega }^2)$. Ions of charge $Ze$ give a further contribution ${\mathbf{P}}_i=-nZ{e}^2\mathbf{E}/(m_i{\omega }^2)$, smaller by $Z{m}_e/m_i$. Thus heavy ions move negligibly at these electron-wave frequencies. Using $\mathbf{D}={\epsilon }_0\mathbf{E}+\mathbf{P}$ gives the cold collisionless plasma dielectric response,For this model ${\mu }_r\simeq 1$, so $n_{\mathrm{refr}}^2={\epsilon }_r$ and $c^2{k}^2={\omega }^2-{\omega }_p^2$ for the transverse electromagnetic wave.

The vertical cutoff and density. Below the plasma frequency, the squared refractive index is negative: the wave is an evanescent wave instead of propagating through a layer hundreds of kilometres thick. The observed $5\mathrm{MHz}$ cutoff therefore identifies $f_p=5.0\times 10^6\mathrm{Hz}$, with ${\omega }_p=2\pi f_p$. ConsequentlyThis uses ${\epsilon }_0=8.854\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$, $m_e=9.109\times 10^{-31}\mathrm{kg}$ and $e=1.602\times 10^{-19}\mathrm{C}$. The density is also $3.10\times 10^5\mathrm{c}{\mathrm{m}}^{-3}$.

The transmission elevation. At $f=8\mathrm{MHz}$, the ionosphere has refractive indexIf $\alpha$ is elevation above horizontal, incidence at the layer is ${\theta }_i=\pi /2-\alpha$. Avoiding total internal reflection requires $\sin {\theta }_i=\cos \alpha <n_i$. The limiting elevation is thereforeStrictly, it must point above this limiting angle to carry upward power through the finite layer: at equality the transmitted wave is grazing. With a suitable satellite position, the stated orbital altitude imposes no additional angular constraint in this flat-layer approximation.

Polarization-selective ground reflection. The ray's incidence plane is vertical. Horizontal electric polarization is s polarization; the other transverse polarization lies in the vertical incidence plane and is p polarization. Reflection of s but not p identifies the Brewster angle, not total internal reflection, which would reflect both in this lossless isotropic model. ThusFor a reflecting height $h=200\mathrm{km}$, the horizontal range consists of two equal legs. The Brewster reflection from an ionospheric interface givesAt this angle ${\theta }_B$ is below the critical incidence angle, so a transmitted ray also exists; the detected horizontal component is an ordinary partial reflection. The s-polarized reflected power fraction is approximately $[(1-n_i^2)/(1+n_i^2){]}^2\simeq 0.059$, whereas the ideal p-polarized reflected fraction is zero. These results use the question's sharp planar interface model and exclude upper-interface reflections.

Snell's law is $n_1\sin {\theta }_1=n_2\sin {\theta }_2$, with angles measured from the interface normal. The Huygens-Fresnel principle explains it by constructing the transmitted wavefront from secondary wavelets: during the same time interval, distances travelled in the two media are proportional to $c/n_1$ and $c/n_2$. The geometry of the advancing plane wavefront then gives $\sin {\theta }_1/\sin {\theta }_2=(c/n_1)/(c/n_2)=n_2/n_1$.

For incidence from index $n$ into vacuum, the transmitted angle is $\alpha =\arcsin (n\sin \theta )$. The ray bends away from the normal, so its positive deviation isFor $0<\theta <{\theta }_c=\arcsin (1/n)$,Thus the graph rises monotonically from zero to a finite endpoint with a vertical tangent:At the endpoint the transmitted ray grazes the interface. Beyond the critical angle for total internal reflection, no propagating transmitted ray exists and total internal reflection occurs.

Inside a prism, the two ray angles with the face normals sum to its apex angle. For the equilateral cross-section, they are $\theta$ and $\pi /3-\theta$. The entry deviation equals the corresponding exit deviation under ray reversal. Adding the two changes of direction therefore givesThis is symmetric about $\theta =\pi /6$. Symmetry alone gives a stationary point; to establish that it is a minimum, differentiate once more:within the transmitted-ray interval. The sum is therefore strictly convex, andThese expressions require $1<n<2$ for a non-grazing path through both faces; $n=2$ is a grazing limit, and $n>2$ admits no such two-face transmitted path. More explicitly, both internal angles must be at most ${\theta }_c$, which is possible only if $2{\theta }_c\ge \pi /3$.

Atmospheric hexagonal ice crystals can present pairs of faces with a $60^{\circ }$ angle, acting as small prisms. Their orientations turn the minimum-deviation cone around the Moon into an optical halo. No transmitted ray of this family is deflected less than the minimum deviation of a prism. Moreover, near the minimum, $\Delta -{\Delta }_{\min }$ is quadratic in the departure from the symmetric ray, so a range of incident directions is compressed into a small outgoing angular range. This caustic makes the inner rim sharp and bright, with the ray density decreasing farther outside. The common 22-degree halo has an angular radius near $22^{\circ }$ about either the Sun or Moon, hence a diameter near $44^{\circ }$.^[ref]

Refraction changes the direction of a transmitted wave when it crosses between media with different phase velocities. For light at a planar interface, Snell's law relates the angles measured from the normal vector through the refractive indices. At normal incidence the direction remains unchanged even though the wavelength and phase velocity can change. The worked example under geometrical optics distinguishes transmission from total internal reflection.

For light, refraction explains why a submerged object looks nearer the surface and why an optical lens can form an image. A wavefront crossing a boundary must retain a consistent wave phase along that boundary: its normal spacing changes with phase velocity, and oblique light rays change direction. Normal incidence separates these two effects especially clearly.

Take stationary, transparent, homogeneous media with isotropy, linearity of their optical response, and real positive refractive indices, at one chosen frequency $f$. With $c$ the vacuum speed of light and ${\lambda }_0=c/f$ the vacuum wavelength, medium $j$ hasThe frequency is unchanged at a stationary boundary; the phase velocity, wavelength and wavenumber can change. Here $n_j$ is a phase refractive index, not the group refractive index. If there is optical dispersion, evaluate $n_j$ at this same frequency on both sides.^[ref]

For an original idealized air–water example, stipulate $n_{\mathrm{a}}=1$ and $n_{\mathrm{w}}=4/3$, with vacuum wavelength $600\mathrm{nm}$. Then $f=c/(600\mathrm{nm})\approx 4.99654\times 10^{14}\mathrm{Hz}$, the water phase velocity is $3c/4$, and its wavelength is $450\mathrm{nm}$. At an air angle of incidence $30^{\circ }$, Snell's law gives a water angle of refraction $\arcsin (3/8)\approx 22.0243^{\circ }$. These stipulated values illustrate the relations; actual refractive indices depend on material conditions and frequency. By reversibility of an optical ray, the same path emerging from water has outgoing air angle $30^{\circ }$.

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.

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]