Question

The speed of light in a medium of index of refraction n is v=ds/dt=c/n. Then the time of transit from A to B is

Answer 1

\[t = \int\limits_{A}^{B}dt=c^{-1}\int\limits_{A}^{B}n ds.\]

By Fermat's principle, t is stationary. If the path consists of two straight line segments with n constant over each segment, then \[\int\limits_{A}^{B}n ds = n_{1}d_{1}+n_{2}d_{2.}\]
and the problem can be done by ordinary calculus. Thus solve the following problem:

Derive Snell's law of refraction: \[n_{1}\sin \theta_{1}=n_{2}\sin \theta _{2.}\]

Answer 2

\[t = \frac{D}{v} = \frac{D_1}{\frac{c}{n_1}} + \frac{D_2}{\frac{c}{n_2}}\]
\[= \frac{n_1}{c}\sqrt{x^2 + y_1^2} + \frac{n_2}{c}\sqrt{(x_2-x)^2 + y_2^2}\]

\[\frac{dt}{dx} = \frac{n_1}{c} \frac{x}{\sqrt{x^2+ y_1^2}} - \frac{n_2}{c} \frac{x_2-x}{\sqrt{(x_2-x)^2 + y_2^2}} = 0\]

\[n_1 \frac{x}{\sqrt{x^2 + y_1^2}} = n_2 \frac{x_2 - x}{\sqrt{(x_2-x)^2 + y_2^2}} \]

look a drawing

\[sin \theta = \frac{x}{\sqrt{x^2 + y_1^2}} \]

\[sin \phi = \frac{x}{\sqrt{(x_2-x)^2 + y_1^2}} \]