 Hello, before I go on to the next module allow me to correct a small mistake that I made Let me just do it on the board. I found that k1 squared equal to k0 squared minus 4 pi rho b k1 squared by k0 squared will be equal to 1 minus 4 pi rho b by k0 squared so n is equal to k1 in the medium k1 by k0 equal to and k0 in the vacuum it is given by twice pi by lambda square so it goes to 1 minus 4 pi rho b divided by 4 pi square lambda square and then in square is this equal to 1 minus lambda square by pi rho b this is what I want n squared and this is a small I said this is small and 1 minus epsilon to the power half is equal to 1 minus half epsilon Applying that I get n equal to 1 minus lambda square by 2 lambda square by 2 pi rho b Can I go back there? This is what I obtained from Assuming the K changes as the Newton travels in the medium energy was zero and here I assumed the energy is E1 or E prime now I also showed you that the refractive index depends in case of x-rays it depends on electron density in case of neutrons let me emphasize it depends on coherent scattering length density plus for a magnetized medium it depends on magnetic magnetics scattering length which is added or subtracted to the nuclear part depending on whether the medium is polarized and the neutron polarization is parallel to it or anti-parallel to it Now quickly I'll tell you polarized Newton and extra neutral reflectometry as I told you they are parallel techniques complementary techniques and it depends on neutron nucleus interaction in case of polarized Newton and unpolarized Newton reflectometry and for polarized Newton reflectometry with unpaired magnetic moment so that means Take a simple example. Let me take iron. It's a 3d group Magnetic material so every iron has got a Unfilled shell which giving is giving rights to his magnetic moment So with that is the neutron can interact with the nucleus with with be coherent plus it also interacts with the Unpaired electron So we get information on number density through coherent scattering length density and also magnetic moment density from polarized Newton reflectometry in case of extra reflectometry it is extra in the charge cloud Interaction and information on density through electron scattering length density. So that means on the same sample I can use coherent scattering density in case of Neutrons to get the density of the medium an electron scattering length density I can translate it to physical density and This is a unique Technique that using these two. I will show you later. I can actually obtain the density of medium Which has got two elements or a density of a binary phase binary? Two component medium So this is a quick comparison between polarized Newton reflectometry and extra reflectometry now Let me just try to figure out what should be the reflectivity of a medium and What should be the critical angle for a medium that I discussed earlier now? This is what I want to show you that for Specular reflectometry you can see there is an incoming energy and intensity incident beam There's a reflected beam and there's a transmitted beam in the medium This is what we did even in schools when you did about talked about refraction now here Because the medium has less than one reflectivity then it has got a total external reflection And the critical angle for reflection So I'll just derive it for you. So in Refractive index is given by my If I consider this Angle from here From the surface and not with the normal then it will be cos theta theta 1 or theta i theta 1 theta 2 theta 2 cos theta 2 So n is given by cos theta 1 by cos theta 2 is exactly same if I had measured the angle from it You have been sin theta 1 by sin theta 2 now n equal to cos theta 1 by cos theta 2 and When this angle is equal to critical angle then cos theta 2 becomes 0 The beam starts propagating Exactly along the interface in that case cos theta 2 theta 2 equal to 0 and Theta 1 I call it theta 1 equal to critical angle theta c when that happens then n is equal to cos theta c So cos theta c and equal to 1 minus lambda square by 2 pi rho b 1 neutron and cos theta c under the assumption that theta c is Almost equal to 0 it is 1 minus theta c square by 2 Equal to 1 minus lambda square by 2 pi Rho b so from here theta c or the Equal to lambda Rho b by pi so Please note this thing the critical angle is decided by the Scattering by the scattering Scattering length Density Now this answers our question that How we can find out density of a medium? By measurement of reflection angle so one is that measurement of the critical angle theta c So this is what so now theta c x-rays is With the same method I can write either lambda equal lambda square root of this is Classical electron radius Electron density by pi square root of that and theta c neutron is Scattering length density of neutrons divided by pi so the critical angle depends on electron density in case of x-rays and Coherent scattering density in case of neutrons, but so far in this expression I have not added magnetism otherwise It will come as b coherent plus or minus b of m Now how do I evaluate the? Reflectivity at an interface so in case of x-rays it should be as we did in our Bachelor's degree that the electric field in case of x-rays e and De by dz should be continuous in case of matter wave matter wave like neutrons It should be the wave function psi and It's derivative d psi by dz y. Let me just show you So I'm talking about an interface in which reflection is taking place specular reflection So I've got a incident beam e to the power i kz i kz I have got a reflect. I have considered that it is Amplitude is unique then if R is the Reflectivity coefficient you have got R to the minus i kz And I have got going in the medium t to the power i This is since I'm considering a z component only I'll let i q1z Explain to you I Q2 I q1z and in the medium I Q2z why so let me explain to you This is the incident beam This is the outgoing beam. I have just shifted it. So for this reflection This is the z component This is the incident beam The reflected beam I shifted there So this is the Q In the medium and same things true for the scattered beam now I call it q1 which is in vacuum in case of Reflectometry in sales sales log the Along the plane component they remain continuous now only component of wave vector which is changing is The z component which I'm calling it q1 here and cute in the medium and I'm Translating this propagation as a propagation only in z direction Because that is the only vector which is changing from one medium to another. So now I have got incident beam e to the power I q1 z plus Reflected beam to the power minus I q1z why why I minus I q1z because This beam is I have taken this beam is propagating in plus z direction If that is the case then this beam is propagating in minus z direction It's propagating in this relation the z component only x component xy component. I am not bothered So this is a if this is a plus z Direction, this is a minus z direction. So the incident beam unit vector unit amplitude Reflection amplitude and this is the propagation and for the transmitted it is t e to the power I q2z So now I have got the incident beam e to the power I q1z plus e to the power minus I q1z and I have To match it with t e to the power I q2z So if I consider At z equal to zero then one There are plus r should be equal to t. This is equation one If I consider d psi by dz d psi by dz You can see it comes to I q1 Plus r minus I I I q1 r so I q1 e to the power I q1z minus I q1 e to the power minus I q1 z One r here equal to t I q2 e to the power I q2z To solve this equation. I have put my origin At the interface when z equal to zero that gave me one It gave me at z equal to zero one plus r equal to t and the other part is Here if I put z equal to zero I will Cancel from both the side So it will come to q1 minus q1 r equal to q2t So this is from continuity of psi Across the interface I have put the interface at z equal to zero z equal to zero and this is d psi by dz at z equal to zero If I solve these two equations, it's very easy to do what I get r equal to q1 minus q2 upon q1 Plus q2 this is a reflection amplitude And this is what I have written here And this is very easy if you write q1 equal to 4 pi by lambda sin theta And all those expressions you will get this expression for r and Reflected intensity is equal to r into r star What a reflected intensity So this is the reflected intensity for an interface Between vacuum And a medium of infinite thickness Infinite thickness Thickness So and the reflectivity pattern that I will get in q space Basically, I will be studying reflectivity as a function of angle If it's a reactor source Or as a function of wavelength as well as angle if it is a spallation neutron source But we will be probing the q space and the reflected intensity and I just show you the experimental result for a highly polished silicon surface This is how it looks like Please you need to digest this that since there's a critical angle As I showed you earlier that up to a critical angle The reflected intensity is one And beyond that the reflected intensity falls And actually when The q is very large it falls as 1 by q to the power 4 Why I will come to it later So it falls rapidly We're at large q and up to a certain critical angle Theta c It is one now this critical angle is a signature of the density of the medium because that critical angle is given by lambda root over rho b by pi Or in case of x-rays lambda root over rho e Re by pi So x-rays will give me The electron density if I measure it with neutrons it will give me scattering length density in this case it is x-rays And this tells you what is the density of the medium From whose interface we get this reflectivity pattern With this I stop today I'll carry on in the next module next lecture With discussions on reflectivity