 So today we will do a little bit of discussion of cavity dumping. Now this is more for the sake of being historically correct and more for the sake of completeness of knowledge because most of the time when we do ultrafast studies we use titanium sapphire lasers and unfortunately titanium sapphire lasers cannot be cavity dumped. Cavity dumping was a very very popular technique when die lasers were used. It is not a technique by which you produce ultra short lasers per se. What you do is you take the pulses out. There is some narrowing down of laser pulses also but more importantly you can take the pulses out of a particularly you can take it out of a die laser cavity and is attractive because as you will see cavity dumping is a technique in which the laser is always on. We are also going to discuss Q-switching after this. We have already discussed mode locking. Cavity dumping is like mode locking where the in the sense that the laser is always on. As you see in Q-switching the laser is not always on and we will talk very briefly about pulse picking which we use in our lab. As I have said earlier also it is a wasteful technique. We threw away most of the energy. We use only one-tenth of it or one-twentieth of it. In cavity dumping when we are not taking the pulse out actually the power of the laser goes up in that time interval. That is why it is such a good technique. It is unfortunate that we cannot use it with the femtosecond lasers we have now. So in cavity dumping once again we use an acoustic modulator. We have talked about mode locking. How you do mode locking using acoustic modulator? That time we said that you have to use it. You have to use the AUM, AOM in Raman Nath regime where L was much much less than lambda square by 2 pi small lambda. Where capital lambda is the wavelength of sound wave small lambda is the wavelength of light. In case of cavity dumping I think we have discussed it in the previous module as well. In case of cavity dumping we have to operate this AOM in the so called Bragg regime. And Bragg regime means L the thickness of the medium is much much greater than capital lambda square by 2 pi lambda. So essentially you end up using you have to use much higher frequency of acoustic wave. L has to be greater than capital lambda square divided by something. So capital lambda would better be small. If capital lambda is to be small the corresponding frequency has to be large. So we will see what kind of frequency is typically used in this technique. Secondly it helps if you have a longer interaction length. And here what happens is when you work in this regime all the higher order refractions are eliminated. Only the 0th order and 1st order rays are sustained. Omega plus or minus capital omega if you write in angular frequency. Sometimes it is written as omega 0 plus minus capital omega. So for now this is what we will consider. We will consider that here when light falls on the Bragg cell part of it goes straight undeflected and it does not interact with the medium. Part of it gets diffracted and it gets frequency modulated by capital omega. This is what we have discussed in the last meeting as well. Now what you do is, in fact we discussed this as well in the last meeting but we went too fast and I think nobody understood anything. So let us see if we understand it better if we make animations. So this is a part of the laser cavity that we are showing. As we had said earlier it is important to remember that there is no output coupler here. All mirrors are highly refracting and we are showing only one end of the cavity. So this is M1, this is M2 and the laser beam goes this way. Somewhere here we have the gain medium and somewhere here we have the other high reflector mirror. Have you understood the cavity? Have you understood what the cavity is? This is M1, this is M2. What is shown there? Then the laser comes from M2 this way. There is a light. This is where the active medium is and this is where the other reflector is. And there might be other mirrors in between. But the crux of the matter is that this is one mirror, the end mirror. M1 is the starting mirror. M2 is the other curved mirror that is important for our discussion. So this is the simplest cavity that you can think. M1, AOM, M2, then the gain medium, Dijet or whatever it is, say M3. This is the cavity. So it is a closed cavity or high reflectors. So you can think that if you pump this gain medium here there will be a laser. There will be a laser beam in the cavity but it will never come out. The question is how do we take it out? We take it out by cavity dumping and how does that help? So before showing you the beam path and all, one thing that I did not say earlier and it is important to say otherwise it is very difficult to follow what we are saying is that this bragg cell here is kept at the focus of M2 and at the center of M1. Focus of M2 that means any light hitting M2, a parallel beam of light hitting M2, if it is normal to the tangent at the point of hitting then it will get focused on to the bragg cell. And since the bragg cell is at the center of, if you draw a line from the center of the bragg cell to the surface of M1 that will be the radius of curvature of M1. What happens then? Let us see if you remember any geometrical optics. I have a curved mirror, spherical mirror. I draw a ray from the center. The ray goes and hits that curved mirror. Where does it go afterwards? That is for focus. Yes? No, no I am saying at the center. For a concave mirror usually focal length is half of the radius, is not it? I am saying it is not at f. It is at 2f. So this is one of the main reasons why we did not understand last day. Remember those ray diagrams that we used to draw? This is a curved mirror. Something like this comes it will get focused. A ray that comes through the focus will, yeah, now from 2f center the ray comes and hits the curved mirror. Where does it go? It retraces its path. Why? Because if this is the, this is a radius then the tangent here, right, the radius is perpendicular to the tangent, is not it? So it is like normal incidence. What happens for any mirror in case of normal incidence? The light retraces its path, is not it? So for a beam that originates at the center of the mirror, all the rays will retrace their paths. I did not say it in as many words in the last day, whereas that is why it was a little confusing and also I showed everything together. Today we are going to show it step by step. Have you understood now what is going to happen with light falls? So then let me draw this. This here is the interactivity beam. Let us see it is coming from the die jet, gain jet. Now for M2 the Bragg cell is at the focus. So this is what the path of the right will be, all right? Now what I have shown is, I have shown the part that goes straight. The part that has the part of the beam that has not interacted with the radio frequency. Will this be the only thing or will there be something else? You follow me, right? What I am saying? E0, interactivity beam, E0 is the, you can say electric field associated with it. That falls on M2, gets focused on the, and I am talking about only one ray. So it goes through the Bragg cell. One part of it goes straight. Is there another part of it that does not go straight? And you can refer to the diagram above. We are working within Bragg regime. So won't we get another beam that will get diffracted like this? What did we say? That you will get the 0th order and the first order beam. Second, third, what we used in modulocking Ramanranth regime, we had omega 0, omega 0 plus minus, capital omega plus minus 2 capital omega plus minus 3 capital omega so on and so forth. Here there is no n omega, only omega, capital omega, omega plus or minus capital omega depending on the angle of incidence, okay. Here we have written, well we will see what we have written. Are we okay? So let me call this, let us say that the electric field associated with this beam that with the first order beam is E1. And let us say the electric field associated with the beam that went through straight is E2, okay. Now let us consider E1 first. After hitting M1, where will E1 go? It will retrace its path, right, because the Bragg cell is at exactly the center of M1. So it will retrace its path, okay. And again it is going through the Bragg cell. So part of it will go straight, we will call that E1 dash. Part of it again we will get bent. And when it gets bent, it will be made by the exactly the same angle. So now it will take the path that you would get if you produce this E2 line back to M2, is that correct? Yeah. So we will call this E1 double dash. Have you understood so far? Now in the direction of E1 double dash is that the only beam that is there or is there something else? What we have said so far is E0 falls on the Bragg cell after reflection in M2, gets divided into two parts, E2 and E1. Then E1 gets reflected by M1, comes back to the Bragg cell. The part that goes straight, we call it E1 dash. The part that bends, we call it E1 double dash. Now think in the, so what will be the fate of this ray that I have drawn associated with E1 double dash? After hitting M2, where will it go? It will go back along the direction of the intra-cavity beam, is it not? So it will be sent back to the cavity. But will only E1 double dash be sent to the cavity or is there something else that we have not considered yet? Exactly. E2 also hits M1 and retraces its path, goes back to the Bragg cell. Then what happens? A part of it goes straight and by the convention that we have used so far, we call it E2 double dash. A part of it bends, but before that, well I forgot the sequence of animation, sorry. So now the field that is sent back to the cavity is E1 double dash plus E2 double dash, understood? So that is what I was talking about. A part of it goes back to the cavity. Now to complete the other part, what will the other part be? E2 after hitting M1 has come back to the Bragg cell. The part that has gone straight, we have called it E2 double dash. The part that bends, we can call it E2 dash. So in this direction, we have a total field of E1 dash plus E2 dash. And now see what has happened. When the beam came from the dijet and hit M2, there was only one spot. But since Bragg cell has performed a diffraction, the beams that come back create an additional spot which is associated with the field E1 dash plus E2 dash. Where will that go? That does not go back into the cavity, does it? Let us see it goes in this direction. So in whichever direction it goes, you keep a prism and the prism acts as a retroreflector. You do not even have to keep a prism. You can keep two mirrors or you can keep this retroreflector that we use whatever it is. Conventional prisms were used in die lasers and this is the beam that comes out. So see even though this laser has no output coupler, by using cavity damper, you can take a part of the beam out. Part of the beam that goes out is of the field associated with it is E1 dash plus E2 dash. But interestingly for every such cycle, E1 double dash plus E2 double dash goes back to the cavity. So this is what I was saying. Even while you take the light out, the laser beam inside the cavity grows. That is why you get a good power out of the laser. Have you understood now? So even if you stop here, I think qualitatively we are good. We have understood how it works. But we will not stop here. We will give you a little bit of a feeling of what kind of numbers we can think of and why it is that during cavity dumping, the repetition rate gets modulated. Why does it that it gets cut down? In the cavity dumped synchronously modulok die laser that I used as a PhD student, repetition rate would typically be cut down by a factor of 10. It would go down to one-tenth of what the repetition rate was initially. And that was important because we did TCSPC with it. So you could not work with 78 MHz kind of repetition rate which is intrinsic of the modulok output of the pump laser. Now again I will ask the question that you did not answer. This is what it is and this is the diagram that I had shown you last day. Now I hope you understand what it means. Now let us try to see what kind of expressions we get. We want to know what is the expression of E1 dash plus E2 dash. So to start with we will write E0 as A0 e to the power i omega t. Now I want to ask you something. You want to say something about E0 or A0? What does it look like? It looks like it is a constant. Is it really a constant? Do not forget what kind of laser we are talking about. We are talking about a synchronously pumped modulok laser. Cavity ramping is only take the thing out. Even if I do not do cavity ramping, it is modulok nevertheless. So E0 is actually a pulse. E0 is an electric field associated with a pulse. So it actually has a little more complicated structure. E0 is not just a constant. That also is something more. We know already the expression of pulses that you get out of modulok systems. So that is the real expression. Here we are writing this only because otherwise expressions become too large. And you will see the implication of it once we are done with this discussion. All right. Output field is E1 dash plus E2 dash. So question is what is E1 dash, what is E2 dash? Let us see if we can figure that out. To start with that, let us first consider that this might look a little scary because we are chemists after all. We are not very good mathematicians perhaps. But this much of, we will not even try to solve it. We will tell you how it begins and then we will tell you where it ends. Now in the Bragg cell also there is this periodic field that you have because you have applied a sound wave. Let us say it is ESS for sound equal to AS e to the power i capital omega t plus phi s. What is capital omega? What is omega in this kind of expression? When I write for light E0 equal to A0 e to the power i omega t, what is that omega? Angular frequency of light. When I write it for sound, ES equal to AS e to the power i capital omega t plus phi s. Capital omega is the angular frequency of sound, that is all. And what is phi s? It is a phase of sound wave. So you might remember those of you have worked with the Pulse speaker particularly you play around with the phase, do not you? So that it is that phase. There are discussions in which they have not considered phase. I am going to share some paper with you. They have actually simpler expressions but phase is a very real thing. We even have a knob to control it. So that is why we will keep it for now. So after the first pass E0 light with associated electric field E0 passes through the Bragg cell the first time. Then what happens? It gets split into E1 and E2. So E1 turns out to be E0 multiplied by square root of eta multiplied by e to the power i omega minus capital omega t. This could have been omega plus capital omega t also multiplied by e to the power i pi by 2 minus phi s multiplied by e to the power i phi 1. Of course you can write this expression in several different ways is not it? Because it is a product of some three exponential functions. So you can add up everything. You can group them differently. This is how it is done in the book. So we will go by this. Let us see what every factor means. E0 we have talked about already. What is eta? Eta is the ratio of intensity of the diffracted beam divided by intensity of the input beam. So this is ratio of intensities. This here we are talking about field E1. What is the relationship between electric field and intensity? What is the ratio? Which one is the square of which one? Intensity is square of electric field. So therefore it is not difficult to understand that electric field is square root of intensity. That is where this square root comes from. So square root of eta no problem I hope. What is the next factor? e to the power i omega minus capital omega into t. Where does this come from? Because you have a light wave with angular frequency small omega. You have sound wave with angular frequency capital omega. They are interacting. E1 is a diffracted beam. Do not forget. So in diffracted beam you are going to have interaction between the two. That is where that omega minus capital omega comes from. Next one is something that will take axiomatically. Where does this pi by 2 come from all of a sudden? Does anybody know? There is no pi so far out of the blue. Where have I got it from? Well this comes because the interaction is between light wave and sound wave. What kind of wave is light? Electricity is fine. I am only talking about the electron electric part. Transverse wave. And what kind of wave is sound wave? Longitudinal. So whenever there is an interaction between longitudinal and transverse wave that pi by 2 term comes in. So these are all very well known from physical optics and all. And then this phi s we have kept and then last one e to the power i phi 1 that you have to consider. Again for a simplistic picture you can set it to be 0. Experimentally you can play around and make all the phi is 0. If there is no phase difference between anything. Perhaps you can try and do that. But this is the general expression. The point is this. Your light is travelling through some glass quartz. So some phase difference will come in. That is why that e to the power i phi 1 comes. Then what about e 2? Now knowing the expression of e 1 let us try to guess what the expression of e 2 will be. E 0 will be there. Then will we have root eta or will we have something else? Remember e 2 is the electric field associated with the light that has gone through straight. No interaction with sound. No. The capital omega comes later. Right now I am only talking about the root over eta. You are right. Well your answer is the correct answer to the next question that I was going to ask. It is right that e to the power i omega t that is what we will get. We will not get capital omega there. But can we come back to the correct answer of the question I have asked? What about root eta? Will it be root eta or will it be something else? No diffraction. But see eta has to be there in some form or the other because after all e 0 is getting split. It is getting split into the diffracted beam. The beam is getting split into the diffracted beam and not diffracting. Intensity of diffracted beam is idiff as we have written. What is the intensity of the beam that has gone through straight? i n minus idiff. So what is 1 minus eta? 1 minus eta is i in minus idiff divided by i in. So here instead of eta we are going to have square root of 1 minus eta. And then we have e to the power i omega t minus phi 2. The phase difference will come. But this is what we have considered so far. This is a Bragg cell. e 0 has hit m 2, gone to the Bragg cell. One part has gone straight. One part has been diffracted. We have not got to e 1 dashed yet. We have to consider e 1 dashed. To do that what we have to consider is there is another pass in the other direction. So e 1 and e 2 both hit m 1 and come back. So this is e 1, hits the mirror, comes back, goes to the Bragg cell again. e 2 hits m 1, comes back, goes to the Bragg cell again. And now for the sake of gravity we are not going to try to work out what is the expression of the field for light that goes back. Let us see if we can get say e 1 dashed. What is the light that goes out? e 1 dashed, is not it? Yeah. So after the second passage what will e 1 dashed be? I say e 1 dashed is which one? The beam that goes straight or gets diffracted. No, e 1 originated from diffraction that is okay. But that was the first passage towards m 1. While going towards m 2, e 1 dashed is the part of e 1 that goes straight. So now see whatever was the original expression for e 1 that will be there. But then something like eta over square root of eta or some square root of 1 minus eta has to come. What will come here? Square root of eta or square root of 1 minus eta? It is going straight. It is not being diffracted this time. Square root of 1 minus eta will come here okay. This is what will happen and you can have some additional, you can have some additional phase difference that comes in okay. That is how you get this carry expression okay. So people who work in optics or physics students and all for them it is very easy. They do not have to do it so slowly. Look at it they understand everything. But then we might not. So it is better to go a little slower. Similarly, you can work out yourself. e 2 dash gets a very, now we have something we like. We have symmetry. We have 2 expressions e 1 dashed and e 2 dashed which look more or less the same. So that is why see earlier it was multiplied by square root of 1 minus eta already. So now it is multiplied by square root of eta okay. The next part of the math is not difficult just manipulation. You do it and this is the final result I am going to show you. Well we get an expression for e 1 dash plus e 2 double dash sorry e 1 dash plus e 2 dash. But then finally what we care about is intensity right. Mod square of that intensity is going to be intensity of the output beam is that right. Intensity of the output beam this is the expression we get for that intensity of the output beam okay. We stop here and we continue from here in the next module.