 So, here we here is where we stopped right we said that intensity is A star into A A star and we said that provided the result first we said that when you lock n number of modes together so that their phase difference is not a function of time then you get pulsed operation repetition rate is not a function of number of pulses that are locked pulse width is that is an important take home message. So, repetition rate is 12 by C all that is important there is what is L, L tells you what the repetition rate will be right but pulse width that is where number of modes coming. So, now let us see how so we start our somewhat sketchy discussion of modulocking I apologize for the quality of slides slides that will uploaded will hopefully be better I tried to do things quickly and this is another moral lesson that you do when you try to do things quickly they do not work out to be very good ok. So, as we said in the previous module we do not really have plane waves we have damped oscillation but then eventually the way we go from time domain to frequency domain is Fourier transformation and the beauty of Fourier transformation is that to get the correct information about frequencies it does not matter if you take a plane wave or a damped wave because they get separated. So, that makes the mathematics a little easier otherwise we will have another damping factor here. So, we are going to work with plane waves where we will say that this electric field is a function of time and is given by E0 the amplitude maximum amplitude into e to the power i omega t what is omega yes angular frequency omega is angular frequency what is the relationship between omega and nu frequency and angular frequency 2 pi now just all you have to know is whether to multiply by 2 pi and divide by 2 pi it is not very difficult right I think we know it ok. So, now what I am saying is I have locked these modes together and I am saying the number of modes that I am working with capital N is equal to 2n plus 1. So, I am working with an odd number of modes because the math becomes little easier as you will see in the next step if I have an odd number of modes it does not matter even if it is an even number of modes you can conveniently neglect one right because remember how many modes are we going to have 5000, 6000, 10,000, 20,000 something like that ok. So, it does not matter whether it is 20,532 or 20,531. So, we formulate using 2n plus 1 number of modes ok. So, here we are saying that from one mode to the next there is a difference in frequencies right and we have worked out the expression earlier what is delta nu delta nu remember delta nu what is delta nu it is difference between the frequency of nth and the n plus 1th longitudinal mode what is it what is the expression C by 2L it is going to be handy remember do not forget that is very important C by 2L which is inverse of round trip time you told me right C by 2L. So, what is going to be delta omega what is delta omega going to be you know delta nu delta nu is C by 2L what will delta omega be that factor of 2 pi will come right we are going to use this. So, do not get confused. So, this is what we are writing is 0 into e to the power i omega 0 plus k delta omega q ok t what is k is a number that varies from minus n to plus n. So, now you already see why 1 2n plus 1 number of modes it becomes nice symmetric symmetrically distributed about a mean about central position not mean central position ok plus we are saying k into delta phi q what is this delta phi q phase difference and that is independent of time let us remember that that is very important. So, how do I expand it I can well e to the power a plus b is equal to e to the power a multiplied by e to the power b. So, I can use that and I can take this e 0 e to the power i omega 0 t outside the summation and then inside I am left with e to the power i k delta omega q t plus k delta phi q I have to do the summation. And now the advantage of summing here it is not showing, but advantage of summing k from minus n to plus n is that I know what the answer is. So, these summations I mean I am sure we might have done many of these in 11 12 maybe in BC or something right the summations are all nicely worked out and it is known that sum over k equal to minus n to n e to the power i k alpha is equal to 2 multiplied by sum over k equal to 0 to n cos k alpha minus 1. And then this you might think one summation is leading to another summation. So, what is the big deal the big deal is that this summation is also known as you are going to see, but let us first write it this is what it will be you have this outside the bracket inside the summation of exponential terms becomes summation of cosines. And also sum over this domain changes right it was minus n to plus n now it is going to be k equal to 0 to n. So, if you took only capital N then we would have this trouble you cannot even write n by 2. So, that is why we want 2 n plus 1 modes ok. So, now all we need to know to go further is what is the summation of cosine terms this is the summation of cosine terms does it look very nice, but that is what it is what can we do right. So, I am going to rush through the next steps you can do the math if you want it is not all that difficult. So, this is what we substitute we substitute for the summation and then we put alpha equal to delta omega qt plus delta phi q. So, in terms of alpha this is the expression that we get ok. You get something like E0 e to the power i omega t and you understand that the moment finally we do not want to work with field right we want to work with intensity. The moment we want to work with intensity this e to the power i omega t is just not going to be there because we will multiplied by its complex conjugate that term is not going to appear. So, eventually we do not have to worry about it we do have to worry about this sin 2 n plus 1 alpha by 2 divided by sin alpha by 2 this term ok. So, let us write it in a little easier form what is 2 n plus 1 2 into small n plus 1 what is that 2 n plus 1 what is that number the answer is there in the slide here total number of longitudinal mode side. So, the other way we write it is capital N. So, now we have the job of the small 2 n is small n is done let us write it in terms of capital N and this is what we get and here on the quality of slide also the starts getting a little better ok. So, we get E equal to E0 e to the power i omega 0 t multiplied by sin capital N multiplied by sum of the sum of delta omega and delta phi divided by 2 yes. So, you have to consider that all at all frequencies the contribution to the equitability set that will be 0. Yes. But equitability. So, as I said eventually it will not matter because you are going to do Fourier transformation right you can add on the amplitudes at that time that is true but then there is a way out there ok. So, this is what we end up with the form of function is something like if you are talking about intensity sin n x by sin x whole square ok what is x is this delta omega t plus delta phi divided by 2 and plot of sin n x by x sin x I encourage you to do it using whatever graph plotting software you want to use you see this is what you get and this is a kind of function that we encounter everywhere whenever we use optic. One thing we should not forget is that this is a periodic function I have drawn only one but it is not going to be one it is going to repeat. So, you are going to get this and this is what I was saying that picture we drew earlier was not completely correct in the sense that you do not just get a pulse you get like this small you can call after pulse if you want ok. But the issue is even this picture is drawn using a comparatively smaller number of smaller values of capital N when we really work with ultra short pulses this duration is so small that in the same scale it is very difficult to see capital T and T P you get to say T P only when you zoom in to into one pulse ok. So, this is what it is. So, do not forget what is the x axis what is the y axis here. So, this is what you get now I have given you the answer already. Now, let us see if we can deny it. So, first let us worry about repetition rate what is the meaning of repetition rate we can say it is time between 2 maxima where will the first maximum occur sin square n x divided by sin square x where will let have maximum value of course, for that we need to work out limits and so it is difficult right because it is some it might sound like there is a discontinuity when x equal to 0 but it is a removable discontinuity right this point here if you think that point is going to be actually 0 is not it 0 by 0 kind of situation not 0 sorry 0 by 0 is what you will get but that discontinuity is something that you can remove. So, first maximum will occur where delta omega q T 1 plus delta phi q equal to 0 where will the next one occur second maximum. I will give you the answer when it is 2 pi does it make sense right ok just plug this into this expression and you will get the answer at 0 and at 2 pi. So, now if I ask you what is capital T capital T you can get by capital T is basically T 2 minus T 1 ok and since it is a periodic function this separation will be same for any pair of consecutive pulses you take right. So, what is T 2 minus T 1 the subtract delta omega q is constant so what do you get if you subtract one from the other you get delta omega q multiplied by T 2 minus T 1 and now see delta phi q would cancel each other the moment you take a difference that is exactly what will not happen if delta phi is also a function of time right you can do this only because delta phi is not a function of time and it is constant everywhere. So, what do you get delta omega q multiplied by T 2 minus T 1 is equal to 2 pi and what is that T 2 minus T 1 that is capital T time between 2 pulses this one occurs at time T 2 this one occurs at time T 1 ok x axis is time multiplied by something right from there you can always convert to time. So, T 2 minus T 1 is equal to capital T. So, I can do it like this difference in time capital T is equal to T 2 minus 1. So, you get delta omega q multiplied by capital T is equal to 2 pi. So, what is T 2 pi divided by delta omega q right now again we come back to what we promised is going to be useful later on later on time has come what is delta omega q delta is difference omega is angular frequency. So, it is difference in angular frequency between what and what 2 consecutive longitudinal modes very good and you already know some quantity associated with it we know delta nu and what is the relationship between delta nu and delta omega we know that already right. So, what we can say is we can write something like this 2 pi divided by 2 pi into delta nu q because delta omega is equal to 2 pi into delta nu when you go from frequency to angular frequency that factor of 2 pi comes. So, you are left with 1 by delta nu q what is delta nu q yes. So, now from there you get 2 L by C right. So, we see that this repetition rate that we get has nothing to do with how many modes are locked that is the nice thing that comes out because otherwise it would have been a problem how do you decide how many modes are available for you and you decide one thing that is there is what is the spectral bandwidth say to start with let us say your spontaneous emission spectrum that is independent of anything else spontaneous emission then remember we had drawn a laser threshold where half width was delta lambda right. So, where that laser threshold will be who determines that if I can somehow increase decrease the laser threshold then I will have more modes that will get locked are you following what I am saying this figure this here is your laser action threshold okay. Now what I am saying is my first thing that is important is which substance what is your active medium that would better have a broad enough spontaneous emission band okay then what I am trying to say is here I have placed the laser action threshold here is it in any way possible to push it further down remember laser action means you should have stimulated emission is there any way for a given substance that I can increase the stimulated emission bandwidth how can I do it exactly pump harder if you pump harder then this threshold can actually go down okay there is more molecules will be excited and so on and so forth do not forget the stimulated emission is like a bimolecular reaction right bimolecular elementary reaction between light and matter. So, you could think of doing that okay so that way you can get more longitudinal modes and you can try to get shorter pulse that is why in a tricepher laser you want to pump with a high energy you want to pump if possible by 5 watt instead of 3 watt you want to pump if possible by 8 watt instead of 5 watt to get shorter and shorter pulse of course there are other issues damages an issue so you have to take care of those but the problem is if the repetition rate also depended on number of modes they would have been an issue is not it then you will you do not want the repetition rate to change otherwise synchronizing electronics everything will change so it is good that repetition rate is just 2 L by C nothing else matters how hard you pump nothing not so for pulse duration okay so for pulse duration it is not very difficult to understand that what happens let us start thinking a little qualitatively let us go back to this analogy of people running around a field. So, when they come together that is the pulse maximum right if you want a pulse of short duration then they have to scatter faster okay now if you have more and more people running with different speeds will you agree with me that scattering is more efficient there yeah in fact if you look at this interferograms where you mix many waves you will see that more than number of waves sharper is the falloff right that is why you need more pulses to get a shorter pulse duration that is a qualitative understanding now let us see what we get from here okay how do I define pulse duration as we are see practically we always talk about full width at half maximum and when we finish this discussion today we will be talking about full width at half maximum but for now it is easier for us if we consider all the normal modes that are there otherwise again pulse shape and all that will come if you want to consider all the longitudinal modes that are there you want to go from here to here right. So, let us see what is the time about this big maximum where your intensity becomes 0 before and after okay difference between these two times is going to give me pulse duration remember this pulse duration is not full width at half maximum. So, if this is 200 because again that means 200 femtosecond then full width at half maximum is perhaps 100 perhaps less depending on the shape of the pulse okay and this is the shape of the pulse also becomes important right if you want to go full width do you agree with me that ratio of full width at half maximum and full width that is a 10th maximum is not going to be the same for a Gaussian pulse and a Lorentzian pulse right will be different. So, right now we do not worry about all that we just talk about the pulse duration. So, when will they become 0 when the numerator is equal to 0 and the denominator is not. So, that condition is sin n delta phi t plus delta omega t plus delta phi by 2 equal to 0 okay this is one condition where sin x equal to 0 when when x equal to 0 next time when does it become 0 yeah pi okay. So, I can put time as t1 I can say this is t1 for t1 this thing is equal to 0 for t2 the next one it is equal to pi. So, again from here you can find out t2 minus t1 very easily and that is going to be tp the pulse duration yeah. So, do it this is what you get tp equal to t2 minus t1 just subtract what will you get you will get 2 pi divided by n delta n delta omega is that right yeah just simple algebra right. Again delta omega what does by now we are sensitized to this issue whenever we see delta omega what should we do and here this is 2 pi divided by delta omega it is even more tempting go over to delta nu and you know the expression for delta nu it turns out to be 2l by nc. So, just this n is different so we are vindicated you need more number of pulses so that this more number of not pulses sorry more number of longitudinal modes because it this n comes in the denominator okay larger denominator smaller tp that is what you want right. But then we might as well have stopped here if you wanted to have only theoretical but finally we want to go over to something that is more tangible who is going to count the number of longitudinal modes it may not be all that easy okay but there may be something else that is easier. So, let us use another expression and you might have forgotten because we have talked about it I think couple of modules earlier remember this in fact we talked about it at the beginning of the previous module as a recap we worked out that number of total number of longitudinal modes capital N is given by 4l delta lambda divided by lambda 0 square okay. So, might as well substitute this so when you substitute what you get you get pulse duration is equal to lambda 0 square divided by 2c delta lambda okay. What is delta lambda in the treatment that we have followed delta lambda is half width that half maximum of the spontaneous emission band. But what is more important for us is that it is the half width at the base of the stimulated emission band half width at base yeah okay. So, now we have something tangible right we can measure spectrum when we measure spectrum we know what lambda 0 is we know what delta lambda is so it is not very difficult to figure out what the pulse width is going to be that is why in our lab we do not bother about trying to measure the pulse width in time domain. As I said shortly we are going to discuss how to do it see what the problem is we are talking about short pulses 100 femtosecond 50 femtosecond 6 femtosecond how will you measure you cannot see it on an oscilloscope directly okay. So, you have to apply some kind of a trick and the trick that is applied invariably depends on some instrument that was made long long ago okay. Let me digress a little bit I love digression what is the speed of like is that large is that small how do you know it is correct who told you that that is the speed of light I mean textbooks told you but how do they know somebody must have measured it right. So, if I want to measure what is the speed of somodipto it is very easy I can do it using perhaps my wrist watch who what is the wrist watch or what is the stop watch that allowed somebody to measure the speed of light again everybody has studied in the 19, 11, 12 sometime in physics Michelson interferometer the moment I say it you will know it is just that you might know it as Michelson interferometer Michelson interferometer right. So, in Michelson interferometer if you remember monochromatic light was used and there was a relationship from which you could find out the from path length. So, basically what Michelson interferometer does is it degenerates a map like we generate a map of the fluorescence decay that takes place in femtosecond by monitoring second harmonic intensity as a function of delay time. So, Michelson interferometer also generates a map okay and it sort of stretches out does not stretch out it we have sense if it stretches out and creates a wave whose wavelength is correlated with the wavelength that falls on it okay there is a relationship it is not very difficult to work out. So, that is how speed of light was determined. So, now once you know the speed of light you can determine small times using Michelson interferometer kind of arrangement we will talk about that but it is a little cumbersome and I do not know how many chemistry utter first labs in India at all have what is called an autocorrelator it is called an autocorrelator the instrument by which you measure the pulse width is called an autocorrelator and we are happily working in this field without ever having had one why because this is our saving place Tp is 4L delta lambda divided by lambda 0 square. So, if delta lambda because if delta lambda is of some value in principle we can calculate what the pulse width is right if you do not want to calculate what the pulse width is we can get some qualitative value we know from our experience that for a 100 femtosecond pulse laser your full width at half maximum has to be 12 nanometer or so. For the amplifier which is shorter pulse laser full width at half maximum has to be more okay this is a relationship that one can use and work happily in an ultra first lab without ever having to buy an autocorrelator fine. But then it may not be as simple as I have proposed it to be because do not forget what this delta lambda is it is the basal width right one thing that we have not considered so far at all is the spectral shape that is actually important. So, let us close today's discussion why talking about it. So, what we are essentially talking about is transform limited pulses and I have goofed up a little bit on this slide unfortunately bear with me see what are we saying we are saying that a broader spread in wavelength and therefore energy corresponds to a shorter spread in time okay that is essentially something that is absolutely correlated with uncertainty principle once again. Greater uncertainty in time is associated with lesser uncertainty in energy and vice versa okay. So, these are called transform limited pulses of course this is the best case scenario you might have a bandwidth full width half maximum of 12 nanometer but not 100 femtosecond pulse if something else is wrong right now we are talking about the best case scenario. So, let us see why they are called transform limited pulses let us start talking about a Gaussian pulse in time domain okay this is a Gaussian function if I want to know its spectrum if I want to know what that pulse looks like in frequency domain what do we do we do Fourier transformation and when you do Fourier transformation what kind of function do you get another Gaussian okay and this thing that I had by mistake written earlier is that full width half maximum delta omega full width half maximum that is given by this kind of function okay what is the spread in time for this Gaussian function in time the spread is given by 2 tau square root of ln 2 alright now if I multiply that 2 together full width half maximum in time domain multiplied by full width half maximum in frequency domain what do I get I get 0.441 so all this tau and everything cancel of ln 2 is a number okay. So, a transform limited pulse which is Gaussian is going to obey this so if your pulse is Gaussian then it is very nice again no need of an auto coordinator the only thing to remember is you should talk about delta nu not delta lambda okay lambda is an inverse scale right that is why that lambda 0 square factor comes so delta t full width half maximum in time multiplied by full width half maximum in frequency gives 0.441 for Gaussian for a transform limited pulse so what is that why do people buy auto correlators in the first place so far I have told you that we are happily working without an auto correlator without having to spend those few lakhs of rupees that are required to buy it or build it why do people buy at all they buy it because not all pulses are transform limited pulses this is sort of the theoretical limit okay and then let us now close by showing you something else this number 0.441 that we got why did we get 0.441 and 0.8 because we worked with Gaussian function okay so similarly this has been worked out for other pulses as well and these are the numbers forget about this for now for a square pulse the product has to be 1 for a diffractive pulse 0.86 Gaussian we have discussed already for hyperbolic sake these are all different shapes right and there is no guarantee that one shape that always you are going to work with Gaussian kind of pulses Lorentzian 0.221 see Gaussian and Lorentzian we are very familiar with see the differences one is half of the other exponential two sided exponential the product is 0.142 right so depending on what kind of pulse you have what kind of spectrum you have in principle you can work out the pulse width if you just look at the spectrum provided you are working with transform limited pulses okay so that is what we wanted to discuss today next let us see how to achieve we have talked about modulocking we have talked about the theory but how do you do it that is what we are going to discuss in the next few models