## anonymous one year ago 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

1. anonymous

$t = \int\limits_{A}^{B}dt=c^{-1}\int\limits_{A}^{B}n ds.$ 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: $n_{1}\sin \theta_{1}=n_{2}\sin \theta _{2.}$

2. IrishBoy123

|dw:1441126765935:dw| $t = \frac{D}{v} = \frac{D_1}{\frac{c}{n_1}} + \frac{D_2}{\frac{c}{n_2}}$ $= \frac{n_1}{c}\sqrt{x^2 + y_1^2} + \frac{n_2}{c}\sqrt{(x_2-x)^2 + y_2^2}$ $\frac{dt}{dx} = \frac{n_1}{c} \frac{x}{\sqrt{x^2+ y_1^2}} - \frac{n_2}{c} \frac{x_2-x}{\sqrt{(x_2-x)^2 + y_2^2}} = 0$ $n_1 \frac{x}{\sqrt{x^2 + y_1^2}} = n_2 \frac{x_2 - x}{\sqrt{(x_2-x)^2 + y_2^2}}$ look a drawing $sin \theta = \frac{x}{\sqrt{x^2 + y_1^2}}$ $sin \phi = \frac{x}{\sqrt{(x_2-x)^2 + y_1^2}}$

3. IrishBoy123

|dw:1441127568535:dw|

4. IrishBoy123

NB TYPO $sin \phi = \frac{x_x - x}{\sqrt{(x_2-x)^2 + y_2^2}}$

5. anonymous

That helped so much thank you!