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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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    There is a very detailed proof, using Ptolemy's Theorem, in Ivan Niven's bwautiful book *Maxima and Minima Without Calculus.*2012-06-04

2 Answers 2

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Perhaps this will help, if you are looking at a non Calculus approach.enter image description here Consider two parallel rays $A$ and $B$ coming through the medium $1$ (say air) to the medium $2$ (say water). Upon arrival at the interface $\mathcal{L}$ between the two media (air and water), they continue their parallel course in the directions $U$ and $V$ respectively.

Let us assume that at time $t=0$, light ray $A$ arrives at the interface $\mathcal{L}$ at point $C$, while ray $B$ is still shy of the surface by a distance $PD$. $B$ travels at the speed $v_{1}=\frac{c}{n_{1}}$ and arrives at $D$ in $t$ seconds. During this time interval, ray $A$ continues its journey through the medium $2$ at a speed $v_{2}=\frac{c}{n_{2}}$ and reaches the point $Q$.

We can formulate the rest, geometrically (looking at the parallel lines) from the figure. Let $x$ denote the distance between $C$ and $D$. \begin{eqnarray*} x \sin\left(\theta_{i}\right) &=& PD \\ &=& v_{1} t \\ &=& \frac{c}{n_{1}} t \\ x \sin\left(\theta_{r}\right) &=& CQ \\ &=& v_{2} t \\ &=& \frac{c}{n_{2}} t \end{eqnarray*}

Thus,

\begin{eqnarray*} n_{1} \sin\left(\theta_{i}\right) &=& \frac{c}{x} t \\ n_{2} \sin\left(\theta_{r}\right) &=& \frac{c}{x} t \end{eqnarray*}

Re arranging this will take us to the Snell's law as we know. \begin{eqnarray*} \frac{n_{2} }{n_{1}} &=& \frac{\sin\left(\theta_{i}\right) }{ \sin\left(\theta_{r}\right)} \end{eqnarray*}

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    This is far simpler than the use of Fermat principle. +12017-10-30
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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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    This is not a purely geometric solution. This is the solution that the poster explicitly requested something different from.2016-10-09