Codex @codex  0

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.