 In this material characterization course, in the last class we just looked at the some of the fundamental properties of x-rays as a waves and then we also looked at the kind of relations that is we talked about phase relations and then and we also talked about how the x-ray waves are represent represented mathematically in the exponential component form as well as a trigonometric form and then finally how if somebody want to use it these expression for arriving at an intensity what kind of expressions we will look at it. So in continuation with that discussion we also slightly looked at a diffraction. I said that most of the phase relations we talk about in order to appreciate the phenomenon diffraction. Today I will just briefly discuss with some of the animation diagrams to illustrate how the diffraction can be appreciated using the phase relations between the waves. So I will just start with this animation where you have let us assume this is a plane polarized wave that means their electric vector is in the same plane of the drawing and then I am going to consider two waves like what is depicted here and then what you are now seeing is that wave is propagating in particular path and then in reaching the wave front B B dash and it starts from the wave front A A dash and what we are now interested here is just look at this wave and for the hypothetical I mean I mean situation you imagine that this wave is splitted into two waves like 2A and then 3 and these waves will have the amplitude one half of this wave 1 this is just an assumption and also you assume that these two waves are in the same phase and first I would like you to compare the 2A and then 3 these two waves are traveling in two different path the wave 3 is traveling in the straight line but wave 2 is traveling in a very different path as compared to wave 3. In fact the amplitude as I mentioned amplitude of these two waves 2A and then 3A should be half of this though this is not quite obvious from this schematic diagram please assume that these two waves are having of the amplitude of this that means we are splitting this wave into two for an imaginary experiment. So now you see that the wave travels this path and then arrives at the wave front B B dash and here also you can see that the wave which travels in a straight line arrives at the wave front B B dash you see these since these two waves have traveled in a two different path length and there you can see that the the wave where they enter the wave front B B dash they are not the same you can see that at this intersection you can see the wave the amplitude is 0 on the other hand the 2A wave where it try to rejoin the the wave 3 where the intersection point you have the amplitude is maximum. So you have very clear demonstration here when the path difference is there and there will be an impact on the amplitude or there will be a change in the amplitude like this you can see I will repeat since the wave travels in the straight line and where it enters the B B prime wave front the wavelength is I mean the amplitude is 0 here you can see that intersection since the wave has taken a different path and it causes a path difference in this case the the wavelength is I would say that the wavelength is enlarged to the quarter wavelength we can assume that so the amplitude is maximum here here the amplitude is 0. We can also assume suppose if you allow the travel I mean allow the wave to travel like this for example in 2B case where your wavelength is the I I would say that the wavelength difference if it is instead of quarter wavelength here if it is half a wavelength then the two waves will be out of the phase by half half wavelength the phase I mean phase shift between the 2A and 3A is about a quarter wavelength. So you can see that similarly if you allow this 2C to travel in a different path and then the wavelength difference is between the the third wave and the 2C wave it is one wavelength completely then the this wave and this wave will be out of the out of phase by one full wavelength but in this time the wavelength though it is out of phase by one full wavelength and it is completely indistinguishable because it has got both I mean you can see that amplitude they are in this same phase. So certain things are very clear here you can just note down few points the two conclusions may be drawn from this illustration differences in the length of the path travel lead to differences in the phase the introduction of phase differences produces change in amplitude. So I think these two points are very clear you can see in this case every wave is is traveling in a different path assuming that the their wavelength is different at least I have shown at this intersection the amplitudes are quite different if they follow a different different travel path. So the greater the path difference the greater the difference in the phase since the path difference measured in wavelengths exactly equals the phase difference also measured in wavelengths. So this is again a very important point though we have we have come across this in the light optical system also and we are reinforcing that understanding here with this illustration. So differences in the path length of various rays arise quite naturally when considering how a crystal diffracts x-rays. So finally we are interested in looking at the the scattering of x-rays by a crystal system and then there we are going to connect this concept to a diffraction. So we need to understand how this path difference will contribute to the the condition for a diffraction that is the ultimate game. So you can see that a diffracted beam may be defined as a beam composed of large number of scattered rays mutually reinforcing one another. So this is a kind of a definition for the a diffracted beam but then how do we qualify this? How do we qualify this statement? How do we visualize this statement? And then and we can come back to the statement and then say this is what it is. So for that what I will do is you look at this schematic again it is an animation. I will play it what you are now seeing is a diffraction of x-rays by a crystal. So okay we will stop here and then we will go one by one. So assume that this is the 2D representation of a three-dimensional crystal A, B, C are different planes and then you have the perfectly monochromatic x-rays are falling on this surface and this is a angle of incident theta and this is angle of reflection theta. So this theta what we are seeing here in x-ray diffraction is slightly different from what we talk about in general light optics where the the angle of incidence and angle of reflection is always with respect to the surface normal to the whatever the angle of incidence and angle of reflection we talk in light optics is with respect to the surface normal of the plane surface plane. But here it is slightly different it is this theta is an angle of incident and angle of reflection is considered in this fashion and that is one point and assume that this is the wave front x x prime and then after the reflection it goes to the goes to the y y prime the wave front and this is the normal we are trying to take a normal to this the first x-ray path that is 1 k 1 prime this is the way we are now going to talk about and now let us see I will put another ray second ray which that is 2 L again this one 2 prime and then you have 3 and 4 and 5 and so on it will keep on going. So now we will talk about the atoms sitting in that top surface plane and it is reflection and how it is contributing to the diffraction intensity for example this is we will talk about the the the second ray little later we will talk about the first plane now suppose if I introduce one more ray like this and then which again comes and hits the atom P and then it reflects and then you can designate this as 1a. So now you have 2 rays 1 k 1 prime and 1a P 1a prime that is 2 rays which are trying to diffract from the the first plane. So now we have to see whether these 2 rays have any path difference we were just talking about a path difference just now. So to to understand that we will just do this the path difference of 1 and 1 prime and 1a and 1a prime between the wave friends x x prime and y y prime is equal to q k minus P r is equal to p k cos theta minus p k cos theta is equal to 0. So let us go back and then look at this q k minus p r q k is there and this is p r and this is nothing but your p k cos theta for both cases whether it is q k or p p r is equal to p k cos theta. So in this case both the the path difference is 0. It is not just this 2 rays any ray any number of ray which comes and hits at the plane 1 will have will not make any path difference like this. So what is the significance? Similarly the rays scattered by all the atoms in the first plane in a direction parallel to 1 prime or in phase and add their contribution to the diffracted beam. So all this will be in phase and then they will contribute to the diffraction. We are still we are yet to qualify the diffraction we are now saying that these 2 rays will have the their phase I mean they will have the same phase. So that means I am also saying that any number of rays which falls on this first plane will have their phase in the same orientation we can say that and the important point is if all this rays reinforce with each other with the the same phase and then they contribute to the diffraction intensity that is what we just made a statement here. A diffracted beam may be defined as the beam composed of large number of scattered rays mutually reinforcing one another. So this is now qualified for the first line first line of atoms. Please remember whatever the schematic is shown here it is shown assuming that these rays are diffracting. In fact the moment the x rays hit on the atoms p and k it scatters all over the direction all the direction it scatters but only the 1a prime and 1 prime are shown as a diffracted beam. Please you have to understand the diagram. We are assuming that only these two waves contributes to the diffraction intensity it is not that I mean other scattering rays are not there they are there they are not in the phase they are not in the same phase that is the that is the meaning you have to look at it. So now what about the other rays which is coming just below this plane you have because we are talking about a three-dimensional crystal lattice and we are only representing with the 2D I mean lattice and you will have an infinite number of planes here and above and so on. So now we will see what is the condition where the ray 2 will have to get it reinforced and contribute to the diffraction intensity. So let us talk about the ray 2 L and 2 prime and if you look at the path difference by drawing a normal m and then n and then we can look at ray 1 and 2 for example are scattered by atoms k and L and the path difference for rays 1 k 1 prime and 2 L 2 prime is m L plus L n is equal to D prime sin theta plus D prime sin theta. So what is that? So these planes are separated by the distance D prime and the path difference is m L and L n this is nothing but D sin theta. So that is how it is measured. So this is also the path difference for the overlapping rays scattered by s and p in the direction shown. Since in this direction there is no path difference between the rays scattered by s and L or p and k. So let us go back and see s and L, p and k. So in the same line there is no path difference similarly there is no path difference in this but when it crosses the next plane there is a path difference but that is measured like m L plus L n here. Scattered rays 1 prime and 2 prime will be completely in phase if this path difference is equal to whole number of n of wavelengths or if n lambda is equal to 2 D prime sin theta. So in order for this second ray to be in phase with the rays which is diffracted from the first plane and there is a condition. This is a condition which exists. So the condition is the path difference should be the integral multiple of wavelength that is n lambda is equal to 2 D sin theta. Now let us now play the other okay. If it is going to satisfy this condition like if the path difference is going to be the integral multiples of lambda they are going to get reinforced with the plane I mean the rays which is coming out of the first plane and second plane and third plane and so on. It is keep on going to reinforce provided if this condition is met. Similarly you can imagine 2A and 2A prime which is going to follow this condition again. So like that you can keep on imagining any line which comes and hits any atom here and then they will follow a condition like this or this two situations in either case you will have the reinforcement of the phase will be there and then they will contribute to the diffraction intensity. So it states the essential condition which must be met if the diffraction is to occur. So this is the condition for the diffraction which is popularly known as Bragg's law and we will now see what are the some useful remarks in considering the above schematic. It is helpful to distinguish three scattering modes by atoms arranged randomly in space as in a monotone gas this scattering occurs in all directions and it is weak. Intensities add by atoms arranged periodically in space as in the perfect crystal. In a very few directions these satisfying Bragg law the scattering is strong and it is called diffraction. Amplitudes add. In most directions those not satisfying Bragg law there is no scattering because the scattered rays cancel one another. So you have to just recall the phase relations what we have just gone through. You see that if the path difference is different then that has got a significant influence on its amplitude and if it is out of phase by quarter wavelength or half wavelength we have seen that it is completely you know cancelling that intensity if it is of opposite sign. So if the Bragg condition is satisfied you will have the rays which is diffracting will be on the same phase and then they will reinforce and contribute to the diffraction intensity. Not all the scattering waves will do this because it has got some specific angle theta by which only it will happen it is called Bragg angle. So we will see the significance of this further. Two geometrical facts are worth mentioning. The incident beam the normal to the diffraction plane and the diffracted beam are always co-planar. The angle between the diffracted beam and the transmitted beam is always 2 theta. This is known as diffraction angle and it is this angle rather than theta which is usually measured experimentally. We will talk about this 2 theta angle when we go to the x-ray diffractometer in the laboratory and we will discuss about the significance of this 2 theta versus theta. Let us look at the another important remarks. Diffraction occurs only when the wavelength of the wave motion is of the same order of magnitude as the repeat distance between scattering centers. This requirement follows from Bragg's law since sin theta cannot exceed unity. Now this is very important. So that clearly implies n lambda by 2d prime is equal to sin theta less than 1. Therefore, n lambda must be less than 2d prime. For diffraction the smallest value of n is 1. n equal to 0 corresponds to the beam diffracted in the same direction as the transmitted beam. It cannot be observed. The condition for diffraction at any observable angle 2 theta is lambda less than 2d prime. So, what is the significance of this statement? For most crystals planes of d prime is of the order of 3 angstrom or less which means that lambda cannot exceed about 6 angstrom. A crystal could not be possibly diffract uv radiation measuring for example of wavelength about 500 angstroms. So, that is the implication. The d spacing and lambda are not compatible. On the other hand if the lambda is very small the diffraction angles requires very specialized equipment. So, we can rewrite this Bragg law like this lambda equal to 2d prime by n times sin theta. Since the coefficient of lambda is unity here a reflection of any order can be considered as the first order from planes real or fictitious spaced at a distance 1 by n of the previous spacing. So, this turns out to be a real convenient. So, the d is equal to d prime by n and we can write a general form lambda is equal to 2d sin theta. We can demonstrate this usage of this kind of writing this Bragg law by considering a second order 1 0 0 reflection for a simple cubic substance. So, what I will now do is I will try to illustrate this on the blackboard. See what I have just drawn here is you have 1 0 0 plane in a cubic crystal where you have the i plane and i plus 1 plane and you have this incoming rays and this is a diffracting rays and they are separated by the. Now, we will suppose if we consider the second order reflection from this that means these 2 planes are out of phase by a 2 lambda. So, that means suppose if there were some scatterers if you assume in the i plus half plane they will scatter the atoms sitting on i and i plus 1 atoms by lambda. You have the that this scatter I mean whether you have the atom here or not and the diffracted intensity from this plane will with respect to i and i plus 1 will have the phase difference of a lambda. Here it is 2 lambda. So, what we can write is in general nth order diffraction. So, considering this the equivalence of a second order reflection from 1 0 0 and the first order reflection from 2 0 0 we can make a general statement. In general nth order diffraction from hkl with the spacing d prime may be considered as a first order diffraction from n h nk nl with spacing d is equal to d prime by n. So, with that we will be able to understand the kind of diffracted intensity which is coming from these crystal planes will be realized. So, what I will do is I will just stop here and then we will continue this the diffractions of the x-rays and its relation with a reciprocal lattice concept in the next class. Thank you.