 Hi, I'm Zor. Welcome to Unisor Education. Today we will talk about convex lenses. Convex lenses. And these lectures, usually as all other lectures, are part of the course. Course presented on Unisor.com. And this particular course is Physics for Teens. And right now we are in the part of this course which is called Waves. So I suggest you to watch this lecture from the website from Unisor.com because every lecture has textual notes which basically are like a textbook. So it's very beneficial, I think, not only to listen to the lecture and watch it, but also to read whatever the notes are. Well, some people are just absorbing information better in this case. Now the same website, Unisor.com, contains prerequisite course called Math for Teens, which I strongly recommend you to study, or some equivalent amount of knowledge should be accumulated because Math is absolutely mandatory for studying physics. Now the website and all the courses are completely free. There are no advertisements, no strings attached, so welcome to go. Now speaking about convex lenses, let me start from something which we have learned before when we were talking about prismatic lenses. Well, when we discussed that, we basically talked about this particular kind of prism. When the light goes here, now it continues perpendicularly to the top surface and here it actually goes at the angle because if this is, let's say, glass and this is air, the speed of light in the glass is slower. So because of this difference in speed between air and glass, there is this type of a refraction and it's related to the Fermat's principle of least time. So to come from this point to this point, apparently it's not the direct line, but it's this line which actually is faster because it's a longer time. If you go to this trajectory, the light will go longer within the slower part of the distance. So there is also the law of refraction. So if you go and draw a perpendicular to the surface of refraction, this is perpendicular. So this angle is incident and this is refraction. So the law of refraction states that there is such an equation where ni is c divided by... Okay, this is the speed of light in vacuum and this is speed of light in this particular place. Now in this particular case, incident is within the glass and refraction is within air. Now we have... oh, I'm sorry, it's a sign here. So we have derived this particular law of refraction from the principle of least time. Okay, so this is given. I will be just using this as is. And now as the first step towards convex lenses, I will assume not just a single prism like this, but a double prism. Double prism would be with two angles. So this would be one angle and this would be another angle. Something like this. Now I have used this type of a drawing in my notes and I would like to use the same letters as over there. So that would be letter y. Okay, so this is pqo. This was one light coming here. Point A1 goes to this, to B1 and refracted to this, point C. Another light falls on another surface. So we have two surfaces and also refracted to this point. That would be A2, B2. Now this would be angle alpha 1 and this would be angle alpha 2. So these two angles define the prism, double prism in this particular case. Now I'm interested in these two rays of light to come to the same point. And I would like to find out how can I make it using basically the location. Let's say this would be x1 and this would be x2. So from the previous lecture about the prism, if it's a normal single prism, the further I am from the point P, the further will be refracted ray of light across the y-axis. This is y-axis and this is x-axis. Now if I have double prism, now this distance can actually make the projection, the refracted ray to go into this point. But since this plane is at different angle than this plane, I can actually choose a point here in the direction such that it will go into exactly the same point. Now as you understand, I'm planning to gradually increase the number of different angles to come to the smooth convex lens. But let's do it step by step. Okay, so we have only two. Okay, now considering we have two, now how can we use this particular thing? We have to draw a perpendicular to the surface, this one, which is m1 and 1, and perpendicular to this m2 to m2. Now, first of all, what we can say is that let's continue this and continue this as if there is no refraction. Now, this is an incident light and this is refracted light. So this difference, I can call it delta 1, is the deviation of the refracted light from the original. So it's not an incidence or a refraction angle, it's an angle of deviation. This is the original direction and this is the deviated. Same thing here, this is delta 2. Now, let's just think about the following. If this angle is alpha between this and this plane. Now, this ray of light is perpendicular to this line, right? And this is normal to this surface. So this is also angle alpha, right? Because these two angles are correspondingly perpendicular sides. So this angle is actually the incidence angle, right? Between the light and the normal to the surface. Now, this angle is also alpha because these are vertical. This one m1, b1, a1, and n1, b1, d1. They are vertical. Some of these, alpha 1, delta 1 plus alpha 1, this is alpha 1, is an angle of refraction, right? From the normal to the refracted light. So I can say that this is theta r. And now I can use this thing, nI times sine alpha 1 equals n2 sine of alpha 1 plus delta 1. So this is basically the same principle just in different angles. And instead of incidence and refraction, I'm using incidence, which is actually an angle between the refracted surface of this particular double prism and horizontal line. And again, refraction I split into two. The same angle of incidence plus deviation from the original direction. Now, let's say I would like all these two lights to come to the point at distance f, right? Now, how can I actually calculate this distance f in terms of the distance x and angles? Well, this distance f is equal to this minus this. Now, this is equal to, let's just think about it. Let's consider this triangle. This one and this one. Okay, this is for 1 and this is for 2. So this one is x1, this one is x2. So let's talk about the one first. So the same distance f can be calculated as a distance between these two points and these two points. Now, these two points is, this is also alpha 1, right? Because it's all the same thing, parallel lines. So it's x1. This divided by this is tangent, so it's x1 times tangent, alpha 1. Now, this one is, I can use this angle. Which angle should I use? This angle. So if this is delta 1, then this is delta 1. It's too parallel and the cutting line between them. So that would be x from this triangle. x, this divided by this would be cotangents. So it would be x1 times cotangent of delta 1, right? So again, from this triangle, considering this is delta 1 and this is delta 1. This is x1 and I would like to know this length. So this length would be this divided by this is a cotangent. So it's x times cotangent. This is a big length. And the small length is x times tangent alpha. So the difference between them, which means x1 minus x1 tangent alpha 1 is f. But exactly the same thing I can say about the second point. Just instead of this, I have to consider this piece. So from this I will subtract this and it will be exactly the same thing. x2 cotangent delta 2 minus x2 tangent alpha 2. That's exactly the same f. So what my point is that if my x1 and x2 satisfy these conditions, now alpha is known angle. It's an angle between this surface and the horizontal. Alpha 2 is also known. And how can I find delta in this particular case? Well, delta I can find from here using alpha. So what is it? Delta is equal to... Well, let's start with sine of delta 1. Sine of alpha 1 plus delta 1 is equal to n1 divided by n2. Alpha alpha 1, right? From here my alpha 1 plus delta 1 equals arc sine of ni divided by nr sine alpha 1. And delta 1 is equal to minus alpha 1. So we know alpha 1 as the angle of the surface and that's how we calculate delta 1. Knowing alpha 1 and delta 1, we can actually find what exactly xi is supposed to be equal, f divided by cotangent delta 1 minus tangent alpha 1. This is exactly the location of the light which will be refracted into this point where f is given. Now, the same thing I can say about the x2. The x2 is also defined, properly defined. So it's either or. Either I can find, from given prism, I can find points which will project to this or refract it to this. Or, for given lights, I can actually find out what kind of angles I should have to basically project to refract my lights into one point. But this is for double prism only, right? Now, on my way to convex, what's my next step? Well, my next step is, instead of two refracting planes of this prism, I will use many. So let's call it multi-prism. So what is multi-prism? Multi-prism is something very similar. But instead of this, you say this, this, this, this. Consider this to be straight lines. So this represents a prism with surfaces at different angles. So what happens here? These are horizontal lines. These are angles, alpha 1, alpha 2, alpha 3, alpha 4. But they are more and more inclined towards the vertical. What happens? Now, if I have a straight prism, then the further my light comes, the further it's projected here. But if these planes are at different angles to each other, greater and greater angle. This, this and this, like here. Then, if it's a greater angle, it will turn the light more towards the previous points. Again, for each particular shape of the prism, I can make, I can find actually the angles. Knowing these angles, I can find the location x1, x2 and x3, which will be refracted to the same point. So this is one point. And for each of these, I can actually draw exactly the same thing. I will have alpha 1. From alpha 1 I can calculate my delta 1 deviation of this particular. This is delta 1. Now this is delta 2. This is delta 3. This is delta 4. And from them, I can find the location of how far from the y axis my ray is supposed to go to concentrate in one point. So again, why did they do it? Because this is yet another step towards the curve. Now, what is the curve? It will always be represented as the infinite number of infinitesimal segments which are inscribed into this curve, right? So let's talk about convex lens. Convex lens is basically a body which is represented in its simplest form. So it's one-sided convex. Let's say you have some kind of a curve. So if this is y, this is x. And the curve is representing by some function y of x. Now if I have this, now this is perpendicular and this is some kind of a graph of the function y is equal to y of x. If I will rotate this around the y axis, I will have a solid body with a curved bottom and flat top, right? So how can I actually use this to reflect parallel rays of light into one point? Well, I have to really think about what kind of a curve is this so that these angles and how each angle actually is determined, well, angle is determined very similar to this. If this is a sequence of straight lines, straight segments, then this segment represents basically an angle. Now if I have a curved, obviously I have to go by tangential line. So let me just make a bigger picture and you will see what it is. So whenever light comes to some kind of a curved surface, how is it refracted? How can I use the law of refraction? Well, apparently I can assume that this is actually a very small but straight flat surface and the tangential line to my curve actually represents the direction of this. So I can basically say that in this particular case the refraction happens according to the same laws as if this is a straight line, which means I have to draw a perpendicular and this is angle of incidence and this would be angle of refraction. So that's basically how it is. So now if I'm talking about the convex lens, so let me just go to another picture, convex lens and that's my picture. So again, this is my y-axis, this is my x-axis and this is my lens. Now I'm discussing everything in the two-dimensional case because obviously when this thing is rotating, the symmetry of the whole rotational symmetry of the whole picture actually allows me to consider just one ray of light and how it's refracted to this point. So it's all because of symmetry. I'm considering the cutting plane section of this three-dimensional figure which is drawn by the plane coming through the y-axis and the ray which is parallel. So within this plane everything, basically all the events happens in this plane including the refracted light. So I can consider only the two-dimensional case in this section plane. Now this is PQ. This is my point where the light goes, it goes to point B and then it goes down. So let's say this is point D, this is F, this is... Now I need a tangential line which is RS and perpendicular to E2 line which is MN and this is point C and distance F, this distance is X. Now I will actually use exactly the same logic as before. So this is my incidence angle and as you see this angle is equal to the angle my tangential line forms within X-axis. So if this is theta i and this is alpha, alpha is equal to theta i. Now why? Because this is perpendicular to this and X-axis is perpendicular to the ray of light which is parallel to Y. So these two angles M, D, A and angle which my tangential line makes with X-axis they are exactly the same. So I will use the alpha. Now this one is my refraction from the normal refracted light but I can actually represent it as two angles delta which is deviation C, B, F, deviation from the original direction and this one is also alpha because it's vertical to this one. So I have exactly the same thing and I can use the same equation instead of using indices I can just use without indexes. I can use it for one particular ray. Now let me see now. It will be the same thing actually. So from this particular equation let me just state right up front how can I find angles alpha and delta? Well alpha is relatively simple thing. Alpha is the angle which my tangential line makes with an X-axis which means that tangent alpha is equal to Y first derivative at point X. Right? Because if you remember from the calculus that the tangent of the tangential line angle with X-axis is the first derivative of the function at this particular point. This is X, this is OD is also X, right? So I know alpha. Now knowing alpha I can find my delta from here, right? Delta here. I mean it looks ugly to tell you the truth but that's what it is. So I know delta. Since I know delta and I know alpha I can actually use these two things to find out some kind of equation similar to whatever I used before. Okay, so how can I do it? So angle CBN, this is my refracted angle is alpha plus delta, okay. Now if we will, let's call this point E. Okay, now the OD which is sum of OE plus ED this is my X, right? This is my X. Now I can represent this X as sum of OE and ED, right? So let's find them separately using the angles and distance X, okay? So if I will multiply F by tangent delta now this is delta, this is also delta, right? So if I multiply F by tangent of this angle I will get OE. So OE is F times tangent delta. Now as far as ED is concerned, well let's just think about it. I know this angle is delta, right? So let's consider angle EBD, that's delta. BD I know, this is Y of X, right? This is the value of the function at point X. So if I divide ED by BD I will get a tangent of this angle or if I divide BD by ED I will get a cotangent. So my ED is equal to BD times tangent which is Y of X times tangent of alpha. So this thing X is equal to F plus Y of X times tangent delta. Basically this is it. This is as far as I can actually go right now because right now you're asking what is delta? Well delta is this. What is alpha? Alpha is this. So from this and this I can find delta and from delta I can find and basically this is an equation. Now what kind of an equation? Well first of all it's a differential equation. I mean if you would like to combine whatever I know about the whole thing into one formula that would be ugly again but nevertheless. So this is delta. Alpha is equal to arc tangent of Y first derivative of X, right? If first derivative is tangent of alpha then alpha is our tangent. Now this I should actually put here. So I will have delta is equal to arc sine. You know usually n divided by nr I will put number n. n is a relative index of refraction. This is absolute relative to the vacuum and this is absolute relative to the vacuum. But the ratio is the relative ratio. So arc sine of n times sine of arc tangent of Y of X. 1, 2, 3, 1, 2, 3, 1, 4 minus alpha which is arc tangent of Y of X. And now I have to have tangent of this. Anyway. So if I will put tangent of this substitute you will see that I have only X and Y. And derivative of Y. So that's why it's a differential equation which I don't actually even dare to think about how to solve. So it can be solved numerically using computers and certain algorithms. And it was done in many cases to produce ideal perfect shape of the lens which concentrate all the rays which are falling on it into one point. Now what's interesting is people were thinking about this perfect kind of a lens for a very long time since probably Archimedes and Newton and all the great physicists actually were involved in this. Nobody was quite smart to solve the problem analytically. I mean, somebody like Newton, they didn't really have computers so they could not use the numerical approach. Now we can. And there was actually an interesting problem introduced by Wasserman and Wolf in 1943. What if you have instead of a flat surface on the top and perpendicular rays? This is the simplest thing. But if you have any kind of a surface, the question is how can you make this surface in such a way to correct all these different directions and still concentrate all the rays which are going down into one point? So what's the shape of this surface based on whatever the shape of this surface is? So my particular problem was a particular case when the flat surface is on the top. That's the easiest part. Finally, quite recently, four or five years ago, some physicists from Mexican University actually solved analytically this problem. And they came up with a really very long formula which describes what's the shape of this surface. But if I will write this formula on the board, the board would not be actually big enough to hold it. It's a huge formula. And if you want, you can actually search internet for the problem of Wasserman and Wolf and you will see basically this formula and many articles about how it was derived, etc. So two guys from Mexican University actually, very young guys actually, maybe even students, I don't know, they have derived the whole thing about five years ago. But that was a difficult solution which is definitely outside of the scope of this course. My problem was to introduce you that this can be an ideal surface. None of the simple things like parabola or part of the circle or hyperbola, none of these are perfect. All of them do not have this perfect quality of every parallel light to be concentrated. They will always be somewhere around so it would be a spot, not a point. Now, a little bit more sophisticated, so to speak, convex lens is when you have two surfaces and both are some kind of a convex form. The lens in our eye actually is how it's working. But we will discuss all the different kinds of how the image actually is built in our eye or in some kind of a photo camera. We will discuss it in one of the later lectures. So far, I was just trying to introduce you to a concept of a convex lens and in particular the perfect convex lens which actually concentrates all the parallel lights in one point. All simple like spherical, for instance, convex lines, they are not perfect. So they always have something which is called aberration. Aberration means that the shape is not perfect and spherical is not perfect and parabolical is not perfect. And that's why there is no one particular point but there is a whole section. The further you are, a little bit further will be the point. If this is spherical, for instance, the further it will be, it will be more deviated from one particular point. Okay, that's it for today. I suggest you to read the notes for this lecture. They have much better picture with relatively smooth like a textbook style explanation of all these angles, etc. But in any case, eventually we will come to this huge differential equation which I cannot go any further right now. So thanks very much and good luck.