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Because $s$ is arc length, $|d\mathbf{x}/ds|=1$. Thus $\mathbf{y}=\mathbf{n}/\alpha$, where $\mathbf{n}$ is the unit tangent to the elastic ray. In an isotropic medium this is also the wavefront normal. Consequently $\mathbf{y}$ is the vector slowness, of magnitude $1/\alpha$; locally $\mathbf{y}=\nabla T=\mathbf{k}/\omega$, with $T$ the travel time and $\mathbf{k}$ the wavevector.

For depth-dependent wave speed $\alpha (z)$, the ray equation gives $d{y}_x/ds=d{y}_y/ds=0$. The horizontal component of $d\mathbf{x}/ds$ therefore has a fixed direction, and the elastic ray lies in the vertical plane determined by that direction and its initial point. A vertical elastic ray is the limiting case. If $i$ is the angle to the vertical, its conserved horizontal slowness givesAt a horizontal interface, matching the tangential wavevector at fixed frequency gives the same invariant on both sides: this is Snell's law.

For radial wave speed $\alpha (r)$, take the cross product of position with slowness. The ray equations givesince the second vector is radial. A nonzero constant $\mathbf{L}=\mathbf{x}\times \mathbf{y}$ is perpendicular to every position on the elastic ray, so the ray lies in a fixed plane through the centre. Radial rays obey the same conclusion with a nonunique plane. The spherical elastic ray invariant is thereforewhere $i$ is measured from the radial direction. Across a spherical interface the radius is common and tangential slowness is continuous, so this form of Snell's law survives an interface discontinuity.

In a homogeneous sphere a P wave follows the straight chord between the two surface points. If their principal angular separation is $0\le \Delta \le \pi$, its length is $2R\sin (\Delta /2)$. For constant P wave speed $c_P$,Now put $a=R^{\prime }$ and denote the mantle and core P wave speeds by $c_m>c_c$. Equal incidence and reflection angles put the core-boundary reflection point on the angular bisector. Each mantle leg has length $L$ given by $L^2=R^2+a^2-2Ra\cos (\Delta /2)$. Hence the core-reflected travel time isThe stated domain is essential. At the reflection point the incoming line approaches from the mantle only if $R\cos (\Delta /2)\ge a$. Otherwise it has already crossed the core boundary. The mantle-only direct chord also exists only for $\Delta \le {\Delta }_g$, with $T_D=2R\sin (\Delta /2)/c_m$. In this interval,so the reflected arrival is later. Its intercept is $2(R-a)/c_m$, whereas the direct intercept is zero. At grazing incidence the branches meet at $T=2\sqrt{R^2-a^2}/c_m$, and both have slope $dT/d\Delta =a/c_m$. A continuation of the homogeneous-sphere direct formula beyond this endpoint is not a mantle-only arrival in the modified model.

The core-transmitted elastic rays consist of three straight legs. Their impact distances in mantle and core are $b_m=p{c}_m$ and $b_c=p{c}_c$, respectively. At core entry,Since $c_c<c_m$, the elastic ray bends towards the normal on entry and away from it on exit. The mantle leg subtends $i_m-i_R$ at the centre, and the core chord subtends $\pi -2i_c$. Thus the total angular sweep along the transmitted path is $2(i_m-i_R)+\pi -2i_c$. This sweep need not equal the principal separation if it exceeds $\pi$; the diagram uses a path with sweep below $\pi$.

Let a slowly varying wave have phase $S(\mathbf{x},t)$, wavevector $\mathbf{k}=\nabla S$, and angular frequency $\omega =-{\partial }_{t}S$. The local dispersion relation is the Hamilton-Jacobi equation $\omega =\Omega (\mathbf{k};\mathbf{x},t)$. Equality of mixed derivatives, followed along a curve with velocity $\dot{\mathbf{x}}={\nabla }_{\mathbf{k}}\Omega$, gives the Hamiltonian ray-tracing equationsHere $d/dt={\partial }_t+\dot{\mathbf{x}}\cdot \nabla$ is the total derivative evaluated along a ray, and $\dot{\mathbf{x}}$ is the group velocity.

For $\Omega =kc(z)$, $\omega$, $k_x$, and $k_y$ are conserved. If $\psi$ is the angle between the ray and the $z$-axis, then $k\sin \psi =(k_x^2+k_y^2{)}^{1/2}$ and $k=\omega /c$. Hence Snell's law is

For the ray launched with $\mathbf{k}=k(\cos \varphi ,0,\sin \varphi )$ at $z=0$, this constant is $|\cos \varphi |/c_0$. Thereforewhich proves

For $0<\varphi <\pi /2$, write $a=\cos \varphi$. Since the ray and wavevector are parallel,Near the upper turning point, put $z=z_m-\zeta$. Expansion givesIf $x=x_m$ at the turn, integration yields the local ray pathThe ray is tangent to the line $z=z_m$, turns smoothly, crosses $z=0$, and repeats symmetrically between $-z_m$ and $z_m$. Increasing $\varphi$ decreases $z_m$, while $\varphi \downarrow 0$ sends the turning level outward.

At minimum deviation, ${\theta }_2={\theta }_3=\varphi /2$ and ${\theta }_1={\theta }_4=({\gamma }_{\min }+\varphi )/2$. Substituting in Snell's law givesFor a known prism apex angle, measuring the minimum deviation with monochromatic light determines its refractive index. Repeating the measurement at several wavelengths measures optical dispersion. If the surrounding medium is not air, the measured ratio is relative to that medium's refractive index.

Let the two internal angles be $r={\theta }_2$ and $\varphi -r={\theta }_3$, since the optical prism geometry gives ${\theta }_2+{\theta }_3=\varphi$. By Snell's law, the corresponding external angles are $F(r)$ and $F(\varphi -r)$, where $F(r)=\arcsin (n\sin r)$. The deviation isOn the transmitted branch with $n>1$ and $0<r<\arcsin (1/n)$,Thus $F^{\prime }$ is strictly increasing. The stationary condition ${\gamma }^{\prime }(r)=F^{\prime }(r)-F^{\prime }(\varphi -r)=0$ forces $r=\varphi -r$, and the positive second derivative makes this the unique minimum deviation. ConsequentlyThe argument assumes a transmitted ray is possible; the symmetric internal angles must lie below the critical angle.

Putand let $q_1=nk\cos {\theta }_T$ be the downward vertical wavenumber in the lower half-space. The given form of Snell's law saysWith the $e^{-i\omega t}$ convention, write the incident, reflected, and transmitted plane waves asThe two interface conditions at $z=0$ giveSolving this linear system gives the reflection and transmission coefficients at a scalar-wave interfaceThus the exact fields are

The acute angles follow from part (ii) and the Snell's law relation:The path lengths below and above the interface are respectivelyHence the total distance and minimum travel time areFor a sketch to scale, choose the interface as $y=0$, with $A=(0,-\sqrt{3})$ and $B=(2,1)$. The crossing is $C=(1,0)$ because $d_1\tan \theta =d_2\tan \varphi =1$. The figure uses equal axis scales and shows the two normal-relative angles.

The angles in the PDF are measured from the vertical normal to the interface. The horizontal displacements of the two path segments are $d_1\tan \theta$ and $d_2\tan \varphi$, while their lengths are $d_1\sec \theta$ and $d_2\sec \varphi$. Thus the geometric constraint and travel time areThe Lagrange multiplier equations $\nabla T=\lambda \nabla g$ giveCancelling the positive geometric factors gives $\lambda =2\sin \theta /c=\sqrt{2}\sin \varphi /c$, henceThis is Snell's law, $\sin \theta /(c/2)=\sin \varphi /(c/\sqrt{2})$. The relation depends only on the ratio of the two speeds, so changing $D,d_1,d_2$ leaves the relation unchanged; it changes which angles satisfy it together with the geometric constraint.

For a minimum check, let $u$ be the horizontal distance to the crossing point. ThenandThus the travel time is a strictly convex function of $u$, and its stationary point is the unique global minimum. For positive $D,d_1,d_2$, $T^{\prime }(0)<0<T^{\prime }(D)$, placing the crossing between the two endpoint projections.

The integrand $f=n(x)\sqrt{1+(y^{\prime }{)}^2}$ has no explicit dependence on $y$, so part (i) givesFor positive refractive index and a regular graph with $n(x)>|c|$, squaring and then restoring the sign of $y^{\prime }$ yieldsIt is the continuous-medium version of Snell's law, expressing conserved horizontal optical momentum. The Euler-Lagrange equation establishes stationarity; on a monotone-height segment, convexity of the integrand in $y^{\prime }$ also gives a minimum among graphs with fixed endpoints.

Place the interface at $y=0$, with source $(x_A,h_1)$ and observer $(x_B,-h_2)$ in the two uniform media. Let $(s,0)$ be the crossing point. The optical length isDifferentiating with respect to $s$ and setting the result to zero givesRecognizing the ratios as the sines of the angles measured from the interface normal proves Snell's law:The stationarity condition is the continuity of the tangential optical momentum across the planar interface.

At an air–glass interface, resolve the incident unpolarized light into components perpendicular ($s$) and parallel ($p$) to the plane of incidence. The Fresnel equations give the parallel reflection coefficientAt Brewster's angle, $r_p=0$. Together with Snell's law, this gives ${\theta }_i+{\theta }_t=\pi /2$ and $\tan {\theta }_i=n_2/n_1$. Thus, taking the incident medium to be air,The remaining reflected component has linear polarization perpendicular to the plane of incidence. This assumes a transparent, nonmagnetic glass with negligible absorption.

The wavelengths specified for incidence from vacuum are ${\lambda }_0$. With fixed incident angle ${\theta }_i$, Snell's law gives $n=\sin {\theta }_i/\sin {\theta }_t$ and $v_p=c\sin {\theta }_t/\sin {\theta }_i$. At the central wavelength,Use a central difference, converting the angle change to radians:The group index from refracted-angle dispersion isDirectly inserting the incident vacuum wavelengths into $v_p-\lambda d{v}_p/d\lambda$ would instead give $1.120\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$. That is a plausible intended rough answer, but it mixes wavelength coordinates; the boxed result consistently uses the incident data. The two agree only to first order in a sufficiently small dimensionless dispersion correction.

Let the first ray reflect at the upper surface and the second traverse the film, reflect below, and emerge at a laterally displaced point. Its internal round-trip geometric length is $2d/\cos \theta$, but the external path to a common outgoing wavefront must also be included. The lateral separation is $2d\tan \theta$, whose projection along the external ray is $2d\tan \theta \sin {\theta }_i$. The net optical path difference is thereforeusing Snell's law $\sin {\theta }_i=n\sin \theta$. Multiplying by the vacuum wavenumber givesFor the usual index ordering $1<n<n_2$, including the stated later value $n=1.2$, both reflections are towards higher refractive index and both reverse the electric-field sign. Their relative reflection phase is therefore zero; $\delta$ is also their total relative phase modulo $2\pi$. A different index ordering or polarization-dependent reflection phase would require that additional term.

Eliminate the internal angle using Snell's law:Neighbouring bright fringes differ in phase by $2\pi$. The local angular fringe spacing of a transparent film therefore givesWith $\Delta {\theta }_i=0.4\pi /180=6.981\times 10^{-3}\mathrm{rad}$, ${\theta }_i\simeq 10^{\circ }$, $n=1.2$ and $\lambda =500\mathrm{nm}$,The data specify a local spacing near $10^{\circ }$, so differentiating is appropriate; choosing particular neighbouring fringe angles slightly shifts this approximate result. The phase decreases as incidence angle increases.

In thin-film interference, coherent beams reflected at the two faces of a transparent film overlap. Let the film have refractive index $n_f$, thickness $t$, and internal ray angle ${\theta }_f$, determined by Snell's law. For vacuum wavelength ${\lambda }_0$, the propagation part of the relative phase isAlthough the internal round-trip ray length is $2t/\cos {\theta }_f$, the two emergent rays must be compared on the same outgoing wavefront. Subtracting their external lateral path contribution gives the effective optical path difference $2n_{f}t\cos {\theta }_f$, explaining the cosine rather than its reciprocal.

Reflection also changes phase. At normal incidence in lossless positive-index media, the electric-field Fresnel equations give $r_{ij}=(n_i-n_j)/(n_i+n_j)$: reflection towards higher index reverses sign, whereas reflection towards lower index does not. If the two interfering reflections differ by one such phase reversal, their total relative phase is ${\delta }_{\mathrm{prop}}+\pi$ modulo $2\pi$. Then reflected maxima and minima obeyIf there are zero or two reversals, the conditions interchange. These bright/dark conditions assume overlapping beams with adequate temporal optical coherence; complete cancellation further requires equal amplitudes. In transmitted light, a lossless complete film has complementary intensity variations because reflected plus transmitted power is conserved.

For multiple internal reflections at normal incidence, summing the geometric series gives the exact film reflection amplitudeThis includes both the amplitude mismatch and repeated-beam interference. A normal-incidence antireflection coating on substrate index $n_s>n_0$ can cancel reflection at one wavelength when $n_f=\sqrt{n_0{n}_s}$ and $t={\lambda }_0/(4n_f)$: both interface reflections reverse sign and the film's round-trip propagation adds $\pi$, with equal interface amplitudes. A varying thickness or angle selects different maxima for different wavelengths, producing the colours of transparent films in white light. Broad bandwidth and large incidence angles generally prevent perfect cancellation across all wavelengths.

A wave propagates an organized disturbance through space. What oscillates depends on the medium: displacement in a string or solid, pressure and density in a fluid, electric fields and magnetic fields in an electromagnetic wave, or a quantum wavefunction. The propagation law follows from the restoring mechanism, inertia and relevant boundary conditions. A common linear equation is $u_{tt}=c^2{\nabla }^{2}u$, whose plane waves obey $\omega =c|\mathbf{k}|$. The resulting phase velocity and group velocity coincide, so a narrow wave packet can travel without dispersive spreading. More generally the material supplies a dispersion relation $\omega (\mathbf{k})$; the envelope velocity is ${\nabla }_{\mathbf{k}}\omega$, while constant phase surfaces move at the corresponding phase velocity.

Mechanical waves make the balance of restoring force and inertia particularly visible. A flexible stretched string has $c=\sqrt{T/\mu }$: greater tension stiffens the string, whereas greater linear mass density slows its transverse motion. Small-amplitude acoustic waves in a uniform fluid are longitudinal, with $c_s=\sqrt{K_s/\rho }$, where $K_s$ is the adiabatic bulk modulus. For an ideal gas, $c_s=\sqrt{\gamma p/\rho }$. Pressure changes in rapid sound oscillations are approximately adiabatic because little heat crosses a wavelength in one period. Fluids do not support ordinary bulk shear waves, whereas an elastic solid supports both longitudinal and transverse waves. In an isotropic solid their speeds are $c_L=\sqrt{(K+4G/3)/\rho }$ and $c_T=\sqrt{G/\rho }$, where $G$ is the shear modulus.

A material can be dispersive even when its equilibrium properties are uniform. In a liquid with a free surface, gravity and surface tension act as restoring mechanisms. A simple inviscid capillary-gravity wave haswhere $h$ is depth and ${\gamma }_s$ is surface tension. Long shallow gravity waves have approximately constant speed $\sqrt{gh}$; deep-water gravity waves have $\omega =\sqrt{gk}$ and group velocity half their phase velocity. Short capillary waves have a different dispersion dominated by surface tension. Bending stiffness similarly changes short-wavelength propagation in a string or plate, giving a contribution proportional to $k^4$ in ${\omega }^2$. Different restoring mechanisms can therefore control different wavelength ranges of the same medium.

Electromagnetic waves propagate in vacuum without material displacement, because Maxwell equations couple changing electric and magnetic fields. They are transverse and travel at $c=1/\sqrt{{ϵ}_0{\mu }_0}$. In a simple transparent isotropic material, the phase speed is $1/\sqrt{ϵ\mu }=c/n$, where $n$ is the refractive index. Real material polarization depends on frequency. A Lorentz oscillator model for bound charges explains why permittivity and refractive index become strongly frequency dependent near material resonances, accompanied by absorption. In a birefringent medium, different field polarizations have different propagation constants, permitting wave plates and mode-dependent delays. A electrical conductor additionally supports dissipative currents; its complex wavenumber gives attenuation and, for a good conductor, a skin depth approximately $\sqrt{2/(\mu \sigma \omega )}$. Energy removed from the wave becomes heat rather than disappearing.

At an interface, frequency is conserved for stationary media, and tangential wave phase must match. These conditions give Snell's law for refraction and constrain reflected and transmitted directions. Amplitudes follow from continuity of the appropriate fields or mechanical forces. On strings the relevant mechanical impedance is $\sqrt{T\mu }$; for sound it is related to $\rho c_s$, and for electromagnetic waves to $\sqrt{\mu /ϵ}$. A mismatch gives wave reflection, and a lossless boundary conserves the total incident, reflected and transmitted power. Multiple reflections can reinforce or cancel, as in a Fabry–Pérot etalon. Total internal reflection and other boundary conditions confine waves in waveguides. Transverse modes then become discrete, and even an otherwise nondispersive material develops longitudinal cutoffs and wave dispersion. A membrane strip, for example, obeys ${\omega }_n^2=c_m^2[k^2+(n\pi /a{)}^2]$.

Propagation also involves transport and loss. The Poynting vector gives electromagnetic energy flux; mechanical energy transport is force or pressure times velocity, averaged over a cycle. Wave absorption, viscosity, electrical resistance and imperfect boundaries dissipate energy. Scattering redistributes energy into other directions and modes, while geometric spreading reduces local intensity without necessarily dissipating anything. These mechanisms should be distinguished when interpreting a decreasing measured amplitude. In strongly absorbing or rapidly varying dispersive regimes, the derivative called group velocity is not automatically the speed of a sharp signal front; pulse reshaping and the medium's causal response must be considered.

Finally, quantum matter waves illustrate how the same propagation mathematics can describe a different physical quantity. For a free nonrelativistic particle, $\omega =\hslash k^2/(2m)$, so the group velocity $\hslash k/m$ equals its particle velocity and differs from the phase speed $\hslash k/(2m)$. In a periodic crystal, Bloch's theorem replaces the free dispersion by energy bands. Their slope controls wave-packet velocity, $v_g={\hslash }^{-1}dE/dk$, and their curvature gives band effective mass. Thus mechanical confinement, material response and microscopic periodicity all reshape propagation through their allowed modes and dispersion, while the physical interpretation of the wave and its transported quantities remains essential.

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.

An incident electromagnetic wave generally produces reflected and transmitted waves because the two media have different electromagnetic response. Consider a flat interface between homogeneous, linear, lossless media with permittivities ${ϵ}_j$ and magnetic permeabilities ${\mu }_j$. Their wave speeds and wave impedances areFor each plane electromagnetic wave, $\mathbf{H}=\hat{\mathbf{k}}\times \mathbf{E}/Z_j$. With no free surface charge or current, the interface conditions for electromagnetic fields require continuous tangential $\mathbf{E},\mathbf{H}$ and normal $\mathbf{D},\mathbf{B}$. Equality of the phase at every point on the boundary gives equal angular frequencies and tangential wavevectors. Hence the reflection angle equals the incidence angle and Snell's law iswhere $n_j=c/v_j$ and each angle is measured from the normal.

At normal incidence use the same tangential axis for all electric amplitudes. The two independent boundary equations areSolving gives the amplitude Fresnel equationsFor nonmagnetic media, these become $r=(n_1-n_2)/(n_1+n_2)$ and $t=2n_1/(n_1+n_2)$. Reflection into a medium of larger refractive index therefore reverses the electric amplitude's sign at normal incidence.

The normal Poynting vector determines the reflected and transmitted fractions of incident power. At normal incidence,It is the power fractions, rather than the squared electric amplitudes alone, that add to unity.

At oblique incidence separate transverse electric polarization ($s$, electric field perpendicular to the incidence plane) from transverse magnetic polarization ($p$, electric field in that plane). For nonmagnetic media, writing $c_i=\cos {\theta }_i$ and $c_t=\cos {\theta }_t$, the Fresnel equations areHere choose a common $s$ direction and, for each ray, the $p$ direction as $\hat{\mathbf{s}}\times \hat{\mathbf{k}}$. This makes the reflected $p$ basis reverse its tangential direction at normal incidence, explaining why the displayed $r_p$ then has the opposite sign to $r_s$. Physical reflectance is independent of this sign convention. For either polarization with propagating transmission,For $p$ polarization, zero reflection occurs at the Brewster angle $\tan {\theta }_B=n_2/n_1$. If $n_1>n_2$, incidence above the critical angle for total internal reflection $\sin {\theta }_c=n_2/n_1$ produces an evanescent field in total internal reflection and $\mathcal{R}=1$ for an isolated lossless interface. A lossy second medium requires complex response coefficients and an absorption term in the energy balance.

The Huygens-Fresnel principle constructs a new wavefront as the envelope of secondary wavelets emitted by points on an earlier wavefront. Let the phase velocities in the two media be $v_1=c/n_1$ and $v_2=c/n_2$, with refractive indices evaluated at the light's fixed frequency.

For refraction from secondary wavelets, let point $A$ of the incident wavefront reach the interface first. A second point $B$ reaches the interface at $C$ after time $\Delta t$, so $BC=v_1\Delta t$. During the same interval, the secondary wavelet centered on $A$ advances radius $AD=v_2\Delta t$ into the second medium. The refracted wavefront $CD$ is tangent to this wavelet, hence $AD\perp CD$; also $BC\perp AB$ because a ray is normal to its wavefront in these isotropic media.

Let ${\theta }_1,{\theta }_2$ be the incident and refracted ray angles from the interface normal. They also equal the corresponding acute angles between the wavefronts and the interface. The two right triangles have common hypotenuse $AC$, givingTheir ratio eliminates both $AC$ and $\Delta t$:This is Snell's law. In particular, entering a medium of larger refractive index reduces the ray's angle to the normal. The construction assumes a stationary interface and homogeneous isotropic media with a propagating refracted wave.

Use the original diagram's positive distances: $u$ from the object to the vertex, $d$ from the vertex to the image, and $R$ from the vertex toward the centre of curvature on the outgoing side. Let a paraxial light ray strike at height $Y$. Its incoming inclination is $\alpha \simeq Y/u$ above the optical axis, its outgoing inclination is $\beta \simeq Y/d$ below it, and the inward surface normal is inclined by $\theta \simeq Y/R$ below it. Thus the incidence and refraction angles relative to that normal are respectively $\alpha +\theta$ and $\theta -\beta$.

The local tangent-plane form of Snell's law, using the small-angle approximation, givesSubstitute the three height-to-distance ratios and cancel $Y$:This refraction at a spherical surface result keeps only first-order ray angles. The surface sag shifts the height's axial reference only at higher order. Distances for other object, image or curvature orientations must be signed consistently; the displayed positive-distance form follows the particular orientation in the paper.