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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
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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[t = \int\limits_{A}^{B}dt=c^{1}\int\limits_{A}^{B}n ds.\] 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: \[n_{1}\sin \theta_{1}=n_{2}\sin \theta _{2.}\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1441126765935:dw \[t = \frac{D}{v} = \frac{D_1}{\frac{c}{n_1}} + \frac{D_2}{\frac{c}{n_2}}\] \[= \frac{n_1}{c}\sqrt{x^2 + y_1^2} + \frac{n_2}{c}\sqrt{(x_2x)^2 + y_2^2}\] \[\frac{dt}{dx} = \frac{n_1}{c} \frac{x}{\sqrt{x^2+ y_1^2}}  \frac{n_2}{c} \frac{x_2x}{\sqrt{(x_2x)^2 + y_2^2}} = 0\] \[n_1 \frac{x}{\sqrt{x^2 + y_1^2}} = n_2 \frac{x_2  x}{\sqrt{(x_2x)^2 + y_2^2}} \] look a drawing \[sin \theta = \frac{x}{\sqrt{x^2 + y_1^2}} \] \[sin \phi = \frac{x}{\sqrt{(x_2x)^2 + y_1^2}} \]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1441127568535:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1NB TYPO \[sin \phi = \frac{x_x  x}{\sqrt{(x_2x)^2 + y_2^2}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That helped so much thank you!
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