 Let us begin. Before we start this lecture, you know yesterday I sort of promised that I will bring some models try to explain certain things. So, I could manage to bring some two models. So, let us just discuss some of the structures which we are trying to do sort of talk about it yesterday. Yes, I hope this is you are able to see this is what is generally known as sodium chloride structure. Now, you can see that there are two type of circles or other spheres one is green and another is silver color. I am not very sure whether you can see it very very clearly, but the colors, but you know for example, let us start with this particular thing this is silver, then this is green, then this is silver, this is green. Let us imagine that these are the centers of these ions and let us assume that this is sodium ion and this is chlorine ion, this is sodium ion, this is chlorine ion. Again, I repeat these are only be treated as the centers of these ions because remember that the sodium ion and chlorine ion they are not of equal size. In fact, they are what you will find that chlorine ion is much bigger and sodium ion is comparatively tiny. So, generally all these things are not of equal size. This is only to show the position where they are located the centers of these ions. So, let us assume that this is sodium and this is chlorine, this is sodium and this is chlorine. Now, if sodium would have been equal to chlorine, it means if this would have happened, let us look like this, if this would have happened that this is chlorine, this is sodium, this is chlorine and if they would have all been of equal color, then this would have been a simple cubic structure because you look at this particular thing, this is a simple cube. This I am not sure whether you are able to see it very clearly. I am trying to rotate it to make it clear. So, this would have been a simple cube structure. If these they are all of the same color but they are not of the same color, they are different ions and remember when I were I define a lattice, a Bravais lattice of a particular material, I had to take completely the translational symmetry into consideration. It means every point has to be identical. This point cannot be represented as the same point here because if you are at this particular point on the right hand side is your sodium while if you are sitting on this particular point on the right hand side is chlorine. So, these two points are not equivalent. So, if I have to define a basis, I have to I mean if I have to define a Bravais lattice, I have to define where the environment is exactly identical. Now, you start looking let us say only at the grey ions or let us say only at the green ions. If you let us say look at only the green ions, this is the green ion, let me put it like that. This is the green ion. This is one green ion, this is one green ion. Let us look at silver. I think silver is little more clearer. Let us look at the silver ion. This silver ion, then there is a green ion, then there is a silver ion. If you look at this silver ion, the next silver ion is here. Now, if you look at this silver ion, then you have another silver ion here, then you have another silver ion here, then there is a silver ion So, if you are if you ignore the green things and just look at the silver thing I hope you are able to see this phase center this, this, this, this and this. So, this is a phase centered cubic lattice because now you start at this particular point and you put two atoms in fact what we have decided that if you take for example on this side you put one sodium ion here another chlorine ion here you put one sodium ion here another chlorine ion in respect to every sodium ion you put one chlorine ion in this particular direction here you take this sodium ion you put a chlorine ion. So, you consider only let us say the silver ones then you can see that the environment of all the silver points are exactly identical these silver ions actually form what we call as a phase centered cubic lattice and in this particular case you have to generate the sodium chloride structure by taking this as your lattice and putting two ions one at the origin which is sodium ion another a chlorine ion which I am putting at the distance of a by 2 i plus j plus k which is the body center. So, this is what is a sodium chloride structure which we say can be represented as a FCC lattice with two ions basis one ion let us say sodium ion being put at origin another ion being put at a by 2 i plus j plus k keep on taking every sodium ion as origin and keep on replacing this by this particular basis you will generate what is a sodium chloride structure. So, often there is a confusion that people think that sodium chloride is a simple cubic structure remember it is not had sodium been equal to chlorine which it is not then you could have thought it this is one of the way of explaining it, but if you are looking at the structure this is actually a phase centered cubic structure and in fact when we are discussing x ray diffraction there is typical signature of FCC lattice which you will find when you are taking the x ray diffraction of this particular material. Now, the next thing which I wanted to discuss was about the diamond structure which is also little confusing. So, this is a model of sodium ion it is not exactly the way I would have liked to draw it, but you know probably you can appreciate it. See this is one carbon atom this is another carbon atom remember this is going to be repeated infinitely. So, this you can see is the phase center there will be another one here there will be another one. So, you can see that these carbon ions are forming their phase centers alright. So, these carbon ions are located at this is one this is one this is one then there will be one here then there will be one here. So, these ions amongst themselves are forming the FCC lattice you look at these things if you look only at these things you will find that they are also forming FCC lattice you take this ion then there will be another ion here then there will be another ion here this is one there. So, this will be the phase center of this particular phase then here you will have one one. So, if you look at these things which are located here they themselves would have formed a FCC lattice. But these two lattices are displaced with respect to each other by let us say this much distance. So, they are displaced by let us say put it here this much distance which is one fourth of the body diagonal because you can start with this origin and take somewhere here this will be your body diagonal and it is one fourth of that particular distance. And what I was trying to tell yesterday that if you take this particular unit cell divided into 8 small unit cells I am taking lattice constant a by 2 then I have divided this into 8. Now, you will find that alternate of them is centered. So, if you have this as centered then this one is not centered then this is centered then this one is not centered this one. So, you have formed out of this unit cell 8 smaller cubes I am not very sure whether you can visualize it very very clearly. So, this is the one we have divided into 8 simple smaller cubes which have each age of a by 2. So, you have 8 cubes now out of all these 8 cubes you can see this is centered but this is not centered here this is centered but then this is not centered. So, out of these 8 cubes alternate of them are centered. So, this is another way of looking at the diamond structure that consider a divide this particular face centered cubic unit conventional unit cell into 8 smaller cubes which are each of age a by 2 then center alternate of them. So, this distance will be automatically one fourth the distance of body diagonal. Now, let us come to our normal lecture. So, I start with recapitulating of what we did last time we defined Bravais lattice we defined their primitive vectors and what we mean by primitive vectors also we discussed then we discussed the concept of basis and related Bravais lattice to the actual crystal structure. How we generate ideal crystal structures is starting from a Bravais lattice and then putting on them on each lattice point exactly in identical fashion a basis that will generate a crystal structure. Then finally in our last lecture we discussed what are called closed packed structures. We discuss specifically hexagonal closed packed structure and cubic closed packed structure then we insisted that cubic closed packed structure is nothing but a face centered cubic structure and with face centered cubic lattice we try to show that it is 111 plane which is actually the closed packed plane. So, we try to look at that particular face centered position to find out about that particular closed packing. Now, let us define the two very very important concepts in solid state physics which we are going to start discussing today and they have I mean in their solid state physics we call these are bread and butter you know you start with ABC these are ABC of solid state physics of course when we start discussing about them they appear to be somewhat abstract until you actually start using them and then you realize that eventual the entire solid state physics you cannot go ahead unless you have these two type of these two you know sort of basic concepts. So, the first concept which I want to introduce is the concept of reciprocal lattice. Now, you look at different textbooks introduce the concept of reciprocal in different fashions. A crystallographer would like to introduce this in a different fashion a mathematical physics person would like to introduce in a different fashion or other theoretical physicists would like to discuss in a different fashion the way I have introduced is a purely by brute force without giving any explanation why I am want to define it I am just defining it and then eventually we will try to give some idea of why this particular concept is useful of course because we are not really having a course on solid state physics. If you have a course a proper course on solid state physics or condensed matter physics you will see that this both these concept one reciprocal another concept of Brouhline's own how these concepts are so important for solid state physics. Now, we have just now discussed that any lattice that we talk can be defined in terms of primitive vectors. In fact, we gave for simple cubic body centered cubic lattice and for face centered cubic lattice what are the conventionally regarded as their primitive vectors. So, in principle if we know the primitive vectors we can generate the lattice primitive vectors would generate the lattice. Now, as the name reciprocal suggests that this is also a lattice but in this case what we talk is something which is not of dimension length but dimension of inverse of length. Now, initially this looks somewhat abstract but on the other hand you should realize that if we take a graph paper for example even in high school when we ask students to plot a graph let us say between two quantities for example if you are talking of t square versus l when we are talking about simple pendulum experiment when we keep on varying length and I keep on finding out the time periods then using that graph paper we do not only plot those things which have dimension of length for example I can choose an appropriate scale I can always say that 1 centimeter is equal to let us say 1 second or 1 millimeter is equal to 1 second and draw something which is having a dimension of time on let us say x axis or dimension of time square or dimension of length. Length of course there is no problem because you can always say that you know whatever is the length I can take the actual length but even in that graph paper many times I will write that I will take length 1 meter equal to 1 centimeter. So, I choose an appropriate scale to define to plot a graph. So, it is not very I mean it is quite standard to draw a graph in which one or both the quantities are not having the dimension of length. So, I can imagine exactly the same thing you can imagine that there is a sort of a three dimensional graph paper and I choose an appropriate length scale I choose a particular scale and then I can plot a lattice as I used to plot the lattice in the case of direct lattice where everything was of dimension length I will choose something a particular appropriate unit let us say 1 millimeter is equal to let us say 1 angstrom inverse or whatever it is you want to say and again in this three dimensional graph paper I can look at these points which I will call as lattice only thing I will realize where I am talking of let us say lattice constant of this particular reciprocal lattice that lattice constant will have a dimension of inverse of length alright. So, I hope this should not create any problem because you know confusion because as I said you by taking appropriate scale you can always draw something which does not have a dimension of length on a piece of paper on a graphic paper and that is what precisely I am trying to do. So, now let us talk about reciprocal lattice reciprocal lattice is always reciprocal of a particular direct lattice. So, let us suppose we have a direct lattice like inverse of a number or reciprocal of a number is of that particular number reciprocal of 2 is 1 upon 2 reciprocal of 4 is 1 upon 4 alright. So, you have to know the number then only you can find out what is the reciprocal of that. Similarly, for reciprocal lattice you have to know what is the direct lattice then you can find out what is the reciprocal of that particular lattice. So, let us suppose there is a direct lattice and for that the primitive vectors are A, B and C. So, A, B, C are the primitive vectors of a particular direct lattice and I have to find out the inverse of this particular lattice or rather reciprocal of this particular lattice then I have to find reciprocal of this particular lattice. I have to define what will be my primitive vectors corresponding to reciprocal lattice which is given by capital A, capital B and capital C okay. So, what I will do using these A, B, C which are supposed to be known because I know what is the my direct lattice type I want to find out what is its reciprocal lattice. So, using these I will find out capital A, capital B and capital C once I know capital A, capital B, capital C these will be treated as the primitive vectors of the reciprocal lattice then I again like a direct lattice take a linear combination of these vectors and generate the entire reciprocal lattice in 3 dimensional space okay. So, starting with this A, B, C let us first find out what are the equations which are relating to capital A, capital B, capital C these are given in the following transparency. So, if I know A, B, C first I have to find out A dot B cross C which is actually the volume contained between these 3 vectors A, B, C then capital A is defined as 2 pi remember in the numerator when there is A there is no A here B cross C divided by A dot B cross C. These factors are common in all the B, C so 2 pi there is a 2 pi here there is a 2 pi here A dot B cross C is here A dot B cross C is here A dot B cross C is here okay. Here there is A so there is a B cross C here there is a B so B is missing so you have C cross A remember you have to go in a cyclic case because C cross A is different from A cross C then you say capital C then you have A cross B. So, from B whatever is the next that you put first so after A there is a B so you put B here then cross C then there is a B so you put for C then cross A then there is a C you first put A cross B alright. So, this is the way I define I just like to mention that if you will take some very very old textbooks of x-ray crystallography okay many times you will find that this factor 2 pi is not there okay. So, in the older you know definitions of reciprocal lattice this 2 pi factors were missing but the current trend or current way of representing these things includes these 2 pi factors. You remember when I mean when at least when we are students many times we are not following the assays I mean they rationalize the MK system view of units. So, this 1 upon 4 pi epsilon not you know we started writing much later because that made further equations comparatively simpler. So, now these 2 pi factors are introduced in this particular definition of A capital A B capital A capital B capital C so that the equations that I am going to talk later turn out to be somewhat simpler because that is what we are going to eventually use as far as the entire solid state fix is concerned okay. So, just I mean if you are getting confused with whatever I am saying you ignore whatever I said just now just take these as the definitions of capital A capital B capital C. Now, let us try to find out a simple let us take a direct lattice a simple cubic lattice and trying to find out the reciprocal lattice of this simple cubic lattice. As we have seen yesterday the primitive vectors of a simple cubic lattice is A is equal to A i remember i, j, k are unit vectors along x, y and z directions. So, you have A is equal to A i B is equal to A j C is equal to A k. So, first you have to find out B cross C. So, we find out B cross C B cross C this is cross C. So, A square there is A here there is A here so it will become A square multiplied by j cross k and j cross k is i as you know. So, B cross C will become A square i then for denominator I have to take A dot B cross C so A is A i. So, I have to take dot product of this with A for dot product does not matter whether you take this side or that side because A dot B is same as B dot A. So, once there is another A is you get multiplied by A cube then you have i dot i i dot i will give you one. So, this A dot B cross C is A cube and of course, you could have written this equation directly because you know that A dot B cross C is the volume enclosed by these three vectors and the volume enclosed by these three vectors will be the volume of the cube itself and which has a volume of A cube. So, now I can write what is my first primitive vectors of reciprocal lattice capital A this will be 2 pi as I have said B cross C which is A square i divided by A dot B cross C which is A cube here. So, this gives you 2 pi by A i. So, this is your capital A similarly you could find out capital B then you have to do C cross A find out capital C then you have to do A cross B and then you will find out that this turns out to be exactly equal to 2 pi by A i B will turn out to be equal to 2 pi by A j and C will turn out to be 2 pi by A k. So, this is the way it will turn out to be and as you can see that this particular thing has a dimension of inverse of length because A has a dimension of length this is a lattice constant. So, 2 pi by A this has a dimension of inverse of length this is inverse of length this is inverse of length and these 3 are all equal. So, if you treat this as some constant which is 2 pi by A this will generate a simple cubic lattice which has a lattice constant of 2 pi by A remember this lattice constant has a dimension of inverse of length and that is why it is a in the reciprocal lattice space or reciprocal space where everything is going to be talked in the dimension of inverse of length sometime we will also call it Fourier space, but you know that also is in the dimension of inverse of length. But let us I mean let us stick to this particular thing whatever we are trying to say trying to think somewhat simple. So, you can see that reciprocal lattice of a lattice simple cubic lattice which has a lattice constant of A turns out also to be a simple cubic lattice and its lattice constant turns out to be 2 pi by A. Let us take for one more lattice let us take for a body centered cubic lattice which is little more difficult than this simple cubic lattice. Earlier if you have remember we have said that A B C which are the primitive vectors of this body centered cubic lattice will be given by A by 2 when there is a first one there is a negative sign here A by 2 minus i plus j plus k this is the second one. So, there is a second one negative sign B is equal to A by 2 i minus j plus k this is third. So, there is a third one with the negative sign C is equal to A by 2 i plus j minus k. Now, first thing I take like before we take B cross C if I take B cross C there is A by 2 here there is A by 2 here. So, that will make it A square by 4 then I have to take cross product of this i minus j cross plus k cross i plus j minus k. So, I have to take I mean if there are three terms here there are three terms here. So, you will in principle get nine terms but you will realize that there are three terms will become zero because i cross i is zero j cross j is zero and k cross k is zero. So, first you start with i. So, i cross i is zero then you take i cross j that will become k then you take i cross k which is minus j but there is a negative sign here. So, that makes it plus j. So, you can verify that whatever I have written is alright if you take this particular cross product the outcome that you will get the result that you will get will be A square by 2 j plus k. Now, I have to take A dot B cross C it means I have to take dot product of this with respect to this then what I have done here A dot B cross C. So, you take this dot product of this then j dot j will become equal to one k dot k will become one i dot j will be equal to zero i dot k will be equal to zero j dot j is one k dot k is one. So, this will just give you two. So, the overall A dot B cross C is A cube by 2 which is also understandable because in body centered cubic lattice you have two points on the average. So, therefore, the volume occupied by A dot B cross C which actually will be the volume of the primitive unit cell not the conventional unit cell. This will be half of the volume of the conventional unit cell which is A cube by 2. So, now you can see that capital A turns out to be equal to 2 pi by A j plus k. Now, if you do exactly the same thing for the other thing you will find out that capital A turns out to be this B capital B turns out to be this and capital C turns out to be this. And if you remember these were the type of primitive vectors which we had for the case of FCC lattice. So, actually the reciprocal of a body centered cubic lattice turns out to be a face centered cubic lattice. And with the lattice constant of 4 pi by A remember here whatever was appearing was A by 2 both in BCC and FCC it was A by 2 which was appearing here in simple cubic lattice it was just A appearing here because A i A j A k but when I am talking of body centered cubic lattice it is A by 2 A by 2 A by 2. So, whatever is this factor multiplied by twice will be the lattice constant and that is what you can see here this particular thing has to be multiplied by twice by by 2. So, this turns out to be FCC lattice of a lattice constant 4 by by A. Similarly, you can show that reciprocal of a face centered cubic lattice will be a body centered cubic lattice again of lattice constant 4 by by A. This is one important factor which I always insist on students to remember especially if I am doing a course on condensed matter physics okay telling them very very clearly that you should remember reciprocal of a simple cubic lattice of lattice constant A has a lattice constant 2 pi by A but when you are going a reciprocal of BCC to A becomes FCC the lattice constant becomes 4 pi by A and not 2 pi by A. So, similarly you can show that reciprocal lattice of a face centered cubic lattice will turn out to be like this you can do the calculation exactly in the same fashion which turns out to be a BCC lattice of lattice constant 4 pi by A. Now, there are certain important relationships which you can see if you take A dot A will turn out to be equal to 2 pi all B dot A C dot A will be 0. Similarly, B dot B will be equal to 2 pi C dot C will be equal to 2 pi all other things I am sorry yeah I think there is I should have written A dot B equal to 0 here that I will correct it here because A dot A is equal to pi A dot B should be equal to 0 A dot C should be equal to 0. So, this I will correct it I mean this is the thing wrong with this thing but I am repeating this thing here I should actually written A dot B is equal to 0 but you know before uploading that I will correct it. So, you can see that only A dot A B dot B and C dot C will be equal to 2 pi others will be 0 this can be very easily seen from the definitions themselves. If I talk take A dot B cross C this will give you one so A dot A will be equal to 1. Similarly, if you take B dot B you will get B dot C cross A which you know by vectors that B dot C cross A is exactly same as A dot B cross A so this will give you one. Similarly, if I take C dot C I will get C dot A cross B again you can show that C dot A cross B is A dot B cross C so this will be equal to 1. Now if I take for example, A dot B then you realize this particular thing this particular vector A is perpendicular both to B and C because this is having a direction of B cross C and B cross remember this is scalar. This particular thing is the only vector thing here B cross C will be a direction which will be in a direction which is perpendicular to both B and C. So, if I take A dot B then I am taking a dot product of something which is in the direction of C to something which is in perpendicular direction to C so their dot product will turn out to be equal to 0 because cos 90 is 0. Similarly, A dot C by the same arguments will be equal to 0. So, you can see that all cross dot products turn out to be 0 other than A dot A B dot B C dot C in fact that is one of the ways of calling reciprocal because this almost appears that 2 but multiplied by 1 upon 2 actually should have been equal to 1 but because of this 2 pi inclusion no this becomes equal to 2 pi. Now like we have been used to defining a translation vector in direct lattice as a linear combination of its primitive vectors. So, a translation vector in a direct lattice is defined as n 1 A plus n 2 B plus n 3 C where n 1 n 2 n 3 are integers positive negative or 0. Similarly, we can define a translation vector reciprocal lattice space using these capital A capital B and capital C which is m 1 A plus m 2 B plus m C. So, a translation vector in reciprocal lattice space will be m 1 A plus m 2 B plus m 3 C where m 1 m 2 m 3 are integers positive negative or 0. So, like the position vectors of all the points of the lattice will be generated by these vectors. Similarly, position vectors of all the points in the reciprocal lattice space will be generated by this particular equation. There is another property which I would which is of interest which I will discuss the use of that I will discuss. If I take g dot t it means if I take the dot product which I have defined this is a translation vector in direct lattice this is a translation vector reciprocal lattice. If I take g dot t then I will get m 1 n 1 then A dot A which I have already seen will be equal to 2 pi. Then if I take dot product of this with this will be 0 take dot product of this with this is 0. Similarly, when I take dot product of this with this is 0 when I take b dot b this will be 2 pi. Similarly, c dot c will be equal to 2 pi all others will be 0. So, this you can see that g dot t will be equal to 2 pi m 1 n 1 plus m 2 n 2 plus m 3 n 3 all other will be 0. And this quantity because all m 1 n 1 m 2 n 2 m 3 n 3 are all integers. So, the entire quantity here has to be integer. Therefore, this thing will be equal to 2 pi times n integer n integer. And if I write this e raise power i g dot t that you know you can write as cos of g dot t plus i sin g g by 2. And because g dot t 2 is 2 pi times an integer it is very clearly that e raise power i g dot t turns out to be equal to 1. In fact, many people define reciprocal lattice using this particular concept. Some people define with respect to the equations which I have given you just now that a dot a is equal to 2 pi. They are all equivalent definitions of the reciprocal lattice. Now, we will try to talk about axillary diffraction. Then we will see how this concept of reciprocal lattice is used in the axillary diffraction. First of all, when I am talking of axillary diffraction, the first step will be to consider that diffraction takes place only from the Bravais lattice points. It is very, very important to realize not very, very clearly understood or mentioned. The thing is that we have said lattice points are only points. On these points we are keeping a reasonably big size atoms. And in fact, we have said a better picture is to say that these atoms are touching each other. So, in principle you have electrons or whatever it is existing at a much larger area than these points. But whenever we are talking of Bravais law or for example, Law's law, it is considering the diffraction only from lattice points. The effect of basis is not being considered here. The effect of basis we will consider later. This is very, very important thing to realize. So, let us assume at this particular moment that diffraction is taking place only from the lattice points. In other words, if I want to be more general, I would say that Bravais law and Law's law takes care of diffraction considering only the translational symmetry. What happens within the basis is not being considered in neither in Bravais law nor in Law's law. These things have to be introduced later, which I will do at the end of this particular lecture. That would be one of my job to explain you this particular difference that when we actually start putting atoms, what additional effects does it cause? But at the moment, let us assume that diffraction is taking place only from the lattice points. It is not taking place from everywhere. See, remember if you are talking for example X-rays, X-ray in fact get diffracted from electrons, electron density and you know if you are talking of an atom, electrons could be everywhere. In fact, it depends on their wave function of what are those electrons, their wave functions and there will be some probability of finding electron various points in the space. Wherever there is an electron, in principle electron, the X-rays could get diffracted from that particular point. So, it is never correct picture that X-ray gets diffracted only from lattice point. But as far as Bravais law is concerned, this is only considering diffraction from lattice point. It is only taking care of the translational symmetry aspect. What happens when actually I put a big size atom at this lattice point has to be considered separately. So, let me repeat initially consider diffraction only from Bravais lattice point. The effect of diffraction from basis will be considered later. Then I describe Bravais law which of course I will not derive it because this is considered a very very I mean in the high school we discussed. But I will like to emphasize only on those aspects where I feel that generally with my experience of teaching IIT students, I find that students are generally confused about this particular thing. So, that is the only thing I would like to emphasize. Now, what Bravais law does is, simply it takes care of the lattice by drawing various set of planes. In yesterday's lecture, we have talked about what we mean by lattice planes. These planes must contain a lattice point and a parallel plane will pass through all the lattice points is what we said. So, you can draw various sets of parallel planes inside a lattice. So, what Bravais did considered various sets of planes inside the lattice. Now, consider one of those sets. Now, in order that a Bravais deflection condition or a diffraction condition is satisfied from these set of planes. For different set of planes we have to consider differently. Let us choose one set of plane. If condition of diffraction has to be satisfied, then two conditions have to be made. This is very very important that two conditions have to be both have to be made together. Just satisfying one of the conditions will not cause diffraction. Both the conditions have to be there. First of all, you must have a condition of what we call as Spatula reflection. Spatula reflection actually means that angle of incidence is equal to angle of reflection. Like reflection which happens in the normal mirror for example, of course in this particular case I will just maybe just mention that this theta is not the angle of incidence. So, there is slight difference the way optics people treat. See if you have let us say one plane and there is a beam of light is coming here and it is getting reflected like this. In optics, this is what is called angle of reflection or let me write phi and this is what is called angle of reflection. Both these angles are equal. While if you are having lattice planes here just drawing. So, this is let us draw a line to when the x-ray is getting incident here it is this angle which you call as a glancing angle is treated as theta. So, you should not get confused this theta is actually the glancing angle which is the angle from this particular plane. So, in x-ray diffraction we always take this as theta and not this. So, both these conditions have to be satisfied that you should have 2 d sin theta is equal to lambda where d is the distance between the planes of course theta is the angle which I have just now defined lambda is the wavelength the x-ray wavelength that you are using and n is an integer. So, if this condition is satisfied then only you will get spectral reflection but only in that particular theta corresponding to which angle of incidence is same as angle of reflection. So, you must maintain this particular condition simultaneously. In fact, this is generally a very standard quiz question that what is the difference between x-ray reflection in fact many times we call it x-ray reflection though actually it is x-ray diffraction many times you do not speak very very correctly. So, what is the difference between x-ray reflection and the normal reflection and the basic answer for this thing is that in a normal reflection if angle of incidence is equal to angle of reflection you will get reflection in that particular direction no further condition has to be satisfied but in the case of x-ray diffraction you will get only when 2 d sin theta is equal to l lambda is also satisfied both these conditions have to be satisfied together. Let me just explain this particular thing using drawing some particular planes. Let me just first read this particular thing. So, the thing is that if you are using a single crystal and a monochromatic beam of x-rays then as we will be seeing that sin theta for various type of planes are already fixed if they are already fixed these are already fixed then it is very very unlikely that the condition 2 d sin theta is equal to lambda will be exactly satisfied. So, generally what we do in actual experiment one of these things have to be continuously buried this is what I will be discussing later. So, if you use a single crystal and a monochromatic beam of x-ray the angle of incidence for all the planes is fixed in that case no break break reflection may be observed. There is small probability that you will observe but chances are that you will not observe this also will become very clear. So, let us just draw a 2 dimensional thing to make it somewhat simpler to explain the point which I am trying to say. So, I have drawn various planes in fact this case I have drawn 2 set of planes one is what I am writing with this red thing and this is sort of bluish thing or brownish thing. So, this is one set of planes I have drawn 2 parallel planes like this I have drawn 2 parallel planes like this these are different planes remember this is containing lattice point this also contains lattice point. So, if I take the perpendicular distance between these this will be the value of d for this set of planes for this particular set of planes this will be the value of d. Now, let us suppose my x-ray beam is incident in one particular direction it is like this of course I have drawn third set of planes which is like this which is this particular thing this is also a set of planes all right. Now, if this x-ray the way is coming as far as this red planes are considered the angle of glancing angle is this theta if I take these planes which are vertical planes then corresponding to this this is your glancing angle if I take these brown planes corresponding to them this is glancing angle. So, you can see that as I am drawing different planes theta's for all this set of planes are fixed this is theta for this set of planes this is theta for this set of planes this is the theta for this set of planes. So, this I have drawn 3 sets of planes and you can see that if x-ray beam is fixed then you are seeing that all the angles of this glancing angles corresponding to all these 3 sets are fixed all right. Now, in order that a Bragg diffraction condition is satisfied let us say from this particular plane this particular set of planes then in addition then you must have first of all 2D sin theta is equal to lambda n lambda if it so happens that for this particular plane set of planes 2D sin theta is equal to n lambda is satisfied then you will get a diffracted beam in this particular direction somewhere in this particular direction where this theta will be same as this theta all right. Now, suppose Bragg diffraction condition is satisfied for this set of planes then in that particular case this is the angle of this is the glancing angle then you will get a diffracted beam in this particular direction somewhere here. So, that this is theta if Bragg reflection condition is satisfied for this particular set of planes 2D sin theta is equal to n lambda then this has been the angle of incidence like this then the diffracted beam will come like this somewhere here like this where this angle will be same as this angle. So, you will see that first of all you should have 2D sin theta is equal to n lambda if at all you have 2D sin theta is equal to n lambda then for this particular plane if you have then Bragg reflection beam will come here. So, that this angle becomes equal to this if it gets satisfied with this then Bragg reflection beam will come this particular direction not in that particular direction because it must satisfy the angle of incidence to be equal to angle of reflection from that particular set of plane for which 2D sin theta is equal to n lambda is also satisfied. So, I hope this particular part is clear of Bragg reflection because I have generally found that this particular aspect people are fairly confused. So, that is what I said you need a continuous variation because generally if you fix everything you fix lambda, thetas automatically get fixed. Generally chances are that you may not get any set of plane for this condition is satisfied then older system we had 3 different type of methods by which you used to vary certain thing one was Lawyer's method. Lawyer's method you are not using a monochromatic beam but you are using a continuous beam of x-rays. So, we are using continuous x-rays for one particular lambda you may not be able to satisfy 2D sin theta is equal to n lambda but there are so many lambdas which are continuously varying. So, for some lambda it may get satisfied. So, then here chances of getting Bragg reflection increases. There is another method which is to called rotating crystal where you used to rotate the crystal. So, this theta were continuously being varied. So, at a one particular value of theta 2D sin theta is equal to n lambda will get satisfied and then you will get Bragg reflection condition. Third method which was more common method which was also called Debye-Scherrin method was powder method where you are actually not using single crystal but powdering the sample to very fine thing and therefore all these powders which are all now single crystals they are randomly oriented with respect to each other. So, you essentially get a continuous distribution of theta. So, that was one of the most common methods. But the days that these days that we are using here the standard modern x-ray machines of x-ray x-ray is a very very powerful technique and there are multiple uses of x-rays people use x-ray I mean it is a sort of bread and butter for a large number of material scientists metallurgists earth scientists physicists chemists and there are special type of machines which are used for a specific type of studies. But let me tell you about the most standard x-ray machine which generally we called as theta 2 theta geometry which is the most commonly used method of x-ray diffraction what we do here let me first draw my x-ray let us suppose my x-ray is like this and my sample somewhere here what you do this is what we call as theta. Now, what you keep on doing you keep on rotating this particular thing. So, you start with a very small value of theta and you keep on rotating this particular thing and you always maintain your detector such that this detector along with this mount always maintains this particular condition theta. So, what you normally do you will start let us say from a very small value of theta then you keep on increasing theta slowly. So, you keep on rotating this particular thing when you keep on rotating this particular thing in order that this theta condition has to be maintained this particular theta has to be same as this theta you can see from this particular geometry that detector has to be rotated with twice the angular velocity. So, if this is rotated by theta detector gets rotated by 2 theta this is generally called theta 2 theta geometry thing here important thing to realize here that this condition of specular reflection is satisfied only for this particular set of plane. So, you are always watching this particular plane when you are doing x-ray diffraction by standard theta 2 theta geometry. So, for example, if this is a polycrystalline material then you will only see this particular plane this particular plane what is happening to this plane for that you know specular reflection condition is not at all getting satisfied. So, only those planes which are parallel to this particular thing only those corresponded them this particular condition will be satisfied. So, you will be only be able to see planes which are oriented which are in this particular direction. So, this is what is called theta 2 theta geometry. So, for example, if you have a single crystal which is 1 0 0 oriented in this particular direction it means 1 0 0 direction is like this then you are what you will be seeing only 1 0 0 2 0 0 3 0 0 planes and things like that. So, this is what happens in the standard XRD geometry these days.