You are here
PHYS 201: Fundamentals of Physics II
Lecture 17
 Ray or Geometrical Optics II
Overview
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.
Lecture Chapters

0

2057

2617

3837
Transcript  Audio  Low Bandwidth Video  High Bandwidth Video 

htmlFundamentals of Physics IIPHYS 201  Lecture 17  Ray or Geometrical Optics IIChapter 1: Parabolic and Spherical Mirrors [00:00:00]Professor Ramamurti Shankar: All right, 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 a yes. So if you have to give an elevator presentation of geometrical optics, you’re riding in the elevator with some big shot, you get what, 60 seconds and you have to tell that person everything, what would you say? Any guess? Student: [inaudible] Professor Ramamurti Shankar: 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. What I did show you is that angle of incidence = angle of reflection. That came out. Snell’s law, that came out. But I was about to do mirrors and 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, and the way it works is, you start with 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. Maybe infinity + 1 for one and infinity + 2 for another one. And we know from the time we were children that they’re all the same, so infinity’s not a good reference point. So you draw the gate here, a mental gate, and you say, “I’m going to attrack these guys from this point on. No one’s got any head start. Then I want to see how long any of them takes.” Well, this guy hits that wall, I mean, this mirror here. The mirror is locally vertical, so it will bounce 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 equal 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 will happen, only that distance and that distance are equal. And 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^{2} = 4xf, where f is the focal length. I derived that for you. Then the next question is, it’s not enough to have — then I told you that if you want, you can approximate this by a spherical mirror. You take a sphere or radius R and you slice a portion of it. A sphere is not a parabola, because sphere is going to — put a sphere here, it’s 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? Well, it looks like a parabola whose focal length is related to the radius of that sphere in this fashion, where f is R/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 a 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, then you have to ask yourself, “How long will this beam of light think that the 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 image of an object at infinity. We want an object at a finite distance. We want object at 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 because 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 was 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 will tell you what I’m going to do. I mentioned that to you last time. We want to make sure that when you follow the rules of ray optics that you learned 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 through the focal point. That’s how it’s defined. Here’s another ray that’s a little interesting. Go through the focal point, you’ll come out parallel. How do I know that? I know that because if I run this thing backwards, it’s a parallel ray going to the focal point and ending up here. If this path was the 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 = r, very obvious incidence = reflection, because the mirror is vertical here. I showed you that the third line demands that h_{1}/u is h_{2}/v. But that’s already true from these two rays. Remember, the two rays intersecting here gave the condition 1/u + 1/v is 1/f and they gave this condition. So the third ray which I drew demands the same thing, so it’s consistent. But 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 will take the same time. Because I’ve only taken three special values for the height y. One is y = h_{1}, this beam. One is y = h_{2}, or if you like, h_{2}, this beam. And y = 0, the third one, and they all happen 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: What are u and v? Professor Ramamurti Shankar: u is the distance of the object measured from here, and v is the distance of the image measured from there. I don’t know what the book calls them. I don’t know which book you guys are using, d_{i}and d_{o} 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. All right, so do you 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 are doing it. We’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 take this here. You want to hit that mirror at some height y, you 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. d_{1} is the distance — I’m going to use Pythagoras’ Theorem to calculate it — is that distance, and d_{2} is that distance. So d_{1}depends on y and d_{2} depends on y and when I 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 = 0, = d_{1}(0) + d_{2}(0). That’s my reference signal. That’s the one that goes here, hits the center of that and comes here. You guys with me now? It should be independent of y, therefore it should be equal to the value at any y I like. I’m going to pick the y = 0 as my reference. You find the distance d_{1} + d_{2}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? If I plot — if I make a plot of d_{1} + d_{2}as a function of y, I’m expecting the answer to be flat, independent of y. And this is the reference point, y = 0. But when you do the calculation, first of all, if the graph of d_{1} + d_{2}looked 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, so it’s got 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 d_{1} + d_{2}as the function of y, it’s got some constant, which is the time for y = 0 + some coefficient Ay + some By^{2}+ Cy^{3}. These are all possible things that can happen. If you want 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 it’s the first two guys that vanish and there is no guarantee about the others. But of course, from dimension analysis, when you have a y and a y^{2} and a y^{3}, 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^{2}, this will be divided by u^{3}. It’s an expansion in a small parameter, and only when the parameter is small will this work. So you have to notice two things: in a parabolic mirror an object at infinity will 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^{2}will be 0, but not to arbitrary powers of y. Do you understand? If you have a graph that’s a power series in y, if it begins with y^{2} it will look like this. If the first nonzero term is y to the fourth, it will look very flat, but eventually it will move. The flatter this is, the further out you can pretend nothing depends on y. So we want to see how far we can go. That’s the purpose. It will also tell you the limits of a real mirror for a real object. Any questions on calculation I’m about to do? Okay, before I do the calculation, you’re going to need the following approximation: u^{2} + θ^{2}to the power ½. You’re going to get such combinations all the time. You can write that as u times 1 + θ^{2} over u^{2} to the ½. That’s exact. Yes? Student: [inaudible] Professor Ramamurti Shankar: Oh, R in the top? R/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. You can ask, “What parabola does it approximate?” I showed you that near the origin, it looks like a parabola whose f = R/2. So if you take a sphere of radius R and you slice a portion off it and you paint it with silver so it becomes a mirror, its focal length will be R/2. But it won’t be a perfect parabolic mirror. It’s an approximate parabolic mirror. There are different levels of approximation. A sphere will not focus a parallel beam even if the beam is too wide. A parabola will focus the parallel beam no matter how wide it is, provided the object is at 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, short compared to u and v and so on, it will do a good job. So let’s see how that happens. So I’m going to need this approximation that looks like u. now this thing, I’m going to write as u times 1 + θ^{2}over 2u squared + … θ to the fourth, u to the fourth, etc. I’m going to drop all this. This is the approximate expansion, (1 + x)^{n}is 1 + nx. So that becomes = u + θ^{2}over 2u. So let’s bear that in mind. This is just a mathematical preliminary I’m getting out of the way so we don’t get stuck in that. Yes? Student: [inaudible] Professor Ramamurti Shankar: 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. u, v, f are treated as big numbers. Compared to that, the height h_{1} of the object, height h_{2}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^{2}. You remember that? So it’s proportional to small squared. Remember the parabola equation is x = y^{2}/4f. In fact, let me write it exactly. You can see that f is a number that’s called — you can call these order one numbers, these are small numbers, these are smaller numbers squared. So in all these calculations, I think you have no trouble seeing why x is much smaller than y, right? The shape of the parabola is, 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^{2}, is even smaller. So 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 calculations 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 in the problem and keep them and how to organize our calculation, if you like, as a series in small and smaller squared and smaller cubed, and you keep a few terms. So let’s calculate now the distance, d_{1} and d_{2}. So we’ll have to look at the picture here. Let me just write down what I get. d_{1} + d_{2}is going to look like h_{1}  y whole squared + u  x squared. That is d_{1}. And d_{2} will be h_{2} + y squared + v  x squared. Can you see in that figure in the right that the 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 h_{1}. This height is y, so this side is h_{1}  y, and this side here is u  x, because u brings you from here to the mirror, but this point is to the right by an amount x. That’s this right triangle. Then there’s a second right angled triangle here to find d_{2}. That height is h_{2} + y and that distance is again v  x. This is the point v and this is the point u. And I want that to be same as when the beam goes to this dotted line here. That’s the y = 0 answer. That’s very easy to do Pythagoras on that one. This dotted square is h_{1}^{2} + u^{2}under root, and this dotted square = h_{2}^{2} + v^{2}under root. So this whole thing has got to = h_{1}^{2} + u^{2} + h_{2}^{2} + v^{2}. So the right hand side does not depend on y. Left hand side depends on y. In the best of all worlds, somehow all the y dependents 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, the term quadratic and 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 only paved the way for this. Remember, h_{1} is a small number, u is the big number. When you do this one, you will get u + h_{1}^{2} over 2u. You see that? h^{2}is like θ^{2} in my formula. And this will look like v + h_{2}^{2}over 2v. So the formula I’m using is u^{2 }+ θ^{2}to the½ = u + θ^{2}over 2u + terms involving θ 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. So let me open out everything in the bracket here. I get h_{1}^{2} + y^{2}  2h_{1}y + u^{2}  2ux + x^{2}. Now the second guy is h_{2}^{2} + y^{2 }+ 2h_{2}y + v^{2}  2vx + x^{2}. That’s got to equal the u + v + h_{1}^{2 }over 2u + h_{2}^{2}over 2v. Remember, the right hand side is not really — I’ve put a dot to tell you it’s an approximation. This square root goes 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, a big square, big times small, and small squared. But you’ve got to be a little careful. x is not small; x is small squared. Remember that? x is proportional to y^{2}. This is proportional to y to the fourth. You cannot keep a term to fourth 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 which are quartic in the small. So even though x makes it, x^{2}doesn’t make it in this approximation. Yes? Student: [inaudible] Professor Ramamurti Shankar: Yeah, it can be anywhere it wants. You may think that because it’s under square root, when you take the root of it, it may become x. Is that your worry? Is that what you’re saying? Student: [inaudible] Professor Ramamurti Shankar: The point you’re making is, I see an x^{2} like that, right? But it’s really not x^{2}; it’s really x. That’s certainly true. But you’re adding a whole bunch of numbers and then taking the square root of that. In that sum, these numbers are going to be big compared to this one, whether you take the square root r or not. So it is dominated by the terms under the same square root. That’s why you drop it. 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^{2}. The rest of these guys are my θ^{2}. You understand that? Unfortunately, they’re not next to each other, but all of that is θ^{2}, this is u^{2}. Then the formula tells me, this = u + h_{1}^{2} + y^{2}  2h_{1}y  2ux over u + v + h_{2}^{2} + y^{2} + 2h_{2}y + 2vx divided by v = u + v + h_{1}^{2}over 2u + h_{2}^{2}over 2v. This is what I want to this order. Now I have an x everywhere, but x is not independent. x = y^{2}/4f. So what is 2ux. 2ux will be −y^{2} over 2u over f. Yes? Student: Should the denominator for those be 2u and 2v? Professor Ramamurti Shankar: Yes, thank you. Very good. Okay, therefore — 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: [inaudible] Professor Ramamurti Shankar: It was to the½, but I’ve taken this approximation that when you take it to the power ½. Look at the top right. u^{2} + θ^{2} to the power ½ is u + θ^{2} 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. This is how we organize. We identify quantities which are quantities of importance, and then 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 + v matches u + 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 it functions equal on the two sides, every power of y should match. Of course, we only have two powers here and they should both match. So take the term linear in y from both sides. Oh, let’s get it to one more stop. Do you see this h_{1}^{2}/2u canceled this one? And this h_{2}^{2}/2v cancels that one. So I’ve got nothing left. So everything on the left hand side should add up to 0, because there’s nothing on the right hand side. So what do I have to linear order and y? This guy is linear and y, this guy is linear and y and that’s all there is. So the coefficient of y looks like y times 2h_{1} over 2u + 2h_{2} over 2v and that’s got to vanish. And that simply says h_{1}/u is h_{2}/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, to first order and y. Then I have to do the y^{2} term, understand? You understand if you’ve got numbers on the two sides, big numbers, square of small numbers, linear and y, quadratic and y, everything has to match, order by order. So this is a linear and y term. What is the quadratic and y term? Quadratic and y, look at the coefficients. y^{2} is coefficient ½u. You see that? And this one has got 1 over 2v, coefficient of y^{2}. So I’ve got 1 over 2u + 1 over 2v, but then I’ve got 2ux and 2vx. So what is 2ux, is −y^{2} u over 2f divided by 2u  2 y squared v by 2f divided by 2v = 0. Student: [inaudible] Professor Ramamurti Shankar: Yes, I’m sorry. I’m not being consistent. So let me just call this the coefficient of y squared. Do you agree? It’s the coefficient of the y^{2}. If you want, it’s in front of everything, all of this is multiplying y^{2}. So this has to vanish. Okay, so cancel the 2’s everywhere, you get 1/u + 1/v. This cancels. Let me see. There is no 2 here, because 2ux = y^{2} u over 2f. This 2 is not there, sorry. This is all you have. You get 1 over 2f  another 1 over 2f = 0. Look guys, we know where we are 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 will happen. So the 2’s will all work out. What does this tell you? It tells you 1/u + 1/v is 1/f. So you can ask, 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 for these two privileged rays, but any ray you can think of drawing from here to the mirror and back here. They all take the same time, in 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 will form a perfect image in a world in which you are able to keep the term to leading order in u, to the order h^{2}/u^{2}, and stop. This is an 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. What it will mean, what do you think it means not to form a good image? Rays will cross at different points and you will get a blurred image. If you put a screen, 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. It will 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. Chapter 2: Lenses [00:34:17]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 leaving this in different directions arrive here and all take the same time? It’s clear that this guy is the shortest path. These 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? Yes. Student: [inaudible] Professor Ramamurti Shankar: That’s correct. The answer was, it is true that this path is shorter than one that goes to the top of the lens and goes 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 the same time. So I’m going to do it only for the two extreme rays. Since I’ve done a lot of algebra, I don’t want to keep doing it. So let us say this height is h, height of the lens, and this distance is u, this distance is v. And the refractive index is n of the lens. What that means is, 1 meter in air — I’m sorry. 1 meter in glass is like n meters in the air, 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/n, therefore the time it takes, when you divide the distance by time, 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, 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. And draw a normal here and call that distance delta and call this height h. This is all exaggerated. In reality, delta’s 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. Right? 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 = R  this side. So let’s write the formula this side, which I don’t know, let’s say is l, is R  l. But we also know that l^{2} + h^{2 }= R^{2}. Therefore l^{2} = R^{2} − h^{2}, or l is square root of R^{2} − h^{2}, and by the same trick I did, it is R − h^{2} over 2R. The same θ^{2} formula. So l is shorter than R by an amount h^{2}/2R, so this delta = h^{2}/2R. That’s the formula I’m going to invoke. You will see why I bother 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. So first 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 glass is n inches of air. If you remember that, you’ll be fine. So let’s see how long is that path. That path = square root of u^{2} + h^{2} + square root of v^{2} + h^{2}. By right angle, this right angle triangle, that right angle triangle, very easy to get that. And that’s going to = the effective length going straight through. So let’s look at the effective length in two portions. That’s a segment here of size delta and a segment there of size delta in glass. So let’s look at the distance in the air first. Distance in the air is u  delta on the left + v  delta on the right. That’s the air distance. Now I’ve got to 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, is should be multiplied by n, because you’re traveling at a speed reduced by n in terms of time. It’s the distance factor increase by n. So this becomes u + v + 2 delta times n  1. That’s the right hand side. The left hand side, it’s u + h^{2} over 2u + v + h^{2} over 2v. Again, in the same approximation, in the quantity h_{2}/u_{2}, or h_{2}/v_{2}. There are higher powers of h/v which I’m not keeping in either side. Even here, there are higher powers of h/R that I’m not keeping. So get used to the notion, okay? u, v, f, R, big. h, y, etc. are small. So you cancel the u and the v. Write the formula for delta. Delta = h^{2}/2r. I’m sorry, did I make a mistake here? I guess not. Okay. So you compare the two sides, you find 1/u + 1/v. In other words, drop these 2’s and cancel the h^{2} everywhere. And 1/u + 1/v = 2 times n  1 over R, which I’m free to write as 1/f if I like. In fact, that will be 1/f. How do I know that it’s 1/f, can you tell me? Why should I call this quantity that I get as 1/f? It is what I got, but my question is, what allows me to identify that? Student: [inaudible] Chapter 3: Focal Point [00:43:37]Professor Ramamurti Shankar: 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 will you find it? Haven’t you guys played with ants when you were kid with a magnifying glass? Come on, what’s the focal point? Where a parallel set of rays will converge on the ant. So if you send rays from infinity, if u = 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, therefore 1/v is 1/f. That’s what allows you to identify then this quantity as 1/f. Because in the limit, u goes to infinity, 1/v should equal 1/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 the focal length. But you can also calculate it from first principles by saying, if I made a lens whose faces are cut out of spheres, the two phases, of radius of curvature R, 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 a derivation of the focal length in terms of other given parameters, whereas parallel rays 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. So if you’ve 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: In that drawing of the lens, what is the f? Professor Ramamurti Shankar: Which one? Student: In the drawing of the lens, what is the little f? Professor Ramamurti Shankar: This little f? Student: [inaudible] Professor Ramamurti Shankar: You want to know what f means? Student: Yes. Professor Ramamurti Shankar: It means take this lens and send a beam from infinity. That means it comes in parallel rays. That means u = infinity and 1/v then becomes 1/f. That means v = f. v is the image point, so that f will be here. So that’s what it means. Okay, so I’ve gotten these formulas, spent a lot of time deriving them. I’m assuming you have used them a lot in your younger days. I just 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/u + 1/v is 1/f, just punching in numbers. I will just 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. This is f and 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 I’m just drawing you a few of them. So from now on, we quit Fermat’s Principle of Least Time. We just go back to the usual diagrammatics, having convinced 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 will go through the focal point. That’s the definition. Then you draw a ray through the focal point. That will come out horizontal, and again, why is that true? That’s 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 you draw this and they intersect 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. Now 1/u + 1/v is 1/f. I’m saying if u is at 2f and I believe v is at 2f, I get them to add up to 1/f, so that’s correct. So 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, “Now 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 but more distant than f, something here. So draw that ray, draw it 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. If all the rays from infinity focus here, if you put an object at the focal point and run the lines backwards, they’ll go in parallel, so they’ll never meet. 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. So here it is, same guy. This is f. I want to take an object here. So the first ray I can draw is the parallel ray going through that. First of all, when you do the algebra, you can see some problems. 1/u + 1/v is 1/f, therefore 1/v is 1/f − 1/u. But if u is smaller than f, 1/u is larger than f, and 1/f − 1/u will be negative. That means v itself will be negative. So 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. Previously, you drew an object and you drew a line through the focal point heading towards the mirror, where the focal point is behind you. So what’s the fate of this? Here is what you do. Continue it 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. If 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, where if you really went to the focal point, you’d 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. In fact, this ray seems to come from somewhere in that direction. That ray seems to come out in that direction and 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 it will look like there is a candle behind the mirror, emitting. So 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. Here’s another variation they probably exposed you to in high school. That is, what if I have a mirror like this and I sent some parallel rays into that guy. You agreed this was not a focusing lens. It’s not a focusing mirror, because you can already see, this 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 is forming — sending 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’ll 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? Pardon me? Student: [inaudible] Professor Ramamurti Shankar: Yes, you say f is negative. It will turn out that everything will work out, 1/u + 1/v is still 1/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 5 inches, then everything will work out. 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 Mickey Mouse problem where you have a focusing lens and 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 lens, instead of being a focusing one, is a convex one that causes divergence, or a lens which is causing divergence also is described by f negative. Okay, there’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. Suppose I have a lens like this, 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 same starting point and the same ending point, then there can be many ways to reach. Then you have a focusing effect. But here, there is no end point where they all come. They don’t ever come to the same point. They seem to be coming from here, but the rays never meet. So starting from infinity, you cannot naïvely 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. This is the silver side of the mirror. And I 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 coming from the other side will do. Why is that? Because this light ray had i = 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 coated one way, if you continue the lines to the other side, they’ll still obey i = r. Consequently, take the following problem: take the focusing mirror coated 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 = 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. Therefore rays coming this way will go there, but this one continued, we know 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 that’s a valid 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 you’ve got — look, I don’t want to even try this. I’m just going to focus on the lens in the human eye. Here are the eyebrows and here is my cornea 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. That’s a very operationally well defined thing. If I can look into your cornea, I look at the candles upright 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 we see it in the cornea or do you see it in the brain? There’s a 1:1 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. After the while, the fact that it’s upside down in 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, “Tada!” 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 any more. He already fixed it, but the brain began to reprogram it with respect to the new parameters. You change 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 to 1:1 map. You can flip it any way you like. You can flip it once, or twice or 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 Lens equation. You want the image to be at that distance, and that’s not 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 how do you think that’s going to happen? 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. If they relax, 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, it takes 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 anything that is closer than a certain distance called the near point. And the near point is 25 centimeters, well, approximately. This is not a law of nature, but this number has crept into all the textbooks. So 25 centimeters is all you can do. Chapter 4: Magnifying Lenses [01:03:57]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? Because what we really see with the eye is the following: if 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 will look as big as a pencil 1 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, is 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 — 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 θ, the best you can do, put θ_{0}, is tanθ_{0} = h over this 25 centimeters. If you can come to 15 centimeters, θ_{0} will be even bigger, but you cannot. So for an object of given height h, the smallest distance is 25. The biggest angle is tanθ = h/25. And for small angles, we forget the tan and we say θ_{0} is h/25. That’s the best you can do without a magnifier. So what does a magnifier do? A 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 parallel ray goes through the focal point on the other side. Then, let’s see, a parallel ray going to the focal point here, that’s where you’re going to have to get a virtual image now. Understand? Another ray you can draw is one through the center that goes straight through. If you continue these guys here, you will 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 if this is h_{2} and this is v, the θ you are getting is h_{2}/v, which is h/u. This is your actual height and this is u. So basically what you’ve done is the following: you managed 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, the angle of opening you get with the magnifier is this, so the magnification for angle is θ over θ_{0} and that equals h over u divided by h over 25. That’s 25 times 1/u. 25 is in centimeters. That we’re going to write as 1/f − 1/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 + the absolute value of v. Therefore the gain you get from using the microscope can be written as 25 over f + 25 over the absolute value of v. v is where the image is. Now where do you want this image to be? Of course, the smaller 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 1 + 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, image will go to infinity. Now you might say, “Look, I’ve never seen an object at infinity. What are you talking about?” 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 farther out, it also gets taller, so you don’t lose anything by moving it farther out, as long as it grows proportionately. It is just that distant objects are when the eye muscles have to do no work. In other words, the human eye is designed to focus on objects 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. So the point to remember is the angle of magnification is really given by 25 centimeters divided by the focal length of your lens. So if you have a lens, a focal length 2.5 centimeters, you get an angle or 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 — you don’t know what I’m talking about, but you will in a few years — then that will give you magnification of 3 or 4. 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 until 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? Yes? Student: [inaudible] Professor Ramamurti Shankar: Yes, he saved you. There are some experiments that contradict this. That’s the only reason to change anything, and I’ll tell you what they are next time. [end of transcript] Back to Top 
mp3  mov [100MB]  mov [500MB] 