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Snell's law of refraction can be derived from Fermat's principle that light travels paths that minimize the time using simple calculus. Since Snell's law only involves sines I wonder whether this minimum problem has a simple geometric solution.

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This is treated in "What Is Mathematics" by Courant and Robbins. See the bottom half of this page. –  process91 Jun 4 '12 at 15:14
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@Michael really? I don't have a copy at hand right now, but from the title of the section I would be more inclined to think that the geometric construction relates to Heron's principle, which is the principle of reflection that incidence angle is the same as the reflection angle, and which indeed has a simple geometric proof. I would be very pleasantly surprised if there is in Courant and Robbins a geometric proof of Snell's law of refraction. –  Willie Wong Jun 4 '12 at 15:25
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Huygen's gave a somewhat geometric proof of Snell's law, however, he did not start with Fermat's principle, but rather the assumption that light is a wave, that wave speed equals the product of wave length and frequency, that frequency is invariant across a boundary, and a continuity criterion. I suppose, however, that is not what you are looking for. –  Willie Wong Jun 4 '12 at 15:30
    
@WillieWong Ah you're right - I read the OP's question too quickly. They mention Snell's law at the end of that section, but don't derive it geometrically. –  process91 Jun 4 '12 at 15:32
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There is a very detailed proof, using Ptolemy's Theorem, in Ivan Niven's bwautiful book Maxima and Minima Without Calculus. –  André Nicolas Jun 4 '12 at 15:38

1 Answer 1

Let upper half plane be a medium with refractive index $n_1$and the lower half plane be another medium with refractive index $n_2$. Let $(a_1,b_1)$and $(a_2,b_2)$ be the starting point and end point of light traveled, while $(x,0)$ is the point of refraction.

Time of light traveled $$t=\frac {n_1}c\sqrt{(x-a_1)^2+b_1^2}+\frac {n_2}c\sqrt{(a_2-x)^2+b_2^2}$$ where $c$ is the speed of light

By Fermat's principle, we need to minimize $t$ $$\frac{dt}{dx}=0$$$$\frac {n_1}c \frac x{\sqrt{(x-a_1)^2+b_1^2}}+\frac {n_2}c \frac {-x}{\sqrt{(a_2-x)^2+b_2^2}}=0$$$$n_1\frac x{\sqrt{(x-a_1)^2+b_1^2}}=n_2\frac x{\sqrt{(a_2-x)^2+b_2^2}}$$ note that $\frac x{\sqrt{(x-a_1)^2+b_1^2}}=\sin\theta_1$ and $\frac x{\sqrt{(a_2-x)^2+b_2^2}}=\sin\theta_2$, proved.

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