 Hi, I'm Zor. Welcome to Inizor Education. We continue talking about properties of light, and today we will talk about effect which is called a refraction. Now, we are familiar with reflection when the light goes to some border between two different substances and then reflects back into the space where it came from. Refraction is when the ray of light actually penetrates the border between these two substances and goes into another substance. For example, light goes from air to water, so the surface of the water is the boundary. And obviously, we will assume that this surface is always flat. Very nice, ideal plane. Now, this lecture is part of the course, Physics for Teens, presented on Unizor.com. I suggest you to watch the lecture from the website because right at the place where the lecture is presented, there is a note, there are some very detailed notes, like a textbook basically. So, whatever the formulas I'm writing on the board are there and the pictures are also there, probably much better. Now, the website is completely free. There are no advertisements, no financial strings attached. There is also a prerequisite course called Mass for Teens. Mass is mandatory to know at the level of calculus, basically, at least. So, you have to know what the derivatives, what the vectors are, the integrals, obviously, differential equations. I mean, all these relatively standard and classical parts of the mass, you have to know before you engage in real studying of physics. Okay, so let's get to the reflection. Now, when I was talking about reflection, I mentioned something which was called Principle of Least Time. Well, this principle was introduced by Fermat, Pierre Fermat, a very talented French mathematician of the 17th century. And the principle is that light goes to a trajectory which assures the minimum time. So, if you have two points and it's supposed to go from one to another, well, if it's a uniform substance between them, like vacuum, it goes along the straight line. So, only the ray, well, if this source of light emits lights in all the different directions, then the ray which goes directly to the point where we are actually trying to sense it, that's the only ray which goes. All others are going in all other directions. Now, if it's not a uniform substance, if it's like in our cases of reflection or refraction, we have two different substances and there is a border between them, then it all depends. Now, as far as reflection, we have actually constructed the way how the light goes to the surface of reflection and to the point where we expect it to be. And we had the law of reflection that the angle of incidence is equal to angle of reflection. So, if this is a mirror, this is source and this is recipient of light, then we have to have such a point here where these two angles are equal. This is angle of incidence and this is angle of refraction. This is angle of the ray with the normal to a surface. Now, very important, these two angles are equal and don't forget that the light doesn't really leave this upper part. So, it's always within the uniform environment. Now, let's consider refraction. So, again, this is the border, let's say this is air and this is water. Now, here is the light, source of light and here is a recipient of light. Well, obviously we can say that maybe the light goes straight. Well, maybe, but let's just think about it. The speed of light in air is higher than in water. So, water is where the light is slower than in the air. Maybe if I will make the way from light to the water a little longer, but make the way from the surface of the water to a recipient of light a little shorter, maybe I will win something. So, I'm making longer where the light goes faster and I'm making shorter where the light goes slower. So, maybe I will minimize the time by finding such a point. So, this is the goal. And obviously this point should, the position of this point should depend on the differences in the speed of light. For instance, if this is air and this is air and there is nothing basically, there is no border basically, right? Obviously it goes along the straight line because that would be, you know, the speed is the same. So, you divide distance by time that would be definitely the shortest time because this will definitely be longer distance and if the speed is the same then the time would be definitely longer. But if the difference, if there is a difference between the speeds, again, increasing one piece segment and decreasing another, increasing where it's faster and decreasing where it's slower might help to minimize the time. Let's calculate. I mean, everything goes to plane mass. Okay. So, let me just simplify it. So, let's say I have these coordinates. This is x, this is y. Now, I'm using the plane here for this picture. Basically, it's a three-dimensional situation, right? So, in theory, this border is a plane and this is a z-axis and this is x and y. However, from certain relatively simple considerations, we can come up with a conclusion that everything actually happens in one plane and that's the plane where I'm actually considering right now. So, it's easier and again to transfer this particular explanation to a 3G, three-dimensional, is really not very difficult. So, I think it will just, if I will simplify it to a two-dimensional picture, it will be much more understandable. So, let's say we have a point B with coordinates A and B, this. And we have point Q with coordinates CD. Now, I'm looking for some point X, X zero, right? On the border. And this is how the light travels. I will try to find such point X to minimize the time. Okay. Well, time contains actually time spent in the air and time spent in the water. By the way, I'm mentioning air and water. Obviously, we understand it can be vacuum and glass or whatever else. It doesn't really matter. So, how can I calculate the total time? Well, I have to know the speeds. So, let's say this is speed V1 and this is speed of the light V2. And V1, we assume to be greater than V2. Okay. This is my assumption and I know the speeds. So, consider that I know the speed of light and the light in the air and in the water. So, I know these numbers. Now, I know A, B and CD as well. And I would like to find out this point X, or at least some kind of a necessary condition for this point to minimize the distance. Sorry, to minimize the time, not the distance. Well, let's make a normal here. This is perpendicular to the X axis. So, this will be my angle theta incident and this will be my angle theta refraction. I'll just lower this. So, fine. Let's just calculate the time. So, first of all, the distance from P to R is square root of X minus A square plus B square. That's the distance, right? This is X. This is A. So, this is X minus A and this is B. Now, the time equals this. You have to divide distance by speed, right? Now, obviously, I assume that within the air, speed always constant and is equal to B1, which by the way is slower than vacuum. Vacuum is the highest speed. Now, analogously, T2 affects the time which light needs to cover the distance from R to Q, from R to Q. This distance is, well, this is C and this is D, right? So, it's, again, X minus C square plus B square divided by D2. By the way, X minus C or C minus X doesn't really matter because it's a square. Just for uniformity, that's the general formula in this particular case. Okay. So, I know the time. How can I minimize the time by X? Well, this is a function. Function has minimum and we know from calculus to find the minimum. We have to find derivative and find the root when derivative is equal to zero. Okay, fine. Let's do it. Derivative, well, first of all, this is the constant, V1 and V2. Derivative of square root is this. But this under the square root, we have some inner function. So, we have to first make the derivative of the square root, which would be 1 over 2 square root of X minus A square plus B square times derivative of the inner function. Inner function is, this is a constant, so it can be ignored. This is a square, so the derivative of this would be 2 times X minus A. Obviously, we can cancel this. Okay. So, that's the derivative of this square root and we also have V1. So, this is T of X first. So, this is the first time derivative by X and that's what we have. Well, analogously, T2 of X is equal to, okay, I will drop tools. We don't really need them. X minus C square plus D square times V2. So, this is the time, this is derivative of the time. Now, the total time is their sum. So, I have to basically add those guys and that would mean derivative of the total time. Sum of these two. Okay. I don't like these square roots, it's kind of cumbersome. But I would like to do, I would like to use the angles of incidence and refraction to simplify these expressions. So, how can we do it? Very simple. Consider, for instance, the top one. So, this is B, this is X minus A. This is point B. Now, this is my angle of incidence. Now, obviously, B divided by X minus A, B divided by X minus A. Well, let's do it the other way around. X minus A by B. X minus A divided by B is tangent of theta i. Right? This is X minus A. This is B. This is angle. So, this calculus divided by this calculus gives me the tangents from which B is equal to X minus A divided by tangent theta i. Let's think about what this expression actually is. Well, it's X minus A square plus B square. B square is also X minus A square divided by tangent square. Well, instead of tangent, I would have sine square divided by cosine square, if you don't mind. So, sine divided by cosine, cosine goes up. Now, why did I do it? Very simple. Because I can take X minus A square out. What I have? 1 plus cosine square divided by sine square. Common denominator is sine square. And it will be sine square plus cosine square, which is 1. So, the whole thing is equal to X minus A square divided by sine square of theta i. And now I can extract square root from this, right? That was my purpose. So, that would be much simpler expression. So, what do I have? I have square root of X minus A square divided by sine square of theta i. That would be my square root, right? What it's equal to? Well, now these two angles are acute angles. So, square root from the sine would be the same sine. Square root of X minus A square depends on where actually X is. If X is greater than A, that would be X minus A. If, in case of a C, for instance, I have exactly the same kind, if X would be smaller than C, then I should have C minus A. So, to make it right, I have to do it this way, right? So, this is basically this square root. So, the whole expression for T1 of X derivative is X minus A divided by V1. And this square root, which is this, X minus A, and sine goes up. So, that is my first component of my derivative. Now, the second component would be exactly the same. So, T2 would be equal to X minus C times sine theta r divided by V2 and absolute value of X minus C. Now, I have to do something with X minus A divided by X minus A by absolute value. Well, again, for the case for the A, X is greater. So, X minus A absolute value is equal to X minus A. Absolute value of X minus C, on the other hand. Since C is greater, it would be C minus A. Which means, if I will divide X minus A here and X minus A here, I will just cancel them out. So, this is equal to sine theta I divided by V1. Now, here, this is X minus C, and this is C minus A. So, it would be minus sine of theta r divided by V2. And again, my, that's supposed to be, my total derivative is a sum of these two. So, sum of these two, this minus this, should be equal to zero, which means sine theta I divided by V1 should be equal to sine theta r divided by V2. This is a necessary condition for optimization of the time. So, the time would be minimum if angles of incidence and refraction are related like speeds of light in the corresponding. Equivalent to this would be five output V1 divided by V2. That's sine theta I divided by sine theta r. So, sines of two angles, incidence and refraction, should relate exactly like speeds. This is a necessary condition for optimizing. So, whenever V1 is greater than V2, like in this case, the corresponding angle should be greater. So, theta I would be greater than theta r. Incidence angle would be greater than refraction. If situation is reversed, for instance, this is a light and this is a recipient, then obviously the angle of incidence would be smaller than. So, it all depends on relationship between the speeds of light. That's very important. Back to reflection. Angles were equal, incidence and refraction. Why? Because it's the same substance. The light did not leave this particular area which is above the mirror. In this particular case, light penetrates and now the speed is different. Speed is air and air is greater than water. So, that's why we have difference in angles. But that's the most important necessary condition for light. I would like to determine, for instance, what is the angle of reflection? You're just knowing angle of incidence. You can find angle of reflection from this particular equation. So, this sine is equal to this sine divided by V1 multiplied by V2. So, very simple. Now, there is another kind of terminology. V1 and V2 are speeds of lights, which are very, very large numbers, as you understand. Now, people would like actually to deal with smaller numbers, so what they do, they introduce number which is called refraction index, which is speed of light, the real speed of light in vacuum, divided by speed of light in this particular substance. So, this is obviously much, it's a small number. Obviously, it's greater than one, because speed of light is greater than speed in any substance, but not by much. Maybe it's one and a half, maybe it's two, something like this. I think the highest number is like 2.6. So, the speed of light in some substance, I think diamonds have something like 2.5. So, the speed of light in the diamond is 2.5 times less than the speed of light in vacuum. Okay, so using this, I can say that V1 is equal to C divided by N1, and V2 is equal to C divided by N2. So, N1 is refraction index for, in this case, air and N2 vacuum. Now, what happens if I divide V1 to V2? Well, C will cancel out, so I will have N2 divided by N1 is equal to sin A theta sin theta r. Or again, if you wish, N1 times sin theta i is equal to N2 times sin theta r. So, using refraction indices, this equation looks maybe a little simpler maybe than this one. So, you don't deal with real speeds, which is like 200,000 something kilometers per second, whatever. This is dimensionless number and the refraction index. And sin is also dimensionless number, so it's kind of easier maybe to deal with this. But again, in this case, you have to know the speed of light in the substance. In this case, you have to have refraction index of that substance, which is this ratio. Same thing. Now, let's analyze a little bit the whole thing. So, let me start with this equation, which I have derived the last one. N1 times sin theta i is equal to N2 times sin theta r. What if my ray of light is perpendicular to the surface of water in this particular case? So, the angle of incidence is equal to zero. When pr is perpendicular, it goes right on the normal to the surface of water. So, the angle is zero. Sin is equal to zero. What follows? This sin is equal to zero. There is no other way. To make this equation true, if this is zero, N2 is definitely not zero. So, this sin should be equal to zero and N2 should be equal to zero, which means that vertical light, let's say from air to water or from vacuum to glass or whatever, the vertical normal to the surface which separates these two substances, it does not change the direction. It changed the speed, obviously, but it does not change the direction. So, perpendicular light to a surface goes perpendicularly. It does not change direction like in this particular case. So, that's one interesting property which we can derive from this. Another interesting property is when, let's say, reverse. So, let's say we have light here with a slower speed and it goes from water to the air in this particular case and as you see the angle is increasing. This speed is smaller and this speed is greater. So, this angle is smaller and this angle is greater. Okay, fine. Obviously, again, if we start from Q on the perpendicular, it will be perpendicular, that's okay. But then, let's just increase this angle. So, as this angle is increasing, this angle is increasing more, right, because the angle should be greater. Now, there will be some point here when this angle of incidence would correspond to 90-degree angle of refraction, when the sine is equal to 1. So, when this sine was equal to 1, when n1 divided by n2 times sine i is equal to 1, right? If this is equal to 1, then this divided by this should be equal to 1. Or if you wish, sine of theta 1 is equal to n2 divided by n1. Now, n2 is less, right, than n1. So, am I right? Okay, so my point is that at certain angle of incidence, my angle of refraction would be equal to 90-degree, pi over 2. Now, this is called a critical angle, because as soon as we will increase it just a little bit more, this angle cannot go, actually, this way. So, it will be reflected. This is called a total internal reflection. So, if light goes from a lower speed to the higher speed, then there is some kind of a value, critical value of this angle of incidence. Now, this is angle of incidence, this is angle of reflection in this particular case, right? So, there will be such an angle when the light goes from a lower speed when the light will go exactly at 90-degree, but a little bit more than that, and light will be completely reflected. And it will actually obey the laws of reflection in this case, because as soon as we will increase this angle a little bit more, this will go exactly equal refraction angle to incidence angle. So, that's very interesting. Now, this total internal reflection effect is used, actually, and used quite efficiently. Now, obviously, we all heard about fiber optics, right? So, it's a very narrow, let's say, cylinder of glass, if you wish. And if there is a source of light here, since this is very narrow, this angle of incidence is very close to 90-degree. So, it's really large, it's almost like here, right? And definitely, total internal reflection works in this case, so the light doesn't leave this glass cylinder. It's reflected and goes this way. Within this glass, so if we have sufficiently clear glass, it will go the very long way. So, that's what's very important. So, the fiber optic is one of the applications of this particular effect of total internal reflection. Another very important application of this same thing is something which we all are very much familiar with. If this is the light, and this is some kind of a very transparent substance, like a glass, for instance, the light goes here, and again, here it's perpendicular to the surface, so it goes in into a lower speed. Now, this angle, if this angle is sufficiently large, I mean, I can actually do it larger than that, I can do it this way. So, the perpendicular would be really large. It will internally reflect here, and same thing here, and then go out. So, that's how we can actually reverse the direction of light. And if this thing is made of the material which really has a low speed of light, and that's, for instance, the diamond, this would actually be very sparkling, because any ray of light which is directed to this would be reflected back. It does not absorb light into itself. It will internally reflect it back to wherever the source is, and that's why the diamonds are so sparkling. The light goes from the top, then it has this internal reflection, total internal reflection, and goes back to the top. And that's why we have such beautiful sparkling diamonds. Obviously, it depends on how clear is the diamond and some other properties. The color, for instance. So, you can actually change the color, because if the diamond is, let's say, yellow, the white light comes in, and the yellow comes out. And again, it's probably changing. So, it's very interesting to actually affect this total internal reflection. And it all depends on the fact that the speed of light in the substance where the source is lower than the speed of light where recipient looks at it. And that's why there is a total internal reflection. Well, that's it for today. I suggest you to look at this lecture on the Unisor.com. We have to go to courses called Physics for Teams. Then there is a part called Waves. And on the next screen, the menu will be you will choose properties of light, and you will find a reflection among different properties of light. Now, why? Because there is a very detailed explanation of whatever I was talking about today, with a much nicer picture, obviously. So I do suggest you to read that material. That's it. Thank you very much, and good luck.