Refraction Of Light

Hey, guys. In this video, we're gonna talk about something called The Refraction of Light, which is the change in the lights angle as it passes from one boundary to another. All right, let's get to it. Remember, guys that at a boundary Lykken do two things it can reflect off of that boundary, or it can transmit. It can transmit through the boundary and make it into the new medium. And remember, that light in general does a combination of the two. Okay. When light crosses a boundary into a new medium, it turns out that it will change its direction. Remember that at a boundary, we always measure angles from the normal to that boundary. So if this is state a one, then the light is gonna come out at some new angle. Fate of two and data one is not necessarily going to equal data to In general, they will not. Okay, remember that in each medium, light still travels in a straight line. Those lines just aren't parallel because the ray has changed direction. Okay, Now, the reason that the light changes it's direction is due to one thing and one thing alone. Light travels at different speeds in different media. Okay, in a vacuum, light travels the fastest it will ever travel. The speed of light in a vacuum is a fundamental constant of the universe. But in a medium light will always travel more slowly than in a vacuum. Okay, remember that a fundamental fact about waves in general, all waves is that their speeds air determined by properties of the medium and the type of wave in this case, since they're all light, is electromagnetic waves. The speed is determined simply by the properties of the medium, the property of the medium that tells us how fast light is going to travel in it is known as the index of refraction. Okay, this is a very important word. Sometimes it's called the refractive index, just so you don't have to call it the index of refraction. Okay. And the index of refraction off any medium is always defined the same way. It's defined as the speed of light in a vacuum divided by the speed of light in that medium. Okay, Now, remember, because light always travels slower in a medium than it does in a vacuum, the index of refraction is always going to be bigger than one. The speed of light in a vacuum is always gonna be larger than the speed of light in any medium. So the index of refraction is always gonna be bigger than one. The two indices of refraction that you need to remember are the index of refraction for a vacuum, which is obviously one right. How fast does light travel in a vacuum? C C, divided by C is one. But for air, the index of refraction is 1.29 It's so so close toe one that we always just say it's simply one. Okay, so the index of refraction of air is the same as the refractive index off vacuum, Which is what? Okay, now I'm going to talk about an analogy to explain why refraction occurs. Okay, I'm gonna minimize myself so we can all see this figure. You can think about refraction like a barrel rolling on a smooth surface to rough surface. So imagine this barrel right here is rolling on smooth, fresh pavement, okay, and it's rolling very, very quickly. And then it encounters some grass, which is rough, and it slows the barrel down when the barrel is rolling entirely on the smooth surface, all of the points along the barrel are moving at the same speed, so the barrel rolls in a straight line. But once the edge of the barrel hits the grass, that edge is now moving more slowly than the opposite edge. Okay, now that this is rotating more slowly than this edge is, it hooks the barrel right. This barrel hooks inward, and it causes it to change direction once the barrel is entirely on the rough surface. Once again, all points on that barrel are moving at the same speed, so it moves in the straight in a single direction. Again, it moves straight again. But since it hooked when it transitioned from the pavement to the grass, it is now moving in a new direction. Okay, this is the idea with refraction. It's due solely to the fact that as you transition from one boundary to another, the speed that you're traveling with changes okay and the speed in the case of the barrel change once the barrel hit the rougher grass and that caused it tow hook inwards and move along the grass at a different angle. Okay, let's do an example to further illustrate this. We want to explain refraction using Hagen's principle. Okay, Before I do refraction, though, I want to talk about what Hagen's principal says about a single ray of light moving through a vacuum. So imagine that that boundary wasn't there. Let me minimize myself for this because I'm going to be drawing over where I am right now. All right, so this Ray is coming in and there's a boundary right here that separates the media on the left side. On the right side, I'm still indicating where the boundary is, but I'm not sorry. Let me redraw that really quickly. I wanted to end at where the imaginary bounder is. I've indicated where the boundary is, but I don't actually have a boundary here. In reality, we're still looking at this light ray without changing the media that it's in. So remember how to do these problems. We're going to draw wave fronts as they approach the boundary. So there's a way front at one time, after a certain amount of time, There it is, right after another amount of time. There it is, and there it is and then it hits the imaginary boundary, and then it hits it again, and then it hits it again. And remember that this was the first time that it hit that boundary. Okay, actually, let me write that over here. This is the first time that it hit that boundary right here. This is the second time that it hit that boundary. And right here, this is the third time that it hit that boundary. So that third point was the most recent time hit. So it produces the smallest Ray. I'm sorry. The smallest wave lit the green dot is the second time that it hits. So it produces the second smallest or the second largest wave lit Because there's only three of them. The blue is the most recent Sorry, the earliest time that it hit. So it's had the longest time to propagate the wave lit. So it has the largest wave lit. And what ends up happening is that this wave front is gonna be parallel to all of those other wave fronts. And this ray is going to go through undisturbed, right? That's exactly what we expect. Because as it passes through no boundary right This is no boundary right here. I drew it as an imaginary boundary, but in fact it's traveling in the same medium because there's no boundary. It shouldn't change direction. It should absolutely move in the same direction. Now let's see what happens when we encounter a boundary where the speed of the light does change. So once again, here's away front. Then a little time later, another way front, little time after that, another way front, then away from hits the boundary. Then time after that, another way front hits and then it hits again. Okay, so it hits at those three places the first places here. The second place is here, and the third place that hit is here. So the most recent encounter is at the third point. The earliest encounters that the first point This is no different than every time we've applied Hagen's principle. Now what's the difference between the situation on the left with the boundary and the situation right here with no boundary on the left side. In the new medium, light travels more slowly than when you don't change media. Okay, because light travels more slowly. These wave let's don't travel as far. So that third wave lit the red one is gonna be smaller over here than it waas right here. The green wave lit is also gonna be smaller over on the left. Then it was here. And finally the BlueWave lit too, is gonna be smaller on the left than it is on the right. And that's simply due to the fact that light travels more slowly in this new medium. So what happens? The new wave front is at a different angle, then the old wavefront, because it's at a different angle. Now, when I draw my ray, my ray is moving at a new angle. Okay, remember that this process, which is called refraction, is due solely solely to the fact that the speed of light is different between these two boundaries. Okay, that wraps up our discussion on refraction. Thanks for watching guys.

Hey, guys, in this video, we're gonna talk about something called Snell's Law. Snell's law is the mathematical law that governs refraction across a boundary. Okay, let's get to it. Remember that when light transmits through a boundary into a different medium, the angle of its path must change. Okay, What I have here is an image that shows one medium indicated by an index of refraction in one and another medium indicated by an index of refraction in two. So the light travels at different speeds in these two media, so long as in one and in two don't equal. If they do equal, then they travel at the same speed and no refraction occurs. Remember that refraction depends on Lee upon a change in speed. So what we have is some incident angle fate, a one and some refracted angle faded to the question is how did those angles relate to one another? Well, Snell's law gives us this angle of refraction, and it says that the number of the index of refraction times sine of the angle, is conserved across the boundary. So in time, sign of data on the left side is also equal to n times Sign of Fada on the right side. Okay, so that quantity the index or faction times sine of the angle is conserved across the boundary. All right, now, this is where our definition of angles becomes very, very, very important. Your angle, which appears here and here, has to be measured from the normal. Fada has to be measured from the normal. Okay, Okay. It has to be measured from the normal notice. These two angles are both measured with respect to this normal line that I drew. This is the normal. Okay, if you measure that angle, let's say with respect to the surface of the boundary, like so that is not going to give you the right answer. Your answer will absolutely unequivocally be wrong every single time you measure the angle with respect to the boundary, do not do this. Always measure it with respect to the normal. Okay, let's do a quick example. Ah, Light ray is incident on an air water boundary at 33 degrees to the normal, with the reflected with the refractive index of water being If the light ray passes from air toe water. What is the refracted angle. What if the ray were to pass from water to air? So we have two different scenarios. We have air, toe, water and then we have water to air. So let's start with air to water. We know the incident angle 33 degrees to the normal, which is always how we have to measure the angles. And we know the index of refraction of water 1. We're going from air to water. So theta one, the angle the incident angle is 33 degrees in one is our index of refraction. Of our initial medium in two is the index of refraction off the medium were passing into which in this case, is water right there toe water 1.33 and then theta two is are unknown. The only thing here that we're missing is the index of refraction of our original medium, which is air. Always remember, the index of refraction of air is one. Okay, so now we have enough to use Snell's law to solve this problem. Snell's law says in one signed theta one equals in two signed theater too. So I need to solve for sign of data to so rewriting this I get sign of fated to is in one over into sign of fate of one which is in one. The incident medium is just one into the medium that we're passing into, which is water is 1.33 and the incident angle was 33 degrees. Okay. And this is equal to 0.41 which means using the arc sine function on our calculators. That this angle is simply 24 degrees, right? Right relative to the normal in the second bound Sorry in the second medium, which in this case is water. All right, that's the first part. What about water to air now the media are flipped, right? We're going from water to air, but the incident angle is still the same. We're still incident at 33 degrees from the normal. So theta one is still 33 degrees. But now, in one, the refractive index of our original medium is now that of water 1.33 and in to the index of refraction for are transmitted medium is that of air, which is one. And we're looking for that refracted angle. So it's the same set up. Let me minimize myself. I'm already sorry. I already have the equation I need to use, so I'm just gonna write that out. Sign of faded. Two equals in one over in to sign of fatal one, which is now 13 3/1. Right. The industries of refraction are flipped, but the incident angle is still 33 degrees and is holding equals about 0.72 So if we use the ark sign on our calculator, we find that the refracted angle is 46 degrees from the normal. Okay, always from the normal. All right, guys, that wraps up our discussion on Snell's law. Thanks for watching.

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