 Suppose that we have a field like something like this linear phase and this all goes then to a given point here. So I have a constant times a linear phase like this and that gives me the Fourier transform something that is very localized like a delta function here. So what is the phase on this side? It's not constant because it's a linear phase because the rays are tilted, so the phase is different here than here than here, increased linear, the wave fronts are perpendicular to the rays. So what happens if I take the complex conjugate of this? The phase instead of increasing decreases let's say, which means that we reverse the direction of the field. So if I change the, if I complex conjugate this field, if the field is, say, here the phase is more advanced than here, but if I reverse it, it's going to be backwards. So actually this is going to go in the other direction and therefore the light is going to end up not here, but where? Here at the mirror image, which is what this is telling us, rather than the, there's a minus sign here in the argument telling us that the light is, this is going to move to this side. So again there's an easy physical interpretation for this property. The Fourier transform of the complex conjugate is a complex conjugate evaluated at a minus sign. What this tells us, by the way, is if the field is real, if f equals f star, then f tilde is f star tilde minus. That means that the real part is symmetric and the imaginary part is anti-symmetric. But for the current topic, this is not so important. For other topics, this is really important. Okay, we can skip that one. So to summarize, we went through, there is a list of the properties. So parsable, that this integral is the same as this integral. And in fact you can do it with two different functions, f star g is the same as f tilde star g tilde. They don't have to be the same function. Then the shift phase thing, that if I shift a function, I cause a phase in the Fourier transform, or if I put a linear phase in the function, I cause a shift in the Fourier transform. Scaling, if I make my function narrower or wider, I make the Fourier transform the opposite. Derivatives turn into multiplication by the variable to the n times a constant and the other way around. So if a convolution goes into a product or a product goes into convolution, then we have this uncertainty relation and finally this property of a complex conjugate. So before I move on to physical optics applications of these properties, well, first are there questions about any of these? No? Okay. So the examples, you already played with these three. So now we can see how we would do this one, just by giving that we have that, we just use the shift property here. In this one, what property would I use? Scaling and shifting. In this one, the convolution. So the Fourier transform of rect is something called sync. So the Fourier transform of rect convolved with rect is what? Sync times sync or sync squared. So any function convolved with itself, its Fourier transform is just the square of the Fourier transform of the function. And then here we did the Gaussian and then x times the Gaussian, we just use the formula for the derivative of the result. So it's all very easy. So you can play with this later on your own. So very quickly in preparation for what I'm going to do next, is there a question? Okay. What happens in two dimensions? Someone was asking, how about bigger dimensions? Well, two is very important for Fourier optics. I'm going to stick with two. We're not going to keep going to three, four, but you could. So now I'm going to use a convention of x underscore, meaning a two vector, vector x comma y. So two component vector. And new underline underscore is news of x comma news of y. It's two vectors as well. And the dot product, of course, between this is this times the first one plus the second times the second. And all the same properties hold. So you can define the Fourier transform the same way. Well, first convolution into dimensions is the same thing, but we have to integrate over both x and y. So it's the same definition, but just in x and y. The delta function, this is interesting. So it's a delta of a vector. It has two arguments and now its units are one over x squared. It's the product of the inverse of each argument because we have to integrate it over both variables to get one. The Fourier transform is defined the same way except because x and new are vectors now. This has to be the dot product, meaning x times news of x plus y times news of y. And same with the inverse Fourier transform. And all the properties look very similar. They use the vector versions of these. Notice here, for example, x. Now x is a vector. So it's a vector multiplying the function. This has to be a vector. And the vector is the vector derivative. It is the gradient of the function. So gradients go to the variable and the other way around. And the uncertainty happens to lose a factor of two. That's the only effect. Now, in optics, we like things that have rotational symmetry. So if I have Fourier transform in two dimensions like this, f tilde of new underscore equal the double integral from minus infinity to infinity of f of x e to the minus i 2 pi x dot new dx dy. What if this function here only depended on how did I call it rho? So it's just a function of rho where rho is the square root of x squared plus y squared. So rather than having x and y, we play with rho and phi theta. So I can change variables here. But what matters then is that x dot new is now x is rho times what? Cosine theta comma rho sine theta dot. The same thing for new. I'm going to say this is new at the radius time cosine phi. I'm going to call the angle in the other space comma new sine phi. And what is this equal to? Rho new in front of everything. And then cosine cosine plus sine sine. We'll remember this one. Don't look at the screen. Turns out that it is minus. Which makes sense because it only depends on the angle between the two vectors. It should not depend on the sum. If I have this is x and this is rho, it can only depend on this angle. And that's theta minus rho theta minus phi. So I can make that substitution. Now this is an integral from 0 to infinity and integral from 0 to 2 pi f rho of rho. If our function happens to have rotational symmetry, it's independent of phi. If it is not independent of phi, we're in trouble. But if it is independent of phi, then we can do this. And then we have e to the minus i 2 pi rho new cosine theta minus phi rho v phi v rho. I can bring this integral in phi here. And I can do that in closed form. It gives me something called a Bessel function. So this turns out, how many of you have never heard of a Bessel function? Okay. So a Bessel function is, well I guess I can just define it that way. So I can, let me write this like this. Integral from 0 to infinity. And then I'm going to put this up. I'm going to put f rho of rho. And then integral from 0 to 2 pi e to the minus i 2 pi rho new cosine of theta minus phi d theta rho v rho. And this thing here happens to be 2 pi times something called the Bessel function of order 0. So there's a discrete number of Bessel functions. And it's evaluated at whatever multiplies this thing. So it's 2 pi rho. So this now becomes 2 pi 0 to infinity f rho of rho j 0 2 pi rho new d rho. And this is a form of the Fourier transform. We're only one integral, even though we're in two dimensions. And this is sometimes called the Hankel transform. It's just another way of writing the Fourier transform if we have rotational symmetries given by this. And the inverse Hankel transform looks the same way. Okay. So while I get the next thing ready, why don't you try this? So evaluate the Hankel transform of these three functions. When f is a delta, when f is 1 for rho less than a given a and 0 outside. And forget about the third one. You can do the third one at home. Just try the first two. And I'll give you a few minutes. Does anyone finish one? Okay, let me do the first one. And then I'll do the second one. So if f rho of rho is delta of rho minus a f tilde of nu, it turns out that it only depends on nu, is 2 pi 0 to infinity. Then I just put this delta rho minus a j 0 2 pi rho new d rho. And what is this equal to? So what does this delta function do? It sets rho equal to a everywhere and then it removes the integral. So this just gives me 2 pi j 0 2 pi rho, sorry, a nu times nu. And I'm doing these examples because they're going to have an importance in imaging. What about the second one? Sorry? Yes, I was just checking that you were paying attention. Okay, what happens? So this is what is called a CIRC function or a CAKE function sometimes. Because if I plot this function, how does it look? It's one inside of a circle and zero outside. So if I were to draw it in 3D, it would be like this sitting on zero. So it's like a cake with radius one. And it represents very well what happens when we have an aperture, which we're going to have in our lens system. So if I insert this function into the integral, what effect does it have? The integral goes from word to word now. Zero to a because after that it turns off. So zero to a and then that's f and then we have j 0 of 2 pi rho. And what do I do now? If I only knew the properties of vessel functions. Good? So I'm going to use that. I call this u prime. Then I need to complete this u prime here. So I need to put a 2 pi nu. So I'm going to leave some room to put whatever I need there. So I have j 0 u prime. And then this rho to become a u prime needs what? A 2 pi which has there a nu. So I'm going to put a nu here. And then the that's u prime. What about the row? It also needs a 2 pi nu. So I'm going to put a 1 over 2 pi nu squared u prime. And this goes from 0 to where? What is u prime when rho is equal to a? Pi a nu. And now this is nu squared. And if I use that formula. This becomes u which is this guy here. Pi a nu times j 1 the first vessel function. This is the vector of the zero vessel function of pi a nu. And this I can cancel the 2 pi. I can cancel the nu. This is a j 1 2 pi a nu divided by it is the diffraction pattern that we're going to be using. And this is the diffraction pattern if we use an annular pupil that has been proposed for some applications. Sometimes you want an annular pupil, someone you want to solid pupil. The third example is what is called an apodized pupil or one form of an apodized pupil. Which is you don't want the pupil to drop from 1 to 0 immediately but you want it to go gradually. It's apodratic. And that will remove in this function some of the oscillations. So turns out it gives you a j 2. So yeah, it's wider but it doesn't oscillate so fast. Just like this one because this one is very discontinuous. This one is the narrowest one but it oscillates a lot. This one is a bit wider but it oscillates less and the other one. We can plot them later. What is missing? No because there were two. We needed one for rho and one for d rho. So this one participated with rho let's say in turning into u prime. But d rho needed another one and we had to compensate by putting one downstairs. So we needed two pi's. One for this and one for this. And we only had one. So we had to put these downstairs. It's what, sorry? Yeah but when I plot them later it oscillates less. And these oscillations for some applications are a bad thing. So that's what this term apodization means. Means that you go rather than having a pupil that is sharp. You have something that goes to zero. And the effect that that has on the Fourier transform is that instead of having something like that that goes like this and oscillates a little bit. You get something that is slightly wider but it doesn't oscillate. Pot, what does this mean in etymologically? When you go to the podiatrist it's your feet. So they think of this as the feet. Like these oscillations. So it's removing the feet. Literally means removing the feet. So it's to remove these oscillations, side oscillations. So you do it by sacrificing some energy because you're having an aperture that is less and less transmissive as you go to the edges. And because you're making your function narrower, your Fourier transform is going to be wider in certainty. But it's not oscillatory. So it depends on the application. Sometimes you don't care about these feet, these oscillations, these side lobes. Sometimes you really care about them. The best apodization is if you try to do something like a Gaussian that drops like a Gaussian then it drops quickly and it's pretty much a Gaussian without any oscillations. So the last part of this note is to do with the discrete Fourier transform, which is how we approximate a Fourier transform with the computer. And it turns out that there's a discrete operation which uses sums instead of integrals that does a good approximation at a Fourier transform. I'm going to go very quickly through this because I want to move on to physical optics. So this is a sum from 0 to sum number n minus 1. That means there are n elements in this sum. And it has an exact inverse transformation. Looks the same. You normalize by 1 over square root of n and instead of a minus you have a plus. And if you want to approximate a Fourier transform with this discrete Fourier transform because we want this to go from minus infinity to infinity when we approximate an integral, we need to take the second part of this sum because this is periodic and put it at the beginning. So we want something more like this from here to here. And then when we let n be very large it approximates minus infinity to infinity or minus something large to something large. So when you're doing this with data what you have to do is here. So you discretize your data. You have a function like that one. Then you discretize it. And then if this is the origin I need to take this old here and move it here. So you have this is the center, the origin, this, and then this part here. If you use Mathematica you have to do that yourself and I use Mathematica. If you use Matlab, Matlab does it for you so you don't have to worry. And the interesting thing is you can see that when you approximate a Fourier transform with a discrete Fourier transform like this, the discrete Fourier transform looks like the Fourier transform divided by some normalization factor but it is sampled at values m divided by n times delta x. Where delta x is the spacing of your samples in the initial function and n is the number of steps. So if I have my function here and let me come back, let me go down, sorry. So I have my initial function but of course to enter this I cannot use the continuum of points. I take a delta x being a spacing. I take these spacings and I take the values. That spacing is delta x. So if I multiply delta x by n what do I get? So I have something like this and then I do this spacing is delta x and then I have m points. So delta x times n is what? It's the size of the interval that I'm using because it's how many steps times the size of each step. So what this tells me is that if I sample that way then the Fourier transform is going to be sampled at spacings that are the inverse of n times delta x that is the total range. And this, what that means is that if I want, if I have something I want to then increase my sampling here. I want to make it more densely sampled. I should not make my initial function more densely sampled. Making my function more densely sampled will add values to the sites. Will not add values, increase the sampling. If I want to increase the sampling instead I should add values to the sites. So I have this at given spacing and then I need to keep going this way and keep going this way. And if the function finishes that means put zeros. That's what it's called padding with zeros. So you put a bunch of zeros here, a bunch of zeros here and then when you do your Fourier transform your result is going to be more nicely sampled. So to increase resolution here you add zeros to the sites here and the other way around. If you want to add zeros to the sites here you need to increase your sampling space. And that is important because it turns out if you don't sample well enough this is not going to die by the end of it gets here and this is not going to die out by the end it gets here. And if it doesn't finish whatever doesn't fit here comes around and comes to the other end and overlaps with this. This operation is sort of defined on a cylinder. So whatever spills out comes on the other side. It's like when you're playing Nintendo or something and you go out one side of the screen you come out the other side of the screen. So you want to avoid that for that you need to do more sampling. We'll see this numerically later. Okay so let me now, any questions? I'll show this with a lot of examples later but I want to do some physical optics before. So we see all this together with the physical optics. Yes. Ah yes sorry. So this delta x is not the same as that delta x. It's the same letters but it means something different. On that context delta x means spacing. On here means the width of the function. So I should have called that little delta x or something just to differentiate. Yes and I wrote an article on this on the uncertainty relation for the DFT. So there is a difference in the uncertainty relation. It's a bit more complicated but it has a nice interpretation but I don't have time to cover that. But I actually wrote a paper about it so I can give it to you. The uncertainty relation for the discretized version involves n. So the right hand side of the uncertainty relation goes like 1 over n, the number of points. Okay so let's change topics for a little bit. So who has heard of Maxwell's equations? Who has not heard of Maxwell's equations? Good. So why are these called Maxwell's equations? Did Maxwell come up with them? No. So why does he get credit? It was Ampère, Faraday, Gauss. And in fact there were a lot of other people, Prisley, Coulomb, Cavendish, that came up with many of these laws and didn't get the credit. What did Maxwell do? He fixed one of them. So here I'm giving incomplete versions. I'm missing here the charges and the currents. And this equation did not have this term. Maxwell realized that this term was missing. So when you have charges and currents there are other terms here and another term here. So for the sake of this class I'm going to skip them. Oddly enough I have to teach my class to my students in Rochester at 5 in another classroom with Skype. And I'm going to go through this. It's going to be the same lecture. I'm just going to go into more detail there with the other term. You'll probably have enough lectures but when Professor Imrana finishes here if you want to go to the next room I'll be in the next room. What Maxwell did was find this and these are laws that tell you the relation between electricity and magnetism. If you have a circuit with current that interacts with a magnet and if you move a magnet through a loop you induce a current and all these things. And also the repulsion between charges, etc. If you write that in differential form it gives you this. Now the other extremely important thing that Maxwell did is if you take for example this one and you take the curl and you use these properties that at some point you probably knew that it's the curl of a curl it turns out to reduce to the gradient of the divergence minus a laplacian and because the divergence of E from this one is zero it gives you the minus a laplacian then here you take the curl of this you swap the order between the time derivative and the curl and then the curl here you use this other one you substitute it and at the end you get an equation like this which is the wave equation it's an equation for a wave that is travelling at a velocity equal to one over the square root of whatever is sitting here and that was the most, I've been one of the most wonderful times in the history of physics because people knew about light and there was something to travel very fast they measured the velocity of light to some precision Poucault and Fissot they used this clever trick do you know of these experiments? so they had a wheel, the first version was with a wheel like this with a lot of slits and then you had a candle here and then a mirror several miles away I don't know how they did this and you could spin this very fast you look through this hole here in the mirror the flame coming through this hole here so you could see the light then you start spinning it and you start spinning so fast that the light going through here by the time it comes back it's no longer hitting this slit but it's hitting this blocking thing so you don't see the light again it's much faster and then you start seeing it again through the next one and with that you can measure the speed of light with some accuracy then the version of Poucault was more precise he used I think a barrel like this with mirrors on the sides and it was, so these were mirrors you were looking this way let's say and the candle was here so the light came in here came back here like that and it spins again so he could do it with more precision and found that it was ballpark what the true velocity of light is but when they combined Maxwell and others combined these equations to find this they found that mu zero times epsilon zero which were found from experiments with electricity and magnetism nothing to do with light this gives you a wave equation whose velocity is the velocity of light and that's amazing not only did they unify electricity and magnetism but also optics light is an electromagnetic wave and it satisfies that that's a remarkable, remarkable thing sometimes we forget about how amazing that is so once we have this wave equation second derivative the Laplacian of E minus let me then write one over c squared the velocity of light the derivative with respect to time of E is equal to zero and what happens if the field is monochromatic if I assume that the field is monochromatic monochromatic means only one so I can say E is some u of only the spatial position times something that depends on time which is E to the minus i omega t or minus i to pi nu t if we want so then if I take this derivative with respect to this one what do I get? this is independent of time so that doesn't change so on this side I get the Laplacian of u of r then minus what happens when I take two derivatives here minus and minus gives me plus and I get an omega squared divided by c squared and then just u of r both are multiplied by the exponential but then I divide by the exponential and it's gone and this is called the Helmholtz equation we call k is equal to what? omega over c so I can call this k squared and it's also equal to what else? two pi over lambda, the wavelength so I can write this as the Laplacian of u r plus k squared u equals 0 and the electric field is a vector but for much of what we're going to do we're going to forget about it we're going to use it as scalar just one number and what that means is we take one component and we use it as a representative however when do you need to use the full vector? when you're doing optics with polarization when you care about the polarization of light then you have to worry about the different components if not, no so for now I'm going to forget the components so I have an equation like this and that's the equation that we're going to solve to model propagation of say, laser light is that clear? so it's only one component then we can put back the many components they're all solutions of this equation and they satisfy the divergence condition can anyone tell me a solution of this equation? a very simple solution the simplest you can find exponential so you let me say I'm going to symbolize that of r is equal to an exponential because I know that if I take a second derivative this, when I take a second derivative has to give me something negative so that it cancels this positive thing so therefore to get a negative second derivative I need to put i i here and then I need something that gives me something like a square but this is in three-dimensional space so I'm going to put r to make this general but in the exponent I need to multiply this by something to make it a scalar I cannot take the exponential of a vector so I should do k vector dot r and this is called a why is it a plane wave? suppose that let me think of this as z so here's where you know who does optics and who doesn't in all other fields of physics the z-axis points up in optics the z-axis points to the right and I always tell students the first law of optics is light goes from left to right if light is going from right to left then optics doesn't work unless you're doing microscopy then it can go down so what happens if the vector k goes in this direction so let's think of this dot product suppose that I'm at a point here r what is this dot product equal to? it has to do with the projection of this point onto the direction of this point that means that this point will make this dot product take some value what about this other point here? will the dot product be a bigger value or the same value? the same the cosine is smaller but this distance is the angle is smaller so cosine is larger but this distance is smaller and those cancel out turns out the only thing that matters is if I take this vector that is fixed I look at all the points that are projected on a perpendicular onto this they all give me the same a dot r the dot product is a projection so at this point and at this point the wave takes the same value so and of course this is in one line but also out of the board so in the whole plane the wave takes the same value if I come to another point here that's a different value but this is the same as this and all those over planes that are perpendicular to the k vector so for this k all this plane is constant and then this other plane is constant etc that's why we call it a plane wave and I remove the time dependence but as I let time go these would move and travel at the speed of light so this we call a plane wave now what happens if I take this plane wave what is the solution of that and you told me this but we haven't checked there's more to it than just saying that it is so what happens if I take the derivative of this so what is the second derivative of this u k of r so when I take a gradient I bring down i times the vector k then when I take the second derivative because this is dot with the first derivative I bring out another i and another k and the two k's are dots with each other so I get minus k dot k times u I could write this as magnitude of k but I will not for a reason that will be obvious so if I substitute this in here what do I get let me write it as the exponential sorry e to the i if I substitute this I get minus k dot k the exponential plus little k squared the exponential equals 0 can I cancel anything the exponential and then I'm left with k equals little k squared so that's the condition for this to work so k dot k is equal to so you might say that means that the length of this wave vector k is what this is the wave vector this is the origin here what is the vector from here the magnitude of that is little k with some catch so let me write k as 2 pi divided by lambda times a vector u or this is of course little k what what happens then to this equation k dot k is little k squared dot u equal little k squared can I cancel this and this is one so u is what type of vector a unit vector so the solutions then are plane waves I can write as e to the i 2 pi divided by lambda unit vector u dot r is another way of writing it but u x sorry u dot u if I write it in terms of the components how does it look y squared plus u c squared and this is equal to 1 that means that of these 3 components how many are independent how many do I need to specify only 2 the third one I can find from this expression and I always choose it so that the wave goes from left to right so I want u to be greater than u so u z is given by what if I solve this equation square root of 1 x squared plus u y squared let's say I have an uncertainty here whether I use the plus or the minus results here but the plus result is a wave that goes from left to right and the other one is one that goes from right to left and I'm going to assume that all my light goes from left to right so I just always choose that one but there's a catch, another catch here I could also find solutions where this is imaginary if u x square plus u y squared are bigger than 1 then this thing here can be imaginary and that's something that I don't know if I have time but we'll see it is something called an evanescent wave how many of you have heard of a vanescent wave okay so there are two types of plane waves the traveling plane waves or homogenous plane waves and then there's the evanescent waves some illustrations of this so the plane wave I can write then as e to the i 2 pi u x divided by lambda times x plus u y divided by lambda times y and let me write this as e to the i I'm going to separate the last one u z divided by lambda where this depends on u x and u y if I write a vector nu like this as u x divided by x u y divided by y sorry by lambda I can rewrite this expression in a form that's going to start to look familiar e to the i 2 pi so I'm going to write n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n wave traveling in the direction of the C axis from left to right. And in fact, I'm going to see how it evolves when time goes by by a quarter of a cycle. If this moves by a quarter of a cycle, that is, those wave fronts are advancing a little bit. Now let me come back to it again. This case of X is, think of it as use of X, and let's forget about use of Y. We're only in one plane. If I change use of X, what happens is I'm going from a vector that goes like this, where this is the C direction, to a vector with X components, so this is X. So that means that the plane wave is now going to go in that direction. If I keep going, it's, I'm just increasing the X component, but that comes at the cost of the C component having to be shorter because this magnitude is fixed. And then I reach a point where this is flat like that, and then that means I'm already here. The X component is one. You would think, I cannot make it any bigger than one because the vector's already going like that. Well, I can force it to be bigger than one at the cost of the C component being now imaginary. So if I keep going, this starts to oscillate faster, but now I have e to the i something times something imaginary. I have an i here, and I have an i here. i times i give me what? Minus. Minus. And e to the minus something times z gives me something that in the C direction drops the case. So this is the evanescent wave. It's a wave that oscillates faster than the wave can, the traveling wave can, at the cost that in the other direction it has to drop. And these waves don't travel very far. They die very soon. As we will see tomorrow, we can decompose any field, any image by using Fourier theory in terms of these waves. This is like the superposition in the Fourier expansion, but in the Fourier expansion we need all values of nu. Some of those values of nu correspond to traveling waves, some correspond to evanescent waves, and those will not travel far. They will decay very quickly. And that is another way of seeing the fundamental resolution limit, because the information in those waves doesn't make it to our detectors, it's gone forever. The only way to see it is with what's called near field microscopy, which is you don't use lenses, you use a tip that you bring right next to the object, and then you detect those waves before they decay. But if you don't do that, they're gone forever. And this is the same if we go in the other direction. So this is going up. If I go down, then I also get into the evanescent wave. Now one thing that I'm plotting here that is going to be interesting for what we're going to do next is the red line is the slice in this direction of this wave. So it has the same period. So it's maximum here because this is maximum here, maximum here because it's maximum here is the slice. So if I just look at the slice here, it's like a sinusoidal whose frequency varies slowly. The more close to the C direction the wave is going, the more slowly that varies. And if I tilt this, it starts oscillating faster over the initial plane. And then to oscillate very fast, we have to go into the evanescent waves. The green line is the imaginary part. The red part is only the real part, which is what I'm plotting here. What does the imaginary part tell us? So the real part is telling us what that wave is doing, let's say now. The imaginary part tells you what the wave is going to be doing in a quarter of a cycle. So it's sort of looking forward into the near future. So for example, this is at the initial time. But if I propagate in time, quarter of a cycle, then I get into now this wave corresponds to the green, not to the red. So the imaginary part tells you where you're going in a quarter of a period. And it serves to distinguish between what's going, say, up from what's going down because the relative position between the red and green line reverses even for the same period. And I should pass the baton. So tomorrow what we're going to do is take these ideas, these plane waves, put it together with Fourier theory, and see how things, how wave fields propagate in space. It turns out that this part is going to provide the Fourier transform. And this is how this tells you what happens to each Fourier component as it propagates. This part is essentially the transfer function of free space. And because free space is a linear shift invariant system, it's described by a transfer function which is precisely this. So we will be doing numerical modeling of propagation of wave fields and several things using this idea of expressing any field as a superposition of plane waves. And then each plane wave is like a Fourier component. And then the longitudinal part is the transfer function, which is oscillatory for traveling waves and is real and decaying for evanescent waves. So I'll put these two pieces together tomorrow. Anna, you want to take over? So now Professor Consortini is going to give a short description of what she's going to be doing in the lab. Do you want me to put the presentation? Okay. Okay. They're a rotated version of. No. Yeah, please raise. You can start and I will raise for you. I am not speaking loud enough. Is this okay? This also should go here. Okay. We don't know. So as I thought before, there are three kinds of the laboratories. The microscope laboratory, which is in the lab, in the lab up. The computer laboratory, which is in a room nearby here, the other room, and the diffraction laboratory, which is nearby the secretariat room. Our groups are made of 10 people each time, but for the diffraction laboratory, 10 people together are too many. So I have decided to divide those 10 people in five to subgroups of five people each. So we will have roughly one hour and a half for each subgroup. So the first five people of a group corresponding to this laboratory will come first, and the second five people, and the second group of five people will become later. Let's say one hour and a half later, but if you need a little bit more time, maybe we can go a little bit further, which is likely. The experiments we will do, although they are in the laboratory of diffraction, are not exactly all diffraction experiments. So we will start with the diffraction and Fourier transforms. This is the first group of experiments, one experiment on evanescent waves, and several demonstrations of the composition of light by diffraction grates. There is an important point that I want to establish from the very beginning. Our eyes are sensitive to energy, not to the field, and of course everybody knows that optics is a field, amplitude and phase, but we never see amplitude and phase. We always see only energy, which means proportionality, but amplitude square. So what we see, for instance, is not a airy function, is the model square of the airy function. To describe the diffraction from an aperture, we will consider different apertures and we will examine the region where the field is diffracted. Starting from an impinging plane wave, which is typically an Illuminaeon laser, we have or an opening, or a wire, or something else, and on the other side of the screen we have a region of the diffracted field. The region of the diffracted field is typically divided into two regions, Fresnel region, which is far from the aperture with respect to the wavelengths, but not far enough to be considered at infinity. Then we go further, in principle, to the infinity, we have the diffraction region. We will experience diffraction of radiation Illuminaeon laser, wavelength 632 or 33, if you want to, a number, one round, by wires of different dimensions, sleets of different widths, and by circular aperture of different radius. Diffraction gives rise to evanescent waves also, but the experiment on evanescent waves will be made in different ways. In the diffraction, especially from, we will consider mostly in our theoretical part, with reference to theoretical part of opening, sleet and circular aperture, it's important the angular dependence of the aperture, because this is related to the resolving power, and so we will check the angular dependence by measuring the width of the diffraction of the pattern at infinity in different cases. In particular, we will consider two different circular apertures, and we will compare the width of the diffraction pattern with the width of the aperture, and we will confirm this way that from the point of view of the optics, we know that as smaller is the hole, and larger is the diffracted field, the angular, and from the point of view of the frene of the transformer, Fourier transform, this confirms what you already know that the smallest is the width of the function, the largest is the width of the transformer. By the way, from the circular apertures, considerations on the solving power for instance of the telescopes can be made, while the case of the microscope requires a little bit more attention, but this end, the microscope, the solving power is developed in the corresponding laboratory where AB formula is most important, so we will not see anything about this, and we will confirm by looking at this diffracted field that a function rect has a transform, a sinc, of course we will see a sinc square, and the sinc has a function, transfer function, and a refunction, which is a vessel function of a given argument, divided by argument, again the square. If the border of the aperture, if the aperture is the border of the converging lens, the transform, which in the theory of diffraction from the aperture goes to the infinity in the Fraunhofer region, by the lens is transferred to the focal plane, and therefore in the focal plane we can see that a lens operates the Fourier transform of the field on its aperture, not the field from an object which reaches the aperture of the lens, and this is the basis, sometimes people say that the lens operates the Fourier transform, but we must be careful, it's not the lens that operates the Fourier transform, it's the border of the lens that diffraction takes place every time there is an abrupt change of the amplitude, and this is made by the border. As a consequence the phase adjusts in a suitable way, but I don't want to enter in this point. The fact that in the focus of an aperture, of a lens, in the focal plane of a lens, one has the Fourier transform of the field of the aperture is the base of the elaboration of images, and because one can utilize a convolution theorem to elaborate amplitude and phase in this case, images which are on the entrance of a lens, in the front entrance of the lens, of course this is not what we will do, but I like to mention this. Now we make an experiment on evanescent waves. As Professor Miguel showed you, a evanescent wave is a solution of Maxwell equation where a wave propagates along a plane, let's say, flow along a plane, but decreases exponentially from the surface. In my case, the propagation direction in the plane is not x, it's the opposite as here, but it's clear. I will show you an equation in a while. Those waves cannot exist without the surface. They were generated and along which they propagated. Difficult cases are diffraction because at the border of the diffraction in the plane of a diffracting aperture, you have evanescent propagating along the aperture, and as it was already said, these are lost because they cannot be collected by common lenses. If you want to collect them, you have to go near the surface on a kind of a microscopy which is called the near field microscopy. But evanescent waves are present in many phenomena. In addition to diffraction, one has diffracted evanescent waves in the total refraction, such as in prisms when we reach a total refraction or in fibers. In fibers, light propagates, modes propagates in the fibers just because total refraction allows guided propagation. This is the geometry of our experiment, in any case, to understand. This part we have, VPT is the complex amplitude that was already mentioned by Professor Alonso. This, you see, there are two parts. One, A, is a constant of no importance, and then an exponential depending on X. And this part altogether is the amplitude. And then we have another part with the I, which is the phase. So we see, looking at this part, that the phase propagation takes place in the zeta direction. That means along this surface. This is the surface. But in principle, the phase is very rare because you can go even here with the phase. But the amplitude decreases exponentially immediately as soon as you move a little bit from the first surface. Because note that K is 2 pi over lambda, and therefore, as lambda is very small, as soon as X reaches values some multiples of the wavelength, this exponential goes rapidly to zero. So the experiment on evanescent waves is of this kind. We produce an evanescent wave by sending a laser beam on a prism. Oh, by the way, this morning, for instance, I prepared just the opposite direction. The laser beam coming here, going here and here. So instead of looking at evanescent wave over this surface, we will see this on one of the others. But this is of no importance. Then we collect the what happens here? A laser beam, which is real, go inside the prism and remains real. And then it is reflected by total reflection. So in principle, on this side of the prism, there should be nothing. On the contrary, there is the evanescent field which propagates along the surface. This phenomenon has reciprocity. That means that if you have a field, an evanescent field on a surface created for some reason, on the other side of the surface, you have a real field reciprocity. So in the fiber for reciprocity, by approaching the fiber near the surface where the evanescent wave propagates, the evanescent wave enters the fiber or better. There is a coupling between the evanescent wave of the prism and the lateral surface of the fiber. And inside the fiber, you produce a real field. And in this way, to the end of the fiber, we will see the light. So we will see the light entering the prism, going back on the other side. Here nothing, but as soon as we put the fiber, we see on top of the end of the fiber, the light. And this will be done with a green laser. Then the third group of experiments is the composition of light by diffraction gratings. Which is the basis of the spectroscopy. If we consider a diffraction grating linear, that means we have a number of parallel opening, let's say, or they can be reflecting instead of we can have transmission and reflection. They can be reflecting or transmitting. Here the example is in the transmitting case and we will mostly make experiments with the transmitting case, but in some cases also for reflection. An impinging light in this example is red, impinges on the grating. And as you probably all of you know, from the grating a number of spectra directions where the diffracted field, all the diffracted fields waves interfere constitutively. Do you know who knows those things or better? Who doesn't know about diffraction grating? Everybody of you knows about diffraction gratings? Yes? Okay. So it's not difficult to see, but thinking for instance of in Fresnel diffraction Orgen's Fresnel principle, if we go, we consider this opening very small and we consider light going out from this and we consider race, we have a constructive diffraction when the different in the path which is here can be seen, this part here is a multiple of an entire multiple of the wavelength. The multiple can be positive or negative value, but when apart from the case when everything is parallel and in case we don't see the dependence on the wavelength. And therefore we can separate the light of a given color from another just by looking at the diffracted field from the grating. This is almost all and we will use linear gratings of different periods. One of the best is 600 and we will see the composition from different sources going from lasers and leads. And in case of the lasers we will have two different lasers, one red and one green, both are a laser, but the green one of course works with a different wavelength and we can see the relationship between the wavelengths of the two lasers by looking at the difference in separation of the diffracted beams of the same order. So for instance we will see that the green and red go in different directions. I will show also something produced by a two dimensional grating. A two dimensional grating is something like this where there are lines in horizontal in two directions, one normal to the others. And this is all. The University of Florence gave us the open lab of the University some diffraction gratings. The electrosynchrotone is supplying, this is an error, some lasers and the experiments in the laboratory are organized by Imrana, Professor Imrana, Dr. Mutual Dinalov from Eletra and myself. Okay, this is all. Oh, I forget to say something. I have recovered from lectures given in 1993, which means 24 years ago, some, no, a lecture on evanescent waves. But of course in order to follow clearly this lecture you need to know what happens before plane waves. So I will applaud in the paper that we will put in the lectures. I will give a link to this in such a way if someone is interested in seeing how these things are developed, this will be maybe can be useful. It's very old, but it's still the same. The diffraction is always the same. Okay, this is okay. Thank you. Thank you to you. Now this can be answered.