17. Ray or Geometrical Optics II
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Shankar, Ramamurti
Ray diagrams are used to investigate the behavior of light incident on mirrors and lenses. The principle of least time is used to show that all rays from an object in front of a concave mirror focus on the image point if they are not too far from the axis. The experiments describing the breakdown of geometric optics are discussed.
 Okay, I'm hoping that this interlude into optics will give you a chance to catch up and be on top of the class. Okay? I take that to be yes. So, if you have to give an elevator presentation of geometrical optics, okay, right in the elevator, it's a big shot, you get what, you have to give a chance and you have to tell that person everything, what would you say? Any guess? Right. Principle of least time covers everything. And I thought I would do that because I'm sure you guys, most of you have seen this in one form or another, and I had to do something slightly different, and that was to show you that it all came from least time. Well, I haven't shown the whole thing yet. I did show you that angle of incidence equals angle of reflection. That came out. A Snell's law, that came out. But I was about to do mirrors, and in the case of mirrors, I said, if you want a parallel set of rays from infinity to focus on one point, then what do you need to do? First of all, focusing itself is a very special case. Focusing means many light rays going in different directions all end up in the same place. That means they all must satisfy the principle of least time. How can all of them be the least time? Because they all take the same time. Therefore, there's more than one way for a light beam to go from start to finish, and that's when you get focusing. This is the first example of that. The way it worked is, you start this guy here. First of all, the rays are coming from infinity. If you're going to keep track of how long that takes, you're going to get infinity for everybody. There'd be infinity plus 1 for 1, infinity plus 2 for another one, and we know from the time we have children that they're all the same. So infinity is not a good reference point. So you draw the gate here, the mental gate, and you say, I'm going to track these guys from this point on. I'm going to discard any head start. Then I want to see how long any of them takes. Well, this guy hits that wall. I mean, there's a mirror here. The mirror is locally vertical, so it bounds right back, and this focal point is where I want everyone to end up. So this ray does that, loops around and comes back here. How much time does it have? Well, it has the time to go here, plus the time to go an angle of the circle distance F on the other side. That means this light ray, which began here, has the same amount of time. But for this mirror here, it's got the time to go there. But you want it to hit the mirror and come here in the same time. That means it'll happen only if that distance and that distance are equal. This must be true for any ray parallel to the axis. That means your mirror is a curve with the property that points on it, have the same distance from the focal point and from this line here. And there's a name for that curve, and that is the parabola, and satisfies the equation y squared equals 4xF, where F is the focal length. I derived that for you. Then the next question is, it's not enough to have, but I told you that if you want, you can approximate this by a spherical meter. You take a sphere of radius r and you slice a portion of it, but a sphere is not a parabola because sphere is going to, put a sphere here, is going to start deviating from the parabola. But you can fool this beam of light if you don't go too far from the axis. In that region, it coincides with the parabola. And you say how big a parabola, but it looks like a parabola is related to the radius of that sphere in this fashion, or F is r over 2. That's called a spherical mirror. So whenever you do this, you should be aware of all the limitations. A parabolic mirror will focus the parallel beam of light no matter how wide it is measured from the axis, because I've cooked this up to work for any height. If you approximate it with a spherical mirror, you have to ask yourself how long will this beam of light think that this sphere is a parabola? Well, you can see as you start moving off, it's going to catch on. So you cannot go too far off the axis. So that's an approximation. And for a spherical mirror, if you send a parallel beam of light, if it's too thick in the transverse direction, it won't focus here. It will focus for small thickness, small deviation from the axis. But now our goal in making mirrors is not only to form the mirror of an object at infinity, but you want an object at a finite distance. You want an object at a finite distance to form an image, and that's what we were talking about. So let's go back to that problem. Okay, I'm going to be writing some equations today, and I want to request people in the last row, if you cannot read something, you have to tell me if the font is small. Guys, in the front row, if you cannot read something, if the handwriting is bad, you've got to tell me. People in the last row may not know why they cannot read it. Maybe the handwriting, maybe the font. But between the two of you, you've got to keep me on the lookout for that. Okay, because I did realize last time I was drawing more and more and more lines in the same figure. It's hard to read. All right, so I'm going to write some equations. I'm trying to plan my blackboard strategy here. But first I'll tell you what I'm going to do. I mentioned that to you last time. I'm going to make sure that when you follow the rules of ray optics that you learn in high school, what do you learn in high school? You say, well, take any object here. You draw a bunch of rays, and you know where they're going to intersect because the parallel ray goes to the focal point. That's how it's defined. Here's another ray that's a little interesting. Go through the focal point, it'll come out parallel. How do I know that? I know that because if I run this thing backwards, I'll have a parallel ray going to the focal point and ending up here. If this path was a path of least time, the reversed one should also be a path of least time. So these two meet here. So we say that's where the image is. But that's not enough. I told you two lines will always meet. You've got to draw more lines, and here is one more that I drew. The one that hits the center here, that's i equal to r, very obvious incidence equals reflection, and the mirror is vertical here. I showed you that the third line demands that h1 over u is h2 over v, but that's already true from these two rays. Remember the two rays intersecting here gave the condition 1 over u plus 1 over v is 1 over f, and they gave this condition. So the third ray, which I drew, demands the same thing, so it's consistent. If you really want to nail this down, you want to be able to prove that no matter what height you pick for the interception of the mirror here, you'll take the same time. Because I've only taken three special values for the height y. 1 is y equal to h1, this beam. 1 is y equal to h2, or if you like, minus h2, this beam, and y equal to 0, then you have to take the same value for the y, and they all have to agree. I want to take an arbitrary y and hope that the answer will not depend on y. The time it takes to travel for any y should be the same. And that's where we will find it is not exactly the same. Yes? Student is asked what are the distance from the object? Oh, u is the distance from the object measured from here, and u is the distance from the object measured from here. I don't know what the book calls them. I don't know which book you guys are using, Di and Do and I and O. There's a lot of different symbols you can use. So you're free to use any symbol you want in your problem set, as long as you explain to the TA what you're using. Unfortunately, the only thing everyone agrees on is that F stands for focal length. That's universal. These are random. You have to understand what I'm going to do, because it's going to take some time and it's not worth doing it unless you know why we're doing it. You're trying to verify that the time it takes to go from here to the mirror and over here is the same for every conceivable ray for any height y. And I'm telling you, so what would you like to show? You would like to show the following. You want to make this here. You want to hit that mirror at some height y. Let me make it bigger. So these are all drawn as very big so that you can follow them. But the understanding is they are not really that big compared to the dimensions of the mirror. So it goes there and it does something and it comes here. So this is the point x, y on the parabola that it hits. d1 is the distance, I'm going to use Pythagoras theorem to calculate it, is that distance and d2 is that distance. So d1 depends on y and d2 depends on y. I want to add them up. That's got to be independent of y. That's got to be the same for all y. In particular, it's got to equal the value when y is equal to 0. It's equal to d10 plus d20. That's my reference signal. That's the one that goes here. It hits the center of that and comes here. You guys with me now? It's going to be independent of y, therefore it should be equal the value of any y I like. I'm going to pick the y equal to 0 as my reference. You find the distance d1 plus d2 when y is 0, you move up and down, calculate it and it shouldn't depend on y. So when I do the calculation, what do I really hope will happen? I make a plot of d1 plus d2 as a function of y. I am expecting the answer to be flat, independent of y. And this is the reference point, y equal to 0. So if you do the calculation, first of all, if the graph of d1 plus d2 look like this, that's no good. Because that means if you take y positive, you take a longer time than y negative. Obviously, that y and that y are not paths of least time, because one is clearly taking less time than the other one. So you don't want anything growing linearly in y, you want to be quadratic at best. But then, even if it's quadratic, you go far out, you can see that the answer depends on y. So what I will show you is that the graph, if you plot it as a function of y, if you plot d1 plus d2 as a function of y, it's got some constant, which is the time for y equal to 0, plus some coefficient Ay plus some dy squared plus Cy cube. These are all possible things can happen. If you want a perfect focusing for any value of y, all of these guys have to vanish, because it cannot depend on y. But we will find out if the first two guys have vanished, and there is no guarantee about the others. So, of course, from dimension analysis, when you have a y and a y squared and a y cube, for them to all have the same dimensions, they will be divided by some numbers. So this y will be divided by, let's say, u, this will be divided by u squared, this will be divided by u cubed. It's an expansion in a small parameter, and only when the parameter is small will this work. So we have to notice two things. In a parabolic mirror, an object at infinity will not form a perfect image, no matter how wide the beam is. But an object at a finite distance will not form a perfect image if the object is too tall. If the height y becomes too much, then you don't get a good image. But you get a pretty good image because the possible dependence on y and y squared will be 0, but not to arbitrary powers of y. You understand? If you have a graph that's a power series in y, it begins with y squared, it will look like this. If the first non-zero term is y to the fourth, it will look very flat. But eventually it will move. The flatter this is, further out you can pretend, nothing depends on y. So we want to see how far we can go. That's the purpose. It'll also tell you the limits of a real mirror for a real object. Any questions on the calculation I'm about to do? Okay, before I do the calculation, you're going to need the following approximation. u squared plus theta squared at the power 1 half. You're going to get such combinations all the time. You can write that as u times 1 plus theta squared over u squared to the 1 half. That's exact. Yes? Student is asking, what is r? r in the top r over 2. r stands for the radius of the sphere from which you cut out that piece. Everybody understands that? We had a parabola here and I'm going to approximate it by taking a sphere of some radius r. And that sphere, if you look at this portion here, it looks like a parabola. And you can ask, what parabola does it approximate? I showed you that near the origin it looks like a parabola whose f is equal to r over 2. So if you take a sphere of radius r and you slice a portion of it and you paint it with silver so it becomes a mirror, its focal length will be r over 2. But it won't be a perfect parabolic mirror. It's an approximate parabolic mirror. There are different levels of approximation. The sphere will not focus a parallel beam even if the beam is too wide. Parabola will focus the parallel beam no matter how wide it is. Provided the object is infinity. If the object comes to a finite distance, I'm saying even the parabolic mirror will not give you a perfect image. But if it's very short, very short compared to u and b and so on, it'll do a good job. So let's see how that happens. So I'm going to need this approximation that looks like this. Now this thing, I'm going to write as u times 1 plus theta square over 2u squared plus dot, dot, dot, theta to the fourth, u to the fourth, etc. I'm going to drop all this. This is the approximate expansion. 1 plus x to the n is 1 plus nx. So that becomes equal to u plus theta square over 2u. This is just a mathematical preliminary. I'm getting out of the way so we don't get stuck in that. Yeah? Student is asking if the u is the square root of the parallel equation. It was squared here, but when I multiplied by this u, it got unsquared. So in this calculation, I mentioned to you some numbers are big and some numbers are small. I said that near the end of the class. So the u, v, f are treated as big numbers. Compared to that, the height h1 of the object, height h2 of the image, and y, the height at which the light hits the mirror, they're called small, and the coordinate x on that mirror where you hit it is proportional to y squared. Remember that, so it's proportional to small squared. Remember the parabolic equation is x is equal to y squared over 4f. In fact, let me write it exactly. You can see that f is a number that's called, you can call this order of 1 numbers, these are small numbers, these are smaller numbers squared. So in all these calculations, I think you have no trouble seeing y, x is much smaller than y, but the tip of the parabola is that if you climb a lot of height, you don't move a lot in the x direction because it's quadratic. And the square of a small number is even smaller. So if y is small, x, which is y squared, is even smaller. I want to do this because no matter what field of life you go into, well, not every field, but in most fields where you have to do some kind of calculation, whether it's in economics or biology or whatever, you should know how to do approximate calculations. I'm trying to show you how to do a systematic approximate calculation. Most of the calculation people do are approximate. In fact, every calculation ever done by any physicist is approximate because we're neglecting something. You show me one result that you swear is accurate. It is not. Everything has got corrections. So we should learn how to find the big numbers and the problem and keep them and how to organize our calculation. So let's calculate like a series in small and smaller square and smaller cubed and keep a few terms. So let's calculate now the distance d1 and d2. So you'll have to look at the picture here. So let me just write down what I get. d1 plus d2 is going to look like h1 minus y whole square plus u minus x square. That is d1. And d2 will be h2 plus y square plus v minus x square. Can you see in that figure in the right? In that right angle triangle, maybe I'll bring it down one more time to show you. In fact, this is what we really need. So let's keep it here. This whole thing is h1. This height is y. So this side is h1 minus y. And this side here is u minus x because u brings you from here to the mirror, but this point is to the right by an angle triangle. That's this right triangle. Then there's a second right angle triangle here to find d2. That height is h2 plus y, and that distance is again v minus x. This is the point v. This is the point u. And I want that to be same as when the beam goes to this dotted line here. Now, that's one thing I haven't summed up anywhere. I wasn't summing up the current lines up. So this whole bottle here goes to the right. Now, that looks similar to that one. That proves very similar to aー afflictive. So aー, a, but that's very easy to do pi tyhehe over that one. This dotted square is h1 square plus u square under root, and this dotted square is equal to h2 square plus v square under root. So, the right-hand side does not depend on y. The left-hand side depends on y. In the best of all worlds, somehow all the y dependence will magically cancel, but it's not going to happen. I can tell you right now. What you will find is that the term linear in y will cancel, but the y dependence will be quadratic in y can be made to cancel, and that's all you can do. So, let's now look at these two things. Let me look at the right-hand side. It's very easy to do this one. I already paved the way for this. Remember, h1 is a small number. u is the big number. When you do this one, you will get u plus h1 squared over 2u. You see that? h squared is like theta squared in my formula. It's going to look like v plus h2 squared over 2v. So, the formula I'm using is u squared plus theta squared to the 1 half is equal to u plus theta squared over 2u plus terms involving theta to the fourth and so on, which I'm not keeping. So, the right-hand side is very easy to calculate. It's that. So, let's look at the left-hand side now. Let me open out everything in the bracket here. I get h1 squared plus y squared minus 2h1y plus u squared minus 2ux plus x squared. Now, the second guy is h2 squared plus y squared plus 2h2y plus v squared minus 2vx plus x squared. That's got to be equal to u plus v plus h1 squared over 2u plus h2 squared over 2v. Remember, the right-hand side is not really neat. I've put a dot to tell you it's an approximation. This square would go on forever. I've kept the leading first leading term. The first leading term happens to be quadratic in h. Quadratic in the small quantity. So, to be consistent, I should look at the left-hand side also to the same accuracy. So, let's look at the left-hand side here. This is quadratic in the small guy. Quadratic in the small guy. Small times small, big square, big times small, big times small square but you've got to be a little careful. x is not small. x is small square. Remember that x is proportional to y squared. This is proportional to y to the 4th. You cannot keep a term to 4th order in the small quantity on the left-hand side if you don't keep a similar term on the right-hand side. So, you systematically drop things with your quadratic in the small. So, even though x makes it, x squared doesn't make it in this approximation. Yep? Student is answering the question. Yeah, it can be anywhere it wants. You may think that because it's an under-square root, when you take the root of it, it may become x. Is that your worry? But is that what you're saying? I mean, the point you're making is I see an x squared like that, right? But it's really not x squared, it's really x. That's certainly true. But you're adding a whole bunch of numbers, then taking the square root of that. So, if you take that sum, these numbers are going to be big compared to this one, whether you take the square root or not. Okay? So, it is dominated by the other terms under the same square root. That's why you drop. Okay, so this fellow is out. Here's what I want to do now. I'm going to write the first thing. So, let's decide what to call what. This is my u squared. This is my x squared. This is my theta squared. You understand that? Unfortunately, they're not next to each other, but all of that is theta squared, and this is u squared. Then, the formula tells me, this is equal to u plus h1 square plus y square minus 2h1y minus 2udx y squared plus 2h2y plus, let me see, minus 2vx divided by v is equal to u plus v plus h1 square over 2u plus h2 square over 2v. That's what I want to this order. Now, I have an x everywhere, but x is not independent. x is equal to y squared over 4f. So, what is minus 2ux? Minus 2ux will be minus y squared over 2u over f. Yep? Student is not going to go to the teacher. Yes, thank you. Very good. Okay, therefore, let me do the, so let's compare everything. The big numbers have to cancel the big numbers, otherwise you've got big problems. You guys have a question? Student is asking if the linear error number is equal to 1.5? Oh, yeah, it was to the 1.5, but I've taken this approximation that when you take it to the 1.5, look at the top right. u squared plus theta squared to the power of 1.5 is u plus theta squared over 2u. Okay? Yes, so I've done this a lot of times, but I want you to follow this, because this is telling you how to organize the calculation. So, this is how we organize you. Identify quantities which are quantities of importance, and you can do this in smaller and smaller quantities, and you balance the equation starting with the big numbers, then go to the smaller numbers. So, the big numbers better match. So, that tells you u plus v matches u plus v. That's fine. Now, take the terms proportional to y. They have to separately match. When you have a function of y, then if the function is equal to y, then you have to do this. So, every power of y should match. Of course, we only have two powers here. They should both match. So, take the term linear and y from both sides. Oh, let's get rid of one more stuff. You see that this h1 squared over 2u cancels this one, and this h2 squared over 2v cancels that one. So, I got nothing left. So, everything on the left-hand side should add up to 0. So, you have to linear order in y. This guy is linear in y, right? This guy is linear in y, and that's all that is. So, the coefficient of y looks like y times minus 2h1 over 2u plus 2h2 over 2v, and that's got to vanish. So, that simply says h1 over u is h2 over v. So, this is what you get by usual geometric optics arguments. I showed you last time with similar triangles, but it comes from the principle of least time by demanding that the function, the time doesn't depend on y, the first order in y. Then I have to do the y squared term. You understand? You understand? If you've got numbers on the two sides, big numbers, square or small numbers, then the, I mean, linear in y, quadratic in y, everything has to match order by order. So, this is linear in y term. What is the quadratic in y term? Quadratic in y, look at the coefficients. y squared is coefficient 1 over 2u. You see that? So, this one has got 1 over 2v coefficient of y squared. So, I've got 1 over 2u plus 1 over 2v, but then I've got minus 2ux and minus 2vx. So, what is minus 2ux is minus y squared u over 2f divided by 2u minus 2y squared v by 2f divided by 2v equal to 0. Yes, I'm sorry. I'm not being consistent. So, let me just call this the coefficient of y squared. You agree? It's the coefficient of the y squared. If you want it, it's in front of everything. So, this is the coefficient of this multiplying y squared. So, this has to vanish. Okay, so cancel the 2s everywhere. You get 1 over u plus 1 over v. This cancels. Let me see. There is no 2 here because 2ux is equal to y squared u over 2f. It's not there, sorry. This is all you have. So, you get minus 1 over 2f minus another 1 over 2f equal to 0. Look, guys, we know where we're going, right? So, it's going to happen. It's happened for the last 300 years and I'm not going to change it. So, it'll happen. So, the 2s will all work out. What does this tell you? It tells you 1 over u plus 1 over v is 1 over f. So, you can ask, you know, what's the point of doing this exercise? This tells you that because of the principle of least time, the time it takes to go not only to this two privilege rays, but any ray you can think of drawing from here to the mirror and back here. They all take the same time. The approximation in which the height over u, for example, to the fourth power is neglected. Basically, anything that looks like small over big to the fourth power and higher powers are neglected. So, it's not going to form a perfect image for everything. It'll form a perfect image in a world in which you're able to keep the term to leading order in u to the order h squared over u squared and stop. It's going to be a very approximate thing. So, for it to successfully work, you've got to keep your object and the image all to be much smaller than u and v and f. That's the rule. If you take any parabolic mirror and you have an object which is as tall as the focal length, it won't form a good image. It'll be, what it'll mean then, what do you think it means? Not to form a good image. The rays will cross at different points and you get a blurred image. If you put a candle here, if that's a candle, you can actually put a real screen here and there'll be a glow on the screen, image of the candle upside down. But if the, and it'll be a very sharp image, if the candle's not too tall, once it gets taller, image will get blurred because the rays will not meet at the same point. All right. I want to do one other thing, then I'm going to stop with the principle of least time. That has to do with lenses. You saw how a mirror does a job. Let's take a lens. Let's say we have the following lens. And I have an object which is infinitesimally tall to begin with. And I wanted to form an image somewhere on the other side. So you can ask yourself, how can the light rays, weaving this in different directions, arrive here and all take the same time? It's clear that this guy is the shortest path. The lenses are taking longer, and yet they have to be competitive with the one in the middle. So how do you think that happens? How do you think they will take the same time? Yep? Student's question is the light is lower than the actual mirror lens and that's shorter. That's correct. Do you understand the answer was it is true that this path is shorter than once that says goes to the top of the lens and comes down here, but this is through glass. It's like saying, I have a lake here shaped like this, and I tell you to run so that you go from here to here in the least time. It's not at all clear that heading straight for the lake is the best thing to do because you've got to swim. This person doesn't have to swim at all, but has to go further. You can imagine somehow cooking things up so that they all take longer to survive. So I'm going to do it only for the two extreme rays. You can, since I've done a lot of algebra, I don't want to keep doing it. So let us say this height is h, the height of the lens. This distance is u, this distance is v. And the refractive index is n of the lens. What it means is 1 meter in air, I'm sorry, 1 meter in glass is like n meters in the air. Do you understand in terms of travel time? Everybody with me, that's what it means to say it goes slower. So a meter of lens is worth n meters of air. That's the consequence of the velocity of light. We're trying to find equal times, so when it travels here, it travels with a velocity c over n. Therefore, the time it takes, when you divide the distance by the distance of the lens, will be n times as big. Okay, then there's one other result I'm going to need, so I don't want to interrupt you later, which is the following. I'll take this glass here, I mean this face of the lens, and let's say it is coming from a sphere of radius r. So that's the center of the sphere. So I'm going to draw a normal here and call that distance delta and call this height h. This is all exaggerated. In reality, delta is a very small number. I'm going to need an approximate formula for delta in terms of h and r. So since it's a sphere, that distance is also r. This is a sphere of radius r, from which I've cut out that portion, which is this face of the lens. The lens ends here, but the mental sphere can keep going like that. So what can you say about this distance delta? You can say the following delta is equal to r minus this side. So let's write the formula. The other side, which I don't know, let's say is l, is r minus l. But we also know that l squared plus h squared equal to r squared. Therefore, l squared is equal to r squared minus h squared, and l squared is equal to r squared minus h squared. And by the same trick I did, it is r minus h squared over 2r, the same theta squared formula. So l is shorter than r by an amount h squared over 2r. So this delta is equal to h squared over 2r. So this is the formula I'm going to invoke. You will see why I bothered to prove that in a minute. So you should now think about what equation you will write down. Okay? I'm going to give you a few seconds to think about what two things should be equal for the condition that everything takes the same amount of time to go from start to finish. And since I'm going to find the time it takes to go on that straight line, by the way, since I'm going to divide by velocity everywhere to get time, distance is all you need, except here you've got to remember 1 inch of glasses, n inches of air. If you remember that, you'll be fine. So let's see how long is that path. That path is equal to square root of u squared plus h squared over 2r. Right? By right angle, this right angle triangle, that right angle triangle, very easy to get that. That's going to be equal to the effective length going straight through. So let's look at the effective length in two portions. There's a segment here of size delta and a segment there of size delta in glass. So let's look at the time in the distance in the air first. Distance in the air is u minus delta on the left, plus b minus delta on the right. That's the air distance. Now I've got to take, add to that, 2 delta times n. That's what you have to remember. That's the only non-trivial part of this, is that this distance 2 delta, it should be multiplied by n, because you're traveling at a speed reduced by n, and the distance of time is a distance factor increased by n. So this becomes u plus v plus 2 delta times n minus 1. That's the right hand side. In the left hand side, it's u plus h squared over 2u plus v plus h squared over 2v. So you're doing the same approximation in the quantity h squared over u squared or h squared over v squared. There are higher powers of h over v, which I'm not keeping on either side. Even here, there are higher powers of h over r that I'm not keeping. So get used to the notion, okay? Uv, Fr, big, h, y, etc. are small. So you have to cancel the u and the v. Write the formula for delta. Delta was equal to h squared over 2r times. I'm sorry, did I make a mistake here? I guess not. Okay, so you compare the two sides. You find 1 over u plus 1 over v. In other words, drop these 2s and cancel the h every h squared every where 1 over u plus 1 over v is equal to 2 times n minus 1 over r, and 3 to write as 1 over f, if I like. In fact, that will be 1 over f. How do I know that's 1 over f? Can you tell me? Why should I call this quantity that I get as 1 over f? It is what I got, but my question is, what allows me to identify that? Yes? Student's question is, what is the location of the image and use the location of the object? Remember, v is not the focal point. v is the location of the image and u is the location of the object. But what do you want the focal point to be? What's the definition of the focal point? How is it defined? If you had a lens and I said, what's the focal point? How is it defined? Haven't you guys played with ants when you were kids with a magnifying glass? Come on, what's the focal point? Where a paddled set of rays will converge on the ant. So, if you send a rays from infinity, if u is equal to infinity, v should be the focal point. If u is infinity, you drop this guy, then v, the image should be at the focal point. So, if u is 1 over v, then u is 1 over f. That's what allows you to identify, then, this quantity is 1 over f, because in the limit u goes infinity, 1 over v should equal 1 over f, and therefore, this is what it is. But this is a calculation of the focal length. See, there are two things. One can define focal length operationally by saying take a lens, send a parallel beam, see where it all focuses. That's a focal length. But you can also calculate it from first principles by saying, if a lens whose faces are cut out of spheres, the two faces of radius of curvature are, and the refractive index of the glass is n, then I can actually calculate the focal length in terms of the radius of curvature and n. So, one is the derivation of the focal length in terms of other given parameters. Whereas, a paddled ray's focusing at f is the definition of the focal point. It doesn't tell you how to calculate it for a lens. But this will tell you. If you have got a focal length that is too big or too small, you can vary it by varying either n or by varying r. You can take different materials, or you can take different radii of curvature. Question? Yes? Student is asking, what is the length of the lens? Which one? Student is asking, what is the length of the lens? This little f? Student is asking, what is the length of the lens? You want to know what f means? It means take this lens and send a beam from infinity. That means it comes in parallel rays. That means u is equal to infinity, then 1 over v then becomes just 1 over f. That means v is equal to f. That means the image point, so that f will be here. So that's what it means. I've gotten these formulas. I spent a lot of time deriving them. I'm assuming you have used them a lot in your younger days. I want to give you a few examples, but I don't want to do all of them, because you can do endless calculations involving 1 over u plus 1 over v is 1 over f. It's punching in numbers. I just will tell you a couple of things to bear in mind, then I want to sort of wrap it up. So let's take some examples. Let's take this focusing mirror now. So this is f. This is 2f. So let's see what happens when I start with an object here. You should be able to sort of draw the ray diagrams and just drawing a few of them. So from now on, we quit the Fermat's principle of least time. We just go back to the usual diagramatics. We can't convince ourselves it all follows from the principle of least time. So you don't want to go back to that every time. So what will be the situation here? You need only two rays to draw. So one ray, you draw a horizontal ray, and you know that'll go through the focal point. That's the definition. Then you draw a ray through the focal point, that'll come out horizontal, and again, why is that true? Because if you sent a parallel ray at this point, it will go to the focal point, and if it ends up here, then the reverse is also true. Any ray that's possible, then the time-reversed version of that's also possible, because if one takes the least time to go from A to B, other takes the least time to go from B to A. That's how we draw this, and the intersects somewhere here. In fact, one can show that if you're to the right of 2f, the image will be to the left of 2f. If you're exactly at 2f, you can show the image will also be at 2f. Let's see why. 1 over U plus 1 over B is 1 over f. I'm saying if U is at 2f, and I believe V is at 2f, I get them to add up to 1 over f, so that's correct. 2f is the point where the image and the object cross. See, objects at infinity focus on f. As you move the object to find a distance, the image moves away and they cross when you put it at 2f. Then you can, of course, go closer than 2f, and you can ask, you know, what image will I get? So maybe I'll draw one more picture for that one. So this is closer than 2f, closer than 2f, more distant than f, something here. So draw that ray, draw a parallel, draw that ray, draw that, and they will meet like this. And finally, if the object is at f, the image will be at infinity, because that's just the reverse of the other picture. So all the rays from infinity focus here. If you put an object at the focal point and run the lines backwards, they'll go out parallel, so they'll never meet. So they're trying to meet at infinity. But of course, you can also take an object even closer to the mirror than the distance f. That's what I want to talk about now. What if you go even closer than f? So let's try that. Let's try that. So here it is. Same guy. This is f, and I want to take an object here. So the first ray I can draw is the parallel ray going through that. So when you do the algebra, you can see some problems. 1 over u plus 1 over v is 1 over f. Therefore, 1 over v is 1 over f minus 1 over u. But if u is smaller than f, 1 over u is larger than f, and 1 over f minus 1 over u will be negative. That means v itself will be negative. v is the location of the image, and it's negative, which means on the other side of the mirror, you can ask, how can light go to the other side? It doesn't go to the other side. But you will see that it looks like something is behind the mirror on the other side. So let's see why that happens. You will have some difficulty drawing the second ray. Remember, the second ray, you draw through the focal point, and you say it comes out parallel. If I draw a ray through the focal point, now I've got to go backwards. Obviously, you drew an object, and you drew a line through the focal point heading towards the mirror, but the focal point is behind you. So what's the fate of this? Here is what you do. Continue backwards, and I claim it will come out like this. Why do I know that is correct? Because I know the reverse of that is correct. In other words, a parallel ray going like this will end up at the focal point. Therefore, this is to guide the eye. You want to know what happens with one particular ray. Join it to the focal point, hit the mirror, and come out horizontal. That's how you do the ray diagram. So this is a little delicate thing, but if you really went to the focal point, you never hit the mirror. You just do that to get the direction, but you continue the other way, and you come out parallel. And the logic is, if I draw it backwards, I know that's correct. But now look what's happening. These rays are not focusing anywhere. If this ray seems to come from somewhere in that direction, that ray seems to come from that direction, so the two of them meet here. That means to a person looking from here, from this side with the eye, the two rays, when you extend them backwards, seem to emanate from this point. So that will look like that as a candle behind the mirror emitting. That's a virtual image. So virtual image is when there is no real image, but everything seems to come from a point behind that. Okay, here's another variation. They probably exposed you to a high school. So this is what if I have a mirror like this, and I send some parallel rays into that guy. You agree this was not a focusing lens. It's not a focusing mirror, because you can already see that the light ray will come here, take off like that. That ray will go right back. This will come here, take off like this. So it is not a converging lens. But what is true is that if you extrapolated all these lines backwards, they seem to emanate from this point. You understand? If you look from the other side, if you have a sun and the sun setting parallel rays, the image of the sun will be here, it will be behind the mirror. It will all seem to be coming from here. So instead of the lines actually meeting at one point, they will seem to come from one point. So that's what happens. But how do you do the calculations in this problem? Does anybody know? What's the mirror equation with this problem? Student is asking if you can write this one over v. You say f is negative. It will turn out that everything will work out. 1 over u plus 1 over v is still 1 over f. But if you want, you can write this 1 over mod f, because this will be a problem where the focal length, if that is 5 inches, f should be treated as minus 5 inches. Similarly, a lens which looks like this is a diverging lens. If you send a parallel beam, it will spread out like that. But to a person on the other side, everything will seem to come from a point here. You can do that lens again by using focal length to be negative. So u and v and f are positive for a certain standard amikimos problem, where you have a focusing lens and you have u bigger than f, then v is positive and everything is positive. But you can then move around in this world of parameters where v becomes negative. That means it's a virtual image. u is never negative because you put the object where you like. And you can also have the focal length being negative if the focal length is negative. So instead of being a focusing one, it's a convex one that causes divergence. Or a lens which is causing divergence also is described by f negative. Okay, that's one interesting point. You can take it or leave it. So relax. Don't kill yourself over this, but I thought I should mention it to you, which is the following. Okay? And people tell you it's going to all seem to come from this point. How do we know that's true? You cannot use the principle of least time to argue. Principle of least time says if you have the same starting point, the same ending point, then that can be many ways to reach. Then you have a focusing effect. And here there is no end point where they all come. They don't never come to the same point. They seem to be coming from here, but the rays never meet. So starting from infinity, you cannot naively use the principle of least time. But I'll tell you why this is still going to be the correct answer. Take any mirror surface you like, some part of a mirror. That's the normal to that mirror. Imagine that a light ray does that. It's the silver side of the mirror. And ask you, what if I silver this side and shine light from the other side, what will happen? I claim the answer is that if you continue this ray here and continue that ray there, then that's what light rays come from the other side will do. Why is that? Because this light ray had i equal to r. You know that. But the normal to this surface is the same normal on the other side, and by opposite angles being equal, this is i and this is r. Therefore, in any mirror, quoted one way, if you continue the lines to the other side, they'll still obey i equal to r. Consequently, take the following problem. Take the focusing mirror, quoted like this. We know these lines will come and join here. Then use the result I showed you. That will then show you that the mirror image of that on the other side will also obey i equal to r. Because if you draw the normal here, if these two angles are equal, those two angles will be equal. Therefore, I know this will lead to that and that will match this one. The result of the ray is coming this way. It will go there, but this one continues. We know it goes to F. That's the reason why if you know the fate of one kind of mirror, you can find the fate if you do it on the other face. And this is the construction that tells you why if you shine light from the right, they will all seem to come from this point. That's something optional. You can take it or leave it, but I wanted to tell you where that result comes from. If you just do the algebra, everything will work out if F is negative, but you need to know why you need to know the picture. Okay, so there's one final gadget I want to talk about, and that's the human eye. So the human eye is pretty impressive when you consider the following situation. So here I got the, look, I don't want to even try this. I'm just going to focus on the lens and the human eye. Okay? So here are the eyebrows, and here is my corneas over there, and you put some object here. It goes and forms an inverted image somewhere here. You must know that the image of everything you see is actually physically upside down in your brain. Did you know that? Or you did not know that? But that's a very operationally well-defined thing. So if you look into your cornea, and look at the candles up right here, the candle flame is here, the flame of the candle will be here, like that. But it doesn't seem to bother us, because where do you really see something? It's not very clear. Do you see it in the cornea or do you see it in the brain? That's a one-to-one correlation between what's registered in the cornea and what I run into in my real life. I'm walking around, I bump into the upside down table, and after a while the fact it's upside down, the cornea is not relevant. I know if I take two steps I'm going to hit this object. The brain has learned how to translate that into what you will encounter. In fact, I'm told in some bizarre experiment they put glasses on people so that everything got inverted one more time, and after a few days those guys were just fine. So the eye has got a lot of software. It's not purely hardware but software. I learned it the hard way when I had an eye operation, and the doctor pulled out the stuff and said, Ta-da! And I couldn't see anything. So I was in a panic. My doctor didn't seem particularly worried. And I said, Are you worried? He said, No, no, you'll be fine in a few days. Now I've heard that before, so I didn't believe that lie. But slowly I got better and better. So it's not that my eye was operated anymore. He already fixed it, but the brain began to reprogram it with respect to the new parameters. He changed the parameters of the eye and he reprogrammed it, so the brain had to correlate everything that it saw with everything that's registered in the brain. So you just need a one-to-one map. You can flip it any way you like. You can flip it once or twice, any number of times. After a while you will learn how to use it. But that's not the only impressive thing. If you have an object here, you go to the Lency equation, you want the image to be at that distance, and that's non-negotiable. The distance from your lens to your retina is fixed. So no matter what you have, as u varies, you want v to be fixed to obey this equation. So that's the amazing thing that's going to happen. Yep. Pardon me? You've got to change f. You've got to change the focal length of the lens. And that's the amazing thing about the human lens, is that it's not made out of glass, it's made out of some jelly-like stuff, and there are some muscles pulling it. And if the muscles pull it, it will become longer and thinner. It will have one focal length. And if it doesn't, it will have another focal length. So as I look at an object, which is very far, that's the normal, relaxed state of the eye, as the object comes closer, you take a certain effort to focus on the object to keep reading whatever is coming near you. And that's the process in which your eye is trying to change its focal length. But it cannot do more than some amount. The eye cannot focus with anything that is closer than a distance called the near point. And the near point is 25 centimeters. Well, approximately. This is not a law of nature, but this number is crept into all the textbooks. So 25 centimeters, all you can do. You see, if you didn't have the limitation, you don't need microscopes. You want to see a piece of bacteria just pull the sucker really close to your eye. Why doesn't it work? What we really see with the eye is the following. This is your eye. What the eye cares about is the angle subtended by the object with respect to your eye. We all know that. So if you have the little bacteria guy here, you can see it just as well if you can bring it that closely. That'll look as big as a pencil one meter away, except the eye cannot accommodate that. That's why you need a magnifying lens. And that's the last thing I'm going to tell you, the principle of the magnifying lens. So here is the deal. The eye really cares about the angle subtended by some object, but there is a limit to how close you can bring it. So if you have an object of height h, and you don't have any lens, you just have your eye and you're looking at it, then if that's the height h, and that's the angle theta, the best you can do for theta naught is tan theta naught equal to h over this 25 centimeters. If you can come to 15 centimeters, theta naught will be even bigger, but you cannot. If the object is given height h, the smallest distance is 25, the biggest angle is tan theta equals h over 25, and for small angles, we forget the tan, and if you say theta naught is h over 25. That's the best you can do without a magnifier. So what does a magnifier do? Magnifier, this is the focal length here, you put the object here, somewhere there, and you try to form the image for that one. So the paddled ray goes to the focal point on the other side, then let's see, a paddled ray going to the focal point here, that's where you're going to have to get a virtual image now. Understand another way you can draw is one through the center that goes straight through. If you continue these guys here, you'll find there's an image here, a virtual image. In other words, again, what I'm telling you is the rays will not actually meet, but they will seem to be coming from a point here. So this is h2 and this is v. The theta you're getting is h2 over v, which is h over u. This is your actual height and this is u. So basically what you've done is the following. You're going to have to bring the object closer to your eye than the near point if you want. That's because the actual image seems to be much further where you can see it. So the magnification you get, the angular opening you get with the magnifier is this. So the magnification for angle is theta over theta over v, and that's equal to h over u divided by h over 25. That's 25 times 1 over u. 25 is in centimeters. So you're going to write as 1 over f minus 1 over v. Now you've got to be careful here. v is a negative number because you're forming a virtual image. So I'm going to rewrite this as plus the absolute value of v. So the value of v, or the gain you get from using the microscope, can be written as 25 over f plus 25 over the absolute value of v. v is where the image is. Now where do you want this image to be? Of course, smaller is the v, better your magnification is, but v cannot be any smaller than 25. The best you can do with v is 25, in which case you will get 25 over f as your magnification. But if you're a jeweler working with jewels all the time, and you want to see your piece of diamond, you don't want the image to be at the near point of your eye, because you're really struggling to see it, because that's the best your eye can accommodate. It's okay for a few seconds, but it hurts. Ideally, you want to see the object at infinity. Therefore, typically you pick v to be infinity. That means you put the thing you're looking at right at the focal point. If you put it at the focal point, the image will go to infinity. Now you might say, look, I've never seen an object at infinity. What are you talking about? Right? It turns out that infinity could be 100 meters, or 10 meters is good enough for a lens. It's far enough so that it looks to be infinite. How can you see something at infinity? The answer is, as it goes further out, it also gets taller, so you don't lose anything by moving it further out, as long as it grows proportionately. So you can't guess that distant objects are when the eye muscles have to do no work. In other words, the human eye is designed to focus on object at infinity. It can manage if you bring it closer by those muscles working, but if you don't want the muscles to be tired, you want to have the image at infinity. So you take this little thing you're looking at, and you use the lens to create an image at infinity, and again, forgetting the extra 1, most people don't bother with the extra 1, but the point to remember is the angular magnification is really given by 25 centimeters divided by the focal length of your lens. So if you have a lens of focal length 2.5 centimeters, you get an angular magnification of 10. Most of the time, the magnification you get is 3 or 4 or 5. Any of these things you buy in a supermarket to read something, but you don't know what I'm talking about, but you will in a few years. So that will give you a magnification of 3 or 4. Okay? All right, so this is the end of geometrical optics, and next time I'm going to tell you why you want to change the rules of the game. But before you go, I won't let you go till you tell me why I would want to change anything. Why would anybody change this? The minute you answer me, you go to lunch. What's the reason I would change all this? Yep? Yes, he saved you. There are some experiments that contradict this. That's the only reason to change anything, and that's really what they are next time.
 Okay, I'm hoping that this interlude into optics will give  you a chance to catch up and be on top of the class.  Okay?  I take that to be yes.  So, if you have to give an elevator presentation of  geometrical optics, okay, right in the elevator,  it's a big shot, you get what,  you have to give a chance and you have to tell that person  everything, what would you say?  Any guess?  Right.  Principle of least time covers everything.  And I thought I would do that because I'm sure you guys,  most of you have seen this in one form or another,  and I had to do something slightly different,  and that was to show you that it all came from least time.  Well, I haven't shown the whole thing yet.  I did show you that angle of incidence equals angle of  reflection.  That came out.  A Snell's law, that came out.  But I was about to do mirrors, and in the case of mirrors,  I said, if you want a parallel set of rays from infinity to  focus on one point, then what do you need to do?  First of all, focusing itself is a very special case.  Focusing means many light rays going in different directions  all end up in the same place.  That means they all must satisfy the principle of least  time.  How can all of them be the least time?  Because they all take the same time.  Therefore, there's more than one way for a light beam to go  from start to finish, and that's when you get  focusing.  This is the first example of that.  The way it worked is, you start this guy here.  First of all, the rays are coming from infinity.  If you're going to keep track of how long that takes,  you're going to get infinity for everybody.  There'd be infinity plus 1 for 1, infinity plus 2 for another  one, and we know from the time we have children that they're all  the same.  So infinity is not a good reference point.  So you draw the gate here, the mental gate,  and you say, I'm going to track these guys from this point on.  I'm going to discard any head start.  Then I want to see how long any of them takes.  Well, this guy hits that wall.  I mean, there's a mirror here.  The mirror is locally vertical, so it bounds right back,  and this focal point is where I want everyone to end up.  So this ray does that, loops around and comes back here.  How much time does it have?  Well, it has the time to go here, plus the time to go an  angle of the circle distance F on the other side.  That means this light ray, which began here,  has the same amount of time.  But for this mirror here, it's got the time to go there.  But you want it to hit the mirror and come here in the same  time.  That means it'll happen only if that distance and that distance  are equal.  This must be true for any ray parallel to the axis.  That means your mirror is a curve with the property that  points on it, have the same distance from the  focal point and from this line here.  And there's a name for that curve, and that is the parabola,  and satisfies the equation y squared equals 4xF,  where F is the focal length.  I derived that for you.  Then the next question is, it's not enough to have,  but I told you that if you want, you can approximate this by  a spherical meter.  You take a sphere of radius r and you slice a portion of it,  but a sphere is not a parabola because sphere is going to,  put a sphere here, is going to start deviating from  the parabola.  But you can fool this beam of light if you don't go too far  from the axis.  In that region, it coincides with the parabola.  And you say how big a parabola, but it looks like a parabola  is related to the radius of that sphere in this fashion,  or F is r over 2.  That's called a spherical mirror.  So whenever you do this, you should be aware of all the  limitations.  A parabolic mirror will focus the parallel beam of light no  matter how wide it is measured from the axis,  because I've cooked this up to work for any height.  If you approximate it with a spherical mirror,  you have to ask yourself how long will this beam of light  think that this sphere is a parabola?  Well, you can see as you start moving off,  it's going to catch on.  So you cannot go too far off the axis.  So that's an approximation.  And for a spherical mirror, if you send a parallel beam of  light, if it's too thick in the transverse direction,  it won't focus here.  It will focus for small thickness, small deviation from the  axis.  But now our goal in making mirrors is not only to form the  mirror of an object at infinity, but you want an object at a  finite distance.  You want an object at a finite distance to form an image,  and that's what we were talking about.  So let's go back to that problem.  Okay, I'm going to be writing some equations today,  and I want to request people in the last row,  if you cannot read something, you have to tell me if the  font is small.  Guys, in the front row, if you cannot read something,  if the handwriting is bad, you've got to tell me.  People in the last row may not know why they cannot read it.  Maybe the handwriting, maybe the font.  But between the two of you, you've got to keep me on the  lookout for that.  Okay, because I did realize last time I was drawing more and  more and more lines in the same figure.  It's hard to read.  All right, so I'm going to write some equations.  I'm trying to plan my blackboard strategy here.  But first I'll tell you what I'm going to do.  I mentioned that to you last time.  I'm going to make sure that when you follow the rules of ray  optics that you learn in high school,  what do you learn in high school?  You say, well, take any object here.  You draw a bunch of rays, and you know where they're going  to intersect because the parallel ray goes to the focal point.  That's how it's defined.  Here's another ray that's a little interesting.  Go through the focal point, it'll come out parallel.  How do I know that?  I know that because if I run this thing backwards,  I'll have a parallel ray going to the focal point and ending up  here.  If this path was a path of least time,  the reversed one should also be a path of least time.  So these two meet here.  So we say that's where the image is.  But that's not enough.  I told you two lines will always meet.  You've got to draw more lines, and here is one more that I  drew.  The one that hits the center here, that's i equal to r,  very obvious incidence equals reflection,  and the mirror is vertical here.  I showed you that the third line demands that h1 over u is h2  over v, but that's already true from these two  rays.  Remember the two rays intersecting here gave the  condition 1 over u plus 1 over v is 1 over f,  and they gave this condition.  So the third ray, which I drew,  demands the same thing, so it's consistent.  If you really want to nail this down,  you want to be able to prove that no matter what height you  pick for the interception of the mirror here,  you'll take the same time.  Because I've only taken three special values for the height y.  1 is y equal to h1, this beam.  1 is y equal to h2, or if you like,  minus h2, this beam, and y equal to 0,  then you have to take the same value for the y,  and they all have to agree.  I want to take an arbitrary y and hope that the answer will not  depend on y.  The time it takes to travel for any y should be the same.  And that's where we will find it is not exactly the same.  Yes?  Student is asked what are the distance from the object?  Oh, u is the distance from the object measured from here,  and u is the distance from the object measured from here.  I don't know what the book calls them.  I don't know which book you guys are using,  Di and Do and I and O.  There's a lot of different symbols you can use.  So you're free to use any symbol you want in your problem set,  as long as you explain to the TA what you're using.  Unfortunately, the only thing everyone agrees on is that F  stands for focal length.  That's universal.  These are random.  You have to understand what I'm going to do,  because it's going to take some time and it's not worth doing  it unless you know why we're doing it.  You're trying to verify that the time it takes to go from here  to the mirror and over here is the same for every conceivable  ray for any height y.  And I'm telling you, so what would you like to show?  You would like to show the following.  You want to make this here.  You want to hit that mirror at some height y.  Let me make it bigger.  So these are all drawn as very big so that you can follow them.  But the understanding is they are not really that big compared  to the dimensions of the mirror.  So it goes there and it does something and it comes here.  So this is the point x, y on the parabola that it hits.  d1 is the distance, I'm going to use Pythagoras theorem to  calculate it, is that distance and d2 is that distance.  So d1 depends on y and d2 depends on y.  I want to add them up.  That's got to be independent of y.  That's got to be the same for all y.  In particular, it's got to equal the value when y is equal to  0.  It's equal to d10 plus d20.  That's my reference signal.  That's the one that goes here.  It hits the center of that and comes here.  You guys with me now?  It's going to be independent of y, therefore it should be equal  the value of any y I like.  I'm going to pick the y equal to 0 as my reference.  You find the distance d1 plus d2 when y is 0,  you move up and down, calculate it and it shouldn't  depend on y.  So when I do the calculation, what do I really hope will  happen?  I make a plot of d1 plus d2 as a function of y.  I am expecting the answer to be flat, independent of y.  And this is the reference point, y equal to 0.  So if you do the calculation, first of all,  if the graph of d1 plus d2 look like this,  that's no good.  Because that means if you take y positive,  you take a longer time than y negative.  Obviously, that y and that y are not paths of least time,  because one is clearly taking less time than the other one.  So you don't want anything growing linearly in y,  you want to be quadratic at best.  But then, even if it's quadratic, you go far out,  you can see that the answer depends on y.  So what I will show you is that the graph,  if you plot it as a function of y, if you plot d1 plus d2 as  a function of y, it's got some constant,  which is the time for y equal to 0,  plus some coefficient Ay plus some dy squared plus Cy cube.  These are all possible things can happen.  If you want a perfect focusing for any value of y,  all of these guys have to vanish, because it cannot depend on y.  But we will find out if the first two guys have vanished,  and there is no guarantee about the others.  So, of course, from dimension analysis,  when you have a y and a y squared and a y cube,  for them to all have the same dimensions,  they will be divided by some numbers.  So this y will be divided by, let's say, u,  this will be divided by u squared,  this will be divided by u cubed.  It's an expansion in a small parameter,  and only when the parameter is small will this work.  So we have to notice two things.  In a parabolic mirror, an object at infinity  will not form a perfect image, no matter how wide the beam is.  But an object at a finite distance will not form a perfect  image if the object is too tall.  If the height y becomes too much, then you don't get a good  image.  But you get a pretty good image because the possible  dependence on y and y squared will be 0,  but not to arbitrary powers of y.  You understand?  If you have a graph that's a power series in y,  it begins with y squared, it will look like this.  If the first non-zero term is y to the fourth,  it will look very flat.  But eventually it will move.  The flatter this is, further out you can pretend,  nothing depends on y.  So we want to see how far we can go.  That's the purpose.  It'll also tell you the limits of a real mirror for a real  object.  Any questions on the calculation I'm about to do?  Okay, before I do the calculation, you're going to  need the following approximation.  u squared plus theta squared at the power 1 half.  You're going to get such combinations all the time.  You can write that as u times 1 plus theta squared over u squared  to the 1 half.  That's exact.  Yes?  Student is asking, what is r?  r in the top r over 2.  r stands for the radius of the sphere from which you cut out  that piece.  Everybody understands that?  We had a parabola here and I'm going to approximate it by  taking a sphere of some radius r.  And that sphere, if you look at this portion here,  it looks like a parabola.  And you can ask, what parabola does it approximate?  I showed you that near the origin it looks like a parabola  whose f is equal to r over 2.  So if you take a sphere of radius r and you slice a portion  of it and you paint it with silver so it becomes a mirror,  its focal length will be r over 2.  But it won't be a perfect parabolic mirror.  It's an approximate parabolic mirror.  There are different levels of approximation.  The sphere will not focus a parallel beam even if the  beam is too wide.  Parabola will focus the parallel beam no matter how wide it  is.  Provided the object is infinity.  If the object comes to a finite distance,  I'm saying even the parabolic mirror will not give you a  perfect image.  But if it's very short, very short compared to u and b and  so on, it'll do a good job.  So let's see how that happens.  So I'm going to need this approximation that looks like  this.  Now this thing, I'm going to write as u times  1 plus theta square over 2u squared plus  dot, dot, dot, theta to the fourth,  u to the fourth, etc.  I'm going to drop all this.  This is the approximate expansion.  1 plus x to the n is 1 plus nx.  So that becomes equal to u plus theta square over 2u.  This is just a mathematical preliminary.  I'm getting out of the way so we don't get stuck in that.  Yeah?  Student is asking if the u is the  square root of the parallel equation.  It was squared here, but when I multiplied by this u,  it got unsquared.  So in this calculation, I mentioned to you some numbers  are big and some numbers are small.  I said that near the end of the class.  So the u, v, f are treated as big numbers.  Compared to that, the height h1 of the object,  height h2 of the image, and y, the height at which the light  hits the mirror, they're called small,  and the coordinate x on that mirror where you hit it is  proportional to y squared.  Remember that, so it's proportional to small  squared.  Remember the parabolic equation is x is equal to y squared over  4f.  In fact, let me write it exactly.  You can see that f is a number that's called,  you can call this order of 1 numbers,  these are small numbers, these are smaller numbers squared.  So in all these calculations, I think you have no trouble  seeing y, x is much smaller than y,  but the tip of the parabola is that if you climb a lot of  height, you don't move a lot in the x direction because it's  quadratic.  And the square of a small number is even smaller.  So if y is small, x, which is y squared,  is even smaller.  I want to do this because no matter what field of life you go  into, well, not every field, but in most fields where you have  to do some kind of calculation, whether it's in economics or  biology or whatever, you should know how to do  approximate calculations.  I'm trying to show you how to do a systematic approximate  calculation.  Most of the calculation people do are approximate.  In fact, every calculation ever done by any physicist is  approximate because we're neglecting something.  You show me one result that you swear is accurate.  It is not.  Everything has got corrections.  So we should learn how to find the big numbers and the problem  and keep them and how to organize our calculation.  So let's calculate like a series in small and smaller  square and smaller cubed and keep a few terms.  So let's calculate now the distance d1 and d2.  So you'll have to look at the picture here.  So let me just write down what I get.  d1 plus d2 is going to look like  h1 minus y whole square plus u minus x square.  That is d1.  And d2 will be h2 plus y square plus v minus x square.  Can you see in that figure in the right?  In that right angle triangle, maybe I'll bring it down one  more time to show you.  In fact, this is what we really need.  So let's keep it here.  This whole thing is h1.  This height is y.  So this side is h1 minus y.  And this side here is u minus x because u brings you from here  to the mirror, but this point is to the right by an  angle triangle.  That's this right triangle.  Then there's a second right angle triangle here to find d2.  That height is h2 plus y, and that distance is again v  minus x.  This is the point v.  This is the point u.  And I want that to be same as when the beam goes to this  dotted line here.  Now, that's one thing I haven't summed up anywhere.  I wasn't summing up the current lines up.  So this whole bottle here goes to the right.  Now, that looks similar to that one.  That proves very similar to aー afflictive.  So aー, a, but that's very easy to do pi  tyhehe over that one.  This dotted square is h1 square plus u square under root,  and this dotted square is equal to h2 square plus v square  under root.  So, the right-hand side does not depend on y.  The left-hand side depends on y.  In the best of all worlds, somehow all the y dependence  will magically cancel, but it's not going to happen.  I can tell you right now.  What you will find is that the term linear in y will cancel,  but the y dependence will be quadratic in y can be made to  cancel, and that's all you can do.  So, let's now look at these two things.  Let me look at the right-hand side.  It's very easy to do this one.  I already paved the way for this.  Remember, h1 is a small number.  u is the big number.  When you do this one, you will get u plus h1 squared  over 2u.  You see that?  h squared is like theta squared in my formula.  It's going to look like v plus h2 squared over 2v.  So, the formula I'm using is u squared plus theta squared to  the 1 half is equal to u plus theta squared over 2u plus  terms involving theta to the fourth and so on,  which I'm not keeping.  So, the right-hand side is very easy to calculate.  It's that.  So, let's look at the left-hand side now.  Let me open out everything in the bracket here.  I get h1 squared plus y squared minus 2h1y  plus u squared minus 2ux plus x squared.  Now, the second guy is h2 squared plus y squared  plus 2h2y plus v squared minus 2vx plus x squared.  That's got to be equal to u plus v plus h1 squared over 2u  plus h2 squared over 2v.  Remember, the right-hand side is not really neat.  I've put a dot to tell you it's an approximation.  This square would go on forever.  I've kept the leading first leading term.  The first leading term happens to be quadratic in h.  Quadratic in the small quantity.  So, to be consistent, I should look at the left-hand  side also to the same accuracy.  So, let's look at the left-hand side here.  This is quadratic in the small guy.  Quadratic in the small guy.  Small times small, big square, big times small,  big times small square but you've got to be a little careful.  x is not small.  x is small square.  Remember that x is proportional to y squared.  This is proportional to y to the 4th.  You cannot keep a term to 4th order in the small quantity on  the left-hand side if you don't keep a similar term on the  right-hand side.  So, you systematically drop things with your quadratic in  the small.  So, even though x makes it, x squared doesn't make it in  this approximation.  Yep?  Student is answering the question.  Yeah, it can be anywhere it wants.  You may think that because it's an under-square root,  when you take the root of it, it may become x.  Is that your worry?  But is that what you're saying?  I mean, the point you're making is I see an x squared like that,  right?  But it's really not x squared, it's really x.  That's certainly true.  But you're adding a whole bunch of numbers, then taking the  square root of that.  So, if you take that sum, these numbers are going to be  big compared to this one, whether you take the square  root or not.  Okay?  So, it is dominated by the other terms under the same square  root.  That's why you drop.  Okay, so this fellow is out.  Here's what I want to do now.  I'm going to write the first thing.  So, let's decide what to call what.  This is my u squared.  This is my x squared.  This is my theta squared.  You understand that?  Unfortunately, they're not next to each other,  but all of that is theta squared, and this is u squared.  Then, the formula tells me, this is equal to u plus h1  square plus y square minus 2h1y minus 2udx  y squared plus 2h2y plus, let me see,  minus 2vx divided by v is equal to u plus v  plus h1 square over 2u plus h2 square over 2v.  That's what I want to this order.  Now, I have an x everywhere, but x is not independent.  x is equal to y squared over 4f.  So, what is minus 2ux?  Minus 2ux will be minus y squared over 2u over f.  Yep?  Student is not going to go to the teacher.  Yes, thank you.  Very good.  Okay, therefore, let me do the, so let's compare everything.  The big numbers have to cancel the big numbers,  otherwise you've got big problems.  You guys have a question?  Student is asking if the linear error number is equal to  1.5?  Oh, yeah, it was to the 1.5,  but I've taken this approximation that when you take it to the  1.5, look at the top right.  u squared plus theta squared to the power of 1.5 is u plus  theta squared over 2u.  Okay?  Yes, so I've done this a lot of times, but I want you to follow  this, because this is telling you how to organize the  calculation.  So, this is how we organize you.  Identify quantities which are quantities of importance,  and you can do this in smaller and smaller quantities,  and you balance the equation starting with the big numbers,  then go to the smaller numbers.  So, the big numbers better match.  So, that tells you u plus v matches u plus v.  That's fine.  Now, take the terms proportional to y.  They have to separately match.  When you have a function of y, then if the function is equal  to y, then you have to do this.  So, every power of y should match.  Of course, we only have two powers here.  They should both match.  So, take the term linear and y from both sides.  Oh, let's get rid of one more stuff.  You see that this h1 squared over 2u cancels this one,  and this h2 squared over 2v cancels that one.  So, I got nothing left.  So, everything on the left-hand side should add up to 0.  So, you have to linear order in y.  This guy is linear in y, right?  This guy is linear in y, and that's all that is.  So, the coefficient of y looks like y times minus 2h1 over 2u  plus 2h2 over 2v, and that's got to vanish.  So, that simply says h1 over u is h2 over v.  So, this is what you get by usual geometric optics arguments.  I showed you last time with similar triangles,  but it comes from the principle of least time by demanding that  the function, the time doesn't depend on y,  the first order in y.  Then I have to do the y squared term.  You understand?  You understand?  If you've got numbers on the two sides, big numbers,  square or small numbers, then the, I mean,  linear in y, quadratic in y,  everything has to match order by order.  So, this is linear in y term.  What is the quadratic in y term?  Quadratic in y, look at the coefficients.  y squared is coefficient 1 over 2u.  You see that?  So, this one has got 1 over 2v coefficient of y squared.  So, I've got 1 over 2u plus 1 over 2v,  but then I've got minus 2ux and minus 2vx.  So, what is minus 2ux is minus y squared u over 2f  divided by 2u minus 2y squared v by 2f divided by 2v  equal to 0.  Yes, I'm sorry.  I'm not being consistent.  So, let me just call this the coefficient of y squared.  You agree?  It's the coefficient of the y squared.  If you want it, it's in front of everything.  So, this is the coefficient of this multiplying y squared.  So, this has to vanish.  Okay, so cancel the 2s everywhere.  You get 1 over u plus 1 over v.  This cancels.  Let me see.  There is no 2 here because 2ux is equal to y squared u over  2f.  It's not there, sorry.  This is all you have.  So, you get minus 1 over 2f minus another 1 over 2f  equal to 0.  Look, guys, we know where we're going, right?  So, it's going to happen.  It's happened for the last 300 years and I'm not going to  change it.  So, it'll happen.  So, the 2s will all work out.  What does this tell you?  It tells you 1 over u plus 1 over v is 1 over f.  So, you can ask, you know, what's the point of doing  this exercise?  This tells you that because of the principle of least time,  the time it takes to go not only to this two privilege  rays, but any ray you can think of drawing from here to the  mirror and back here.  They all take the same time.  The approximation in which the height over u,  for example, to the fourth power is neglected.  Basically, anything that looks like small over big to the  fourth power and higher powers are neglected.  So, it's not going to form a perfect image for everything.  It'll form a perfect image in a world in which you're able to  keep the term to leading order in u to the order h squared over u  squared and stop.  It's going to be a very approximate thing.  So, for it to successfully work, you've got to keep your  object and the image all to be much smaller than u and v and  f.  That's the rule.  If you take any parabolic mirror and you have an object  which is as tall as the focal length, it won't form a good  image.  It'll be, what it'll mean then, what do you think it means?  Not to form a good image.  The rays will cross at different points and you get a blurred  image.  If you put a candle here, if that's a candle,  you can actually put a real screen here and there'll be a  glow on the screen, image of the candle upside down.  But if the, and it'll be a very sharp image,  if the candle's not too tall, once it gets taller,  image will get blurred because the rays will not meet at the  same point.  All right.  I want to do one other thing, then I'm going to stop with  the principle of least time.  That has to do with lenses.  You saw how a mirror does a job.  Let's take a lens.  Let's say we have the following lens.  And I have an object which is infinitesimally tall to begin  with.  And I wanted to form an image somewhere on the other side.  So you can ask yourself, how can the light rays,  weaving this in different directions,  arrive here and all take the same time?  It's clear that this guy is the shortest path.  The lenses are taking longer, and yet they have to be  competitive with the one in the middle.  So how do you think that happens?  How do you think they will take the same time?  Yep?  Student's question is the light is lower than the  actual mirror lens and that's shorter.  That's correct.  Do you understand the answer was it is true that this path is  shorter than once that says goes to the top of the lens and  comes down here, but this is through glass.  It's like saying, I have a lake here shaped like this,  and I tell you to run so that you go from here to here in the  least time.  It's not at all clear that heading straight for the lake is  the best thing to do because you've got to swim.  This person doesn't have to swim at all,  but has to go further.  You can imagine somehow cooking things up so that they all take  longer to survive.  So I'm going to do it only for the two extreme rays.  You can, since I've done a lot of algebra,  I don't want to keep doing it.  So let us say this height is h, the height of the lens.  This distance is u, this distance is v.  And the refractive index is n of the lens.  What it means is 1 meter in air, I'm sorry,  1 meter in glass is like n meters in the air.  Do you understand in terms of travel time?  Everybody with me, that's what it means to say it  goes slower.  So a meter of lens is worth n meters of air.  That's the consequence of the velocity of light.  We're trying to find equal times, so when it travels here,  it travels with a velocity c over n.  Therefore, the time it takes, when you divide the distance by  the distance of the lens, will be n times as big.  Okay, then there's one other result I'm going to need,  so I don't want to interrupt you later,  which is the following.  I'll take this glass here, I mean this face of the lens,  and let's say it is coming from a sphere of radius r.  So that's the center of the sphere.  So I'm going to draw a normal here and call that distance  delta and call this height h.  This is all exaggerated.  In reality, delta is a very small number.  I'm going to need an approximate formula for delta in terms of  h and r.  So since it's a sphere, that distance is also r.  This is a sphere of radius r, from which I've cut out that  portion, which is this face of the lens.  The lens ends here, but the mental sphere can keep  going like that.  So what can you say about this distance delta?  You can say the following delta is equal to r minus this side.  So let's write the formula.  The other side, which I don't know, let's say is l,  is r minus l.  But we also know that l squared plus h squared equal to r  squared.  Therefore, l squared is equal to r squared minus h squared,  and l squared is equal to r squared minus h squared.  And by the same trick I did, it is r minus h squared over 2r,  the same theta squared formula.  So l is shorter than r by an amount h squared over 2r.  So this delta is equal to h squared over 2r.  So this is the formula I'm going to invoke.  You will see why I bothered to prove that in a minute.  So you should now think about what equation you will write  down.  Okay?  I'm going to give you a few seconds to think about what two  things should be equal for the condition that everything takes  the same amount of time to go from start to finish.  And since I'm going to find the time it takes to go on that  straight line, by the way, since I'm going to divide by  velocity everywhere to get time, distance is all you need,  except here you've got to remember 1 inch of glasses,  n inches of air.  If you remember that, you'll be fine.  So let's see how long is that path.  That path is equal to square root of u squared plus h squared  over 2r.  Right?  By right angle, this right angle triangle,  that right angle triangle, very easy to get that.  That's going to be equal to the effective length going  straight through.  So let's look at the effective length in two portions.  There's a segment here of size delta and a segment there of  size delta in glass.  So let's look at the time in the distance in the air first.  Distance in the air is u minus delta on the left,  plus b minus delta on the right.  That's the air distance.  Now I've got to take, add to that, 2 delta times n.  That's what you have to remember.  That's the only non-trivial part of this,  is that this distance 2 delta, it should be multiplied by n,  because you're traveling at a speed reduced by n,  and the distance of time is a distance factor increased by n.  So this becomes u plus v plus 2 delta times n minus 1.  That's the right hand side.  In the left hand side, it's u plus h squared over 2u  plus v plus h squared over 2v.  So you're doing the same approximation in the quantity h  squared over u squared or h squared over v squared.  There are higher powers of h over v, which I'm not keeping on  either side.  Even here, there are higher powers of h over r that I'm not  keeping.  So get used to the notion, okay?  Uv, Fr, big, h, y, etc.  are small.  So you have to cancel the u and the v.  Write the formula for delta.  Delta was equal to h squared over 2r times.  I'm sorry, did I make a mistake here?  I guess not.  Okay, so you compare the two sides.  You find 1 over u plus 1 over v.  In other words, drop these 2s and cancel the h  every h squared every where 1 over u plus 1 over v is equal  to 2 times n minus 1 over r,  and 3 to write as 1 over f, if I like.  In fact, that will be 1 over f.  How do I know that's 1 over f?  Can you tell me?  Why should I call this quantity that I get as 1 over f?  It is what I got, but my question is,  what allows me to identify that?  Yes?  Student's question is, what is the  location of the image and use the location of the object?  Remember, v is not the focal point.  v is the location of the image and u is the location of the  object.  But what do you want the focal point to be?  What's the definition of the focal point?  How is it defined?  If you had a lens and I said, what's the focal point?  How is it defined?  Haven't you guys played with ants when you were kids with a  magnifying glass?  Come on, what's the focal point?  Where a paddled set of rays will converge on the ant.  So, if you send a rays from infinity, if u is equal to  infinity, v should be the focal point.  If u is infinity, you drop this guy,  then v, the image should be at the focal point.  So, if u is 1 over v, then u is 1 over f.  That's what allows you to identify, then,  this quantity is 1 over f, because in the limit u goes  infinity, 1 over v should equal 1 over f,  and therefore, this is what it is.  But this is a calculation of the focal length.  See, there are two things.  One can define focal length operationally by saying take a  lens, send a parallel beam, see where it all focuses.  That's a focal length.  But you can also calculate it from first principles by saying,  if a lens whose faces are cut out of spheres,  the two faces of radius of curvature are,  and the refractive index of the glass is n,  then I can actually calculate the focal length in terms of the  radius of curvature and n.  So, one is the derivation of the focal length in terms of  other given parameters.  Whereas, a paddled ray's focusing at f is the definition of  the focal point.  It doesn't tell you how to calculate it for a lens.  But this will tell you.  If you have got a focal length that is too big or too small,  you can vary it by varying either n or by varying r.  You can take different materials, or you can take  different radii of curvature.  Question? Yes?  Student is asking, what is the length of the lens?  Which one?  Student is asking, what is the length of the lens?  This little f?  Student is asking, what is the length of the lens?  You want to know what f means?  It means take this lens and send a beam from infinity.  That means it comes in parallel rays.  That means u is equal to infinity, then 1 over v then  becomes just 1 over f.  That means v is equal to f.  That means the image point, so that f will be here.  So that's what it means.  I've gotten these formulas.  I spent a lot of time deriving them.  I'm assuming you have used them a lot in your younger days.  I want to give you a few examples, but I don't want to do all  of them, because you can do endless calculations involving  1 over u plus 1 over v is 1 over f.  It's punching in numbers.  I just will tell you a couple of things to bear in mind,  then I want to sort of wrap it up.  So let's take some examples.  Let's take this focusing mirror now.  So this is f.  This is 2f.  So let's see what happens when I start with an object here.  You should be able to sort of draw the ray diagrams and  just drawing a few of them.  So from now on, we quit the Fermat's principle of  least time.  We just go back to the usual diagramatics.  We can't convince ourselves it all follows from the principle  of least time.  So you don't want to go back to that every time.  So what will be the situation here?  You need only two rays to draw.  So one ray, you draw a horizontal ray,  and you know that'll go through the focal point.  That's the definition.  Then you draw a ray through the focal point,  that'll come out horizontal, and again, why is that true?  Because if you sent a parallel ray at this point,  it will go to the focal point, and if it ends up here,  then the reverse is also true.  Any ray that's possible, then the time-reversed version  of that's also possible, because if one takes the least  time to go from A to B, other takes the least time to go  from B to A.  That's how we draw this, and the intersects somewhere  here.  In fact, one can show that if you're to the right of 2f,  the image will be to the left of 2f.  If you're exactly at 2f, you can show the image will  also be at 2f.  Let's see why.  1 over U plus 1 over B is 1 over f.  I'm saying if U is at 2f, and I believe V is at 2f,  I get them to add up to 1 over f, so that's correct.  2f is the point where the image and the object cross.  See, objects at infinity focus on f.  As you move the object to find a distance, the image moves away  and they cross when you put it at 2f.  Then you can, of course, go closer than 2f,  and you can ask, you know, what image will I get?  So maybe I'll draw one more picture for that one.  So this is closer than 2f, closer than 2f,  more distant than f, something here.  So draw that ray, draw a parallel,  draw that ray, draw that, and they will meet like this.  And finally, if the object is at f,  the image will be at infinity, because that's just the  reverse of the other picture.  So all the rays from infinity focus here.  If you put an object at the focal point and run the lines  backwards, they'll go out parallel,  so they'll never meet.  So they're trying to meet at infinity.  But of course, you can also take an object even  closer to the mirror than the distance f.  That's what I want to talk about now.  What if you go even closer than f?  So let's try that.  Let's try that.  So here it is.  Same guy.  This is f, and I want to take an object here.  So the first ray I can draw is the parallel ray going  through that.  So when you do the algebra, you can see some problems.  1 over u plus 1 over v is 1 over f.  Therefore, 1 over v is 1 over f minus 1 over u.  But if u is smaller than f, 1 over u is larger than f,  and 1 over f minus 1 over u will be negative.  That means v itself will be negative.  v is the location of the image, and it's negative,  which means on the other side of the mirror,  you can ask, how can light go to the other side?  It doesn't go to the other side.  But you will see that it looks like something is behind the  mirror on the other side.  So let's see why that happens.  You will have some difficulty drawing the second ray.  Remember, the second ray, you draw through the focal point,  and you say it comes out parallel.  If I draw a ray through the focal point,  now I've got to go backwards.  Obviously, you drew an object, and you drew a line through  the focal point heading towards the mirror,  but the focal point is behind you.  So what's the fate of this?  Here is what you do.  Continue backwards, and I claim it will come out like this.  Why do I know that is correct?  Because I know the reverse of that is correct.  In other words, a parallel ray going like this  will end up at the focal point.  Therefore, this is to guide the eye.  You want to know what happens with one particular ray.  Join it to the focal point, hit the mirror,  and come out horizontal.  That's how you do the ray diagram.  So this is a little delicate thing,  but if you really went to the focal point,  you never hit the mirror.  You just do that to get the direction,  but you continue the other way, and you come out parallel.  And the logic is, if I draw it backwards,  I know that's correct.  But now look what's happening.  These rays are not focusing anywhere.  If this ray seems to come from somewhere in that direction,  that ray seems to come from that direction,  so the two of them meet here.  That means to a person looking from here,  from this side with the eye, the two rays,  when you extend them backwards, seem to emanate from this  point.  So that will look like that as a candle behind the mirror  emitting.  That's a virtual image.  So virtual image is when there is no real image,  but everything seems to come from a point behind that.  Okay, here's another variation.  They probably exposed you to a high school.  So this is what if I have a mirror like this,  and I send some parallel rays into that guy.  You agree this was not a focusing lens.  It's not a focusing mirror, because you can already see  that the light ray will come here, take off like that.  That ray will go right back.  This will come here, take off like this.  So it is not a converging lens.  But what is true is that if you extrapolated all these lines  backwards, they seem to emanate from this point.  You understand?  If you look from the other side, if you have a sun and the sun  setting parallel rays, the image of the sun will be  here, it will be behind the mirror.  It will all seem to be coming from here.  So instead of the lines actually meeting at one point,  they will seem to come from one point.  So that's what happens.  But how do you do the calculations in this problem?  Does anybody know?  What's the mirror equation with this problem?  Student is asking if you can write this one over  v.  You say f is negative.  It will turn out that everything will work out.  1 over u plus 1 over v is still 1 over f.  But if you want, you can write this 1 over mod f,  because this will be a problem where the focal length,  if that is 5 inches, f should be treated as  minus 5 inches.  Similarly, a lens which looks like this is a  diverging lens.  If you send a parallel beam, it will spread out like that.  But to a person on the other side, everything will seem to  come from a point here.  You can do that lens again by using focal length to be  negative.  So u and v and f are positive for a certain standard  amikimos problem, where you have a focusing lens and you have  u bigger than f, then v is positive and  everything is positive.  But you can then move around in this world of parameters where  v becomes negative.  That means it's a virtual image.  u is never negative because you put the object where you like.  And you can also have the focal length being negative if the  focal length is negative.  So instead of being a focusing one,  it's a convex one that causes divergence.  Or a lens which is causing divergence also is described by  f negative.  Okay, that's one interesting point.  You can take it or leave it.  So relax.  Don't kill yourself over this, but I thought I should  mention it to you, which is the following.  Okay?  And people tell you it's going to all seem to come from this  point.  How do we know that's true?  You cannot use the principle of least time to argue.  Principle of least time says if you have the same starting point,  the same ending point, then that can be many ways to  reach.  Then you have a focusing effect.  And here there is no end point where they all come.  They don't never come to the same point.  They seem to be coming from here, but the rays never meet.  So starting from infinity, you cannot naively use the  principle of least time.  But I'll tell you why this is still going to be the correct  answer.  Take any mirror surface you like, some part of a mirror.  That's the normal to that mirror.  Imagine that a light ray does that.  It's the silver side of the mirror.  And ask you, what if I silver this side and shine light from  the other side, what will happen?  I claim the answer is that if you continue this ray here and  continue that ray there, then that's what light rays  come from the other side will do.  Why is that?  Because this light ray had i equal to r.  You know that.  But the normal to this surface is the same normal on the other  side, and by opposite angles being equal, this is i and this  is r.  Therefore, in any mirror, quoted one way,  if you continue the lines to the other side,  they'll still obey i equal to r.  Consequently, take the following problem.  Take the focusing mirror, quoted like this.  We know these lines will come and join here.  Then use the result I showed you.  That will then show you that the mirror image of that on the  other side will also obey i equal to r.  Because if you draw the normal here, if these two angles are  equal, those two angles will be equal.  Therefore, I know this will lead to that and that will match  this one.  The result of the ray is coming this way.  It will go there, but this one continues.  We know it goes to F.  That's the reason why if you know the fate of one kind of  mirror, you can find the fate if you do it on the other face.  And this is the construction that tells you why if you shine  light from the right, they will all seem to come  from this point.  That's something optional.  You can take it or leave it, but I wanted to tell you where  that result comes from.  If you just do the algebra, everything will work out if F  is negative, but you need to know why  you need to know the picture.  Okay, so there's one final gadget I want to talk about,  and that's the human eye.  So the human eye is pretty impressive when you consider the  following situation.  So here I got the, look, I don't want to even try this.  I'm just going to focus on the lens and the human eye.  Okay?  So here are the eyebrows, and here is my corneas over there,  and you put some object here.  It goes and forms an inverted image somewhere here.  You must know that the image of everything you see is actually  physically upside down in your brain.  Did you know that?  Or you did not know that?  But that's a very operationally well-defined thing.  So if you look into your cornea, and look at the candles up  right here, the candle flame is here, the flame of the candle  will be here, like that.  But it doesn't seem to bother us, because where do you really see  something?  It's not very clear.  Do you see it in the cornea or do you see it in the brain?  That's a one-to-one correlation between what's registered in  the cornea and what I run into in my real life.  I'm walking around, I bump into the upside down table,  and after a while the fact it's upside down,  the cornea is not relevant.  I know if I take two steps I'm going to hit this object.  The brain has learned how to translate that into what you  will encounter.  In fact, I'm told in some bizarre experiment they put  glasses on people so that everything got inverted one  more time, and after a few days those  guys were just fine.  So the eye has got a lot of software.  It's not purely hardware but software.  I learned it the hard way when I had an eye operation,  and the doctor pulled out the stuff and said,  Ta-da!  And I couldn't see anything.  So I was in a panic.  My doctor didn't seem particularly worried.  And I said, Are you worried?  He said, No, no, you'll be fine in a few days.  Now I've heard that before, so I didn't believe that lie.  But slowly I got better and better.  So it's not that my eye was operated anymore.  He already fixed it, but the brain began to  reprogram it with respect to the new parameters.  He changed the parameters of the eye and he reprogrammed it,  so the brain had to correlate everything that it saw with  everything that's registered in the brain.  So you just need a one-to-one map.  You can flip it any way you like.  You can flip it once or twice, any number of times.  After a while you will learn how to use it.  But that's not the only impressive thing.  If you have an object here, you go to the Lency  equation, you want the image to be at that distance,  and that's non-negotiable.  The distance from your lens to your retina is fixed.  So no matter what you have, as u varies,  you want v to be fixed to obey this equation.  So that's the amazing thing that's going to happen.  Yep.  Pardon me?  You've got to change f.  You've got to change the focal length of the lens.  And that's the amazing thing about the human lens,  is that it's not made out of glass, it's made out of some  jelly-like stuff, and there are some muscles pulling it.  And if the muscles pull it, it will become longer and thinner.  It will have one focal length.  And if it doesn't, it will have another focal length.  So as I look at an object, which is very far,  that's the normal, relaxed state of the eye,  as the object comes closer, you take a certain effort to  focus on the object to keep reading whatever is coming near  you.  And that's the process in which your eye is trying to change  its focal length.  But it cannot do more than some amount.  The eye cannot focus with anything that is closer than a  distance called the near point.  And the near point is 25 centimeters.  Well, approximately.  This is not a law of nature, but this number is crept into  all the textbooks.  So 25 centimeters, all you can do.  You see, if you didn't have the limitation,  you don't need microscopes.  You want to see a piece of bacteria just pull the sucker  really close to your eye.  Why doesn't it work?  What we really see with the eye is the following.  This is your eye.  What the eye cares about is the angle subtended by the object  with respect to your eye.  We all know that.  So if you have the little bacteria guy here,  you can see it just as well if you can bring it that closely.  That'll look as big as a pencil one meter away,  except the eye cannot accommodate that.  That's why you need a magnifying lens.  And that's the last thing I'm going to tell you,  the principle of the magnifying lens.  So here is the deal.  The eye really cares about the angle subtended by some object,  but there is a limit to how close you can bring it.  So if you have an object of height h,  and you don't have any lens, you just have your eye and  you're looking at it, then if that's the height h,  and that's the angle theta, the best you can do for theta  naught is tan theta naught equal to h over this 25  centimeters.  If you can come to 15 centimeters, theta naught will be even  bigger, but you cannot.  If the object is given height h, the smallest distance is 25,  the biggest angle is tan theta equals h over 25,  and for small angles, we forget the tan,  and if you say theta naught is h over 25.  That's the best you can do without a magnifier.  So what does a magnifier do?  Magnifier, this is the focal length here,  you put the object here, somewhere there,  and you try to form the image for that one.  So the paddled ray goes to the focal point on the other side,  then let's see, a paddled ray going to the focal point here,  that's where you're going to have to get a virtual image now.  Understand another way you can draw is one through the center  that goes straight through.  If you continue these guys here, you'll find there's an image  here, a virtual image.  In other words, again, what I'm telling you is the  rays will not actually meet, but they will seem to be coming  from a point here.  So this is h2 and this is v.  The theta you're getting is h2 over v,  which is h over u.  This is your actual height and this is u.  So basically what you've done is the following.  You're going to have to bring the object closer to your eye  than the near point if you want.  That's because the actual image seems to be much further where  you can see it.  So the magnification you get, the angular opening you get  with the magnifier is this.  So the magnification for angle is theta over theta  over v, and that's equal to h over u divided by h over 25.  That's 25 times 1 over u.  25 is in centimeters.  So you're going to write as 1 over f minus 1 over v.  Now you've got to be careful here.  v is a negative number because you're forming a virtual image.  So I'm going to rewrite this as plus the absolute value of v.  So the value of v, or the gain you get from using the  microscope, can be written as 25 over f  plus 25 over the absolute value of v.  v is where the image is.  Now where do you want this image to be?  Of course, smaller is the v, better your magnification is,  but v cannot be any smaller than 25.  The best you can do with v is 25, in which case you will get  25 over f as your magnification.  But if you're a jeweler working with jewels all the time,  and you want to see your piece of diamond,  you don't want the image to be at the near point of your eye,  because you're really struggling to see it,  because that's the best your eye can accommodate.  It's okay for a few seconds, but it hurts.  Ideally, you want to see the object at infinity.  Therefore, typically you pick v to be infinity.  That means you put the thing you're looking at right at the  focal point.  If you put it at the focal point, the image will go to  infinity.  Now you might say, look, I've never seen an object  at infinity.  What are you talking about?  Right?  It turns out that infinity could be 100 meters,  or 10 meters is good enough for a lens.  It's far enough so that it looks to be infinite.  How can you see something at infinity?  The answer is, as it goes further out,  it also gets taller, so you don't lose anything by  moving it further out, as long as it grows proportionately.  So you can't guess that distant objects are when the eye  muscles have to do no work.  In other words, the human eye is designed to focus on object  at infinity.  It can manage if you bring it closer by those muscles working,  but if you don't want the muscles to be tired,  you want to have the image at infinity.  So you take this little thing you're looking at,  and you use the lens to create an image at infinity,  and again, forgetting the extra 1,  most people don't bother with the extra 1,  but the point to remember is the angular magnification is  really given by 25 centimeters divided by the focal length of  your lens.  So if you have a lens of focal length 2.5 centimeters,  you get an angular magnification of 10.  Most of the time, the magnification you get is  3 or 4 or 5.  Any of these things you buy in a supermarket to read something,  but you don't know what I'm talking about,  but you will in a few years.  So that will give you a magnification of 3 or 4.  Okay?  All right, so this is the end of geometrical optics,  and next time I'm going to tell you why you want to change the  rules of the game.  But before you go, I won't let you go till you tell me why I  would want to change anything.  Why would anybody change this?  The minute you answer me, you go to lunch.  What's the reason I would change all this?  Yep?  Yes, he saved you.  There are some experiments that contradict this.  That's the only reason to change anything,  and that's really what they are next time.
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