Suppose light ray passing through a medium with refractive index $n=n(y)$. In the case of an inhomogeneous medium in which $n$ varies continuously in the $y$-direction, We have curved rays that satisfy Snell's law in the form:

$$n\cos\psi=\mathrm{constant}$$ where the angle $\psi$ is the slope of the tangent to the path.

Fermat's principle states that

The actual path taken by a light ray between two fixed points makes the travel time of the ray stationary.

So that $$T[\mathcal{P}]=c^{-1}\int_{\mathcal{P}}nds$$ which reduce to (in present case) $$T[y]=c^{-1}\int_{x_0}^{x_1}dx \ n(1+\dot{y}^2)^{1/2}$$ with the help of Euler-Lagrange's Equation $$\frac{n}{(1+\dot{y}^2)^{1/2}}=\mathrm{constant}$$

And on writing $\dot{y}=\tan\psi$, this gives snell's law.

Question: If I put $\dot{y}=0 \Rightarrow y=$ constant that is not extremals and therefore not rays. But since such a ray would experience a constant value of $n$, How does the ray know that it must bend?

  • 1
    $\begingroup$ Please correct your equation -- you are missing a $dx$ -- and define your notation, $\dot y = dy/dx$. I know it seams obvious to you, but for most physicists a dot means time derivative. $\endgroup$
    – Semoi
    Commented Dec 15, 2020 at 12:49

1 Answer 1


Fermats principle becomes clear, if you start from Huygens description of light. Huygen assumes that

light is a wave, which propagates in such a way that each disturbance generates a 
secondary spherical wave. These secondary waves interfere with each another

Using this picture we don't have a directed light beam. Instead we have a superposition of all possible paths. Thus, your original question "how does the light know which path is shortest" disappears, because now the light does not have to choose a path a priori: It just takes all the possible path and the superposition principle makes sure that correct path is dominant/enhanced.

  • $\begingroup$ I'm not concerned with the Huygens description, Can you explain it with the principle that I have given. $\endgroup$ Commented Dec 15, 2020 at 17:02
  • $\begingroup$ The answer I gave is the best I can do. If this does not help you I am sorry. Maybe somebody else is able to. $\endgroup$
    – Semoi
    Commented Dec 15, 2020 at 20:45
  • $\begingroup$ Thank you for your answer though. $\endgroup$ Commented Dec 15, 2020 at 20:46

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