 I have written it here, this is the expression, only thing is that one that I have written that rho e electron number is equal to ni zi as I wrote there, this is the classical electron radius. Here if I introduce the absorption in a medium, x-rays are absorbed heavily, this is the linear absorption coefficient, then it gets n gets slightly modified, it gets 1 minus delta, but there is an absorption term which is i beta, so this basically e to the power ikx if I take, with this i beta it gives, this gives the absorption in the medium and reduction in the propagated vector. Here it is normally, we know many times we write e to the power ikz for a plane wave propagating in, here it will be nk if you consider in a medium and in that medium if you have absorption then this will give the e to the power minus some terms, it will give the reduction in the wave propagated in a medium, so this is for x-ray. Now let's come to neutron, we have discussed in our lectures in the basics that the v of r potential given by 1 neutron is twice by h square by m delta r into b, it's a constant b coherent, we wrote b coherent basically average scattering length which gives us diffraction and this logic I have used earlier, I am just telling you that the fundamentals of using delta r as the potential spatial extension is because the nuclear dimension is just femtometer and the thermal liter wavelength is around angstrom and b coherent is the coherent scattering length, just to remind you if you remember I wrote it b coherent e to the power i q dot ri i that is equal to scattering vector that is equal to scattering amplitude apart from some constants, I wrote it earlier, so the same b i because now the potential is a delta function but now I am talking in terms of densities, so when I am talking in terms of densities then my v, I wrote v is equal to twice pi h square by m b coherent delta r for one scatterer, for one, now I am talking about a medium with a certain density rho, density means so many scatterers per unit length, I am sorry unit volume then my v I can qualify it as coherent into rho, so this is given now by the density, so now when the neutron approaches this medium it sees this potential, now please consider the fact that I am doing experiments at very low q, low q means the spatial resolution by uncertainty principle is twice pi by q and this delta r allows me to use the density for the medium as the potential, so now we have got a potential which depends on the density, so as the neutron approaches the medium it sees a potential which is given by 2 pi h square by m b coherent rho, so it sees a step potential, now the neutron is travelling in the medium, so this is how it looks like, so with the density I have this potential and the neutron is travelling in the medium, I show the neutron energy greater than this potential otherwise it cannot go through the potential and you have to talk about tunneling but here it is not tunneling, it is the electron energy in the free medium, now I can tend to draw the similarity between x-rays and neutron, now e inside the medium is equal to energy in the vacuum minus the potential, so for a free neutron it is h square k square by 2 m, this is h square k prime square by 2 m minus twice pi h square by m rho into b, I am just taking out the subscript of b coherent, this is what I get, so k square equals to multiply by 2 m by h square k prime square minus 4 pi m h square by h square rho b equal to, sorry I just made a mistake here, here it is the energy in the medium, so k prime square and here it is k0, in the medium the apparent energy is lower compared to because here if you consider 0 this is the energy of the medium outside them when the neutron is outside the medium and this is the energy when it is inside the medium, so it is e0 minus the height of the potential which is v, so I am sorry for this, so let me just write it correctly, so it becomes k prime square equal to k0 square minus 4 pi h square cancel, h square square m by h square m rho b equal to k0 square minus 4 pi rho b, this is what it comes, so now once I write this, so this comes, so now n equal to k1 by k0 that is the definition of refractive index that wave vector inside the medium and divide by wave vector outside the medium, now if I write k1 by k0, I can write it as equal to as I showed you here k0 equal, so I can write again k prime equal to k0 minus, if I consider again this is equal to small then it goes to the power half, so k0 square minus 2 pi sorry this is one second, so open it, so k1 by k0 is given by from here I can just write k1 square by k0 square equal to k1 square, k prime square by k0 square equal to 1 minus 4 pi rho b and then this goes to if I take the square root of k prime by k0 this will be 1 minus 4 pi rho b to the power half which is equal to 1 minus 2 pi rho b and this gives me the refractive index in terms of now I can switch it back switch back to lambda from k prime to k and then I can write because lambda is equal to twice k is equal to twice pi by lambda, so n equal to 1 minus lambda square by 2 pi b coherent rho which is again 1 minus delta, this expression is exactly same as what I got for the x-rays here it was 1 minus lambda square by 2 pi r e rho e here it is 1 minus lambda square by 2 pi b coherent into rho, there it was density of electrons here it is scattering length density multiplied by scattering length, so that gives me the total scattering length per unit volume and n is given by that, so very very similar to what you got for x-rays that, so I tell you once again this is k1 square I just multiply the whole thing with twice m h square twice m by h square I get k1 square equal to k0 square minus this factor and then k1 by k0, k1 square by kz square is 1 minus 2 pi rho b coherent and when I divide it by when I square root it assuming that these are very small value as I said that 1 minus epsilon to the power half is equal to 1 minus half epsilon approximately under the approximation I get the value of refractive index for neutrons, so so far in this expression I have not brought in any magnetism but if I consider b coherent as a coherent scattering length and if I add scattering length which is magnetic in nature then I can tell you that for polarized neutrons if there is this is b coherent which is nuclear in origin now if there is a b magnetic so there is a magnetic moment density in the medium then for and if my sample is polarized and I have got a polarized beam getting reflected from here sorry missed that yeah yes so a polarized beam gets reflected from here then for up polarization and down polarization with respect to the sample magnetization my potential is b coherent plus minus b magnetic so either my potential becomes slightly higher for the plus neutrons or it is slightly lower for minus neutrons so now what does that mean now let me talk in terms of a critical angle then it will become clear what does it mean so first I talked about nuclear potential delta there is a v which gives me a refractive index which is n equal to 1 minus lambda square by 2 pi b coherent rho which is 1 minus delta as I told you this delta is small delta is of the order of 10 to the power minus 6 now I find that this potential height changes if I have a magnetic scattering length in the medium then when I reflect polarized neutrons then neutrons polarized parallel and anti-parallel will see slightly different potentials this is very very important for magnetic neutron reflectometry or polarized neutron reflectometry I will come to it and I will use this result again and again so now let me just sum up what you found common is that whether it is neutrons or x-rays the refractive index is given by 1 minus alpha minus i beta slightly different from delta but this expression this is the this is the part coming from electron density or scattering length density this term is coming from absorption so this is what I have written here that alpha is given by rho lambda square by 2 pi b coherent which I showed you in the previous transparency this is b coherent giving this height but you have plus or minus b magnetic depending on the polarization of the surface I mean polarization of the magnetic thin film and the polarization of the neutron with respect to it and this is beta neutron is an absorption term for neutrons often it is negligible except when you are talking about neutron absorbing medium and reflection from them like cobalt gadolinium cobalt gadolinium they are strong neutron absorbers otherwise this is often zero for almost a very large number of materials in case of x-rays we have a very similar expression but f0 plus f1 I have written here rho lambda alpha x-rays for f0 plus f1 and here f0 comes from sum over ni zi which I wrote earlier here ni zi but the fact is that there is some deviations from this number as you go up in z values so that's why I call it the dispersion term as f2 and beta x-rays exactly same as beta neutron which is given by the absorption coefficient of x-rays and that's large for almost all the medium the b's and f's are the respective scattering length for x-rays and neutrons and r0 here as I used it earlier it's a classical electron radius so both these techniques they can give us density of a thin film I'm just exaggerating the thickness of a film on a substrate density of a film as a function of z if the z axis is the one which is going inside the medium starting from vacuum as we go in this reflected intensity of x-rays and neutrons gives us the rho z values from reflectivity data however I'll tell you in the next module