 Huygens says, light is a wave and gives us a way to figure out how the wavefronts evolve. A wavefront can be thought of as a set of particles which are oscillating in sync with each other. So if this was, say for example, light, then according to Huygens, every point on this wavefront acts as a source for secondary waves. And the envelope of these secondary waves, a common tangent to them, represents a new wavefront. But can we use this to prove laws of refraction? Let's say we have light moving from one medium to another, say air to water and here is our incident wavefront. The moment it hits the surface of the water, every point of the water behaves as a Huygens source and starts producing secondary waves in the water. We are of course neglecting the reflected waves over here. And notice the waves are slower in water compared to that of air. And now when we draw a common tangent to these secondary waves, we get the refracted wavefront. And indeed we see because the waves are traveling slower, the refracted wavefront is bent towards the normal. So now let's see how to draw this step by step on a piece of paper and see if we can prove Snell's law. So here's our air and water. So let me draw a couple of incident rays of light. So I'm going to put my ruler over here and here's my first incident ray. And I'm going to draw two incident rays of light. And here's my second ray. I'm going to draw them parallel to each other. And so basically I'm imagining that the source is far away. You might recall that when the source is far away, the rays of light become parallel to each other. The sources are infinity, you can say. And when that happens, what happens to our wavefront? Remember, when the sources are infinity, the wavefronts are plane wavefronts. And when I'm drawing over here, they're going to be straight lines. And how do I draw my wavefront over here? Well, wavefronts are always perpendicular to the direction of the rays. And so I can now bring in my set square and draw my incident wavefront. So let me use dotted lines or dashed lines so that we can differentiate between rays and wavefront. So here's my incident wavefront. All right. So let me write that as well. This is incident wavefront. And this is going to be perpendicular to our incident rays. And now to prove Snell's law, we have to draw the refracted wavefront. And we can do that using Huygens principle. Huygens says that every point on this can be thought of as a source. And when the incident wavefront hits that point, it's going to start, you can imagine it gets activated and it starts producing waves, secondary waves, spherical secondary waves. Now the waves produced in the same medium are the reflected waves. And we're going to ignore that because we are only interested in the waves in the water. And the important thing, as you will see now, is that the waves in the water are slower than the waves in air. And so now notice as the wavefront goes forward, as the Huygens sources get activated, more and more secondary waves are formed. But notice the secondary waves are slower than the incident waves. And now a common tangent to all these circles will represent our refracted wavefront. And you can immediately see that the refracted wavefront is bent. In fact, it is bent towards the normal. And I hope you can imagine if you had some different media in which the wave was faster than the first media, then the refracted wavefront would have been bent the other way around, away from the normal. Okay, but how do I construct this now? How do I draw this in a piece of paper? Well, I don't need all the circles to draw the tangent. I just need two. So I'll only consider two Huygens sources. One Huygens source over here. The first one that got activated, I'll draw one circle over here. And another Huygens source, which I'll consider is over here. The last one just got activated when the wavefront hit over here. And so it has just started producing the wave so that wave is a point. And so all I have to do then is draw a tangent from this point onto that circle. So let's go ahead and draw that. Okay, but how big should I draw that circle? Well, remember that the radius of that circle represents the distance traveled by the secondary wave. In the time, the incident wavefront went from here to here. So if we are in a denser medium, then the distance traveled would be smaller because the speed is slower. In that case, that radius should be smaller than this distance. But if the second medium was a rarer medium, then that radius should be bigger than this distance. Does that make sense? So in this case, since we are considering a denser medium, water is denser than the air, let's draw a circle which has a radius smaller than this distance. And to do that, I'm going to bring in my compass and I'm going to set the length of the compass to be a little smaller than this distance. And I'll bring that compass over here. And I'm going to make an arc. So here goes. Okay. So this is the secondary wave from this Haigans source. And at this moment, the secondary wave from this Haigans source has just started. So it's just a point. And so my refracted wave will be a common tangent, which is basically a tangent from here to here. So to draw that tangent, let me bring in my ruler. I'm going to move that so that becomes a tangent. It's going to look somewhat like this. Okay. And here we go. Refracted wave front. And again, let me go and write that. This is our refracted wave front. But remember, I have to prove Snell's law, which means I have to draw incident ray and refracted ray because I want to bring in angle of incidence and angle of refraction. So how do I draw the refracted rays now? Well, rays will be perpendicular to the wave front. So I'm going to bring in my set square again. So here's my set square. I've already set it perpendicular to the wave front. And so I'm going to draw one ray from here to here. So here is our first refracted ray. And let me draw a similar one over here as well. This is going to be our second refracted ray. There we go. And again, this is perpendicular. So we're done with the construction. Now let's go ahead and prove Snell's law. To do that, we have to prove sine i by sine r is a constant. So we need to draw angle of incidence and angle of refraction. So for that, I'm going to drop a normal. We can drop a normal over here. So let me, oops, let's, all right. So here's a normal. And let's write down our angles of incidences and refraction. So this is going to be our angle of incidence between the incident ray and the normal. And between the refracted ray and the normal, that's going to be our angle of refraction. And I need to prove that sine i by sine r is a constant. How do I do that? Well, now we're in the geometry world. So let's think mathematically. We're talking about sine i and sine r, trigonometry. So we require right angle triangles. And I can see two right angle triangles over here. So maybe if I can somehow bring these angles into the triangle, if you know what I mean, then I can use sine i, I can calculate sine i and sine r from the triangle and then I can take the ratio and somehow prove that that's really a constant. I know it's a little vague, but I really want you to give this a shot. So why don't you pause the video and see if you can prove this. Okay, if you've tried, let's see, let's try to first bring in the angles into the triangle. Let's concentrate on the incident right angle triangles. Let me dim this. So if you look over here, because this is perpendicular, I know that if this is i, this has to be 90 minus i. This has to be 90. I'm gonna write that over here, 90 minus i. Okay. But if I look at this whole thing, it should also be 90 degrees because normal is perpendicular to our surface. And so if this is 90 minus i, then that means that this should be, this should be i. So let me write that over here. This should be i. So I brought in the angle. Let's do the same for our refracted triangle. Okay, now if we concentrate here, because again, normal is perpendicular to the surface, if this is r, this has to be 90 minus r. But wait, I don't want 90 minus r. Let me write that. This is 90 minus r. 90 minus r. But I want r. I want to bring in r. Well, look at this right angle triangle. Since this is a right angle triangle, these two angles should add up to become 90 because it's already 90. Total should be 180. And so if these two should add up to become 90, if this is 90 minus r, this should be r. There we have it. We have brought in i's and r's into our triangle. Now let's calculate sine i by sine r and see what happens. So if I do sine i, sine i divided by sine r, what do I get? What do I get? So if I look at sine i in this triangle, it's going to be the opposite side divided by the hypotenuse. For sine r, it's going to be the opposite side divided by the hypotenuse. Same hypotenuse. So the hypotenuse cancels out. So that's nice. But the big question now is, how much is this opposite side? How much is this opposite side? How do I figure that out? Well, again, think about it. What does this length represent? That's the distance traveled by the incident wave front in some time from here to here. And that distance traveled I can write in terms of speed because I know that in this equation speed is going to come. I know that. We've already seen that before. Therefore, let me write this in terms of speed. Speed equals distance into time. So distance is speed into time. So if the wave is traveling at a speed v1, let's say, then I can say that, oh, this distance has to be v1 times the time, the time it took to go from here to here. Does that make sense? Distance is speed into time. Similarly, over here, the wave has a different speed. In this case, a slower speed. Let's call that as v2. And the time taken is exactly the same as the time taken to and from here to here. So the time taken will be t itself. And so over here, it's going to be this distance. Over here, it's going to be v2 times t. So our sine i is going to be v1 times t. The opposite side divided by the hypotenuse. I'm just going to call that as hypotenuse because sine i is going to cancel out. And divided by sine r is going to be v2 times t divided by the hypotenuse. And so notice the hypotenuse cancels out because there's the same hypotenuse. The t cancels out and I end up with v1 over v2. Which means that sine i by sine r is equal to the ratio of the speeds, which is a constant for a given media. For air and water, this will be one number. So it doesn't matter what angle of incidence I choose. Sine i by sine r will be that same number. So we have indeed proved Snell's law. Sine i by sine r is a constant. And Heigen also helps us understand what that constant is. It's the ratio of the speeds in the two media. Victory for Heigens. To check your understanding, I highly encourage you to reconstruct and prove Snell's law all by yourself. Do it right now when everything is fresh. And if at any point you get stuck, no worries, just go back and revisit that part of the video. This is the best way to learn new concepts, trying to recall stuff rather than just summarizing.