 So, I stopped at this point as I told you earlier that I can find out the density of the film from the critical angle in general and also the roughness of the film from the fall of the intensity profile. But usually these films are also of finite thickness. So, what about thickness of the film because that is also important parameter and one would like to know. So, then let me go. So, if I have a film of finite thickness, let us look at this schematic. I have a substrate this is a substrate usually one will deposit a film on this of some thickness may be 100 angstrom may be 200 angstrom 1000 angstrom. So, now I have an incident beam which gets reflected from the surface and we know from optics, optics even optics tells us that whenever there is a refractive index mismatch, index mismatch there will be reflection. This is our daily experience even in our windows. Even if the window is transparent you will find that when you look at it carefully you will can see your image a light image because there is a refractive index mismatch between air and the glass you will have reflection. So, same thing is true here the whole phenomenon is taking place at grazing incidence, but the facts are same. So, I have an incident beam which partially gets reflected from the surface and then we have a transmitted beam which gets reflected from the substrate and the film interface here I have got one more reflected beam and these two beams they will interfere they will interfere and that interference pattern will be stamped on the reflectivity plot. So, quickly let me just tell you mathematically the reflectivity is given by a finite thickness film let me just show it like this is a finite thickness film this is the substrate. So, it is given by the d rho z by d z e to the power i q z. The Fourier transfer of the density gradient in the z direction q is the wave vector transfer d z this is what i q. Now, if I have a film of constant density of thickness l let us say then the density gradient we will have two delta functions here this is the film this is the film and this is the density you can see that one delta function will be at the film air interface and the other delta function will be at the film substrate interface. So, basically let me just plot it you might so, if I consider rho z let us say this is a film air interface it goes the film goes then this is a substrate this is the density for the substrate this is the density for the film and this air has got a refractive index 1. If I differentiate this this will look once if I differentiate this d rho by d z versus z if I do this will give a delta function here of certain amplitude and the delta function here there are two delta functions and this l is the thickness of the film. So, in that case the it becomes r f delta 0 plus delta at l e to the power i q z d z. So, ultimately this comes to r f delta at 0 gives me 1 plus 3 power i q l this will be my reflected intensity if I have a film of finite thickness. Now, from here you can see whenever q l is equal to twice pi then this becomes 2 when it is pi it becomes 0 and so, there will be oscillations and whenever there is q l equal to 2 pi you will have a peak in the reflected intensity. So, this is what I want to show you. So, this r equal to r f into 1 plus e to the power i q l because there are two delta functions because the reflections are taking from the top and bottom of the film and there will be oscillations and the oscillation width in q space is given by twice pi by f. So, this so, if I call it delta q because twice pi by l 4 pi by l 6 pi by l they will give oscillations. So, the delta q between the oscillations will be a signature of the thickness of the film. And now I show the example these are experimental examples. So, this is a FEPT film that I had measured and you can see. So, this is a consolidation the reflection from the top and reflection from the bottom of the interface and that gives me in this is an XLR profile this is a polarized neutron profile. So, there are several things that needs to be explained here. One is that in case of XLR you can see the Kiesig oscillation this is a copper K alpha copper K alpha radiation 1.54 angstrom. So, with this we get this oscillation. So, this is this should have been if for an infinite film it should have been a flat one, but here you see these oscillations these are known as Kiesig oscillations. So, this is a schematic of the film of iron platinum where the film is of finite thickness and then this is deposited on silicon wafer. So, silicon substrate and you can see the finite thickness of the film actually gives you this Kiesig oscillations and from the fitting of this whole data you get thickness of the film here. The same film when you use polarized neutron reflectometry one you will have different critical angle for up and for down neutrons plus this same Kiesig oscillations will be visible on the falling edge of the reflected intensity. No Kiesig oscillations below critical angle because below critical angle the film the intensity or the beam the radiation does not penetrate the medium. So, you cannot see any kind of oscillations below critical angle, but above critical angle because your beam is now penetrating the film you have got these two components and these are the polarized beam intensities together with the rough together with the thickness signature as Kiesig oscillations on the falling edge of the reflectivity pattern. So, by fitting this we get magnetic moment density, magnetic moment density, physical density from both XRL and PNR because this gives electro density which can be taken back to physical density, this gives nuclear plus magnetic moment density which can be taken back to physical density. So, I get physical density, I get magnetic moment density only from PNR, I get interface roughness from both of them. So, all these parameters have been extracted for these experiments and usually when you make a film the before you take it to polarized neutron reflectometry it makes sense to do the XRL on the same film provided it is a high C material film and in not a biological film which has got hydrogen as a deuterium material because in that case XRL will not give you anything. So, now I will go one step back. Now, I talked about infinite medium which gives you Fresnel reflectivity, I gave you finite thickness film, finite thickness. Now, what about a film which is actually a repetition of layers, often we study such films. So, that means the film will look somewhat like this, let us say this is a substrate I will call it as will A of certain thickness D1, A then film B not film layer B let me call them layers, layer A layer B of certain thickness D2 then again layer A of D1 thickness then D2 thickness and then I continue like this for some time maybe some 10s or 20 such layers are deposited. So, this structure this thin film structure we will often call them thin film heterostructures in this direction it is a periodic structure periodic structure you will only agree with me it is a periodic structure. So, typical let me say this A can be 100 and strong B can be 50 and strong so on and so forth. So, this is a periodic structure, but the length scales are very different from what you have seen for crystals crystalline materials at atomic level, but this periodic structure is also similar to a crystal I call it pseudo crystal or you can call it a crystal and mesoscopic length scale and if there is a crystalline structure one-dimensional crystalline structure we have seen in our master's day in Kittel that for n number of such layers you will have intensity which is sin square n q by 2 upon n q d by upon sin square q d by 2 and as n goes to infinity this goes to a Bragg peak. So, this synthetic crystal should also give me Bragg peaks in reflectivity profile since this structure is large. So, this Bragg peak will come at low q here I must remind you that I have shown you a crystallographic structure for magnetic flux lattice using small angle neutron scattering that was very similar to a Laue experiment which you do with crystalline materials which are ordered at atomic level, but that Laue pattern was seen at low q region because again vortex lattices are very large in size. I am taking another example here that if I grow such bi layers of periodic nature then I will see a Bragg peaks and I am just showing you data unpolarized neutron reflectometry data from a nickel field and you can clearly see that the critical angle the what you have been discussing so far is very much there, but the key zig oscillations due to finiteness of the field it is there. So, this is coming because this is a periodic bilayer crystal with 10 bilayers that means I have deposited nickel and titanium in this case nickel and titanium one above another and we have got 10 such bilayers. So, the key zig oscillation is here due to the total thickness of the field. So, there are so there is nickel, titanium, nickel, titanium and 10 such bilayers. So, there is a total thickness of the field. So, the key zig oscillation signifies the total thickness of the field, but also because this is a periodic bilayer of periodicity d periodicity d d equal to d 1 plus d 2 if I may say I should also get a Bragg peak satisfying 2 b sin theta equal to n lambda. In this case I can see only one Bragg peak the key zig oscillations in the critical angle. So, from this we can fit and get the structural information including the magnetic moment density in the nickel layers the interface roughnesses and the magnetic structures. So, but now the case is that how do we calculate the reflectivity or how do you fit the reflectivity profile for a periodic bilayer or periodic material or a single film or a bilayer or any thin film structure that I want to study or we have studied. So, I have talked to about Ritwell fitting regarding crystalline structure. So, there in Ritwell refinement we start with the assumed structure of the crystalline solid and then keep refining that till we come to a model which fits my experimental data. Then I discussed to you about liquid and amorphous solid they are also we start with a model structure three dimensional model and then we kept modifying it till we got a good fit with the experimental data. So, basically I am always starting with a model in R space in real space space then find out the experimental data in Q space and for all the experimental points and then I so, this is the model for ith point and then compare it with the experimental data squared them sum over i with some constants and like chi square. So, I always start in the real space produce the data or rather the data that corresponds to the model in Q space and then compare it with the experimental data. I am trying to reduce the error bar here also we go to the similar exercise, but here to generate the reflectivity pattern of a layered structure we use parents formalism. This is a very robust formalism introduced by parent together with the possibly the first most famous reported data which I mentioned earlier of extra reflectometry and the parent formalism can be equally applied to x-rays and neutrons to generate the reflectivity pattern of a model layered structure. So, the model layer structure it looks somewhat like this you have a substrate on which you grow your film can be bilayer or it can be single layer it can be aperiodic it can be trilayer whatever and any layer is signified by its thickness in this case and how do you evaluate the reflectivity profile of a film with n-drawers well briefly introduce you to parent formalism. So, earlier I showed you that if I have only one interface if you remember I used e to the power i qz plus r e to the power minus i qz and t e to the power i q to z for the transparent. Now, I have got a number of layers number of layers. So, I have to do the same exercise layer by layer. So, now let me consider the n-th layer and the n minus one-th layer in this multi-layer step. So, my layer number is going up as I go inside them I should call the thin film hetero structure. So, let me just highlight the n minus one-th and n-th layer. So, here at this interface I have to do exactly what I did earlier for a single layer. So, now let us see I can write if I consider at the center of the layer this is the thickness n minus one is there is the n minus one thickness and n-th layer is d n thickness and at the center of the n-th layer I consider there are two parts of the wave function like this one is psi another one is psi r and n is the suffix talking about the layer that I am discussing. So, for such a setup I have to consider this interface and this interface. Now, let me just talk about a phase factor this phase factor because of the psi and psi n I have assumed at the center of the layer center of the layer. So, at the n-th layer it is here. So, I have to consider now this phase factor this is a in the n minus one-th layer it is the psi at the at here. So, I have to take this psi forward by e to the power minus i q n d by two because it moves d by two this phase factor with respect to the center of this layer changes by e to the power minus i q n d by q n minus one d in minus one by two and that I so a n minus one equal to e to the power minus i q n minus one d n minus one divided by two. So, if I go forward this psi n I have to multiply the forward propagating forward propagating wave function part of wave function there is only z component because x and y component I will ignore here the x z I can consider the reflection plane and I am the things the phase factor is changing only with respect to z. So, if psi n minus one was at the center of the n minus one-th layer I take it forward. So, it becomes a n minus one psi n minus one. Now, with this I need to add the reflection amplitude just I did before. So, here so now, but the reflected beam it has to be brought back. So, it will be e to the power plus e to the power i q n minus one d n minus one by two I will write a n minus one inverse psi n minus one r. Now, consider this layer I am talking about equality of wave functions at this interface. So, this becomes equal to a n inverse because here the forward moving wave function here has to be taken backward. So, I have to do a n minus a n inverse psi n plus a n psi n sorry the reflected part psi n and we will write it clearly in the next page. So, for equality of the wave function a n minus one psi n minus one moving forward from the center of n minus one-th layer by distance d by two to come to the interface plus the reflected beam inside this layer I have to bring it backward at this point I have to bring it backward by d by two. So, it becomes a n minus one inverse psi n minus one r and that should be equal to this is the n minus one-th layer this is the n-th layer. So, from the center now my psi n has to be brought back. So, exactly opposite I have to do and the reflected beam is anyway moving in this direction inside n-th medium. So, this is a n psi n r. So, this is for equality of psi. The equality of d psi by d z is. So, this is equality of psi and just differentiating ones I get from here a n minus one psi n minus one into q n minus one and just because e to the power i q n minus one z it gives you i q n minus one and it gives q n. So, this is from the. So, this is from psi this is from the equality of d psi by d z where this phase factor basically takes care of propagating the beam from the center of the medium to the interfaces. So, with this I have got two equations. Now, I know that the reflectivity at the n minus one and anything beam interface is given by psi n r divided by psi n which is there hidden in this equation. So, now I have to solve these two equations to get psi n minus psi n r upon psi n this is the incident sorry reflected and this is the incident beam that is what I did at last time. So, to solve these two equations I have to just add and subtract them I am not doing the algebra right here rather the results are more interesting results are more interesting. So, I can write down parameter or n minus one n this is given by a n minus one to the power four where a n minus one was e to the power minus i q n minus one d n minus one by two. So, apart from this constant factors to the power four means it will be e to the power twice i this, but in that you have r n minus one n which is n square psi n r by psi n this is the reflection amplitude if I had only n minus one and nth layer because if you remember I derived it for you for a medium which is at the medium infinitely thick medium and air q one was the q parameter in air and q two was the wave vector transfer in the medium. So, here it is wave vector transfer in n minus one layer minus wave vector transfer in the nth layer. So, it is similar to this. So, this is one parameter f in comma n minus one and this is the reflectivity parameter r n minus one comma n, but I have solved it, but I have a problem here you see this is a recursion formula for r where I can go from layer to layer and keep calculating r n minus one n and then by squaring I can get the scattered reflected intensity, but interestingly to get r n minus one n I need r n comma n plus one. So, that means to get the reflected intensity at the interface of n minus one and nth layer I need the reflected amplitude at nth and n plus one nth layer. So, that is the problem because I do not and to get the reflected intensity amplitude at the n and n plus one nth layer I need the reflection amplitude at n plus one and n plus two nth layer. So, how to get rid of this problem and Parat did made an excellent assumption. So, to get n n minus one I get n into n n plus one and to how to terminate this. So, what was assumed by Parat and even today we do the same thing whatever enters the substrate nothing comes back it goes out from the bottom of the substrate. That means with your film going deep you have a substrate which is very thick and whatever enters the substrate nothing goes back that means at this interface I consider r n n plus one equal to zero I get nothing back from the bottom of the substrate and now once I put r n n plus one n is the large n is the last layer and the substrate. The last layer is the nth layer whatever goes inside the substrate nothing comes back once I put it here I can keep going up and keep calculating the reflected intensity there for each and every layer and ultimately I can arrive at the film air interface provided I have made a model in which I have assigned density thickness for each and every layer. So, that means now I have got a handle. So, I have got I can make my sample it can be even a single layer not need not be a multi layer but I can assign I can break it up into layers and I can assign thickness density for this layer even magnetic moment density. So, magnetic moment density for each and every layer and for that I can calculate the thing at the film air interface what we measure actually I take the whole stack and put it in the reflectometry machine instrument and measure the reflected intensity but assuming that r n n plus one equal to zero that means nothing comes back from the substrate and with that I can have a model and I can have a model reflectivity pattern which I can compare with the experiment. So, parrot formalism allows us to calculate the reflectivity profile for a given multi layer stack with density together in magnetic moment density if it is polarized neutron reflectometry that define for the layer and now we can try a model solution and then we go back to the same technique what we do for all other diffraction techniques we try to do a tri square minimization. So, I will take up in the next part the data analysis how we do it in the next lecture.