Past exam of the mathematical methods II course of the University of Oxford / 2014 / 8 / a / Solution

In each layer a minimum-time path is straight because the speed is constant there. The total time as a function of the two crossing positions is

t (x_{1}, x_{2}) = \frac{1 + x _{1}^{2}}{u} + \frac{1 + ( x _{2} - x _{1} ) ^{2}}{v} + \frac{1 + ( X - x _{2} ) ^{2}}{w} .

Setting its two partial derivatives to zero gives

\frac{x _{1}}{u 1 + x _{1}^{2}} = \frac{x _{2} - x _{1}}{v 1 + ( x _{2} - x _{1} ) ^{2}} = \frac{X - x _{2}}{w 1 + ( X - x _{2} ) ^{2}} .

The angles in the held diagram are measured from the vertical interface normals, so these ratios are

\frac{sin α}{u} = \frac{sin β}{v} = \frac{sin θ}{w} .

This is the layered-speed form of Fermat's principle and Snell's law. Signed horizontal displacements can be retained if the endpoint lies on the other side of the origin.

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