I am interested in equations of the form $|\nabla d|= F(x)$, where $F(x)$ is piecewise constant and $d(x) = 0$ on $\Gamma_D$, a subset of the boundary. In particular, like in the figure, one can consider $F(x)$ taking two values, delimited by an interface $\Sigma$ (which can be considered smooth).

Snell's law gives a relation between the incidence angles of a ray passing through the interface: $\sin \theta_+/\sin \theta_- = V_+/V_-$. I cannot find any clear references proving that Snell's law holds for the eikonal equation described above. Therefore I arrive at my question:

  1. Does Snell's law hold across the interface $\Sigma$?
  1. If the answer to the first question is affirmative, can you indicate references where a proof can be found?

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1 Answer 1


Yes, Snell's law holds for any smooth interface, it follows directly from the eikonal equation, see for example section 2.4 of González-Acuña and Chaparro-Romo - Stigmatic Optics. I summarize the derivation:

We substitute into the eikonal equation the definition of a ray, being a unit vector ${\bf v}$ in the direction of $\nabla d$. Then the eikonal equation $|\nabla d|=F$ can be rewritten in vectorial form as $\nabla d=F{\bf v}$, and hence for any closed contour $C$ one has $$\oint_C F\,{\bf v}\cdot {\rm d}{\bf l}=\oint_C\nabla d\cdot {\rm d}{\bf l}=0.$$ If you take a contour as in the figure (with tangential unit vector $\hat{\bf t}$ and infinitesimal perpendicular thickness), you arrive at $$\hat{\bf t}\cdot(V^+{\bf v}_+-V^-{\bf v}_-)=0,$$ which implies Snell's law.

  • 1
    $\begingroup$ Thank you for the reference and for the intuitive explanation! $\endgroup$ Sep 30 at 21:28
  • $\begingroup$ The gradient $\nabla d$ might not be continuous across the interface where the speed changes. Is the contour integral of $\nabla d$ still equal to zero in this case? $\endgroup$ Sep 30 at 21:38
  • $\begingroup$ the component of $\nabla d$ parallel to the interface must be continuous, because $F$ is only discontinuous perpendicularly to the interface, so yes, the contour integral of $\nabla d\cdot{\rm d}{\bf l}$ vanishes. $\endgroup$ Sep 30 at 22:07
  • $\begingroup$ "The component of $\nabla d$ parallel to the interface must be continuous" is equivalent to Snell's law. To bring this argument in justifying the contour integral is zero leads to a circular argument. $\endgroup$ Oct 1 at 20:49
  • $\begingroup$ circular argument? The continuity statement follows from the eikonal equation, that's all I need. I don't see any logical flaw. $\endgroup$ Oct 1 at 21:14

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