 Hi, I'm Zor. Welcome to Unisor Education. Today we will talk about a particular case of reflection. Reflection of the parabolic mirror. Now, this lecture is part of the course Physics 14 presented in Unisor.com. I suggest you to watch this lecture from the website. So go to Unisor.com, choose the course Physics 14, then go to Waves and Properties of Light. And this is one of the properties of the light reflection. That's what we are talking about. The previous lecture was about reflection of the flat surface. And this lecture is about reflection of the curved surface. In this particular case, it's parabolic mirror. Now, this website, Unisor.com, contains the prerequisite course, math for teens, which contains the knowledge which you definitely need to study physics seriously. Now, this is for teens, which means it's basically high school and maybe a little bit after the high school. And it's kind of advanced, but it's still within the reach of everybody. So if you accurately take the course one lecture after another, you will succeed. Now, back to reflection. Now, first of all, we have to address somehow how reflection happens of the curved surface. Now, the previous lecture, as I was saying, was about the flat plane reflection. And we have proven using the Fermat's principle of the least time that reflection of the flat surface is obeying certain laws of reflection. And the primary is that the incidence angle is equal to reflection angle, which we kind of intuitively understand. Now, what about the curved surface? Well, to prove what I'm going to say right now is kind of difficult and definitely beyond the course of this, definitely beyond this course. But again, it's intuitively kind of obvious that if you have some kind of curved surface, I don't even know how to, whatever, if you have a curved surface and you have a beam of light which falls on it, it falls on certain area. And if the surface is smooth, and we are talking only about the smooth surface, then the principle I'm talking about is the reflection of this point on a surface is equivalent to reflection of the tangential plane which is at that particular point, tangential to a surface. So the smooth surface obviously has a tangential plane at any point. So whenever the beam of light, the ray of light falls on the specific point of a surface, this particular neighborhood of that point can be flattened out and it should be considered, instead of it, it should be considered a tangential plane and then the beam will reflect off the tangential plane exactly as we were talking about the previous lecture, with angle of incidence equal to angle of reflection with a normal to the tangential being in the same plane as reflecting and in incident lights, etc. So that's the principle which I'm not going to basically prove in a way I have proven in the previous lecture using the Fermat's principle of the least time that it occurs actually according to these laws of reflection. But if we will consider this particular approach to replace any curved surface with a tangential plane at the point of falling light, then basically we know everything about how it behaves because it's a flat surface, it's a plane, a tangential plane and we know the laws of reflection. So I will use this principle again instead of a surface, curved surface at any point whenever the light falls that we will consider the tangential plane instead of that surface. Okay, now being done with this we will talk about a specific curved surface which is called paraboloid. Now, it has a lot of practical applications which I will be talking about at the very end of this lecture. And that's why I have chosen. And at the same time it's curved so it basically fills the scope which I would like to talk about which is reflection of the curved surfaces. So what is paraboloid? Well, if you will consider XYZ coordinate system and consider the plane XZ only and if you will have a paraboloid here Y is equal to... it's Z is equal to KX squared. So we are talking about this plane, XZ plane and now we will rotate this paraboloid around the Z axis which is the axis of rotation. What will be? Well, there will be some kind of this kind of body, if you wish. So this is a paraboloid which is a result of rotation of a paraboloid like this around the Z axis. Now, the three-dimensional formula for this obviously is for the entire paraboloid. So this is just a paraboloid within XYZ, XZ plane and this is the equation of three-dimensional body called paraboloid. Now, what's interesting about this paraboloid? So let me tell you basically something which will be proven, so to speak. And if you will have a light which goes somewhere here from inside it will hit the wall, it will hit the paraboloid surface and it will reflect and it will hit the Z axis. Now, first of all, why does it hit the Z axis? Well, because of symmetry. If after this reflection it goes without crossing the Z it will basically contradict the principle of rotational symmetry. So we have to have this point on the Z because otherwise, because if we will turn the paraboloid it will not really, it will basically change into itself. So this particular thing also should be kind of changing into itself. But if you miss the Z axis with a reflected light it will not change to itself. So that's basically kind of obvious. Which means that instead of considering paraboloid surface and the tangential plane at this particular point I can just cut the whole thing. If I will cut it along the X-Z plane, what will I have? Well, I will have only a parabolo. I will have the beam of light goes here, reflected here and that would be a tangential line. Instead of tangential surface I will have a tangential line. So I will, because of this rotational symmetry consider the flat picture within the X-Z plane. This is X and this is Z. And I will examine where exactly this point is located. Well, relatively let's say to the original coordinates. Now, why did I choose the paraboloid and paraboloid in this particular case? Because a very interesting thing. If I will go with another beam it will reflect into the same point. So basically that's what's very important. It's very important for applications because if this is the light from, let's say, sun and we have concentrated all these beams into one point we can boil the water here and generate electricity or if we can put a source of light at this point it will reflect as a parallel beam of light and that's how projectors are working. So it's very important this parabolic form of our curve or paraboloid form of actual body which makes this type of application. So that's my goal. To find out how exactly this reflection goes and prove that it will all go into one particular point which is called the focal point. So that's my plan for this lecture. I'm going to prove this particular property of parabola so whenever the beam of light goes at any distance from the z-axis it will still go into the focal point. Okay. So I have explained my plan. Now let's do it. And to do it I need to draw a picture with all these details. It's kind of complicated. So I have this picture on my phone so I will not... okay. So first of all let me just draw the axis of sin. Axis of coordinates. So this is z and this is x. Okay. Now we have to draw a parabola. So parabola would be something like this. Let me use maybe another color light. Okay. Hope it's visible. Okay. This is my parabola. I hope it's visible enough. Okay. Next I have to have a tangential line to this parabola at the point of the beam. Okay. So where is my... My beam would be let's say here. Well let me start from the tangential line. With something like this. This is my tangential line. This is the point where the ray of light comes and this is perpendicular, normal to my line or to tangential line and this is a reflected light. Okay. So let me just continue to do this. Now I need the letters. Okay. So this is c. This is a. This is b. And this is z. And this is... That's it. Okay. Now this is angles. This is incident. This is reflection. They are equal to each other obviously. And incidentally right now, so qi is equal to qr equals, I mean not q, it's theta. So it's theta. It looks like q. That's why I mixed it. It's theta. Greek. So theta. Theta i, theta incidence angle and r, which is reflection angle, are equal to each other. The beam of light is reflected against tangential line, which we were talking about, that reflection against a curved line is the same as reflection of the tangential line at the same point. And the beam of light goes first down to b. This is vertical, parallel to z. And it reflects and goes to the light c. Now cg is parallel to x axis. I need this. Now what I'm going to calculate right now, based on this distance, let's call it a, distance from... distance of the beam of light from the z axis. I will calculate this oc length. If my oc length doesn't depend on a, doesn't depend on how far my ray of light comes to the parabola, that would prove that all parallel to z line beams of light will gather at the same point. So if oc is independent from a, I will have this property proven. So let's just basically start to calculate the oc. Now the parabola should have some kind of an equation, right? So it's z equals to kx squared. That's my parabola. So all I need to know is oc as a function of a, well, and k, obviously. And if that function doesn't depend on a, then I have proven that all lights coming here or here or here or here, they will all gather in one point c. Okay? So what's important is this angle. Let's think about what this angle is. It's between tangential line and horizontal line. What is this? It's angle between vertical line and tangential line. So as you see, angle theta i and this angle, b e a, b e a. They are congruent. Why? Because this is perpendicular to this and this is perpendicular to the normal, right? So again, this e a is perpendicular to e b and e b is perpendicular to normal to the surface. So these two angles are formed by mutually perpendicular lines. So it's something like this. This is one and this is another. So this is perpendicular to this and this is perpendicular to this. So they are congruent. So that's very important. This is also theta. Okay, great. Now let's think. Let's analyze, basically. How can we find out what is Oc? Well, Oc can be found as... Now, Cd is parallel to the x-axis. So Oc can be found from a b minus bd. So that's my analysis. I'm thinking about how can I find Oc? Well, I can find if I know a b and bd. So it's a b minus bd equals a b I know because if this is a and this is parabola, which is this, so a b is equal to k a square minus bd. So k a square is known. Bd I don't know. So how can I find bd? Well, bd can be very much related to angle this one. And Cg. It's Cg times cotangence, right? Cotangence of this angle. Cbd. Angle Cbd. So Cg I know, this is a. Cg is equal to OA because they are parallel and this is the distance of the beam of light from the z-axis. So times cotangence of angle Cbd. Now a is known. So this is a known thing. I have to know what's the angle Cbd. Well, it's easy. Cbd is pi minus 2 theta, right? So angle Cbd equals pi, the whole thing, right? Minus 2 theta. So to know the cotangence of this, I need to know the theta, right? Okay, fine. Now, what's important is that this theta and this theta is equal to this one. So how can I find out this? Well, it's actually easy because this is a tangential line to a parabola. Now we all know the property of the tangential line. Tangential line has basically the tangent of the angle between tangential line and the x-axis is the first derivative of the function, right? This is my function. Now its derivative is equal to 2kx and at point x is equal to 2, my z is equal to 2ka. So this is the tangent of theta. So I know the tangent of theta. All I have to find out is cotangent of angle Cbd, which is pi minus 2 theta. Okay. Now, cotangent of pi minus 2 theta is equal to minus cotangents of 2 theta. This is the plane trigonometry. And again, if you forgot about properties of this, you can always refer to the mass routines, prerequisite course where it's all explained. And now this is cotangent. I don't remember the formula for cotangents. I do remember tangent. Now the tangent of 2 angles is equal to 2 tangent of angle divided by 1 minus tangent square of angle. Now the cotangent is reverse. So that's equal to minus 1 minus tangent square theta divided by 2 tangent theta. So that's minus. And this minus, so let me just put it here. So cotangent of angle Cbd equals to tangent theta, tangent square theta. Minus and this minus, I will change it. Minus 1 divided by 2 tangent theta equals to 2ka. So it's, yes, just use this. Tangent square is 4k square k square minus 1 divided by 2 times 2ka, so it's 4ka. Am I right? Okay. So this is a cotangent of angle Cbd. Cbd, cotangent of this. So if I will multiply it by a, I will have bd. So bd is equal to a times this. Now a will go off the denominator. So I will have 4k, 4k square a square minus 1. So I've got the bd. Now if I will subtract it from the k a square, I will get my oc. So oc is equal to k a square minus this. k a square minus 4k, 4k square a square minus 1. Now the common denominator is obviously 4k. And what do I have? 4k, 4k times k a square 4k square a square minus this. 4k square a square plus 1. k a square goes down. And what do I have? I have oc is equal to 1 over 4k. As you see it's independent of a. A was cancelled everywhere. So what does it mean? It means that this beam at the distance a and this beam at the distance less than a or this beam at the distance more than a from the z. They will all go into the same point. Into the point which is called focal point. So parabola like this has a focal point at the distance 1 over 4k from its bottom. Always. That's a very interesting property of parabola and obviously paraboloid if you will go into a three-dimension. And as I was basically talking in the very beginning there are many different applications. So the obvious application is the you have a flashlight, right? So you have some kind of a source of light in the focal point and this is some kind of a paraboloid mirror and then all the lights from this source of light will go parallel to the z-axis in one direction. So that's how we project light onto something. So the mirror which is around this source of light must be parabolic to really do the job to parallelize the lights. Some of you might think that it's actually part of the sphere. Something like this. And the source of light here. That's wrong. It's not part of the sphere. It's part of the paraboloid. It should be something like this. Then all the lights will go into one direction. Will be reflected into one direction. It's not part of the sphere. It's part of the paraboloid. It's very important. Now what other applications? Well everybody saw the dish antenna. Now dish antenna also has parabolic form. I mean obviously you cannot really expand it to infinity. It has certain boundaries. So you have parabola like this. And obviously lights which go here will not hit the mirror and will be lost. But the greater the height of this parabola the more light will be captured. Or if we are talking about radio signals from satellite you have an antenna dish which is directed to a satellite which has a constant position above the surface of the Earth. So it basically circling the Earth with exactly the same speed. Earth itself is circulating. So you have the same thing. So you have many different rays of radio waves are collected by this dish and reflected towards the focal point where some kind of a receiver is located. So what other applications? Well something which is not really very scientific but whenever somebody doesn't hear a remote sound people are doing something like this. So that's your parabolic dish so to speak. I mean obviously it's not perfect but it helps. The sound goes here and then it focus to the ear. And there are some other applications which we actually can think about. Well that's it. My purpose was to explain how the light is concentrated by parabolic mirror in one focal point just as an example of reflection of the curved surface. That's it. I would suggest you to read these notes for this lecture on this website. The notes contain a much better picture than I have drawn right now. It has certain colors and you can always click on it with the right button and open it in another tab on your browser. It will be bigger and more visible over the details. So you have to go to Unisor.com Physics 14 course and choose the part of the course called waves. And in that part you have properties of light and one of these properties is reflection of the parabolic reflector. That's the name of the lecture. Now I forgot to mention probably that the website is completely free. There are no ads, no strings attached. You don't have to even sign in if you don't want to. If you just want to listen to the lecture take exams for instance. There are exams, there are problems. So this is a good source of knowledge. Good luck. Thank you.