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Problem 71 Hard Difficulty

Let $ v_1 $ be the velocity of light in air and $ v_2 $ the velocity of light in water. According to Fermat's Principle, a ray of light will travel from a point $ A $ in the air to a point $ B $ in the water by path $ ACB $ that minimized the time taken. Show that
$$ \dfrac{\sin \theta_1}{\sin \theta_2} = \dfrac{v_1}{v_2} $$
where $ \theta_1 $ (the angle of incidence) and $ \theta_2 $ (the angle of refraction) are as shown. This equation is known as Snell's Law.

Answer

$\frac{\sin \theta_{1}}{v_{1}}=\frac{\sin \theta_{2}}{v_{2}}$

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Video Transcript

Yeah, were asked to prove Snell's law using optimization. So the background is the one is the velocity of light and air and V two is the velocity of light and water. Was the gay I shouldn't. Yeah, and v two, this is Velocity. Remember hearing the story So really? According to firmas principle, a ray of light will travel from point A in the air trade point B in the water by path acb that minimize the time taken. We have to show that the sign of data one over the science data to equals V one over v two Mm. Where the fate of one is. The angle of incidents in theta two is the angle of refraction. Just kidding. Yeah, as is shown in the figure and exercise 71 of this section. And so this equation, the ratio of the same fate as equals The ratio of the velocities is known as Snell's Law. To prove this free flees the Pythagorean theorem by the Pythagorean theorem. If we use the figure in the book, you have a distance from the point in air A to C or C is the point on the surface of the water is the square root of a squared plus X squared and the time it takes light to travel from A to C. We'll call this T A C. This is this distance traveled, divided by the velocity or the square root of a squared plus X squared over. The one who is it Travels through air. Yes. Now take the derivative of tea with respect to X by using the chain Rule two so t prime a sea of X. That's it. This is one over B one times one over the square root of a squared plus x squared. Yeah. Actually, I'm sorry. This should be X over square root of a squared plus x squared. This is Roy. So me now we'll use trigonometry in the length of a C to find the value of sine of theta one. This is using triangles X over. Get him the square root of a squared plus X squared. Hello. Where the fuck is this goddamn beat? Work here. Now I'll plug sign of data one into t prime of a C that would drive. So we have t prime of a sea of X. This is actually equal to one over V, one times the sine of 3 to 1 which you have to remember in this case, data is a function of X now. Likewise. Bye, Pythagorean theorem. We have the distance from B to C from under the water to the surface. Point is the square root of B squared plus C minus X squared. Sure. And likewise, to find the time it takes to travel from C to B T s, A B C. This is the distance square root of B squared plus C minus, X squared, divided by the velocity of light and water. V two. Yes. Once again, I'll find the derivative T prime BC of x. This is one over V. Two times, right, and negative X minus C or Sorry, Negative. What is this? Yeah, negative. C minus x Yeah, over the square root of B squared plus C minus X squared. Yes. Yeah, so takes care to Yeah, And now again using trigonometry in the length of BC, we find that you'll be sorry. Sign of data to is equal to C minus X over the square root of B squared plus C minus, X squared three and plugging sine of theta two into t prime of BC. We have the t prime of BC of X is one over V. Two times negative sign of data to yes. And that in shoes. Three outfits. Now we have that. Yes. Is this is girl? Oh, hello. Yeah, well, this one is great, because it's like the time from A to B is the same as the time from I'm sorry. The time From A to C. No, it is a time for me to be is the same as the time from A to C plus the time from C to B. But of course, this ordinary equation implies that the differential equation t prime of a B X is equal to t primacy of X plus t prime BC fx as well. So therefore we have that. Yeah. So right, yeah. I want to have some crackers. This is equal to. And then he walks t prime. A sea of X which we found was signed data one over V one smaller, easier and then t prime BC of X's negative. Sign data to over V two and now buy from Mars principle. We want to find the path that minimizes the time taken to want to find the value of theater. I guess you could say the value of X such that t prime of a B equals zero mhm and therefore we obtain the sign of data. One over V one equals the sign of data to over V two, and this is Snell's law.