 Okay. Very good. Thank you very much. Good morning. Okay, so today I'm going to continue talking about Fourier transforms and how they connect to the way waves propagate and then doubt how that relates to things like resolution of images. So just to remind you, in 1D, if I have a function, some function f of x as a function of x, what Fourier theory tells us is that this function is the same as something that is pretty boring plus something that oscillates slowly. Of course, there's a lot of intermediate steps, a continuum of n plus something that oscillates faster, etc. So any function can be constructed by adding things that oscillate at different frequencies. That's what Fourier theory tells us. And they can be cosines or sines or combinations that are these exponentials. In 2D, we have the same thing. Before I go to 2D, what do you think that would happen, say, if these things, for example, these parts start to oscillate very fast? If I block them, what would happen? What would happen to this function? It would be truncated? You miss the sharp detail. Because these are the things that are very fast, they are the ones that contribute to things that are very fast here. Well, these contribute to things that are very slowly. So the fine detail is lost here. And notice that to filter out these fastly oscillating things, one thing you could do is take this, break it into these by doing a Fourier transform, then applying a filter that is multiplying by something, and then inverse Fourier transforming. What is another name for that, which is Fourier transform, multiply, inverse Fourier transform? That's another way you can do that without doing Fourier transforms. You can do a convolution, a blurring. So a blurring would have the same effect of removing some of the frequencies. So this is in one dimension, in two dimensions is the same thing. But now rather than plotting this like this, I'm going to plot the function as some distribution here. And now this is equal to something that is pretty boring, plus something that, also let's say like this, some frequency plus dot plus something that oscillates much faster, etc. But it can be not only in this direction, it can also be in this direction. So something that oscillates like this, sinusoidal, plus something that oscillates much faster. But I can also do in other directions, for example, I don't know something like this, etc. So we have a continuum of possible directions of oscillation and frequencies of oscillation. And the same thing, if we block some of these, now it's a two-dimensional space, we get different things. So to illustrate that, let me use this Mathematica notebook. So these two gentlemen here are Augustin Fresnel, the guy who proposed the mathematics, first proposed rigorously the mathematics of wave propagation of light as a wave. And this is Mr. Fourier. And let me take Mr. Fresnel. And here what I'm showing is on the right is Fresnel, on the left is the Fourier transform of Fresnel. And you see that this looks like a star. There's a lot of detail here, a lot of structure, and then there's these two lines. Can anyone tell me why there's two lines there? It looks like a star. Well, it turns out that here I'm using white as one, let's say, black as zero. And what I had to do to do the Fourier transforms numerically, because I'm going to use those numerical Fourier transforms, the fast Fourier transform that is good for numerics, and I had to do what is called pad with zeros. I'm going to show this in more detail in another animation. So I put a bunch of zeros to the sides. So there's a big square that is the boundary of this picture where it's almost one and then zero. And we have a big boundary between black and white here. Those abrupt changes along lines generate a lot of structure in the Fourier transform in the perpendicular direction. So this edge here gives rise to this line here, like this edge and this edge. Well, this edge and this edge give rise to these lines. So for example, there's not many features that Mr. Fresnel has, but this part of his face probably has to do with a tiny little line that is here in this direction. So the main features about Fresnel himself is all these dots around here. This is an artifact from the square window that I'm using. I'm going to simulate what happens if I were to do a Fourier transform, then block what is outside of a given circle or region, an inverse Fourier transform. So if you look on the left, if I click here, often you cannot see this all the way because there's a circle. I'm multiplying by a circle and it looks more or less the same. If I keep bringing this down, so now it's all in a circle here. Did you notice something? He's starting to get blurry, doesn't he? Exactly. And I'll make an analogy with that. It's like you're not wearing your glasses. The more I close it, the more blurry he gets and until it is really, really almost indistinguishable. So there we see that what's near here gives us the rough parts, the general features while what's out here is the detail. And that's a general property of Fourier transform. I'm going to show you a couple of versions of this. Let's see if this one works better. It's calculating something. And here I'm going to show both Mr. Fresnel and Mr. Fourier. So I have a slider here and if I'm here, Fourier is on this side, Fresnel is on this side. If I go all the way to the other side, they switch. What I'm going to do is I'm going to mix on each side. So on the Fourier space, I'm going to put a circle. And outside of that circle, I'm going to put the Fourier transform of one of them and inside the part of the Fourier transform of the other one. So that at some point, if I'm in the extreme, this is when this circle is very big, so on each side I only have the Fourier transform of one and the other one is on the extreme when it's very small. But if I go somewhere in between, this still looks a lot like one of them. But if I start going down here, at some point something interesting happens. I think this is probably it. So I don't know if you remember who they are, but who's this guy? Fourier, Fourier is the guy with the not so thin guy and Fresnel is the thin guy. Let me come near where you are. So everyone says Fourier is on the left and Fresnel is on the right. Let me go a tiny little bit. Still the same, isn't it? Now squint, do this. We're doing a low pass filter. No, they're not switching yet. At some point, when you do that, they switch. Ah, no, I had to go up, sorry. So now, if you close your eyes a little bit, then all of a sudden this is Fourier and this is Fresnel. Isn't it? Even afford to do a tiny little bit more. I went too far. So why is that? How can we do that change? That when we see it like this, it's one, but when we do this or remove our glasses, it switches. What's happening there? Another way of doing this is let me, if I come here and I grab this, I cannot do that. I'll show you another version of this, but why do we see one of them when we are looking it clearly and then when we look at it without our glasses or doing this, it switches? What happens when we do this? We are blurring things. So we're filtering out the high frequencies. And since the high frequency information contains information about one of them, that is a fine detail that dominates our brain when we see it, but when we remove it, then all we have is the other one. So I'll show you another version of this later that works a bit better. Let's do it now. So do you know this famous musician? So who, what's his name? It's Professor Niemela, who gave the inauguration of this school. So he's a saxophone player. I'm sure you will see him play saxophone in a couple of weeks. So I'm going to use poor Joe as my experiment. So just to illustrate how I use the Fourier transforms, here is Joe. I just got a square around him, and I'm doing the zero padding that I was saying. So I put a lot of zeros around. And here I define the Fourier transform. So I define these commands rearrange and inverse rearrange that puts things in the right order for me to do the numerical Fourier transform. And then I define, because Fourier in Mathematica doesn't do that, I define my version of Fourier as doing this, your arrangement, doing the Fourier transform, and then fixing it again. So just to show how that rearrangement works, if I apply that to poor Joe, I break it into four pieces and reorganize the way that I need it to do the Fourier transform. So that's Joe's Fourier transform. And again, it's dominated by this line and this line because of the frame. Yes, so the discrete Fourier transform, let me come back here. So when we do the discrete Fourier transforms, that's exactly here, that's good. This numerical approximation to the Fourier transform goes from zero to n minus one is a sum. And we have to make it, so once the first value, the central value of the origin, to be the first point. And then it keeps going and then to the left you have to remove it and put it on the right so that you do the Fourier transform. So in order to do numerically an approximation of Fourier transform, n is the number of samples. Yes, yes. So what happens is that to do the Fourier transform, you have your sample function here and you want this to be at the center, in the screen, to be the main, the central point. So you have to break this into two parts and you have to, here it is. You have to take all this half and move it to the other side. So instead of having this function, it looks something like this. So this is the center, here, then you have some zeroes and then the other parts. That's something you have to do to approximate the Fourier transform with a numerical algorithm. That's in one dimension, in two dimensions, let me skip here. You have a function like this and again you have to break it like this, you have to break it like this and then this quadrant has to be here, this quadrant has to be here and like that. So something you have to do, you do that rearrangement, then you Fourier transform with the command and then you undo this arrangement, you put it back together. If you're using MATLAF, it does it for you. I think if you give it the right command. But it's good to know that that happens. That that is a necessary step to do the numerical approximation of the Fourier transform. So here I can do the same thing that did before. I can calculate, I can put a filter around the Fourier transform of Joe to filter different parts of different levels of detail. And so if I remove more and more high frequencies, you start getting at this. And this is very similar as we will see what happens in an imaging system when we have a finite pupil. But when we're using coherent illumination, if this were some sample or something that we were illuminating with a laser and we're using that laser light as the light for imaging because there's full correlation between all the lights coming from different parts, then they interfere. And as a result of interference, we see these fringes here. So these are not curtains. These are interference that we're seeing because of coherent effects. So if I, sorry, they are curtains. But we can see some interference effects, some fringes that form. If we're doing in coherent imaging that we will talk about later, those are not there. So I'm going to show the same effect. And I needed another person playing the saxophone, so I found John Coltrane on the web, a famous saxophone player. And again, we can mix Joe with John Coltrane. So who's that? Coltrane or Joe Niemela? If you look at the face, it looks more like Coltrane, especially if I make this big. But if I make it small, it starts becoming Niemela. So that's Niemela, Niemela, Niemela. Then all of a sudden it starts being Coltrane. So why when I make it small, does it become Joe? What's happening now? Sorry? It's low pass filter because we have our pupils that are, and when I scale it down, the high frequencies become much higher, and they start not being perceptible by us. The high frequencies that have John Coltrane's information are not making it, but I have to make it this very big so that we can start seeing John Coltrane's face. So high frequencies have the fine detail, and low frequencies have the general structure. So we discuss the Fourier structure. Whenever we apply it to a system that's linear shift invariant, the Fourier space has a meaning, and in the case of optics, we know what that is, so that has to do with plane wave propagation. So this is the axis of propagation, Z. This is the transverse axis, X, and I have some initial distribution here. I can break that this way into different waves, so into different frequency components, so I can have a frequency component like this. But now this has a meaning because this, for a given frequency component, is a slice of a plane wave propagating in some direction. The period of the plane wave is something fixed, let's say lambda is this much, so I need to have some sort of plane wave with wave fronts at this distance, such that if I put it here and I rotate it, it fits with that. So it has to go through this point, and then this point, and this point, the maximum how to be here and here and here. Well, I can see if I take that, put it here, and I rotate it, also it's going to click when, let's say, this is traveling in this direction. At this angle, even though this is fixed, lambda, when I do it at that angle, this spacing corresponds to the period of that spatial frequency. On the other hand, if I had something that goes more like this, faster, would correspond to what angle? Higher or smaller? More going more that way or going more that way? The distance is fixed, and I have to take that, put it here and rotate it until it clicks. So there's a maximum here, a maximum here, a maximum here. Is it going at a larger angle or a smaller angle? It's larger. Yeah, remember, larger because this is the axis of propagation, so it has to be something like that. Or if I had something like this, that looks exactly the same. Let me just draw the points here. So if I have something like this, then what angle is that going? The propagation. So the propagation is perpendicular to the wave fronts. So where these are more space is going at this angle. When this is less space, it's more rapid. I have to tilt this more so that these wave fronts fit with these oscillations. This is going very fast, so I have to tilt it a lot. It's almost going perpendicular to this. So let me now turn this around. If I had something that goes directly in that direction, what is the frequency of oscillation here? How fast does this oscillate? Or how slow? If I want a wave that is not going like this, but it's going like that, it doesn't oscillate because it's going like this, the wave fronts are like that, and these never cross this point. So this corresponds to the DC component, no oscillation. So we can see that these frequencies now are each mapped with a different angle of oscillation. Yes. Is that clear? Any questions? One more. What happens if my things are spaced like this? So I have a frequency that is so fast that the spacing between these things is very, very close. The spacing is closer than lambda. At what angle am I propagating? Like this, but if I put this like this, if I put it like this and I put these wave fronts like this, it's not even fine enough to match those points. So what are we talking about then? Those evanescent waves. So when we had this simulation here, that's precisely what I was trying to show if we come here. Let's say I do this. I'm going to show you a different version. So this is the same thing. So this curve that I'm drawing here is the red curve that we see here, and that's the wave, so that's the z direction. But if I tilt it somehow, when it's going at some angle, then this intersection is oscillating slowly. If I tilt it more, it's going at a large angle now, and it's oscillating faster, and then it's going at a very, very large angle. It's like this case here, so it's oscillating much faster. But then if we force it to oscillate faster than that, it cannot do it as a traveling wave. It has to become an evanescent wave that decays. So in this case, it's going in this direction, but it's also decaying exponentially in this direction. And the faster it oscillates, the faster it decays. It's just a case very quick. So those are the waves that just cannot carry any information. And the same if I go in the other direction. So if I go now in the other direction. So now there's a natural meaning for the Fourier transform. Which is, it's just a dry riser. So I have this Fourier variable. I called it nu, and it's just 1 over lambda, use of x, use of y. Where these are the coordinates of a unit vector that tells me in what direction I'm propagating. So if I take my, the blackboard is not big enough to use my arms. I'm going to have to use my pen. So if I put here, I have a circle like this. So this is the Fourier domain. Let's say this is a nu of x, nu of y. Then I can see that if I have, I do the Fourier transform and I have a point here, this point, whatever that is, corresponds to something physical, which is a wave that is traveling in what direction from with respect to here. Well, I just put this pen here and I put it such that the shadow, the projection of the tip is falling into that point. And that tells me the direction in which that component of the light is going. So this corresponds to the light going that way, while this corresponds to the light going in that way. Where should I draw the point of something going in that direction? You tell me when I'm right, yeah, there, something like this. This would, the projection of this would go in this direction. So in this Fourier space, all the interior of the circle would correspond to what? The waves that travel far away, because I can always take my unit vector and put it in a position where the projection this back here gives me a direction of propagation. What is this point here outside? Well, there's nothing I can do with my pen to have it, it's difficult here, because I would need to do the z component of my pen be imaginary and I cannot do that with my pen. So if I could somehow rotate it such that this part of my pen were imaginary, then I could stretch this all the way and that would give me an evanescent wave that is traveling in this direction and it's decaying as I propagate away. So everything outside of this circle is evanescent waves, everything inside of the circle is traveling waves. So in the Fourier domain, there's a clear meaning. We go from, we go to a representation where each point tells you a direction of propagation, particularly if we're inside of the circle. So there's a clear meaning. The Fourier space, did that make sense? So Fourier again has a meaning. So as we say in sounds, because our ears are more or less linear shift invariant, frequency, the Fourier space frequency is stone, something we perceive. In paraxial optics, monochromatic optics propagating, not paraxial, sorry, monochromatic optics, because also free space as some medium of propagation is linear and shift invariant if I move is the same, then the Fourier space has to have a meaning and the meaning is directions of propagation. So this tells me this plane wave going in that direction, this plane wave going in this direction. So let's go back and think about the uncertainty relation that we discussed before. So what happens if I focus a beam? So I have a beam, it's focused like this. It's going in that direction. And if I look at this slice here, let's say, so it's going in the C direction, X direction. So the cross-section here is something like this, some maximum. What is delta X in this case with a point of view of the uncertainty relation? It tells me about what's width. It tells me about the width of the intensity on this. So it tells me about this width here. Delta X is how small that dot is. But what about delta nu? Delta nu, nu now is in one direction, let's say nu X. Let me say nu X is u X divided by lambda. So this is the same as delta X. Delta u sub X divided by lambda. But what is u sub X here? u is a vector that tells you the direction of propagation. So the width in u is the width in direction. So I have a beam that has all a bunch of direction of propagation, directions of propagation, but it's sort of constrained to a cone. So you can think of as having this angle here, let me call this delta theta, and really what I have is, what is the relation between theta and u sub X? I have a sign of the angle, because if I have this going in this direction, so u is 0, I'm here, the angle is 0, and I go to 1 when the angle is 90 degrees. So I have a sign, a projection onto the screen is a sign. So u of X in this sense is the sign of the angle. So this is delta X, delta sine of X, sine theta, sorry, divided by lambda. And if we're in the paraxial approximation where angles are small, sine of theta and theta are more or less the same. So the uncertainty principle tells me that this is greater or equal than 1 over 4 pi. So therefore delta X times delta sine theta is greater than or equal than lambda divided by 4 pi. So what is the meaning of this? Resolution, once more. It's another way of seeing resolution, because the smaller I make my dot, the bigger I have to make the angle, the cone angle of the beam. Conversely, if I want to do laser communications, for example, if I want to send a signal far away, what do I want as small? Which one of these two things I want to be small? The angle. You want a very, very collimated beam. So I need this. What do I need to do for that? Because of this inequality, if I want to make this very small, sorry, here, if I want to make this very small, I have to make this large, because the product cannot be smaller than that. So I have to make my beam big. There is a balance. You cannot have a very thin beam that is very narrow. You just can't. So for a given angle, you can have a minimum size here, and that has to do with resolution as we will see. So if I'm illuminating a spot and I want to make it as small as possible, what are the only two degrees of freedom that I have? So I want this to be small, so what can I do? I can make this big, or I can make this small. I cannot change for pi, so I have to change lambda or this. So I can increase the angle, a bigger cone of light, as we say a bigger numerical aperture, or I can go to smaller wavelengths, and that's why people change from CDs to Blu-ray. That lets you just make using a smaller lamp. That makes sense? Of course, when I make this angle bigger, another thing that you do is there's another characteristic which is the depth of focus, so how this is really a blob here of light. For some applications, we want this to be as small, and this also to be as small as possible, because that's the longitudinal resolution if I want to do 3D image. So for some applications, this is very good to have a very narrow depth of focus. For others, it is not. For example, in photography, sometimes it's good, sometimes it's bad. If I want to take a photo of you guys that is very sharp, I use a camera with a big lens. But then one or a row of you is going to be very sharp, and the people in the row behind and the row in front are going to be blurry. And sometimes that's nice, like if you take a photo of a crowd and you want to highlight the person, like there's a bicycle race or something, and there's the famous racer in the middle, and you want to focus on that person, everyone else is blurry, then that's good. But if you want a photograph that's where everyone looks good, then that's bad. So there's a trade-off between having a very good resolution laterally and a very good resolution longitudinal. Yes, it's a depth of focus. So sometimes you want a long depth of focus, sometimes you want a very short one. Because when you're imaging, for example, a 3D biological sample, you might want to go into the sample and image different depths. And there, if there's some stuff here, because the beam is so extended, you don't see it, especially if you're using non-linear techniques where you focus a lot of light to excite the matter there to emit light in a different wavelength, then you really want to concentrate the intense regions to as small as possible region here. I knew some people that had a project of designing barcode scanners, for example, and that's an interesting trade-off where this plays a role. Because if you... So you have your barcode tags, and there's separation, and you have a laser that scans this, and you want the laser to fit into the dark and light so that it can reach the barcode. So that means that for your laser, at the focal point, you want the spot to be as narrow as possible so you can read the barcode. But if your barcode is going to be used not only in the supermarket, but it's for storing things like in a warehouse or something, then the person in the warehouse is walking with a reader, a gun, and they just point it at the tag and reading. But often you have stacks of boxes. So there's a box here with its tag, then it's one with the tag, then one with the tag. So you don't want to get a letter every time to go and read all the books. You just want to point your gun and get the reading. So that means that while you want delta x to be small, you also want delta theta to be small because you don't want your beam to spread too quickly. So that is the type of design situation where you're fighting against this or you're trying to look for the compromise that lets you over a region of distances have this be good enough while at all distances not only at the center. So because these correspond to plane waves going in different directions, and we have the plane wave e to the i, let me write it as k u dot r. I can write this as e to the i k plus u y y and then I separate the part in c e to the i k u z times z. What we did last time is say, okay, if I separate this way, this looks like the part that I need to define the Fourier transform. This is what matches these oscillations in the initial fields. And this part is how this accumulates an extra phase when I go, say, if I had one of these like this. So initially it reproduces this oscillation here, but as I go far away, this oscillation, if I do another slice, it's going to look the same, but it's shifted. That shift comes from a factor of the phase that I accumulated as propagating from this distance to this distance. And this is precisely the transfer function. So we do the Fourier transform, then we see how much longitudinal phase we accumulated, and then we reconstruct it. That's how we modeled the Fourier transform. Where this u sub c has to be defined as the square root of 1 minus what? Remember, the vector u satisfies this property. It's a unit vector. So this is u x squared plus u y squared plus u z squared. So u z is equal to 1 minus, and we use this when u x squared plus u y squared is less or equal to 1. Otherwise we use i. If this is bigger than this, then we have to use u x squared plus u y squared minus 1 if u x squared plus u y squared greater than 1. So this thing here is then the famous transfer function. If I come back to Joe, where's Joe? Ah, before I do that, let me just mention something. The very question is, am I going too fast? Yes, Anna. Okay, okay. Yeah, that's right. So one important lesson is every scientist you talk to has their own notation. So we have to be fluent in see past the names of the letters and go to the meaning of the letters. Here I have a dream. So in this Fourier space, then this center is the propagating wave. So these are the parts of the signal when we do the Fourier transform that manage to travel far away. Well, all this stuff outside is the stuff with the very high detail that doesn't manage to travel far away because it's a venison waves. So even free space is a low pass filter when we're propagating because the high frequencies are lost. Now, in many cases when we have, for example, a laser pointer or something, suppose that I have a laser pointer going directly into the C direction. The Fourier transform over the transverse plane of that laser pointer that is very directional lives where in this Fourier space. Close to the center. It only has components that are going over a very small range of angles. So it's a very directional thing. So this small region here is called the paraxial region. It's the region of very small angles because I'm only using plane waves that have a very small angle with respect to the C axis, which is coming out of the board. So the paraxial region, we say that Ux and Uy, let's say, much less than 1. And in that case, we can approximate Uz, which is the square root of 1 minus Ux squared minus Uy squared as 1. What is the next correction to this? When we have, let's say, something like 1 minus something very small, square root, anyone remember what? So this is very small. If it's extremely small, then 1 is good. But we need a little bit more. And that's the paraxial approximation. So we just approximate this as that. So the last formula we wrote last time then is if we have the field at x, y, z, how can I model the propagation of this field? I have to go to the Fourier space. So I'm going to start from the left. I have to do the Fourier transform of the field at the plane 0 x, y. When x goes to nu x, y goes to nu y, then what do I do to that? You multiply it by the transfer function. And then what else do I do? The inverse Fourier transform. Where this transfer function is precisely this e to the i k c, this u of c that I mentioned before. And it can be the non-paraxial or the paraxial version. And that's it. That gives us the propagation of the field. So this is the signal and this is the response. So this is the standard linear system where to model propagation, I can do a Fourier transform, multiply by something, inverse Fourier transform. So I'm going to do that here with Joe. I'm going to use the paraxial approximation just for simplicity, but it's just as easy to do the x-axis. So what I have here is I define something that I call a chirp, which is just a table of some of the squares of nu x and nu y. And then the transfer function is e to the i 2 pi times the chirp. Because I had to divide by lambda, I put 2 pi squared. It's 2 pi all squared divided by 2, so it's 2 pi squared. So then Joe propagation is this. So I'm taking Joe Fourier transform, multiplication by the transfer function, inverse Fourier transform. Initially c is 0. So what happens to the transfer function? It's 1. So I'm taking the Fourier transform and the inverse Fourier transform. Mod squared, I recover Joe. But as we start moving away, we see that things start to form. And we get something very... Well, it would look almost exactly like we would do if we had a laser that is nicely collimated. We printed a transparency of Joe, illuminated that transparency, and then put our detector at different distances. The light would diffract like this. So it's Fresnel diffraction. Fresnel diffraction is what happens when you go very far away. And in that case, what you see is just the Fourier transform, which is what Professor Consortini is going to show you in the lab. If you go very, very far, and the logic for this is the following. So if I have c and x, as we said, I have some object. I can break it into waves that are going in different directions. So let's say this one here, and another one that is going here, and another one that is going here, and a continuum. I mix all these waves in such a way that they reproduce what we want here. But as they propagate, they mix in different ways, creating different patterns, which is what we're seeing here. If we go very, very, very far away, you can sort of imagine that at this point, only this wave got there. And at this point, only this wave got there, etc. So very far, we just separate the different Fourier components, and essentially we see the Fourier transform. And what happens here very far is precisely this Frank-Hoffer diffraction, where we see essentially the mod square, the amplitude square of the Fourier transform. But near to the object, we're more into the Fresnel regime, where we see different mixtures of the different waves given different behaviors. So I don't let this slide very far, but if I could keep going, at some point this would become more or less the Fourier transform. Makes sense? So modeling propagation is very easily numerically. You just do a Fourier transform into dimensions, multiplication by a simple thing, inverse Fourier transform. That's it. Now, when we're talking about the systems we had, so if we have a system and I have a signal and a response, I said, way to model this is the Fourier transform then, so I have S tilde, then here the response is just the multiplication of S tilde with the transfer function, and then inverse Fourier transform. This is a diagram that we drew in one of the previous cases. So this is a linear shift invariant system, like free space or many others. But there's a direct way to go from here to here without doing Fourier transforms. So that's the impulse response, which is just a convolution. So I can go here saying that the response is the convolution of the signal with a point spread function. Where the point spread function, the impulse response, I'm getting ahead of myself, the impulse response is related to the transfer function. It's the inverse Fourier transform of the transfer function. It's just the inverse Fourier transform of the transfer function. Well, the same must be true here. The impulse response is the inverse Fourier transform of this transfer function. So even in free propagation, we have that the impulse response of x, y, z is the inverse Fourier transform of nu x going to x, nu y going to y of e to the i k c, this use of c. Where use of c is given by this expression. Now this calculation happens to have a closed form expression that is very hard to get. So most courses just avoid it because it's very difficult to get it. But there's a way to do it. And the result happens to be very intuitive. It gives minus 1 over 2 pi a spherical wave e to the i k square of x squared plus y squared plus c squared divided by x squared plus y squared plus c squared. What is the name of this thing here? A spherical wave. And actually we get the derivative in c of that as it turns out. So you can expand this. It has two terms and oftentimes one is much bigger than the other and you drop that one. But this is the exact result. And there's that this z derivative of the spherical wave is very much like the spherical wave itself times an obliquity factor. And what this tells you is that in this impulse response regime if I have propagation from a field here each point in this field is behaving like a spherical wave emanating from there. So this is here. So if I want to see the field here, this point would get blurred because what was coming out of this point is now a wave that is very extended. Then what is coming out of this point is also getting blurred because it's another very extended wave. What's coming out of this point is another very extended wave. So we have to blur each point in the object with a spherical wave that is coming out from that point. And that's what causes that blurring. That's the image to be like that. What is the name that we always associate with this idea? Huygens. The Huygens principle that each point in the object becomes maybe thought of as a source of a spherical wave coming with it. It's not as active as a spherical wave, it has this extra thing but it's like a spherical wave. So this Huygens principle is fully consistent with this image. This is the impulse response. And the point spread function which is this phase propagation of the plane wave is the transfer function. So when we think of propagation in terms of Fourier transforms we're naturally thinking about plane waves. When we're thinking of propagation in terms of convolutions we're naturally thinking about spherical like waves. And there's a mathematical connection between the two. Okay, yes. No, this one is exact and this one is the hard one. Thank you. If you use the paraxial approximation then it's easy because it's all about integrals with Gaussians and what you get is the paraxial approximation of this one which is where you also expand this square root, this derivative you only keep one term and this one actually you just approximate a C, the one downstairs. So the paraxial approximation is much easier and that's what gives the Fresnel formula. If you use the exact, actually this is called the Rayleigh-Sommerfeld 1 formula. There are two Rayleigh-Sommerfeld formulas. The other one is sort of useless but Rayleigh-Sommerfeld 1 is this one with this propagator and it's exact. The paraxial approximation when you use this before taking this inverse Fourier transform, that's easy and that gives you the Fresnel propagation formula. Okay, so in the last half hour I'm going to leave, stop thinking only about free space and thinking about what happens when you have a more complicated system. Any more questions about this? So what happens if I'm propagating through a system that also has lenses? How would you model this? Yeah, so we know that a lens does a Fourier transform and that's how I'm going to do it for the interest of time now but rigorously what you would do is just start with your field here. You say use the Fourier transform angular spectrum plane wave propagation that we describe. So Fourier transform multiplication by transform function, inverse Fourier transform to take you here. Then you have the field and you know that a lens gives you a face so it doesn't change the amplitude much but it transmits a face to the field. So then once you calculate the field there you multiply it by the face of the lens, the quadratic face. Then you Fourier transform again, you propagate all the way here. You convert back to the field. Then you multiply it by this new face and then you propagate again. So you're sort of jumping between real space and Fourier space. Fourier space is very good for propagating. Real space is very good for applying masks like a lens or an aperture or something. And this is... Well, the extreme of this is what is called a split step method where you go jumping from one to the other. When you're doing, for example, nonlinear propagation where some effects are easy to describe in Fourier space some in real space, you have to be jumping from one to the other. You go to one, you multiply by something, you come to the other when you multiply by something. So free space propagation is easy to go to Fourier space, multiply by transform function and come back. And then a lens is better to just stay with the field and multiply it by a face function. However, that's not what we're going to do. So suppose that I have a system where this is a focal length, so let me use my curly F and a focal length here and then I have another one here and another one here. If I just look at the first one, I know that the properties of lenses from ray optics is that if I come out from this point on the other side of the lens all the rays are going to have what? Parallel rays. And the direction of these rays depends on the direction of this or on the position. If I take this, all the rays are going to come in some direction. If I come here, all the rays are going to go in this direction. So a lens is sort of translating positions into directions which is like doing a Fourier transform because the Fourier transform of position is direction. When you do a Fourier transform of the transfer position of a field, you get a distribution in directions. So going from this plane to this plane, you're sort of exchanging what is position and what is direction. The lens is doing that for you because also if I had directions here, if I have a bunch of rays with the same direction, they are going to go to the same position. So there's this exchange between position and direction and that's why it is a Fourier transforming object. So when I start from here, I can think of a system that forms images because as this so-called 4F system because from a point we go to directions and this takes directions back into positions. And any imaging system is a complicated version of this but this is the canonical picture for thinking of an imaging system if you want. And usually because lenses are not perfect, in this intermediate space we have something called the pupil, something that limits what light gets through the system and what light doesn't. So if I have some object here and I'm going to measure something here, essentially what I have to think, of course this is going to be upside down but if I were to reverse my coordinates on this one, let's say, I can think of this as doing a Fourier transform, multiplication by an aperture and inverse Fourier transform. And roughly that is what any optical system is doing. Of course this focal distance here in practice is going to be very different than this focal distance here and that's why we get magnification and all these things. But just to understand the effects of our solution we can simply understand it through this. So the aperture here is very important because it limits what? It limits the aberration which is going to be the next thing we're going to be discussing but suppose there are no aberrations. The spectrum, if we have the object here, here we have the Fourier transform, it's like this case over here, did I remove it? So if I limit that pupil, the image starts getting worse and worse because remember what I did here is precisely take the Fourier transform, put an aperture that only lets some of the light go through and then inverse Fourier transform. So the resolution of my image is going to be seriously affected by this. Now the key geometrical argument is here what range of directions here would make it through the system? Because different directions here, as we said, correspond to different spatial frequencies, to different fine or short detail, yes? No frequencies. Low frequencies we'll get but can you tell me which one is the limiting one? Yeah, so this one I would say is probably not going to get through the system. This one is probably going to get through the system. There's a geometric way in which you can tell me exactly which one is the last one that is going to get through the system. Yes, and because we're at the focal plane here, let me start from here and walk backwards. So I start with this ray and this is going to bend in what direction? So that's the direction, the limiting direction. Everything going here is going to go through. Everything here is not going to get through. So I can clearly see that there's this cone here that is the one that gets through and that gives me the range of frequencies of the object that are going to be transmitted onto the other hand. What is the name of this, the sign of this angle? It's the numerical aperture and that's going to determine the resolution of the system because once we get this way, our images are not going to be a delta function, they're going to be some pattern of this type and it's a calculation that we did already. So if I have a point here and I have a point here and this forms another image, then sometimes we can distinguish those and sometimes we can't and that sets the resolution. So we can see the point spread function on the detector which is what we measure, but it's sometimes illustrative to think of the point spread function here on the object which is sort of, if I were to reverse things, the size of the blob of light here, the very intense blob of light that if I were to limit this region I would have because by seeing how much that overlaps with other one that gives me the limit of fundamental resolution. That is independent of how much I'm magnifying. So by just looking at this cone of waves that are coming here and how focused they are, that tells me this voxel, if you will, of light is very small or very big and how much it overlaps with the neighbors and that sets up the resolution. I cannot form independent images of something here or something here if their volumes coincide. So we're going to do the calculation of this and this gives us the resolution. So if I flip this and I have a lens and I have this focus field here, how do I calculate this way? So I have all the plane waves are transmitted so I have an integral and I have e to the i k u dot r d ux d uy. So all these plane waves, each one of these arrows is really a plane wave and they're all making it here but this integral should be limited to what? Remember, the u's, ux and uy are the components of the unit vector telling me the direction of the wave. So it's a unit vector telling me the direction is the x component, the y component is out of the board, the c component is the longitudinal one but what I want is that the maximum angle they can form is the numerical aperture. So I need ux squared plus uy squared to be what? Bigger than something or smaller than something? Smaller than what? That's if we would have all a numerical aperture of one. Let me make it easier. I put a square root. This is the na. So the numerical aperture is like the sign of the angle. It's like the transverse component of this. Assuming that the refractive index is one. If the refractive index is not one, then we have to scale things. So this is when we, if we happen to put all waves here such that they're all in phase at this point and they're all in equal amounts, I have to solve this integral and this will tell me as a function of space the field amplitude for this blob of light. This integral cannot be solved in closed form unfortunately. Except, but we can do the transverse slice and the longitudinal slice. And those are good enough to get an idea of what happens. So let's do that. So this then is e to the i q ux times x plus u y times y d u x d u y. Given that the region is circular, do you think it's a good idea to use Cartesian components? Here I'm doing in terms of u x and u y. We should go to polar. It's better to go to polar. So if I say u x comma u y is equal to some u transverse. Let's say times cosine of theta of phi, sorry, u transverse sine phi. Then this becomes the integral of e to the i k. And same with x comma y. Let me say this is cosine and we're using theta already. So let me use, I don't know, curly phi, rho, sine, curly phi. Then what is this dot product? This times this plus this times this gives me u transverse, rho, then cosine times cosine plus sine times sine, cosine of phi minus phi. The phi, the u transverse, the u transverse. And the integral in phi goes from where to where? Zero to 2 pi because this is the angle, it's not this angle, it's the angle around here. And this one here goes from zero to DNA. What is this integral here? A Bessel function, we solved that yesterday? I'm losing track of time now. So this is 2 pi integral from zero to nA, the Bessel function of whatever is sitting here, k, u transverse, rho, u transverse, p, u transverse. And then we use a property of Bessel functions when we solve this. So yesterday we used the method of using properties, today we're going to use the method of laziness and asking Mathematica. So this also works. Not always, Mathematica makes mistakes. I've discovered a few recently. It's too small. Integrate Bessel J for zero comma k, u comma from zero, integrate that in u from zero to nA. This is probably correct but it's not what I want. Why is this doing this? What if I just say u? So are there any mathematicians here? Good, I can tell you a joke. There's a group of physicists, optics people like us that we come all to the winter college and then we go on excursions somewhere in the field. And it turns out that they put us in a hot air balloon and we go flying and all that, but then the wind blows us somewhere in the middle of nowhere and the balloon comes down and we're lost. And there's no one and then we see someone walking and say, excuse me sir, can you tell us where we are? And the guy is carrying like books and has a hat and looks at us and you are in the basket of a balloon. And they say, you're a mathematician, aren't you? Like, yes, how did you know? You gave us an answer that is absolutely correct and completely useless. So Mathematica in this case gave me an answer that is absolutely correct and completely useless. Well, what this gives you, I'll just jump to the result. J1 is proportional, I might miss the constants. J1, K, Na times rho divided by, and this is the famous, normally what we see is not this, but what? Because this function can be positive or negative. This is the field. So what we see is the mod square of this. So what we would see in the detector, if I have the lens, the aperture, the lens here, what I would see is something like this, which is precisely the mod of J1, K, Na, rho divided by K. So in fact, so the plot of that is one of the blue curves here. Now, suppose that we have two objects and each one is forming its point-first function, but they're separated by some amount. Just look at the plot actually on the right. And suppose that each one is being illuminated by a source that is completely unrelated. So the light is, the face of the light coming from one of the objects is not related to the other one. Then what would, should we add the field contribution of each one of them and then take the modulo square, or should we add the intensities of each one? Yeah, you add the intensities because really we should add the fields. But the field of one in time might be going out of phase with the field of the other quickly. So when I add the two fields, then I take the modulo square, I'm going to have the intensity of this, the intensity of this, and a cross term. And that cross term is going to change a lot in time. Something's going to be positive, something's going to be negative, and it's going to average out to zero, so we don't see it. So for incoherent imaging, when we're illuminating with incoherent light or when we have something like fluorescence that the emission from one molecule is incoherent, is not related to the light from the other, then what we should do is add the intensities of these three patterns. And the Rayleigh criterion tells us that we're going to be able to see the sum, something like this, if I see this black curve, which is the sum of those two, I can know that there are two things. And maybe I can even do better than that. Even if they're a bit closer, I can see that there are two things, but it's not so clear, especially when you have noise and pixelation and all these things, it gets difficult. If they're very close, definitely it looks like just one black curve, but if they're very far away, then it's very clear that there are two things. So the Rayleigh criterion is that when the separation is such that the argument of this is 0.609, which means that the maximum of this one coincides with the first zero of this one. Go back to that case, if I just run this again. So here the maximum of this one coincides with the first zero of this one, then it's clear that you can separate them. Closer, it's a bit dangerous, further away, it's safe. So it's not like it works, it works, it works, it doesn't work. It's gradual, but you have to put a boundary somewhere and Rayleigh chose 0.609. Rayleigh or Abbey, they both did it in slightly different ways. So this is the limit on resolution. Now, if the two objects were illuminated by the same laser, for example, and their faces were strongly correlated, then we would have more situations like this on the right. So I have this one and this one, now I add the fields and take the intensity. And there's a, that's worse. You can see that that cross term is hurting us because this is actually, we don't see the two humps here, we just see one. Certainly it's very difficult. Assuming that the two are at certain distance or situations where the face coming from both of them is the same. Now, it could be that the face between them is slightly different and that's what I'm changing here. If I had the face, one with respect to the other, then actually if they're completely out of face, it's better than the incoherent case. This black curve now is, goes even down to 0. So we can really tell very clearly there are two things. So when we change the face of one with respect to the other one, the incoherent case, which we have on the right, doesn't care. It's always the same. But the coherent case goes from being quite bad to quite good. And in fact, if I average all possibilities, that's how I get the figure on the right. It's sort of the average of all possibilities of in-face and out-of-face. Yes. Yeah, the Rayleigh criterion. So we're right at the Rayleigh criterion here. It tells me that there's a separation at which I can say, oh yeah, I can tell there are two. And that's when the separation between, so the separation, x2 minus x1 is greater than 0.609. And this would be divided by k, correct? So k is 2 pi divided by lambda. So this is something in terms of lambda. So the bigger DNA, the smaller this number so that the more I can keep these things separated. This is thought for incoherent illumination. So incoherent. So in the time of Rayleigh and Abbey, because they didn't have lasers, all light was incoherent illumination. So they never thought about coherent illumination. They only thought about incoherent illumination. So they only thought about the case on the right. But now that we have microscopes that can use coherent illumination, we need to care about the two cases. But this is just thought for the incoherent illumination. If we have coherent illumination and we manage to engineer the phase between neighboring things to be opposite, then we can do better, much better. But we cannot be sure. We can also do much worse if they're in phase. And this boundary where this is exactly 6.9 is the one that I'm showing there as a default. So at the moment, that separation corresponds exactly to the boundary between, is the Rayleigh criterion closer, breaks the Rayleigh criterion further away, satisfies the Rayleigh criterion. So as we separate this more, clearly that's better. So as Umberto was saying yesterday, so what I would see, this is just a slice, but I would see a blob and another blob. And they look like this and they overlap. If I take this image and I take it to the photocopy and make it bigger, well, I'm going to get blob and another blob. They're not going to look any more different than this. You cannot get better resolution by just making things bigger because you're also magnifying your point spread functions. So what matters is the size of your point spread function. If you increase your NA, for example, then instead of having, the central points are the same, but instead of having that blob, you might have this blob and this blob. And that's quite good. So that's how we increase resolution by increasing the NA or by doing what to the wavelength, decreasing the wavelength. And those are the only two things that let us make our focus smaller. Okay, we've run out of time. Next time we'll do the other integral, which is what happens longitudinal. Okay? We meet at 11. Does everyone know about what lab you're going to this afternoon? Okay. And that's in the next room here. Yeah? Group three. Number two. I am three. Okay. Okay. Now the key question is, do you all know whether you're one, two, or three? Okay. Good. All right. So for the ones that are coming with me, I don't have a plan. I do this every year. I just tell you what do you want to do and we'll try to do it. So anything that we've learned in any of the lectures here, we'll try to implement it in mathematics. And we'll all learn together. Okay. See you at 11.