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Refraction Through Multiple Boundaries

Patrick Ford
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Hey, guys, let's do an example. Ah, light ray enters a 25 centimeter deep layer of oil at an angle of 37 degrees to the normal. Below. The oil is a layer of water 15 centimeters deep. How far does the light rain move horizontally from the point it enters the oil to the point, it hits the bottom of the water. Okay, this is actually like a very classic Snell's law problem. So the odds are you're gonna see it. Maybe it's a homework problem. Maybe is a test problem, so I would study this problem because it's a very, very common problem. All right, so what's happening here is simply that an incident light Ray is hitting oil right here, so it refracts going through the oil. Now the oil, we're told, is 25 centimeters deep. Right then, that light ray, which is refracted once, already passes through into the third medium. The first medium is air. The second medium is oil. The third medium is water right here, and it refract again. Then it travels centimeters further down. That's the depth of the water at a new refracted angle and eventually hits the bottom of the water. Okay, so it's gonna cross some horizontal distance through both media through the oil. It crosses this horizontal distance, which I'll call Delta X one. And then through the water, it crosses this horizontal distance, which I will call Delta X two. Sorry, that's cut off a little Delta x two, and quite clearly, our total distance traveled. X is just Delta X one, plus Delta X two. All right, so at the boundaries, this boundary and this boundary, we're gonna have Snell's law that we have to apply to figure out the refracted angles. But after that, it's just a geometry problem. Okay, but buckle up. Because this is a long geometry problem. All right, let's talk about the first boundary. This is the air toe oil boundary. What we have is meet all that's solid. We have our incident, ray from air, which has an index of refraction of one hitting this at an angle of 37 degrees. It then passes in tow. Water, which has an index of refraction of sorry passes into oil. First oil sits atop water on index of refraction of Okay, and it's gonna be refracted at some new angle feta to We need to apply Snell's law to find what that new angle is, Snell's law says in one sign. Fatal one equals in to sign Feta to So that means that sign of Fada, too. It's simply in one over in two sine theta one, and that's going to be won over 1.47 times. Sign of the incident angle, which is 37 degrees. Once again, it's measured from the normal. The problem said to the normal. Always make sure that the angles that you plug in the snows law are measured to the normal. Okay, and this whole thing equals about 041 And so our first refracted angle is if we use our calculator 24 degrees. Okay, let me bubble. This that is the angle at which the light travels through the oil. So this angle right here, 27 degrees. Okay, let's talk about the second boundary now. Okay. In the second boundary, we have light that's coming in. At what angle? We don't know that angle, but we do know this angle. That angle is 24 degrees that we found right here. That's just a refracted angle. This angle is the alternate interior angle. So it too is 24 degrees. Okay, so now we can apply Snell's law to find what is the refracted angle, which I'll call Data three inside the water. So in two, once again for oil was 147 in three. The third medium for water is 133 I'm gonna use this exact same equation. I'm just going to use different numbers. Two is now gonna become three, and one is now going to become too. So sign of Fada three is in two over in three sine theta three. This is going to be 147 over. 133 Sign of our incident angle, which is 24 degrees. And it's holding equals 0. Okay. Which means what? That if we use our calculators, they refracted angle into the water is 27 degrees. Okay, so on Lee, slightly off from our refracted angle into the oil. Okay, Now let me separate some area right here. Now we need to do our geometry. Okay? If we look at our figure, we have two triangles. We have a triangle inside the oil and the triangle inside the water. Let's start with the oil. The triangle inside the oil looks like this. This is the incident. Sorry. This is the rate traveling through the oil which has an angle to the normal of degrees. How far horizontally it goes. We called Delta X one. How far down it goes is just the depth of the oil which we said Waas 25 centimeters. So all we have to do is use trigonometry to figure out what Delta X one is. Since we have the opposite and the adjacent edges, we should use tangent. Tangent of 24 degrees is going to be the opposite edge Delta X one divided by the adjacent edge, which is 25 centimeters. So Delta X one is gonna be 25 centimeters times tangent of 24 degrees, which is going to be 11. centimeters. Okay, now what about through the water? But we have another triangle that looks like this. Okay. Where the refracted angles inside the water is 27 degrees. How far it goes horizontally is Delta X two, right? And how far it goes vertically is just the depth of the water, which is 15 centimeters. And the high pot news is, of course, just that Ray that's traveling through the water. So once again we have opposite and adjacent edges. We want to use tangent, so tangent of 27 degrees equals the opposite edge over the adjacent edge. So Delta X two is 15 centimeters times tangent of 27 degrees, which is going to be 76 centimeters. So lastly, putting this all together bumper to bumper bum. We have that X, which was Delta X one, plus Delta X two equals 11 centimeters, plus seven six centimeters, which equals 18 7 centimeters. All right, guys, like I said, this is a classic problem, and it has a butt load of geometry. But the Onley physics involved in it is the application of Snell's law at the two boundaries. Alright, that wraps up this problem. Thanks for watching guys