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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Physics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[t = \int\limits_{A}^{B}dt=c^{-1}\int\limits_{A}^{B}n ds.\] 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: \[n_{1}\sin \theta_{1}=n_{2}\sin \theta _{2.}\]
|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 \[sin \theta = \frac{x}{\sqrt{x^2 + y_1^2}} \] \[sin \phi = \frac{x}{\sqrt{(x_2-x)^2 + y_1^2}} \]
|dw:1441127568535:dw|

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

NB TYPO \[sin \phi = \frac{x_x - x}{\sqrt{(x_2-x)^2 + y_2^2}}\]
That helped so much thank you!

Not the answer you are looking for?

Search for more explanations.

Ask your own question