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If the integrand $L(q_1,\dots ,q_m,q_1^{\prime },\dots ,q_m^{\prime })$ has no explicit dependence on its independent variable $z$, its Euler-Lagrange equations imply the displayed first integral. Indeed, differentiate the left side and cancel the second-derivative terms to obtain ${\sum }_j{q}_j^{\prime }({\partial }_{q_j}L-d({\partial }_{q_j^{\prime }}L)/dz)=0$. This is the several-variable form of the Beltrami identity. Applied to Fermat principle, it expresses conservation of the component of the optical momentum of a ray along a translation-invariant direction.

For height $x$ and horizontal position $y$, Fermat principle in a positive index profile $n(x)=1+\beta x$, $\beta >0$, conserves $n(x)y^{\prime }/\sqrt{1+(y^{\prime }{)}^2}$ on a graph $y(x)$. For positive magnitude $c$ of the horizontal optical momentum, integration gives the displayed catenary. Its minimum height is $x_0=(c-1)/\beta$, at $y=y_0$. Thus $c>1$ produces a smooth turning ray wholly above a road at $x=0$. The two graph branches have opposite signs of $dy/dx$; $x(y)$ continues smoothly through their common horizontal tangent. For a signed negative graph constant, use $|c|$ in the prefactor; $c=0$ gives a vertical ray separately.

For speed $c(y)=y+c_0>0$, Fermat principle and the Beltrami identity yield circular light rays whose centres lie on $y=-c_0$. The ray between $(\pm a,0)$ in the upper half-plane is $y=\sqrt{a^2+c_0^2-x^2}-c_0$.

For a positive refractive index depending only on $x$, Fermat principle applied to $\int \mu (x)\sqrt{1+y^{\prime 2}}dx$ has $y$ as a cyclic coordinate. The Euler-Lagrange equation therefore makes the displayed quantity constant along a stationary ray. When ${\mu }^2>p^2$, its slope is $y^{\prime }=p/\sqrt{{\mu }^2-p^2}$. If $\theta$ is the angle to the $x$ axis, this is conservation of $\mu \sin \theta$, the stratified-medium form of Snell's law. For fixed endpoint values, the integrand has second derivative $\mu (1+y^{\prime 2}{)}^{-3/2}>0$ with respect to $y^{\prime }$. Its strict convexity and the constant momentum show, by the tangent-line inequality and integration, that any such ray is the unique minimum among competing graph paths with the same endpoints.

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].

For a positive refractive index $\mu (r)$ in cylindrical coordinates, a light ray with constant radius $R>0$ and varying axial coordinate has the form $\theta (z)={\theta }_0+\omega z$ when it exists. Fermat principle makes $L=\mu (r)\sqrt{1+r^{\prime 2}+r^2{\theta }^{\prime 2}}$ stationary. Conservation of the momentum conjugate to the cyclic angle makes ${\theta }^{\prime }$ constant along a constant-radius path. The radial Euler-Lagrange equation is thengiving the displayed formula. A nonzero real helix requires $0<-R{\mu }^{\prime }(R)/\mu (R)<1$. If ${\mu }^{\prime }(R)=0$, $\omega =0$ gives a straight axial light ray; a generic radial profile need not support a constant-radius helix. Its helical pitch is $2\pi /|\omega |$. This ray model is useful for a graded-index optical fiber when the refractive index varies smoothly and geometrical optics applies; it does not by itself impose wave-mode quantization or boundary confinement.

A path of light propagation in the geometrical optics approximation. In a smoothly varying propagation speed, Fermat principle makes its travel time stationary.

For a light ray parametrized by arc length $s$, with unit tangent vector $t$ in a smooth positive refractive index $\mu$, the optical momentum is $p=\mu t$. This is the momentum for optical path-length normalization. The Euler-Lagrange equation for Fermat principle, $\int \mu (r)|r^{\prime }(\tau )|d\tau$, gives $d(\mu t)/ds=\nabla \mu$. Therefore an index independent of $z$ conserves $p_z=\mu dz/ds$. In coordinates $r(z),\theta (z)$ this is $\mu /\sqrt{1+r^{\prime 2}+r^2{\theta }^{\prime 2}}$, the first integral supplied by the Beltrami identity for several dependent variables.

For the refractive index $\mu =1/y$ on the upper half-plane, Fermat principle and the Beltrami identity give $1/[y\sqrt{1+(dy/dx{)}^2}]=1/R$. The nonvertical rays are arcs of circles centered on $y=0$; vertical lines are the remaining rays. These are also geodesics of the Poincaré half-plane model. A ray entering the strip $0<y\le 1$ toward decreasing $y$ at angle $\theta$ to the vertical has limiting horizontal displacement $\tan (\theta /2)$, at most one. The line $y=0$ is approached only after infinite optical path length because the refractive index diverges there.

For positive differentiable speed $c(y)$ and a smooth ray written as a graph $y(x)$, the Fermat principle minimizes travel timeA variation vanishing at the endpoints gives the Euler-Lagrange equation for $F=\sqrt{1+y^{\prime 2}}/c$. Its derivatives are $F_y=-c^{\prime }\sqrt{1+y^{\prime 2}}/c^2$ and $F_{y^{\prime }}=y^{\prime }/(c\sqrt{1+y^{\prime 2}})$. Thusand multiplying by $c^2(1+y^{\prime 2}{)}^{3/2}$ givesThe displayed differential equation requires differentiability, beyond mere continuity of $c$. For constant positive speed it reduces to $y^{\prime \prime }=0$, so rays are straight lines; a straight segment minimizes travel time because it has the shortest Euclidean length.

For $c(y)=y$ in the positive-height region, the Beltrami identity gives $F-y^{\prime }{F}_{y^{\prime }}=1/[y\sqrt{1+y^{\prime 2}}]=1/R$. Hence $y^2(1+y^{\prime 2})=R^2$, whose nonvertical solutions are $(x-h{)}^2+y^2=R^2$. Formally imposing the given boundary points fixes $h=1/2$, $R=1/2$, and gives the ray in a linear-speed mediumIt is the upper semicircle of diameter one, with vertical limiting tangents at its endpoints.

There is a necessary endpoint qualification. The speed is zero at the two stated endpoints. For any rectifiable path entering positive height, $ds/y\ge |dy|/y$, and its travel time diverges logarithmically on leaving or approaching height zero. Thus the semicircle is the formal interior ray, but there is no finite-time minimizer between the zero-speed endpoints. For endpoints at height $ϵ>0$ the circular ray has radius $\sqrt{1/4+{ϵ}^2}$ and finite travel time $2\mathrm{arsinh}(1/(2ϵ))$, which tends to infinity as $ϵ\downarrow 0$. This explains the intended semicircle as a limiting hyperbolic geodesic, without asserting finite travel time where $c=0$.

For an integrand $I(y,y^{\prime })$ with no explicit $x$ dependence, the Euler-Lagrange equation is $I_y-d(I_{y^{\prime }})/dx=0$. Differentiate the proposed first integral:Thus $I-y^{\prime }{I}_{y^{\prime }}$ is constant along an extremal, the Beltrami identity.

For the optical path the Fermat principle uses the travel-time integrand $I=e^{-\lambda y}\sqrt{1+y^{\prime 2}}$, in the stated speed units. Its first integral isSet $u=k{e}^{\lambda y}$. Then $u^{\prime 2}={\lambda }^2(1-u^2)$, and integration gives $u=\cos [\lambda (x-x_0)]$ on a positive arch. The symmetric endpoint conditions require $x_0=0$ and $k=\cos \lambda a$. For $\lambda >0$ and $0<\lambda a<\pi /2$, the resulting ray in an exponential-speed medium isIt is above zero inside the endpoints. Substitution also verifies the Euler-Lagrange equation $y^{\prime \prime }=-\lambda (1+y^{\prime 2})$, so the first-integral construction does not accidentally select a spurious constant-height path.

Here $y^{\prime }=-\tan \lambda x$. For $\lambda a$ close to $\pi /2$, the endpoint slopes are steep and the central height $-{\lambda }^{-1}\log \cos \lambda a$ is large. The travel time evaluates directly toIt tends to $2/\lambda$ as the arch height diverges. Positive $\lambda$ is the increasing-speed interpretation needed for a ray arching into $y>0$; for $\lambda <0$ the same algebraic expression bends below that region and is not an admissible interior path. At zero $\lambda$ the limiting straight boundary path has travel time $2a$.

For variations $y+ϵ\eta$ with $\eta (x_1)=\eta (x_2)=0$, integration by parts gives $\delta I={\int }_{x_1}^{x_2}[f_y-(d/dx)f_{y^{\prime }}]\eta dx$. Since this vanishes for every such variation, the Euler-Lagrange equation isWhen $f$ is independent of $y$, it has the first integral $\boxed{\partial f/\partial y^{\prime }=C}$.

For a ray represented as a graph traversed with increasing $x$, the travel time is $T={\int }_{x_1}^{x_2}\sqrt{1+y^{\prime 2}}/c(x)dx$, with $c(x)>0$. By Fermat principle, use $f=\sqrt{1+y^{\prime 2}}/c(x)$. The first integral becomesThe sign of $y^{\prime }$ is constant, and $y_2>y_1$ makes it positive. Write $C=1/k$ with $k>0$ and solve to obtainA finite slope requires $k>c(x)$ in the interior; a limiting vertical tangent is possible at an endpoint.

For $c(x)=ax$ and $k=a{x}_2$, positive speed on the whole interval and $k\ge c(x)$ select $a>0$ and $0<x_1<x_2$. The slope is $x/\sqrt{x_2^2-x^2}$. Integrating backwards from $(x_2,y_2)$ givesso $\boxed{(y_2-y{)}^2=x_2^2-x^2}$, with the branch $y\le y_2$. The endpoint integral is finite despite the vertical tangent. The specified value $k=a{x}_2$ fits the first endpoint precisely when $y_2-y_1=\sqrt{x_2^2-x_1^2}$; for other endpoint data the preceding integral determines a different $k$.

The Euler-Lagrange equation is $f_y-d(f_{y^{\prime }})/dx=0$. Since $f$ has no explicit $x$ dependence,Thus a stationary curve satisfies the Beltrami identity, $f-y^{\prime }{f}_{y^{\prime }}=C$.

By Fermat principle, the light ray makes the travel time $\int \sqrt{1+y^{\prime 2}}/(y+c_0)dx$ stationary. Its Beltrami identity becomesWriting $R=1/C$ and integrating the resulting first-order equation gives $(x-x_0{)}^2+(y+c_0{)}^2=R^2$. Both endpoints have height zero, so subtracting their equations gives $x_0=0$, and then $R^2=a^2+c_0^2$. The arc lying in $y>0$ isIts maximum height is $\sqrt{a^2+c_0^2}-c_0$. The ray rises into the faster region before returning to the boundary; its circle centre is $(0,-c_0)$. The prescribed boundary endpoints are understood by continuity from the open half-plane. This is the circular light ray in a linear speed profile.

First integral. If $F$ has no explicit $x$ dependence, the chain rule givesThe Euler-Lagrange equation makes this zero without requiring division by $y^{\prime }$. Hence the Beltrami identity is

Optical paths. By Fermat principle, a ray makes the optical path length stationary. Within the strip $0<y\le 1$, its functional for a graph $y(x)$ isThe Beltrami identity givesFor a nonvertical segment $k>0$, write $R=1/k$. Then $(y^{\prime }{)}^2=(R^2-y^2)/y^2$, soThus the optical rays in a reciprocal-height refractive index satisfyrestricted to the part in the strip. They are arcs of circles centered on $y=0$. Vertical lines are the additional solutions, obtained as a limiting case or directly from the constant horizontal optical momentum. The same curves are geodesics of the Poincaré half-plane model, whose length element is $ds/y$.

Horizontal displacement. Let the ray enter at $(x_0,1)$, travelling to the right and downward at angle $0<\theta <\pi /2$ to the vertical. The refractive index is continuous at $y=1$, so there is no instantaneous angular jump at entry. The tangent has $(dx/ds,dy/ds)=(\sin \theta ,-\cos \theta )$; hence $y^{\prime }(x_0)=-\cot \theta$. The first integral gives $k=\sin \theta$, or $R=\csc \theta$. The circle derivative $y^{\prime }=-(x-c)/y$ then fixes $c=x_0-\cot \theta$.

As $y\downarrow 0$, the rightward ray approaches $x=c+R$. The total limiting horizontal distance is thereforeFor vertical incidence, $D=0$ by continuity. For a leftward ray the displacement has the opposite sign, with the same magnitude. Since $\tan (\theta /2)$ increases on $[0,\pi /2]$,If entry is required to have a strictly downward component, this is a supremum rather than an attained maximum. A horizontal tangent at entry gives the limiting unit-radius arc, which immediately bends downward in the medium.

The lower boundary is a limit, rather than a point reached in finite optical travel time. On a nonvertical circular ray, $ds/dy=R/\sqrt{R^2-y^2}\to 1$ as $y\downarrow 0$. Thus the optical path length contains a divergent integral proportional to $\int dy/y$. The horizontal distance stays finite even though propagation toward $y=0$ takes infinite time.

Vary $r$ by $r+\epsilon \eta$ with $\eta (a)=\eta (b)=0$. The first variation of the functional isThe boundary term from integration by parts is zero. The fundamental lemma of the calculus of variations therefore gives $\boxed{F_{1,r}-d{F}_{1,r^{\prime }}/dz=0}$. For $I_2$, independently vary $r$ and $\theta$ with zero endpoint variations; the two Euler-Lagrange equations areFor the optical path, Fermat principle makes $\int \mu ds$ stationary. In cylindrical coordinates its integrand is $L=\mu (r,\theta )s$, with $s=\sqrt{1+r^{\prime 2}+r^2{\theta }^{\prime 2}}$. There is no explicit $z$ dependence. The Beltrami identity for several dependent variables givesThe conserved quantity is the axial component of the optical momentum of a ray. As $\mu >0$, squaring and inverting yieldsNow suppose $\mu =\mu (r)$ and seek a path of radius $R>0$. The angle is a cyclic coordinate, so $L_{{\theta }^{\prime }}=\mu R^2{\theta }^{\prime }/\sqrt{1+R^2{\theta }^{\prime 2}}$ is conserved. At fixed radius this is strictly increasing in ${\theta }^{\prime }$, forcing ${\theta }^{\prime }=\omega$ constant. The radial Euler-Lagrange equation reduces toConsequently the helical rays in a radially graded refractive index haveA real nonzero helix exists at such a radius when $0<-R{\mu }^{\prime }(R)/\mu (R)<1$, and its helical pitch is $2\pi /|\omega |$. If ${\mu }^{\prime }(R)=0$, the solution is instead a straight axial light ray. Thus radial dependence permits constant-radius rays for suitable profiles; an arbitrary radial profile need not support a nonzero real helix. This qualification is necessary, for example when the refractive index increases with radius.

For fixed endpoint values, the Euler-Lagrange equation is $f_y-d{f}_{y^{\prime }}/dx=0$. Along a stationary curve, the chain rule givesWhen $f_x=0$, the Beltrami identity therefore takes the requested sign convention

By Fermat principle, the elapsed-time integrand for the given speed is $f=c_0^{-1}\sqrt{1-ky}\sqrt{1+(y^{\prime }{)}^2}$. ThusThe constant is negative for a finite regular ray. Squaring gives $(1-ky)/(1+(y^{\prime }{)}^2)=c_0^2{A}^2$. At a maximum height $y_0$, the slope is zero, so $c_0^2{A}^2=1-k{y}_0>0$. HenceSeparate variables on either side of the turning point. Integration gives the parabolic light rays in a reciprocal-square-root speed profileThe quadratic opens downwards, with its greatest height $y_0$ at $x=x_c$. The positive index condition holds throughout the layer because $kh<1$. Vertical rays are a separate zero-horizontal-momentum case, outside the regular graph representation $y(x)$; they are not returning rays of the type considered below.

Equal endpoint heights at $x=\pm x_0$ force $x_c=0$. Setting $y(\pm x_0)=0$ givesFor nonzero separation, a real returning ray requires $k{x}_0\le 1$ and has candidate turning heightsOnly choices satisfying $0<y_0<h$ are rays wholly within the specified layer. Depending on its thickness there may be two, one, or no admissible choices; at zero discriminant the two algebraic candidates coincide. The entry and exit points at $y=0$ are interpreted as limiting boundary points of the open medium.

For endpoint-preserving variations of a functional, the Euler-Lagrange equation isAlong an extremal, the chain rule gives $d(F-y^{\prime }{F}_{y^{\prime }})/dx=F_x+y^{\prime }[F_y-d{F}_{y^{\prime }}/dx]=F_x$. Thus when $F$ has no explicit $x$ dependence, the Beltrami identity isFor a positive speed profile, the elapsed-time integrand in Fermat principle is $F=\sqrt{1+(y^{\prime }{)}^2}/c(y)$. Its first integral givesIntegrating each monotone-height branch yieldsReal branches require $Ac(y)\le 1$. Turning points are joined in the original graph $y(x)$, even though the inverse derivative is singular there.

For the inverse-height speed profile, this becomes $dx/dy=\pm a/\sqrt{y^2-a^2}$, with $a=A>0$. Integration gives $x-b=\pm a\mathrm{arcosh}(y/a)$, so the light rays are catenariesEqual endpoint heights at opposite $x$ values force $b=0$, since the real hyperbolic cosine is even and strictly increases with the absolute value of its argument. The remaining endpoint condition is $a\cosh (1/a)=\cosh 1$. One exact solution is $a=1$, givingHere $ds/dx=\cosh x$ and $1/c(y)=y=\cosh x$ account for the squared hyperbolic cosine in the time integral.

The stated endpoints also allow a second stationary ray. Set $t=1/a$. The function $\cosh t/t$ decreases and then increases: its derivative has the sign of $t\sinh t-\cosh t$, whose derivative is $t\cosh t>0$. Its unique minimum occurs where $t\tanh t=1$, at $t\simeq 1.199679$. Since the known root $t=1$ lies below this minimum's location, there is exactly one further root, $t\simeq 1.4242930655$. Thus the two stationary catenary rays in an inverse-height speed profile also includeNo extra condition in the question selects uniquely between these rays. The minimum height is a minimum along the chosen ray, not a global travel-time assertion. Indeed, competitors descending vertically to height $\eta >0$, crossing horizontally, and rising again have time ${\cosh }^2(1)-{\eta }^2+2\eta$. As $\eta \downarrow 0$ this tends to ${\cosh }^2(1)\simeq 2.381098$, below both stationary times. Their corners can be smoothed, so Fermat principle here must be understood as stationarity rather than unrestricted global minimization.

By Fermat principle, a light ray makes optical travel time stationary. Multiplication by the constant vacuum speed of light does not change the stationary paths, so the functional can be taken to be the refractive index integrated over arc length. The constant denoted $c$ below is an integration constant, not that vacuum speed.

Assume $c(y)>0$ is differentiable along the ray, with sufficient regularity for the following variation, and fix the endpoints. Fermat principle makes the travel time stationary:For the variational integrand $f=\sqrt{1+y^{\prime 2}}/c$, the Euler-Lagrange equation isExpanding the derivative and multiplying by $c^2(1+y^{\prime 2}{)}^{3/2}$ givesand thereforeEquivalently, the Beltrami identity gives $1/[c(y)\sqrt{1+y^{\prime 2}}]=\text{constant}$. Mere continuity of $c$, as stated in the wording, does not by itself justify the appearance of $c^{\prime }$; the displayed classical ray equation requires this additional differentiability.

Fermat principle states that the optical path length $\mathcal{S}=\int nds$, or equivalently travel time $\mathcal{S}/c_0$, is stationary under small variations of the ray with fixed endpoints. Stationarity need not mean a global minimum.

For a plane ray written as $y(x)$, use $\mathcal{S}[y]=\int F(x,y,y^{\prime })dx$, with $F=n(x,y)\sqrt{1+y^{\prime 2}}$. Its stationary paths obey the Euler-Lagrange equationIt follows by taking the first variation, integrating the term in $\delta y^{\prime }$ by parts, and using zero endpoint variations.