WAVE Optics Physics One Shot 2024-25 | Class 12th Physics Complete topic by Ashu Sir

Welcome everyone to Science and Fun, where teachers teach you with both heart and mind. So today, we are going to talk about a very excellent chapter named Wave Optics. The one-shot for Ray Optics was uploaded earlier, and Wave Optics is a chapter, I don't know why, that students fear a lot. Somewhere, because you see, I'll give you a small example, then I will definitely move on to the chapter. If you have noticed, whenever we used to study about animals in Biology, we could understand better about humans and animals. But whenever it came to studying about plants, we felt a bit strange because we don't know plants very closely, their parts, their components. But when it came to humans, the pancreas came up, and we have also conducted many experiments on Ray Optics. But the problem with Wave Optics is that we are going to talk about the wave nature. I am telling you this further. Therefore, a very big thing in this chapter is that this chapter becomes very boring when I record it live. See, I won't lie to you, I had even started its recording, and I had taught a lot of topics, at least for half an hour. Then I felt that it had become very monotonous here, meaning it felt like I wouldn't be able to give my 100% because this topic is more fun when some students are sitting in front and asking cross-questions. Believe me, you have never seen a one-shot of Wave Optics like this in your life because while teaching, I realized that students ask very peculiar questions, and then explaining them makes this chapter even more interesting. Okay, so that's why I am putting the recording here, offline. I am telling the truth, the recording of half an hour was done, I came with full mood, then I felt that no, I should deliver a good lecture to you all because the exam is very close, and this chapter is already boring. If I teach it like this, it might become even more boring, and I will keep adding some things in between to tell you. So, let's start the chapter first. It's a very good chapter. I probably told you on Day 1 of Ray Optics that Wave Optics is not going to be easy because in Wave Optics, you have never seen most things before. That's why the name of this chapter is Wave Optics, because light had two natures: ray and wave. I had asked you a question, if you remember, when we started Ray Optics, and the question was: If I throw a matter, say a tomato, at the wall, and it sticks, and I throw another tomato, and it sticks, then how many tomatoes will be on the wall? The answer is two tomatoes, because matter plus matter will always give matter. It cannot be that matter plus matter gives no matter. But if I throw light at the wall, and then throw light again at the same point, it is possible that I get darkness. But because you have never seen this, what needs to be done to see this? Are some certain conditions to be created? And because this is not so normal, not seen in daily life, this chapter becomes a bit difficult for students. What are we going to study? In Wave Optics, we are going to talk about the wave nature of light. Broadly, there used to be three topics in Wave Optics: the first topic was Interference, the second topic was Diffraction, and the third topic was Polarization. Now, out of these, Polarization has been deleted from the syllabus. Who has been removed from the syllabus? So now only two topics remain for us. One is Diffraction, which will be our second topic in this chapter, and one is Interference, which is the first topic of this chapter. That is, you can say that two topics are to be studied in this chapter, and the chapter will end. That's it, Interference and Diffraction. This is about the study of nature, and light shows both natures. Who said this? De Broglie. And who proved it? The experiment by Davisson and Germer, which is the topic of the next chapter. Okay, let's start. First, let's talk about what a wavefront is. What is a wavefront? What is a wavefront? Listen. First of all, if you consider this as a light source, consider it a bulb, consider it a star, consider it a candle flame, whatever you wish. There is a light source. Now, imagine how you would show light coming out of it. Imagine how you would show it. Imagine how light comes out of it. If it's a bulb, then you've imagined it. So, I hope everyone has shown light coming out like this. Tell me, that light is coming out perpendicularly, like this, because until now, what study were we doing? Ray Optics, and we were considering light as a ray, that it travels in a straight line. But light is an electromagnetic wave. Light is an electromagnetic wave. Light is an electromagnetic wave, because of which, in reality, light is not coming out like this. In reality, light must be coming out like this, like this, like this. Tell me. It's a different matter that it might be happening at a very small level. It might actually be happening like this, at such a small level, at such a small level, at such a small level, that it will seem to you that it is a straight line. But if I look at it on a larger scale, even if it looks like a straight line, light is actually a wave. So, from these bulbs, waves must be coming out from any light source. What must be coming out? Waves. What must be coming out? Waves. What must be coming out? Waves. Now, waves have a phase. What is it? Which you can call an angle in simple terms. What else can you call it? An angle. Look here. This is a light source. Light came out from here. So, when the wave reached here, let's say its phase came out to be π, 0, π/2, 3π/2, 2π. Electromagnetic waves follow sine and cosine. Both are perpendicular. They have the same phase, they have wavelength. So, when it came out, what was the phase? π. Similarly, when it comes out here, let's say somewhere here, what comes out? π. What comes out here? π. What comes out here? π. π. If I join all these πs, if I join all these πs, will I get a surface? This is a point source. The phase difference here is also π, so it's here too. It's here, so it's here too. It's here, so it's here too. So, if you move anywhere on this entire sphere, the phase you will get will be π. So, a surface that is formed by integrating all those points where the phase is the same, that surface is called a wavefront. That surface is called a wavefront. You must have remembered equipotential surfaces here. Go to hell. An equipotential surface is also something like this, where around a charge, there is a surface where the potential is the same everywhere. Here the potential is not the same, here the phase is the same. So, a surface around a light source where, no matter where you move, the phase will be the same, meaning there will be no phase difference. There will be no phase difference at any point on those surfaces. We call them wavefronts. What do we call them? Wavefronts. So, what definition can be given for a wavefront? A wavefront is defined as the continuous locus, continuous locus means a kind of meeting, an integrated surface of all such particles of the medium which are vibrating in the same phase at any instant. Sir, I don't understand this, is it the phase of light or the particle? It's the same thing. How? There are two types of waves in the world: electromagnetic waves and mechanical waves. Waves that need a medium to travel are called mechanical waves. Those that don't need one. Now, if there is a vacuum, then you all know that we have assumed particles of light, which are called photons. What are they called? Photons. Let's talk about sound first. Is sound a mechanical wave? It needs a medium to travel. So, when I speak, the air between my mouth and your ears vibrates. It vibrates in such a way that it matches the vibrations of my voice and carries the sound to your ears. And when my voice reaches your ears, it vibrates and returns to its place. So, practically speaking, particles also vibrate with sound. Similarly, particles are also vibrating with light, even if you call those particles photons. Sir, photons are imaginary, aren't they? No, not at all. Photons have mass. Do they? Yes, sir. So, a surface around a light source where, let's say light or particles, are in the same phase, is called a wavefront. What is it called? Wavefront. What is it called? Wavefront. Wavefronts are formed for different people. Come. For example, I have a point source here. This is a point source. So, for this point source, will this be a wavefront for me? There will be some phase difference on it. Let's assume the phase difference here is π/2. So, will the phase difference of every point here be π/2? Okay, okay, okay. Similarly, will there be another wavefront ahead of it, brother? And on that wavefront, let's assume the phase difference is π, and there is another wavefront ahead of it, let's assume the phase difference is 3π/2. Whatever it is. So, tell me, how many wavefronts will be around it? A foolish question. Okay. Spherical wavefronts. After this, cylindrical wavefronts. Cylindrical wavefronts will be for whom? For a linear source. Like spherical wavefronts were for a spherical or a point charge. Where did the child come from? Come out of the first chapter. Okay, come. Then who came? Cylindrical wavefronts. Cylindrical wavefronts will be for a linear source of light. What is a linear source of light? Like, isn't there that profile light used during Diwali? Straight light pipes, like lights. Understand it like that. But even in that, you know, profile lights are actually made of small point sources. So, you have to imagine a light that is straight. Now you can say tube light. [Music] You can also say a laser beam. But actually, the origin of a laser beam is a point source. Then it is passed through a concave mirror. Okay, too many complications. For now, understand it as a point source, a small cylinder of light, a wire that is glowing, understand it. So, around them. Okay, one second. We repeatedly say "light" while teaching the chapter, but this is for every EM wave. The chapter is also for microwaves, infrared, ultraviolet, and of course, visible light, X-rays, gamma rays, for all of them. But because we can see it, we will say light repeatedly. Correct? Yes, sir. Now imagine this is a linear source. So, how will it emit light rays? Just like a spherical source emitted in all directions, but a linear source's this point will emit like this. Tell me, yes. It will emit like this. Its lower point, meaning it's emitting outwards. Tell me, yes. So, for it, the wavefront will be like this. Listen. This is a linear source. This is a linear source of light. So, the phase difference here is π/2. So, will it be here too? Here? Here? So, for the lower point, at the same distance, it will also be 5/2. Yes. So, for it, will the wavefronts be cylindrical? Yes. Oh, tell me, yes. So, this entire wavefront, on this, the phase difference is the same, which is let's say π/2. So, will the phase difference on this be, let's say π? So, it will be π on all of them. So, brother, for a linear source, how will the wavefronts be formed? Cylindrical. How will they be formed? Cylindrical. How will they be formed? Cylindrical. Plane wavefronts. You know, a light source that is in the shape of a sheet. Now, how is it in the shape of a sheet? You have seen those large lights installed on false ceilings. Actually, even the round lights installed above are also flat sources, but make it a bit bigger. So, okay, understood. If you open it, you will find many small LEDs inside. Small LEDs. Okay, for now, you can understand it as a sheet source. Now, if, let's say, this is a light source, it's in the shape of a sheet. So, will it emit light like this? Like this? Does anyone come to your mind looking at this? Yes. Just keep it in your mind. So, what will be emitted from this? Tell me, light. What will be emitted? Light. So, if light is emitted from this, then if the phase difference here is π/2, will it be here too? Here too? Here, here, here, here, here, here. At the same distance from the sheet, it will be π/2. So, for a sheet, will the wavefront be a sheet? Sir, for a sheet, the wavefront will be a sheet. Tell me. So, brother, for this, the wavefronts will be formed like this. Okay. On this side too. It's possible that there are infinitely many wavefronts between them. Tell me. They can also be on this side. Correct. Correct. Now, think, how will convex wavefronts and concave wavefronts be formed? The most important question to come in the exam. Look here. Let's assume I have a point source, and here I have placed a convex lens. So, will the light rays coming out of it be like this? And if I have placed it at its focus, then the light coming out will be parallel. Everyone knows this much. So, the wavefronts formed here, will they be of this shape? Because this is, in fact, a point source. Tell me. And for a point source, it is spherical, and a part of a sphere is convex or concave. Okay, children? Yes. So, the shape of the wavefronts formed here will be concave with respect to the light source. What will it be? What will it be? And here, the wavefronts will be like a sheet, i.e., plane wavefronts, which will represent it. So, these are our wavefronts. What is a wavefront? Where, at every point, the light is vibrating, or the light will be, their phase will be the same. So, what do we call such wavefronts? Concave wavefronts. Similarly, think about convex ones. Asking with respect to this side. Here we have a convex lens, and parallel light came. What will happen? Tell me. Will it meet at the focus? So, the light source is somewhere here. Oh, tell me, yes. So, it's meeting. The light source is somewhere here. So, its wavefronts will be what? Sheet, no, plane. Okay. But what do we call these wavefronts? Plane wavefronts. Sheet is the name of the source. Okay. And here, the wavefronts, will their shape be like this? Oh, tell me. So, in fact, these two are the same, but with respect to the light source, its shape is convex. Convex. And here, from the light source, its shape is okay. What? Like this? Can any other shape come? Then it is. Oh, brother, very dangerous. Before moving forward, I have to ask many questions. First of all, okay, we are playing a game. The game is that we have this classroom, this classroom, and there is a very dark environment. Like, right now, if I tell you, many things are white, but imagine it's completely dark, nothing is visible. But here, there is only one light source, one light source, one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, fourteen, fifteen, sixteen. Let's assume these 16 children are sitting in this classroom. So, if I ask these 16 children how many light sources are there in this room, what will they say? One, brother, here it is, one. Right? But if I put a partition here, an opaque object, through which nothing can be seen. So, how many light sources are there for them? And for all of them, there is none. But if I make a hole here in this partition, and make a hole here, and now ask all of them, then how many light sources are there? Brilliant. Two light sources. What difference does it make to them that these two light sources came from the same place? They don't know. They don't know that before they came, there was a partition. And there were holes. So, they thought, oh, how many light sources are here? Two. For them too, how many? Two. For them too, how many? Two. Then the next day, I again put a partition here. Tell me, for the third row, how many light sources are there? Zero light sources. For the second row, two. And for the first row? Okay. Will the first row see light from this side? Tell me. No, brother. Light is going like this, like this, like this. So, for this, here. I made how many holes here? Three. Now tell me, for these people, how many light sources are there? Three light sources. Okay, brother. And if the next day, I put a partition here and make N number of holes, N number of holes. Then how many light sources are there for them? I know the first question in your mind will be, sir, will the intensity of light decrease as it goes there? Who is talking to you about the intensity of light? I am just asking you how many light sources there are. Tell me, how many light sources are there? Absolutely correct. There are N number of sources. So, what game did we play? This game is Huygens' Principle. Huygens' Principle says, listen carefully, its Hindi, because English is very difficult. Huygens' Principle says that if I have a wavefront, what is it? Wavefront. Then every point of it will act as a new source of light. If I have a wavefront, then every point of it will act as a new source of light. Our source was this parent source. And if this was a wavefront, then how many points did I let light pass through from it? So, we thought there were two sources. Okay, there can be many deep things.

In the size of this hole, I can make 10 more holes. Oh, like you asked, who did you do it with? Let's say I did it with that punching machine. So, in the size of a hole made by a punching machine, three or four holes of a compass can be made. A deep thought. Yes sir. So if I did it with a compass, how many light sources would be created in the same area? Three or four. That means a point is an imaginary word, actually. But for now, you should understand these points as small as you can. Understood, beta? What did Mr. Haig say? Every point on a wavefront acts as a new source of light. That is, if, if I have a light source here, beta, this chapter, smart children cannot do it. Still talking, you'll die, you'll cry, it's a very dangerous chapter. It will become more interesting later. Don't take this chapter lightly. You will have to leave it for the board because no one can explain it with as much dedication as someone is explaining it now. You can do as many crash courses as you want, do everything, but the way a teacher explains a chapter for the first time, they never explain it again. Now, understand it alone, with so many examples, no one will explain it. So listen and understand once. You are talking, man. This, beta, is what we have: a light source. Now, another deep thought. Oh, will the light source itself be made of many points? That's why I'm saying this study will become deeper and deeper. We don't have to make it that deep. For us, it is a point source. What is it? A point source. Is this its wavefront? Yes sir. See, there are one, two, three, four points on it. I have taken four. So, will this also act as a source of light? Yes sir. This will also act as a source of light. This will also act as a source of light. This will also act as a source of light. This is its wavefront. So, this is also a source of light. Every point on a wavefront acts as a new source, a fresh source of light. Okay. Yes sir. Okay. Yes sir. Who said this? Mr. Huygens said it. Okay. Mr. Huygens also said that light and any other electromagnetic wave require the same time to travel from one wavefront to another. He said, you will say, "Oh sir, this is a very easy point." No, it's not easy. See, I have made two wavefronts. Wavefront one, wavefront two. So, if light takes one second to go from here to here, it will also take one second to go from here to here. Wasn't this common sense? It wasn't. It seems like common sense to you because you have only seen such wavefronts in your life that are beautiful. Look, I'll show you an ugly wavefront. Look, no, this is fine. Look, this is a wavefront, and this is a wavefront. Listen carefully, sir. When will such wavefronts be formed? We haven't made them yet. Wait, why won't we make them? Listen. Light went from here to here, *thak*, and from here to here. Now tell me, will the time be the same or different? Same. Because what is the law? What does the law say? Light will always take the same time to go from one wavefront to another. You didn't enjoy it here. Here it seemed like common sense. There, you said, "Light has to travel more, and it will still take the same time?" Yes. How will this happen? This is the fun part of the chapter, but for now, we are saying Mr. Huygens said every point on a wavefront acts as a new source of light, and light, any other electromagnetic wave, requires the same time to travel from one wavefront to another, no matter what. Whether the gap between their wavefronts is uniform or non-uniform. Sir, but two things are not understood. First, why will such wavefronts be formed? So far, the wavefronts we have made have equal distances. You are making equal distances. You are making equal distances. You are making equal distances. Because the medium is the same so far. In this case, the medium above is different, and the medium below is different. The medium above is different, and the medium below is different. The medium above is rarer. It is rarer. And the medium below, because of this, the wavefronts have become crooked. Why did they become crooked? I will teach you. They have become crooked. Okay. Now, if they have become crooked, light travels faster in a rarer medium and slower in a denser medium. So, in the same time, it will travel a shorter distance, and that one will travel a longer distance, but the time will be the same. I've given you a hint that this will only be possible when the medium changes. The medium changes. If I say this entire board is air, what is it? Air. What is air? So, such wavefronts cannot be possible. It's impossible. How can the time taken to go from here to here and from here to here be the same? But if the medium is the same, then what is the formula for speed? Distance upon time. Say, die, yes sir. So, what is the formula for distance? Speed into time. If the medium here and here is the same, then the speed will be... Hey, what will be the speed? Same. But what is different? So, what will be different? Which is not possible because what did Mr. Huygens say? It will take the same time to go from one wavefront to another. The next point, the third point, which is also the last, listen carefully. No backward wavefronts are possible. No backward wavefronts are possible. No backward wavefronts are possible. He said that if there is a light source, wavefronts will always form in front and move forward. Backward wavefronts cannot be formed. What cannot be formed? Backward. Okay, first, we need to understand the meaning of backward. This word "backward" itself is very deep, man. Look, beta. Look, I am making a tube light. This is the choke coil. Choke coil. Yes sir. Choke. Yes sir. Starter. Sir, starter. And this tube light. Beautiful, beautiful. Turned it on. It lit up. It lit up. Now tell me, this tube light, this strip, this choke, this tester, oh, starter. If I say, what is backward for this tube light? Okay, true or false. Is this tube light emitting light in both forward and backward directions? I asked you a question. Is this tube light throwing light in both forward and backward directions? How many children say true? Very nice. How many children say false? Shiva, false. How? This side. The back side too. Look, look. Backward means, in English, one meaning of backward is behind. In physics, backward means inside the source, behind the source. It's very difficult to understand this. This tube light is there. This choke is there. And it's stuck on the wall. First, tell me, will this tube light be shining light on the wall too? Yes. On the choke coil too? Yes. And it's also shining on your waist. But both of these are forward. Because forward means you start from the source and move forward. So, if this is a tube light, you are starting from the source and moving forward. You are starting from the source and moving forward. This is not up. This is forward. You are starting from the source and moving forward. You are starting from the source and moving forward. Forward. Not in front of the tube light, but in front of the source. Okay. What is the source of light inside the tube light? Let us suppose it is fluorine gas. What is fluorine gas? So, is light coming out of the fluorine atom or going inside some fluorine atoms? It's coming out, isn't it? Do you know what LED stands for? Light Emitting Diodes. If you have ever looked at an LED with your heart, with love, one leg is longer, one is shorter. But I'm not talking about that. If you have seen it with your heart, you will see something small like this inside the LED. It is called a diode. What is it called? Diode. Right? We will study it in semiconductors. It is a PN junction diode. We don't have to study LEDs. Earlier, LEDs were also in our syllabus. Now, only PN junction is there. But this diode, when current passes through it, this PN junction diode, silicon, germanium, arsenide, what does it release? Light. So, if, if, for example, a silicon atom is releasing light, is it going out of the silicon atom or also inside? It's going out. So, Mr. Huygens said exactly this: light will never create backward wavefronts from the source. It will always create forward wavefronts. What will it create? Forward. And he didn't just say it. He also derived the formula for the display of wavefronts. We don't have to derive its formula, nor will there be any numerical on it. But Mr. Huygens derived its formula and said the formula is 1 + cos phi. What is it? 1 + cos phi. Where phi is the displacement and phi is the phase. Give me any one value of phi. Make y negative. Make y negative. We will teach you for free all year round, and if you take the money, we will return it. It cannot happen. It cannot be negative. Zero. Negative. Oh, it cannot happen. Because the minimum value of cos theta is -1. So, even if you put 180 here, what happens? It becomes -1. And 1 + (-1) = 0. So, the minimum value of y can be zero. That is, wavefronts can stop for some reason. But y negative, meaning y negative, meaning wavefronts cannot come backward. Now, the Huygens principle is very confusing. As you move forward, you will understand it better and better. Because it is like ABCD. Until you use it, you won't understand what is happening. So, as you move forward in the chapter, you will understand. But how do questions come on these topics? How do we practice? This is a comment on every video. So, now it's your turn. If you have reached Wave Optics, it means 70-80% of your syllabus is finished. And we are also making you practice all of this from here. And in my opinion, this is the simplest yet effective book. First of all, it is cost-effective. A sample paper book has more than 45 papers, and its cost, in my opinion, is less than 200 rupees. So, somewhere around 2-3 rupees per paper. The only point is that it is cost-effective. And the best thing is its answers. I believe questions are similar in many books, but answers decide which book is good, how the answers are framed. And here, I think there are very few books that frame very good answers, and Shivdas is one of them. So, you can trust it completely for your board preparation with sample papers. I think you should start now. Go and buy the book. We trust it completely. You will too. Many children have already bought it. Let's continue with the one-shot. Okay. Yes, beta. Here you will find a word called "wavelets." What will you find? Wavelets. What does wavelet mean? When we combine all the secondary sources to form a new wavefront, we give it the name wavelet. So, wavelets are also wavefronts in a way. Let's read the English here. First point: Each point on a wavefront acts as a fresh source of new disturbance, new light, new... I'm saying light again because we understand it with light. You can write it too; marks won't be cut. But what do we actually mean by disturbance here? Wave. What is a wave? A disturbance that carries energy from one point to another. The secondary wavelets spread out in all directions with the speed of light in a given medium. This is probably common sense: the wavefronts will form at the same speed at which light travels. Obviously, right, sir? A new wavefront at any later time is given by the forward envelope. That is, no backward wavefront is possible. No backward wavefront is possible. And they gave a formula for it, which is 1 + cos phi. If you put pi in it, meaning 180, even then it becomes -1. And 1 + (-1) = 0. So, y's minimum value can be zero, meaning wavefronts can stop for some reason. But y negative, meaning y negative, meaning wavefronts cannot come backward. Add a note here, whether you do it as a note or however. Light rays, light rays in different media travel with different speeds of light, and hence, and hence wavefronts, and hence wavefronts will be non-uniform in such cases. Wavefronts will be non-uniform in such cases. Also, light rays are always perpendicular to the wavefront. Light rays are always perpendicular to the wavefront. Two things are written. You have understood the first one, as this has already been discussed. And the second one, you just have to remember it. They are making an observational angle with the light rays. Whichever diagram you see, light is coming out from here too. Obviously. So, they will always make a 90-degree angle. Okay, fine. Tell me once, what are the two biggest things in ray optics? First is angle i = angle r, and the second is sin i / sin r = constant. And these two biggest things of ray optics, where are we going to prove them? In wave optics, of which the first is the laws of reflection. So, our objective is to prove angle i = angle r. I feel like we have been studying angle i = angle r for maybe five or six years. I feel like we've been studying it since seventh grade, maybe more. Say it. But today, for the first time in our lives, we are going to prove it, establish it. I don't know if we will do it today or not, but its diagram, everything is here. A very beautiful derivation. A derivation that requires thinking. A fun derivation. This is a very beautiful reflecting surface we have. You can consider it a mirror. Mirror. Light rays came from here, and another light ray. Both will collide and go away like this. Say it. And this one will also collide and go away like this. Okay. Did I tell you that these will come parallel and go parallel? I just drew it. We have to prove this. We know it happens like this, but we will draw it like this, right? Say, die, beta. Yes sir. These two light rays are coming, and these two are colliding and going away. Understood, right? This is our normal, and this is also our normal. What is this angle? Angle i. And what is this angle? r. This is i. This is i. We basically have to prove that angle i = angle r. I'm removing it now, but our objective is to prove angle i = angle r. Okay, math. What is this angle? 90. Pink, orange, 90. Pink, orange, 90. This angle is 90. Dotted white, 90. Dotted white, 90. 90, 90. This angle is common. This angle is i. Seventh-grade math. 90, 90, common i. Okay? Now, we have to prove i and r. So, I will prove this i and this i. You are also not doing it for the first time in your life. I mean, you are doing this derivation, but you have done this many times. Indirect derivation. Okay. Now, for this, we will choose two triangles. AOC. This is a dirty name. Let's change its name. Let's call it C. Now, if I prove triangle AOC congruent to triangle BCO, then by S.S.S., i and r. So, our objective is now to prove these two triangles congruent. So, if they become congruent, then by S.P.C.T., we will also prove that angle i is equal to angle r. This was given to you as homework. Today, our actual objective is to prove sin i / sin r is constant and angle i = angle r. So, laws of reflection using wave theory. Okay. So, here you can say that in triangle AOC and triangle BCO, the very first thing is OC is equal to OC because it is common. I think everyone must have found this. What is common? I think you must have also found that angle OAE, angle OAE, is equal to angle CBO, each 90 degrees. But the task must have been for the third side. I think it must have been for the third side. Those who have done it, for the third side, I think it must have been. But maybe you didn't find it. And that is the third side. If you have done it, then to find the third side, a little bit of your physics is utilized here. Not chemistry, here it is physics. Oh, not math here. Sorry, what is here? Physics. Now, what is it? Side. Listen carefully. If I tell you that AC is equal to what? To BD. If AC is equal to BD, then how? Let's understand. Look at the red color. Is AC going from one wavefront to another wavefront? Yes. AC is the light ray from the first wavefront to the second wavefront. And if you look, OB is also a light ray from the first wavefront to the second wavefront. That is, it is from one wavefront to another wavefront. And you were taught in the last class that when light travels from one wavefront to another, it takes the same time. So, is it correct to say that if it took time t to go from A to C, then it will also take time t to go from O to B? Are we talking about the same medium here? We are talking about reflection here. So, if the medium above is air, then it is air. If it is water, it is glass, whatever the medium is, it is the same. Which means the speed of light will also be the same. Now, what is the formula for distance? Speed into time. What is the formula for distance? Speed into time. So, if both their speeds and both their times are the same, then will both their distances be the same? Sir. That is, the distance from A to C and from O to B will be the same. Again, the formula for distance is speed into time. So, the time is the same because light is traveling from one wavefront to another. And the speed is the same because the speed of light does not change in a medium. So, accordingly, we will write here that brother, A and B are both equal. T is equal to v, where v is the speed of light and t is time. Now tell me, what do these three sides represent? This is the hypotenuse. This is the right angle. What is this? Side. So, is it correct to say that triangle AOC is congruent to triangle BCO by RHS rule? And if they are congruent, then accordingly, angle AOC should be equal to angle BCO. Angle AEO. Sorry, my bad. Angle AEO. So, here comes which one? So, angle AEO should be equal to angle BCO by S.P.C.T. And AEO is our angle i. And BCO is... Now we have to prove that which one is equal to which one. Children, sin i is equal to what? Sin i. What is the ratio of sin i / sin r? Constant. Reflection on the basis of wave theory. We are going to talk about it. Its diagram will also be quite important like the previous diagram. We are going to study two cases first. Rarer medium to denser. First, we will do denser to rarer. We will work on both. Same derivation. This is our interface. Above, we have a rarer medium. Below, we have a denser medium. Light, and another light ray. This is our normal. This is angle i. The light will bend towards the normal. Away from the normal. Towards. Why? And perhaps it's so obvious that it will be like this. But we don't need it. So, we won't draw it. We can draw it, but everything won't be useful. So, let's leave it. This is our angle r. Okay. A normal will also be formed here. If you look at it, it's also r. Again, not needed. Okay. We are going to draw waves. Is this a wavefront? This is a wavefront. In which medium? Rarer medium. Rarer medium. Now, the wavefront that will form in the denser medium, you have to start from here and extend it to this light ray. Now, imagine how you would draw it. I'm giving you options. Okay, first, second, third. Which one looks best? If I have to draw a wavefront from here, the time it takes for light to travel from here to here should be the same as the time it takes to travel from here to here. But because the medium there is what? Denser. So, light will travel faster. Fast. So, as a result, it will cover more distance in the same time. Suppose it takes 0.1 seconds to come from here. Then here too, it will be 0.1. In 0.1 seconds, light will be able to travel more there because the speed is higher. Here, in 0.1 seconds, it will travel less. So, the distance between the wavefronts here will be less. It will be less. So, these two wavefronts will not be parallel to each other. Understood? They would be parallel if this light continued to go like this. If it continued to go like this, then all these wavefronts would become parallel to each other. So, something like this will form. Let's do some naming. Come on, brother. Let's name it O, A, B, C. That's enough, probably. Okay. Now, what is our objective? To prove sin i / sin r is constant, which is equal to the speed of light in the first medium upon the speed of light in the second medium, which is also known as Snell's Law. We have to prove this. Math says if this angle is i, then this will also be i. We have already discussed this. Listen again. This angle is 90, and this angle is also 90. So, this is common. Let's remove it. So, if this is i, then this angle is also i. Seventh-grade math. Angle sum property. Are you dead? Should I solve it completely? What is this? What is this entire thing? So, this part is 90 - i. What is this? So, this is also i. Because the sum of all three. So, in total, you don't have to write all this. Just know that this angle is i. So, angle OBC is equal to i. Okay. Now, tell me, what will sin i be equal to? Sine is perpendicular upon hypotenuse. So, if this is angle i, and this is 90, then A is perpendicular, and what is the hypotenuse? Also, sin i. Tell me what it is. For sin i, who is perpendicular? O. And hypotenuse? Let's call this first. Let's call this second. Dividing first and second. Sin i / sin i. O cancels O. What do we get? AB / OB. Now, here comes, brother, Huygens. So far, if you look, we haven't used wave theory at all. Now it will be used. What is the time taken to go from A to B? Will it be the same as from O to C? Because it takes the same time to go from wavefront to wavefront. So, here too, you are going from one wavefront to another. Here too, you are going from one wavefront to another. Which means the time will be the same. But what is the speed here? It is v1. What is the speed here? It is v2. What is the formula for distance? Speed into time. So, the distance from here to here will be speed into time. Yes. And the distance from here to here will be speed into time. So, if I want to write AB, it will be speed into time. And if I want to write OB, it will be the speed here into time. But the time is the same. So, T. So, children, sin i / sin i, which is v1 / v2, we have proved it. That is, it took us 10 minutes to solve the two biggest things of ray optics. But we waited so long for this because we were waiting for wave theory. That is, Huygens' principle. Okay. So, where did we go from where to where? Now, where do we go from where to where? Denser to rarer. Light bends towards the normal. Away. This is angle i. This is angle r. I am drawing the first wave here. So, from here to here, no tension. This is the same. The second wavefront, start from here and draw it. But we were saying that in terms of distance, which is the best answer among these? So, this wavefront is formed. Understood? Here, light is in a denser medium, so it will move slowly. Here, in a rarer medium, it will move faster. Because of this, this distance should be greater. In the diagram. Now, let's do the rest of the naming. Come on, brother. Let's name this O. Let's give this A. Let's give this B. Let's give this C. This angle is i, so this will also be i. This is r, so this will also be r. The reason is the same. Everything is the same. The only difference is that in denser, we always show the speed as v2, and in rarer, last time we proved v1 / v2. This time it will be v2 / v1. You should know this much. Come on, same derivation. Let's do it again. What do we have to prove? Tell me. That sin i / sin r is v2 / v1, which is a constant. In exams, generally, prove Snell's Law. In very few cases, it also comes: where to go.

From where to go, from rarer to denser, or from denser to rarer? Does this need to be measured and made, or is it by estimation? The child is asking by estimation. Now, if you have reached this point, the most important thing you would have read by now is angle i = angle r. It's a very important derivation, and the ratio of sin i / sin r is constant. Now, the derivations in the further topics of this chapter are not even in our syllabus. Although I will do those derivations for you, I will keep meeting you in between and telling you. You can skip them if you want, but these two derivations we have just done, proving angle i = angle r and the constant ratio of sin i / r, are now the most important derivations of this chapter. And I bet you, take it or leave it, sorry, that in the board exams, one of these derivations is definitely there in one of the three sets, 101%. So, do it carefully. Second, now we are going to read the Principle of Superposition. You have also read this in 11th class. So, if you know it, you can skip it. But I have taught it completely, and questions can also come on this in the exam. This does not mean it is not in this class. I have taught you extra. No, it is in 12th as well, it was in 11th too. So, if you had studied it in 11th, you can skip it. If you don't know it, you will have to watch it, otherwise you won't understand further. Okay? So, first, look at the Principle of Superposition, then look at constructive and destructive interference, then we will move forward. What is interference? What is interfering? What is it to interfere? To push? What did you say, to meddle? You've used difficult words again. Oh, what does it mean? In the middle, what dirty things are you talking about? Oh, what does interference mean? It means superposition. What does superposition mean? Interference. Okay, listen. A wave has two parts. One is a crest, and one is a trough. And what kind of wave is it? Longitudinal. So, there will be a compression and a rarefaction. When a crest comes over a crest, or a trough comes over a trough, the wave behaves differently. When a crest comes over a trough, it behaves differently. This superposition of two waves on each other, meaning to climb on each other, is called interference of two waves. What is it called? Interference of two waves. Its Hindi translation is that brother, to meddle in each other's lives. That's a bit of an Urdu word, to interfere. Look, the point is, interference can be of two types. Meddling can be of two types. One, someone is meddling in your life and has ruined you. I haven't said anything yet. And one type of meddling is that it makes your life prosperous. There are two types of meddling, son. One, where someone meddled and you got ruined. And one, yes, that's the point. And one type of meddling is where someone comes into your life and makes your life even better. Which interference are you waiting for? But son, the point is, it's not in our hands. I'm talking about waves. The point is, it's not in our hands how two waves will superimpose. That's why we have to study their different cases. What did I tell you on the first day of this chapter, in the first five minutes? If I throw a tomato at a wall and throw another tomato, and they stick, how many tomatoes will be on the wall? Two, because a tomato is a particle. But if I throw light at a wall and throw another light, it might become dark there. First, one light was thrown, it was light. The second light came, it interfered in its life and brought darkness. What did it bring into its life? So, two types of interference are formed in this way. Oh, one is constructive interference. What is it? How to define it? When two waves superimpose with no phase difference. With no phase difference. If two waves are superimposing on each other, and there is interference, and there is no phase difference, then such interference will be called constructive. I am explaining, I am explaining. Look, look. Yes. Oh, crest on crest, he says, crest on crush. Such naughty kids, his mind has different fantasies, and crush on crush. Oh, come here, son. Here we have a wave, y1, and here, son, we have a wave, and this wave is y2. Now, look here. If I draw this wave, you have also done this in 11th, but now our method is going to change a lot. The basic things are the same. Now, if I make this reference line for phase, then here is my second wave. Now, if you notice, its amplitude is less, and its amplitude is... But the point is, if this is creating a crest, then this also... If this is at zero, then this is also at zero. If this is at trough, then... This means their phase is the same. Their phase. If this is zero at zero, and maximum at pi/2, then this is also zero at zero, and at pi/2. Which means both are in the same phase. There is no phase difference between them. And if there is no phase difference between them, and if both these waves superimpose on each other, then what will happen? The crest is saying, if this particle is saying to go up, then this one is also saying to go up. Go up, go up, go up a lot, because both want to displace in the same direction. Tell me, son. So, here, if its amplitude is a1, and its amplitude is a2, then its amplitude will increase to a1 + a2. And it's not just that they are saying it, we have to prove it mathematically too. If its amplitude is a1 and its amplitude is a2, then its amplitude is a1 + a2. That is, in constructive interference, the net amplitude is a1 + a2. We will prove this in a little while. But then comes what kind of interference? Destructive interference, the one that ruins things. I don't know how kids don't know how to use phones. Look, son. This is y1. Okay, if such interference happens, I hit light on the wall, and another light, but in what phase? Constructive. In what phase? So, light plus light will give more light. More light will be received. Two. And this motion graph of two. Tell me, in destructive interference, two waves superimpose with a phase difference of pi. That is, 180 degrees. If the phase difference between two waves is 180, and they superimpose, then destructive interference will occur. Meaning, if its amplitude is a1 and its amplitude is a2, then look, is it crest on crest, or trough? And here, trough on... So, the interference here will be destructive. That is, in this case, the amplitude we will have is a1 - a2, or a2 - a1, depending on which direction you are talking about. Like, as is visible in the diagram here, which one is bigger? a2 is. So, we will do a2 - a1, or a1 - a2. So, it will come from below, because it is more. But their overall amplitude will decrease. And you all must have understood that if the amplitude of these two is the same, then one is to ruin, and one is to completely... Tell me, son. Yes, sir. So, this is completely ruined because their amplitude was the same. Their amplitude was the same. But their phase was opposite. One was a crest, and the other was a trough. But if their amplitude is more, then if this was more, then a1 would have... If a1 was more, then the graph would be from above, like this, right, son? Okay, okay. But how can we prove that in constructive interference this is happening, and in destructive interference this is happening? Let's move on. For this, we will have to talk about the derivation of destructive and constructive interference, and here we will use the Principle of Superposition, which we have also done in which class. If anyone doesn't remember, the Principle of Superposition stated that when we superimpose two waves, the new wave function formed is equal to the algebraic sum of both wave functions. Write it down. According to the Principle of Superposition, according to the Principle of Superposition, the wave function of the wave form after interference, the wave function of the wave form after interference is the algebraic sum of the wave functions of the individual waves. Algebraic sum of the wave functions. Wave functions of individual waves. Let's assume that y1, that is, the first wave, its wave function is a1 sin omega t. Now, you know this, how is it? Again, 11th class. Sine wave, cosine wave, what function do they follow? A sin theta, or A cos theta. And theta is omega, right? And the second one, brother, will be a2 sin omega t + phi. Here, phi is the phase difference. There can be any phase difference between them. It can be zero, 90, 180, or any angle in between. But why have we kept omega the same for both? Because, look, we have kept the frequency of all of them the same. If these two waves are completing, then this one is also completing two waves. Because two waves completely superimpose only when their frequencies are the same. You should know this. Write it down if you don't know. Please. Two waves completely superimpose on one another. Two waves completely superimpose on one another when their frequencies are equal. When frequencies or wavelengths of the waves are equal. When frequency and wavelength are both equal. If this is the case, then we have to take omega the same here. Now, according to the Principle of Superposition, what is the Principle of Superposition? According to the Principle of Superposition, y net will be equal to y1 + y2. It will be. So, y net is y1. What is y1, son? Tell me. a1 sin omega t. And y2 is a2 sin omega t + phi. From here, a1 sin omega t + from here, a2 sin... The formula for a + b is sin a cos b + cos a sin b. This is a 10th-grade formula. It's 11th grade. So, the children of science should write this down, that the formula for sin(a+b) is sin a cos b + cos a sin b. It's an 11th-grade thing. Okay, fine. Now, if this is the case, then y net here becomes a1 sin omega t. And sin omega t. Take sin omega t common from these. It will be a1 + a2 cos phi. And here, let's write cos omega t separately for a bit. And here it becomes a2 sin phi. Now, listen, son. There might be a little problem here, but the rest of the derivation is very simple. Do you remember that we proved in 10th grade that root 3 is irrational? Do you remember, son? We had to prove root 3 is irrational. So, first, we used to assume it was rational. Yes, we used to assume it was rational, then we used to prove that we assumed wrongly, and then it became irrational. Was there such a chapter? I don't know if that chapter was removed or what happened. But I am telling you, it was the most useless chapter, such a horrible chapter, brother. At that time, it was in our syllabus, but it was torture. Because there is no logic in it. Like today, your Euclidean geometry. Did it not come in Euclidean? There were some postulates in Euclidean, do you know? What was the first postulate? Two right angles are equal to each other. You are laughing. It was there. There was a postulate that if one line segment is equal to another, and the second is equal to the third, then the first will be equal to the third. Look, there was a postulate that from one point, infinitely many... These things were discussed. Imagine today, if I tell you, son, my phone has touch. You'll say I'm crazy. But 50 years ago, if you said, I have a wireless phone, it has touch, and video calls can be made. Video calls can be made from home. Only Narayan Narayan could do this. Oh, you know, I saw in that, he used to do this, Narayan Narayan, and an image used to appear in front of him, and he used to talk. We can do this today. If you say this 50 years ago, they will call you crazy. I say today, I am making a phone that will come out of my pocket when I hit the seat tomorrow and say, "Whatever you command, my lord." You are laughing now. Maybe it will be possible. You never know. Look, listen, listen. That was a chapter, generally, because what we are going to do is a repetition of that in a way. So, listen carefully. Here, two things, we, no, not us, but the people who did the derivation, assumed two things, and after assuming, they took the derivation forward and saw that what they assumed was correct. And think, it wouldn't have happened like this. Look, it wouldn't have happened that they assumed this first. Let me tell you. They said, let's assume that a1 + a2 cos phi, a1 + a2 cos phi, is A cos theta. Theta is the new phase that the new wave will have. Oh, say yes, son. And A is the total amplitude. What is it? And they assumed a2 sin phi, a2 sin phi, as A sin theta. And it's not that they assumed this in the first try. Everyone's luck is not like yesterday's, where a dream came and the structure was given. Understand? Some people also work hard in life. So, they must have assumed this for the first time. It wouldn't have come in one go. But they must have assumed things multiple times. But when they assumed this, and after that, when we do what we are going to do next, when the final result comes, we will understand that these two are correct. What is the final result? If you add one sine wave to another sine wave, then either no wave will come, or if it comes, it will be a sine wave. It's not possible that sine and sine add up to a cosine. No, cosine can come. Cosine and sine are the same thing. Just the phase is different. A cosine came, or a parabolic, or an ellipse came. Nothing like that will happen. Okay, if this is the case, then y net becomes sin omega t. It is sin omega t. And here, a1 + a2 cos phi, what am I writing it as? A theta. And a2 sin phi as A sin theta. So, what do we get? Tell me. Here, A sin omega t cos theta + A cos omega t sin theta. Do you feel anything? I'll bring it. I'll bring it. I'm bringing it. sin omega t once. sin omega t cos theta + cos omega t sin theta. Is it correct? Now, is it coming? What formula is this becoming? sin(a+b). So, sin omega t + theta. So, should a third sine wave come by adding two sine waves? Yes, sir. It is coming. Yes, sir. This means they assumed correctly. And to be honest, we are not behind this. We are behind them. We are behind them. Don't forget the purpose. What is the purpose? We need the formula for net amplitude. You haven't got it yet. Now, what are we doing? We are squaring both and adding them, son. It's a difficult topic, right? Even good students don't get it properly. You will have to watch more. Look here. Squaring and adding both sides. Square. a1 + a2 cos phi squared + a2 sin phi squared. And here it will be A cos theta squared and A sin theta squared. Does everyone remember that a1 and a2 were? What were a1 and a2? Amplitude of individual waves. And A is the amplitude of the superimposed wave, which you can also call the final wave. Phi was the phase difference between both waves. And theta is the new phase of the superimposed wave. Open the square. So, a1 squared + a2 squared cos squared phi + 2 a1 a2 cos phi. And here, a2 squared sin squared phi. Here, A squared is taken common, so inside it will remain cos squared theta + sin squared theta. If I take A squared common here, cos squared theta + sin squared theta is 1. Exactly in the same way, from here and from here, what will I take common? a2 squared. So, here it will be a1 squared. Here it will be a2 squared. Inside it will remain cos squared phi + sin squared phi. And here it will be 2 a1 a2 cos phi. And here it has become A squared. We needed to find the value of A, that is, the net amplitude, from here, which will be under the root, because this is a square. It will become under the root. This has become one. So, what will remain? a1 squared + a2 squared + 2 a1 a2 cos phi. Does this formula look familiar to you? Maybe it looks familiar, but it's not at all like you think. That formula was very different. Resultant of two vector addition. That was the magnitude of the new vector obtained by adding two vectors. But first of all, we are not adding two vectors here. Here we are adding two waves, and the new amplitude obtained by adding them is its formula. Here, a1 is the amplitude of the first wave. Let's say one wave. And a2 is the amplitude of the second wave. Phi is the phase difference between them. And A is the amplitude of the total new wave that came. Now, from this formula, we will prove the constructive and destructive interference parts, which is going to be very simple. If we have the formula for A, a1 squared + a2 squared + 2 a1 a2 cos phi. So, according to this, if constructive interference is happening, then we have read that phi should be 0 degrees. Now, if I put 0 in place of phi, then cos 0 is 1. So, it will be a1 squared + a2 squared + 2 a1 a2. Can I do this by applying the identity (a1 + a2) squared? And if this cancels out from here, then the net amplitude will be a1 + a2, which we also discussed in the last class, that if constructive interference is happening, then you will get this net formula. But if, on the other hand, interference is happening, which one? Destructive. According to that, phi will be 180 degrees. Phi will be 180 degrees. Son, were you waiting for rose water? Close the gate downstairs. They must have stopped everyone, the Gulab Jal. They must be waiting for you, son. If it is destructive interference, then phi will be 180. How will it be? Because trough climbs on crest. So, the angle becomes 180 degrees. What is the value of cos 180? -1. So, according to that, it will be a1 squared + a2 squared - 2 a1 a2. And this will become the identity of what, son? (a1 - a2) squared. It cancels out. So, the net amplitude is a1 - a2, and a2 - a1, because it is squared. Reaching up to YDC will take a little time, because now we have to talk about what you see on the wall. Do you see the amplitude of light, or intensity? Intensity. So, a new word has come. What is intensity? Son, intensity is related to the amplitude of any wave. Just like sound has loudness, pitch, and intensity, light also has different strengths: frequency, energy, amplitude, and along with that, intensity. We are talking about intensity now. I hope you remember that intensity is only related to amplitude. Related to what? Yes, we learned this from the 9th-grade sound chapter, that the intensity of a wave is directly proportional to the square of the amplitude. What is it? Yes, that is, for any wave, be it sound, the intensity is directly proportional to the square of the amplitude. So, according to this, the intensity is directly proportional to A squared. And if I want to put an equal sign here, what do I have to put? A constant. Makes sense. Okay, now we have the formula for A net, that is, the square of the net amplitude, which is a1 squared + a2 squared + 2 a1 a2 cos phi. Now, how can I convert this formula of amplitude into intensity? Let's see. It's very easy, son. Can I write A net squared as I net by K? Are you dead? Yes, sir. Can I write a1 squared as i1 by K? Because K is a constant. a2 squared as i2 by... But what do I have to write for a1? Under root i1 by K, because it is a1, and it is a1 squared. And here, what can I write? Under root i2 by K. Along with what is going on? cos phi. Here, it becomes I net by K. Here, it becomes under root i1 by K. Under root i2 by K. And here, it becomes 2 under root i1 by K, under root i2 by K. K cancels out with K. So, from the formula of net amplitude, we have derived the formula for intensity, which is i1 + i2 + 2 sqrt(i1 i2) cos phi. Understood? So, this is the formula for brother, intensity. Now, let's apply constructive and destructive interference to this as well. Come on, let's do it quickly. Do it in your mind. Come on, do this. Solve it in your mind, brother. Everyone. [Music] Did you do it? So, in constructive interference, what must you have put for phi? Zero. And in destructive interference, what did you put for phi? 180. Correct. So, from here, what formula did you get for I net, brother? i1 + i2 + 2 sqrt(i1 i2). And what was the value of cos 0? So, this must be the formula you got. Okay. If the amplitude of both sources is the same, then I hope you understand what will come here. So, here it will be 2i. And here, zero. And if the amplitude is the same, then the intensity will also be the same. So, here it will be 4i. What will it be? Understood? How i + i = 2i, and i + i + 2i = 4i. Okay. And if we talk about here, then the net intensity we will have will be i1 + i2 + 2 sqrt(i1 i2) cos 180. The value of cos 180 is -1. So, what did we get, brother? i1 + i2 - 2 sqrt(i1 i2). Now, if the intensity is the same here too, then it should be zero, because if the amplitude is zero at the same amplitude, then the intensity must also become zero. Understood? So, tell me, if I throw light on the wall, and I get increased total intensity, then it was normal for me. But if the intensity there decreases or becomes zero, then it will be very surprising, because this does not happen in particle nature. Okay? Let's go. Now, let's move on to another topic, and understand it properly. That is, coherent sources. Which sources? Yes, it's a topic that needs a little attention, son. It's oral, but it needs attention. Because if they are coherent, then the phase difference of the waves coming out of them will either be zero, or if there is any phase difference, it will remain constant. For example, let's assume a wave came out from here, and a wave also came out from here. If I look here and look here, if we assume there is no phase difference between them, but here you can see a phase difference has come. Has it come? And look here, you are again in a different phase difference, and here, they have become completely opposite. Look, one is making a crest, and the other is a trough. So, their phase is changing repeatedly. Their phase difference is changing repeatedly. So, these are not which sources? Coherent. Because the difference in their phase is not constant. But if I make a source here, and it is something like this, and there is another source here, and for that, it is something like this, then are these coherent? Yes. Because when this is maximum, this is minimum. So, there is a phase difference of 180 between them. And when this is minimum, this is maximum. How much is it? So, we see here that always between them... Okay, ignore a little bit. Their phase difference is always remaining constant. So, such sources in which either there is no phase difference at all, or if there is, it remains constant, such sources are called which coherent sources? What are they called? Coherent sources. What are they called? Coherent sources. What are they called? Coherent sources. How can coherent sources be made? NCERT says one line, it is also written in the booklet. That line is: Two independent sources can never be coherent. Listen, how? If you have a source here, what is the smallest source in the world in your house? Light. Even if you take the most intense source of light in the world, it is made up of millions of atoms. Like an LED. In your house, the tiniest source of light is the LED. An LED is also made up of millions of atoms. In semiconductors, we will read that an LED is made of semiconductor. Let's suppose germanium. Actually, it's made of an alloy, but let's assume germanium. And let's assume you used even 1 gram of germanium. How much? So, even in 1 gram, there are approximately 10 to the power minus 3 moles. Oh, the molar mass is also assumed to be 1000. Let's assume the molar mass of germanium is 1000. Math is not coming. What is moles? Formula for moles is mass upon molar mass. I am saying, assume it's 1000. It's not 1000. Assume it's 1000. How much did you get? 10 to the power minus 3 moles. Multiply by 10 to the power 23. 10 to the power 20 atoms. Where are the millions? There are no millions here. Millions are nothing compared to this. How many zeros are in millions? So, this is 10 to the power 20. But millions of millions. Understand the point. In such a small source of light, there are so many atoms. Did all of them turn on at the same time? Were they excited? Did they emit light? No. So, brother, forget about big sources like matches, bulbs, lights. Even the tiniest source in your house is made up of millions of atoms. So, then the question comes. Let's imagine there is an atom. What is it? There cannot be a smaller source than an atom. Say, die, die. Let's assume. What is it? Even according to Niels Bohr's theory, an atom changes its energy after 10 to the power minus 3, minus 4 seconds. Even an atom cannot be coherent. Because this atom, this...

What will atoms change every millisecond? Their waves, so how will they become current? So two individual sources, even imagine one atom, which is not practically possible, even then what is it not? Then why are we studying this, and for DSC, which sources do we need? Coherent. The solution is Huygens' principle. What is it? Huygens told you something very nice, if you remember. The first thing he said was that if we assume, then assume, then the study is over. Let's assume, let's assume even 100-watt bulbs, you won't see what you want to see from them. So what sources are needed? Okay, I will tell you. I haven't told you yet why we need a source. I haven't told you that yet. We just need light. Where from? Because two individual sources can never be coherent. So the answer is, don't take two, take one. Don't take two, take one, and what is its wavefront, make two holes in it. Here, say it. This is its wavefront. If this is its wavefront, then will the light particles that are assumed to be here and here be in the same phase? So if light emerges at pi/2 from here, will it also emerge at pi/2 from here? Sir, if it changes, then let it change. This will also change because it is the wavefront of the surface, and if it is the wavefront, then the same phase will remain on both of these. What are coherent sources? Sources in which the phase between the light waves is either zero, or if there is any difference between them, it will remain constant. So, according to the definition, two sources of light which continuously emit light waves of the same frequency and wavelength. Well, we have been talking about this since day one, that if we have to study interference, then the frequency or wavelength must be the same, only then will they superimpose. We have been talking about this since we started studying interference. With a zero or constant phase difference, either the phase difference between them will be zero, or if there is any, it will be constant. And if this is not the case, then those sources will be called incoherent. Then we talked about the fact that two independent sources cannot be coherent. Two different sources can never be coherent. There are some reasons for this. First, light is emitted by individual atoms and not by the bulk of matter acting as a whole. If you imagine a source of light, whether it is a tube light or a bulb, then the bulb is not a source of light. It is made up of many atoms, which is the smallest source of light in the world. As we discussed in our homes, like an LED, if you look at it, it will also be made up of millions and trillions of atoms. And even if you imagine that there is one atom, even then, according to Niels Bohr's theory, its radiation, its energy, its phase will vary somewhere in the range of 10 to the power of -5, -6, -7, -8. According to books, it is around 10 to the power of -8. But you know, this was studied only for hydrogen-like atoms. For everyday atoms, it is still in the range of 10 to the power of -3 to -5. Even if you consider the range of 10 to the power of -3 to -5, you can say that the atom will be changing its frequency, its wavelength, and its phase thousands of times in a second. Therefore, two sources can never be coherent. Our aim is that we will have a screen here, and we will not have a source here. We will have a very small source of light, and we will have a wavefront here, and from that wavefront, we will make very small holes here and here, and they will become two sources for us. What is the YDS? Young's Double Slit. Young's Double Slit Experiment. So we will do it like this, and light will emerge from here. Light, light, look, light is emerging. What is emerging? What is emerging? Light is emerging. We will hit it on a wall, and by hitting it on the wall, we will create an interference pattern here. What will we create? And do you know how the interference pattern will be formed? Look, it will be like this. Look, it will be like this: here will be bright. Bright means constructive interference. Bright means constructive. The amplitude came in the same phase and increased, the intensity increased, so it became bright. What became bright? Then it will be dark. Dark means destructive interference happened. Then it will be bright. Then it will be dark. Then it will be bright. Then it will be bright. Dark, bright, dark, bright, bright, dark. Dark means not darkness. Dark means nothing. Understand it as darkness. Okay, there is no light. Okay. Then this is the center. Then, leaving one, dark, then one bright, then one dark, then one bright, it will continue like this. What will this be called? Fringes. What will it be called? Fringes. What will it be called? Fringes. Now imagine, if all these fringes come very close to each other, will you be able to see these fringes? Will you be able to see these fringes? No. You will just think it is light. You will just think, tell me with your heart, if I made this, would you see that there is darkness in the middle? Let's play in reverse. Let's start from here. Would you see that there is a gap between them? Would you just think it is bright? Because your eyes have the capacity to see two things individually only when there is a certain gap between them. Okay, even if not, if I tear one of the slits and make two holes here and place them in the middle like this, will bright, dark, bright, dark, bright, dark, bright, dark, bright, dark, bright, dark be formed here? Yes. But we won't see it. Why? Because the gap between them is so small. What did we call these just now? Fringes. What did we call them? Fringes. And the gap between them is called fringe width. What is it called? Fringe width. What is it called? Fringe width. This bright, this bright, the gap between them is called fringe width. We will represent it by beta. By beta. And Young, YDS, gave the formula for this fringe width to the world first, which we will also give you. And the formula came as lambda d by d. What came? Lambda d by d. Where lambda is the wavelength of light that you have used. Capital D means the distance between these holes and the screen. The distance between these slits and this screen. And this small d is, brother, the distance between these two slits. Do you want beta to be maximum or minimum? Maximum. Oh, you fool, only then will you see it. If you want to see bright, dark, bright, dark, bright, dark, then what do you want? Beta to be large or small? Fringe width to be large or small? Larger than your eye's capacity or smaller? Larger. So, to increase beta, what will you increase? Lambda. You will increase lambda. The maximum lambda you can use is red color. Now, if you say, I will use radio waves, you won't see them. I'm talking about waves. You won't see them. What you will see is visible. In visible, the largest lambda is of red. Okay, so which light should we use? Red. Second, capital D should be large, should it be small? It should be large. And small d, what should we keep it? Minimum. For that, what will we do? We will make or take two holes very close to each other, as close as possible, just so close that they don't overlap. If we create all these conditions, then maybe we will see fringes. Now we will talk again. If we have to do YDS, then we have to draw the diagram. The most important thing is that the derivation will be done from the diagram. Whose derivation? Not derivation, derivation of fringe width. Whose derivation? Some questions are still pending. Why are we taking coherent sources? And many other questions. How to take light? What is monochromatic? Monochromatic means of one color, because one color has only one lambda. If you take white light, there will be seven colors, seven colors will have different lambdas, different fringes. It won't be fun to see. It will look good, but it won't be fun. Oh, we will do it, leave it. Okay. Here we have brother, who? A source. This is our parent source. And here we have, oh, we are making two sources here. We are making two sources. We have named this brother S1 and this one S2. And this source for us is A. In front, we have placed the screen. This screen is placed very close, very far, very far, or very, very far. As far as possible. And the distance between these two is capital D. Now, we are also making a line from the center of both of these, which is a reference line. O has come. Now listen, brother. Light will emerge from here. Where will it go? To the entire screen. Light will emerge from here. Where will it go? To the entire screen. For now, let's assume that both these lights are meeting at this point. Obviously, the light emerging from here will meet everywhere. But for the diagram, we have to assume one point. So we have assumed this point, that light emerged, emerged, emerged, and met there. We are naming the point where they meet as point P. And again, a reference line from here, which will be useful in the derivation. And a reference line from here as well. Now, if this is D, then all these will also be D. Okay. Now, the distance from P to our O, we are considering this as X. We are considering this as X. Now, if the distance between these two slits is small d, which has already been discussed, then quickly tell me, what will SA be? Don't say, think. Okay, no problem. This whole thing is beta. Everywhere it is X. This whole thing is X. It is small d. And from here to here, it is X. So, you have to subtract from this. So, if this whole thing is D, then this will be D/2. What will this be? D/2. And this whole thing is X. So, from here to here, it will be X - D/2. If this is needed. So, similarly, tell me what BP will be. It will be X + D/2. What will be X + D/2? What will be X? Okay. Now, Young's Double Slit Experiment. Here we are starting. Keep in mind, the derivation of Young's Double Slit Experiment has been removed from our syllabus. But I think it will take me half an hour to explain it. If you want, you can skip that entire half-hour part. I will also increase its speed for you so that your time is not wasted. But if you want to know where this formula came from, then I am putting the derivation here for you. But I am telling you again, its derivation has been removed from our syllabus. Now you only have to understand bright and dark. You will understand bright and dark positions now. You have to find fringes. I am writing here. Now what you have to study is, you only have to study the position of bright fringes, where they will appear. Similarly, you have to study the position of dark fringes, where they will appear. And then you have to study the fringe width, that brother, what is fringe width. And here we will make an intensity graph. We will make an intensity graph. You should know this. Okay. So you should know this. But how we reached here from the derivation is no longer in our syllabus. Okay. Similarly, the next topic, which we will do, we will put the entire YDS setup in water. Its wavelength will change, etc. That will change. All that is also in our syllabus. It will come after this. After this comes diffraction. In diffraction also, it is the same. Position of bright, dark, bright, dark. You have to study. They will be interchanged. I have also given the reason for that. Again, the reason is not in the syllabus. If you understand, fine. Otherwise, leave it. And the derivation of this is also not in the syllabus. Only you. So the chapter has become very easy, hasn't it? The derivation of both has been removed, due to which you will have to study the same things again in diffraction. You will have to study the same things in diffraction as well. Okay. You will have to study in diffraction as well. Where did it go? Yes, here it is. Diffraction. After this, if you look at it, the chapter ends. But I told you in the video, maybe you have heard it by now. If not, it has been discussed further. After this, there is a topic called polarization. Polarization has been removed from our syllabus. But I don't know why NCERT has given its basics. So I have also put the basics here. This is the first topic in our NCERT that has remained in our NCERT but is not written in our syllabus. So there is a lot of controversy about this. I would say, on a safer side, don't get into trouble. Study the basic, basic polarization, which I have taught you. Not much, just the basic, basic. What is unpolarized light? What is polarized? And then finish this chapter quickly. In triangle S1P, in triangle S1P, S1P is a right-angled triangle. So I can apply Pythagoras here, which is S1P squared equals to S1A squared plus AP squared. Okay. Now, what is S1P for now, we don't know. But we know S1A. It is D, capital D, i.e., the distance between the sources and the screen. And we also wrote AP in the last class as (X - D/2) squared. Many children might have asked the question, why are we writing X - D/2? Because we need it in the derivation. Okay. Let's call this the first equation. Also, in triangle S2PB, S2PB is also a right-angled triangle. So here also, the hypotenuse we have, i.e., S2P, its square will be equal to S2B squared plus BP squared. Now, S2P again, what is it for us? We don't know. But this is again, S2B is D for us. And the distance from here to here is (X + D/2) squared. Let's call this the second equation. Now you have to subtract these two. Why we are subtracting, you will also not understand now. In just two minutes, three minutes, you will understand why we are subtracting them. What do we need to find? Do you remember? We need to find X. What do we need to find? And with the help of X, we will find fringe width. Our main focus, main aim, is to find fringe width. So for now, we are subtracting these two. Subtracting the first from the second. We are doing second minus first, basically. So if we do this, then S2P squared minus S1P squared will come. And here, D squared will be cancelled from D squared. Here I will have (X + D/2) squared minus (X - D/2) squared. Here we will apply the identity. S2P minus S1P and S2P plus S1P. Here, if I open the identity, you know that X squared will be cancelled from X squared. (D/2) squared will also be cancelled from (D/2) squared. What will remain here will be 2XD/2. And here also it will be 2XD/2. You can increase the steps if you don't understand what I said. You will understand what is remaining and what is not. Now here we will see S2P minus S1P and here S2P plus S1P. And here, 2 will be cancelled from 2, 2 will be cancelled from 2. So, in total, it will be X. What will it be? X. Now, what we are going to do requires a little attention. So far, it was just math. Now, tell me, are S2P and S1P equal to each other? Looking at the diagram, it doesn't seem so at all. There is a big difference. S2P looks much larger in length compared to S1P, and it is. But this diagram is not like that. This is made to show you. This is the reason why we spent half an hour in the last class explaining why YDS is being done and what its final formula will be. It is probably the first time that before the derivation, we discussed its formula, gave an explanation, and now we are going to derive it because it is necessary. So I told you yesterday that S1 and S2 are actually very, very close to each other, almost overlapping. And is this screen very close or very far? I told you to imagine that capital D is the length of this room, which is approximately 4 meters. And S1, S2, P are approximately 1 millimeter away. Just imagine they are 1 millimeter apart. So, accordingly, if I redraw this diagram, I won't be able to do it on this board. But let's suppose I still have to do it. So, let's say this is S1. First of all, the width I have made S1, both S1 and S2 will fit in this. But okay. Now, look closely. And this is S2. It is even closer than this. I have made it very far. I have made it very far. And this point P, I will make it here. Again, this point P is even more distant than this. Very far. Very far. And if that is the case, then the distance from here is S1P. So, will S2P be the distance measured from here? And now, tell me, are S1P and S2P equal? They look equal now. Although S2P will still be slightly larger than S1P, but we are so used to physics that the difference is not so large that they cannot be considered equal. So here we are considering them equal. But the condition is that S1 and S2 must be very close, and point P, i.e., the screen, must be very far. And this is what is going to happen. Understood? Say yes, sir. Are they equal? Yes. Now the next question, what are both of them equal to? So the answer is, both of them are equal to D. Capital D. Now, what is D? The distance between these. So, the distance from here to here is the distance of the screen. Isn't it, brother? This is D. So tell me, what are both of them? And secondly, I am telling you again, I have made them very close. Imagine that in your house, some whitewashing or plastering work is to be done. So they will come to your house and measure the length and width of the house. Suppose a carpet is to be laid, whatever needs to be done. So when they place the measuring tape here, if they place the measuring tape from here to here, or from here to here, or from here to here, then obviously there will be a difference. But it will be so minute that no one is paying attention to it. Are they paying attention to whether the measuring tape should go there perpendicularly from here? So the answer is no. It can be slightly up or down. It doesn't matter. If this is 4 meters, then this will be 4.00 meters or 4.00 meters. For us, it will be considered approximately four. So, are these two points, S1P and S2P, both equal? Yes. But they are also approximately equal to D. Equal to D. So, if that is the case, then will it become 2D here? I know what your next question will be. I am coming to that. But S2P is approximately equal to S1P, and they are approximately equal to capital D. So, a question is coming to your mind: if they are equal, then S2P - S1P should become zero. And this is not the first time you are doing this. You have understood this in electric potential at any point of a dipole. That brother, if they are equal, then they are equal. If I consider them equal here, then it will become zero, and the derivation I am going to do, I will never be able to find its value. I will ask a math question. If X = A * B / (A - B), and if B is very, very small compared to A, then can I approximately write it as A * B / A? But if I ignore B from the numerator, then it will become zero. The answer will be zero. So, this is not the first time you are learning this. We have done this many times before in physics. So, here also, you have to keep in mind that we have approximately considered S1P and S2P equal, but we have not considered their difference equal. And to be very honest, I am going to tell you about its difference. But understand this much. Now, if that is the case, then S2P minus S1P has come here. It has become 2D. From here, the value of S2P minus S1P has become 2XD/2. 2 has been cancelled from 2. 2 has been cancelled from 2. So, in total, it will be X. What will it be? X. Small d by large D. Let's call this equation A. A very valuable equation. Now, tell me, is S2P minus S1P the path difference of two waves? Path difference means, what is the distance in the path of both waves? And S2P is the distance or displacement that light, the wave, has to cover to go from the second source to point P. And S1P is the distance or displacement it has to cover to go from this source to there. So, what is the difference in their paths? S2P minus S1P. Oh dear, it had to travel 5 centimeters. It had to travel 4.9 centimeters. So, what is the difference in their paths? Just subtract. Say. So, if we look at S2P and S1P, are they path differences? So, the value of S2P minus S1P has also come. And S2P minus S1P can also be called path difference. So, from A and B, from A and B, what is the formula for path difference? It has come as X small d by large D. What has come? X small d by large D. What has come? X d by D. Okay. It doesn't look like we are going to find fringe width yet. We are going to find X. No problem. Is everything understood up to here? Let's understand again. The derivation is very long. It's not even half done yet. Listen. What we did was, first we saw two triangles, S2PB and S1AP. We applied Pythagoras to both. We subtracted them. At last, what did we get? This step is important. S2P minus S2P plus. We have approximately considered it equal to D. Because the sources are very close, and the screen is very far. But we did not consider it equal here because we had to find it. So its value has come here. Now, if we look at S2P minus S1P, it is basically what? It is also path difference. So, we have got the path difference as X d by D. What has come? X d by D. Okay. Now, let's go to 11th class. Which class? Do you remember that Niels Bohr said that the path difference of an electromagnetic radiation is an integral multiple of its wavelength? It is 11th class. Write it somewhere if you have forgotten. According to, according to the theory of radiation, according to, or you can say, you will not bring this to the exam. According to the theory of radiation, the path difference between two EM waves is always, is always equal to, is always equal to an integral multiple, is always equal to an integral multiple of the wavelength of the radiation, of the wavelength of the radiation. Integral multiple of the wavelength of the radiation. Hindi translation: The path difference between two waves is always a multiple of the wavelength. Remember? We proved it. 2 pi r = n lambda. Do you remember anything? If you don't remember, then remember it. But for now, Niels Bohr's theory says that the path difference between two waves is a multiple of what? N lambda. Of what? N lambda. For brights. Who are we talking about? Brights. Who are we talking about? In interference, two things will be formed: bright, dark, bright, dark. Let's talk about brights first. So, according to Niels Bohr's radiation theory, the path difference for brights should be how much? N lambda. How much? N lambda. And the path difference is also equal to X d by D. So, accordingly, n lambda is equal to X d by D. From here, find the value of X. So, what will be the value of X? It will be n lambda d by d. What will it be? n lambda d by d. What will be the value? n d d. What will be the value? n d d. Where the value of n is 0, 1, 2, and so on. You have got the general formula for brights. What have you got? We find the position of brights. What will be their position on the screen? Position of brights. What will be the position of brights? Come, brother. See, is it correct to say that we are talking about which fringe first? Say, brother. Bright fringe. Who are we talking about? Its formula has come. What is it? N lambda d by d. According to this, first we will find the central bright. Central bright means, the one at the center. For that, you have to put the value of n as zero. So, what will be the value of X? Zero. Okay. Now, let's talk about the first bright. When will the first bright be found? For the first bright, what value do we have to put for n? So, the first bright will be found at n lambda d by d. Now, if we talk about the second bright, for the second bright, what will we put in place of n? We will put 2. So, the second bright will be found at 2 lambda d by d. Let's do the third one as well. For the third bright, we will put 3 in place of n. So, it will be 3 lambda d by d. So, the first bright will be found where? Center bright. So, leave the center. Where will the first bright be found? At lambda d by d. Where will the second bright be found? At 2 lambda d by d. Where will the third bright be found? At 3 lambda d by d. Where will the fourth be found? At 4. Where will the 10th bright be found? At 10 lambda d by d. And where will the nth bright be found? At n lambda d by d. So, accordingly, what is the location of the brights? Brother, the location of the brights is 0 lambda d by d, 2 lambda d by d, 3 lambda d by d, and lambda d by d. But here we said that if this is the center, then something like this is happening above it. But will something like this also happen below it? So, if the value of n is 0, 1, 2, 3, 4, then will it be plus minus 1, plus minus 2, plus minus 3, plus minus 4 with respect to the bottom? Yes. If we are talking with respect to the bottom as well, then what value can be put here for n? Plus minus. So, here also, what will come? Plus minus. Here, what will come? Plus minus. So, here, what will come? Plus minus. Plus minus. Plus minus. And on all of these, what will come, brother? Plus and minus. That is what integral multiple means, that we have to multiply it by an integer. Okay. But the main thing is, where will the dark fringes be? True or false. Dark fringes will always be between two bright fringes. Because if there is no dark fringe between two bright fringes, then it is just one bright fringe. Yes, sir. Something to think about. If I say that this is bright, what is it? Bright. And if nothing came between them, then it is just one continuous bright fringe. But if a dark fringe came between them, then it is bright, dark, bright, dark, like this. So, if I say I have two bright fringes, it means that what came between them? Otherwise, why would you say two bright fringes? Say, brother. That is, you are saying that you are getting the first bright fringe at lambda d by d. The second one.

Where is it found? That is, somewhere in between those two, what will there be? Like this, where are you finding the third? So, in between this, what will there be? It will be dark. And to come in between them, to come in between them, you will have to put the value 2n - 1/2. It is math. Did not understand? Come on, it is math. Now let's talk about for dark. Now, where will the darks come? For bright, what path difference did you take? Oh, for brights, what path difference did you take? n lambda. So, for dark, you will have to take n. If I see, what do I want? That I get a dark in between two brights. So, math says, for that, you will have to put this value. Let's understand why I am saying this. Okay, again, you can put plus here, or minus. If you put plus, then you have to remember that you have to start the value of n from zero. But at zero, what have we assumed? Central bright. What have we assumed? So, that is why we will start the value of n from here, from one. For this, you will have to put 2n - 1/2. I am explaining, if you did not understand this, then this is not a formula. This is basic common sense. Half comes between one and two. 3/2 comes between two and three. 5/2 comes between three and four, which will come from this. Put 1 in place of n, 1/2. Put 2 in place of n, 3/2. Put 3 in place of n, 5. Oh, are you not even getting this much? Put 1 in place of n, what came? Put 2, put 3, put 7, put 7, put 10. Just keep putting these. Come, see, if it is like that, then what? Oh, if the path difference for darks is 2n - 1/2 lambda, if it is for dark, then for us, it was also equal to what, brother? x d / d. The path difference for us was also equal to x d / d. From here, find the value of x. So, the value of x came as 2n - 1/2 lambda d / d. This came for us for darks. For whom did it come? Now, if it is like that, then let's find the position of darks. Where will darks be found? Will there be a central dark? No, child, because at the center, what is there? So, the first dark came. For the first dark, what will we put in place of n? One, or say plus minus one. It is very important to write here what you have taken for n: plus minus one, plus minus two, plus minus three. Sir, if I write 0 plus in place of n, then it will be wrong. If you want to start from zero, then the value here will be 2n + 1/2. If you start from zero, but that is very confusing, because there, what will come as your first dark, in that you will put zero in place of n, you will die, ha ha. So, remember, for dark, what to put? 2n - 1/2, and from where to start the value of n? Remember this. Okay, so if we put it, what formula came? 2n - 1/2. So, if you put one, what comes on putting one? Say, lambda d / d's half. But if you put plus minus, then plus minus will come. Second dark will come, brother, second dark. For the second dark, what will you put in place of n? You will put 2. Let's put 2. So, if I put 2 in place of n, what will come? 3/2, 3/2, plus minus 3/2 lambda d / d. And third. Yes, but there is no minus. We don't have to put minus. Minus means we are going to the other side of the origin. That is why, that is why I am saying, what will you put? One. What will you put? Two. What will you put? Three. But the value that is coming, let's say 5/2, that will be in plus minus, right? Because what does plus minus mean? That if this is the center, if this is the center, then all the fringes above it we have considered as plus. All the fringes below it we have considered as minus. Like for the origin, it is here and there in our graph. Yes, so if I take the third dark, then what will come in place of n for us? 3. Now, if 3 comes in place of n, then tell, for the third dark, what happened? Put 3, 5. 5/2 lambda d / d, which can come in either plus or minus. So, if I finalize the position of darks here, then what will be the position of darks? 1/2 lambda d / d, 5, sorry, 3/2 lambda d / d, 5/2 lambda d / d. It will go on like this. Now, we are going to represent this with a graph, but first, bright and dark. If I write the position of brights here, then we will remove it. So, compare, is there always a dark in between two brights, and a bright in between two darks, or not? What was the position of bright? Zero, lambda D / d, 2 lambda D / d, 3 lambda D / d. Now, mathematically check, brother, is there always a bright in between two darks, or a dark in between two brights, or not? Check, it should come anyway, we have put the formula like that.

[Music] Yes, negative direction is representing direction, it would be wrong to say this. It would be wrong to say that negative is representing direction. Negative is representing position. Positive means all the fringes above the center. Negative means all the fringes below the center. Fringes do not have any direction. Done? Okay. Now, if it is, yes, we can also say bright in between two darks, or dark in between two brights. These will come consecutively, child. Come, now we want to see this through a graph. We are calling this. This graph comes a lot in exams: Intensity Distribution Curve for YDSE. We have to make this same curve for diffraction. What does Intensity Distribution for YDSE mean? Look, if this was our screen, then this was its central part, which we had named what? What was it named, brother? And what did we assume? That at the center, what is forming? Bright. I am making a reference line here which is representing maximum intensity. How much will be the maximum intensity and how much will be the minimum intensity, we will also tell you that now. But we had given you the formula for intensity in the last class. What is the formula for intensity? K square. What is it? K square. Okay. Now, if at the center, what is forming? Bright. What is forming? Bright. So, this bright came at the center for us. But just after bright, what will form? Just after bright, what will form? Dark. Then after dark, what will form, brother? Bright. Then what will form? Then what will form? Then what will form? Dark. And this is not happening only on this side. This is also happening on this side: that bright formed, after that dark formed, then what formed? Bright. Then what came? Dark. Then what came? Bright. And then what came? Dark. Now, is the width of everyone, is the width of every bright the same? We will have to see this, because in diffraction, that is, in the next topic, it will not remain equal. First of all, this is the first dark. Where will it be? Where will it be? The answer is, it will be at half of lambda d / d. And on this side, where will it be? At half of minus lambda d / d. Okay, is this our first bright? Where is this first bright? At lambda d / d. Okay, this is also our first bright, but on the opposite side, then who will it be? Minus of lambda d / d. Okay, is this our second dark? Where was the second dark, brother? Minus, sorry, this is plus. 3/2 lambda d / d. And here, our second dark, brother, where will it be? -3/2 lambda d / d. Okay, is this our second bright? Where was the second bright, brother? 2 lambda d / d. Oh, say, yes sir. Is this also our second bright, brother? Where is it? -2 lambda d / d. So, what did we make? Oh, like this, the next dark will come. And this is what we made: the intensity distribution curve. That is, this is representing our intensity. That, brother, look, here, where we have made the positions of brights, there the intensity is highest. And where we have made the positions of darks. Okay, where dark is made, is the intensity zero or less? Why are you saying zero? So, dark means zero? Oh, dark does not mean zero. Dark means destructive interference is happening there. There, the net amplitude is what? A1 - A2. The net intensity is equal to what? I1 + I2 - 2 under root I1 I2. Remember? Dark does not mean darkness. Dark means less light. Understood the point? When will dark be 100% dark? When will it be pure dark? When the amplitude of both waves that have superposed on it is the same. Now, tell me, looking at this diagram, is the amplitude of these two the same? Tell me, looking at these two, is the amplitude of both the same? Yes, because these both are made from the same sources, from slits of the same width. So, in this case, that is, in the case of YDSE, the dark will be 100% dark, and the bright will be a bright of double amplitude. We are doing that too now. But, child, dark means 100% dark, this is not it. In this case, dark is 100% dark, this is the point. Okay, I hope you understood this much, that actually this curve is not like this. You understood? Actually, the curve is like this, right? That is what we made like this. Actually, it is like this. We had a screen here, there were two holes here, YDSE was forming here. So, actually it is like this. But in the exam, we will not make it like this, we will make it like this. But this was forming on the screen in front. Conditions for sustained interference. A question on this will get stuck in MCQ, Assertion Reason, or in any way, that if we want to see the fringes well, what conditions apply? We had read three-four conditions in the last class, and there are more. Come on, before that, before that, let's finish this and talk about fringe width. Do you understand what a fringe is? What are these brights and darks called? But today we are talking about fringe width, which we are representing with beta. What will be its definition, brother? Distance between two consecutive brights or darks. So, how do we find fringe width? Shall I subtract (n-1)th bright from nth bright? Oh, did we read about fringe width in the last class? The distance between two consecutive brights or two consecutive ones is called fringe width. So, tell me yes or no, is this fringe width from here to here? Okay, tell me yes or no, is this also fringe width from here to here? Okay, tell me yes or no, is this also fringe width from here to here? Tell me yes or no, is this also fringe width from here to here? Because this is the distance between bright and dark. Come on, tell me yes or no, is this fringe width from here to here? So, fringe width can be found in any way, because it is the distance between two brights and two consecutive darks. We can find it from darks too, from brights too. Yes, of course. We will find it from brights, brights are a bit positive. So, we will say nth bright, n lambda d / d will be subtracted from n - 1 lambda. The position of brights was lambda d, right? Yes sir. Gaurav, is this coming in the paper? Are you looking at the fans? What are you doing, Ladoo Mut? Are you looking at the fans or sitting in the class? What is in it, child? I feel like putting my hand in it. Oh, man, tell me, what came? n lambda d / d minus n lambda d / d, what happened? Cancel. What happened? Cancel. What came? lambda d / d. What came? And this formula is not a new formula. We talked about it for half an hour in the last class, that the distance between fringes or fringe width always comes out to be lambda d / d. How much does it come? How much does it come? So, this is our fringe width. What is it? Fringe. I told you, there is a chance of one question coming on this, child. What are the conditions? If I want to see good sustained interference. First, two sources should emit light of what? Same frequency and wavelength, so that what can happen? Superposition can happen. What can happen? Superposition. Second comes, brother, for us, that the sources of light should be what? Coherent. Come, now let's talk about this, why we have been saying for two days that our source should be coherent. Although, obtaining coherent sources is how difficult? From different sources, so we take coherence. Our S1 and S2 sources are coherent. Right now, if the phase at this point is pi, and its phase here is 2 pi, then what is the phase difference between them? Pi. And what kind of interference will they make here? Destructive, because the phase difference is pi. But tomorrow, in a fraction of a second, it changes its phase. So, this one is emitting, let's say, 2 pi, but this one emitted what in a fraction of a second here? Now, what is the phase difference between them? Now, what is it making here? Tell me, where 0.001 seconds ago, what was forming? Dark. Now, what is forming? Bright. So, is this a good thing or a bad thing for you? The fringe pattern will change repeatedly. How will you observe it? Again, I am saying, enjoying is a different thing. You want that bright, dark, bright, dark, bright, dark remain where they were, so that you can observe them, see them. And for this, if constructive interference is happening at one point, then what should keep happening there? But for that, what will you have to keep constant? Phase difference. And when will that be possible? When the sources are coherent. So, remember, counter questions to all these will come in the exam. What will happen if sources are not taken? I have told you. Write down homework question: What if we take incoherent sources in YDSE? What if we take incoherent sources in YDSE? Okay, homework. Next, child, what is contrast? Contrast. Do you understand contrast? Yes, child, do you understand contrast? So, where we differentiate between two colors, we need good contrast in the pattern. So, the bright should be bright, but the dark should be pure dark. To do this, what will you do? You will keep the amplitude the same, right? Say, child. For a better contrast between maxima, that is bright, and minima, that is dark, the amplitude of the interfering wave should be equal, and we took that too. Two sources should be narrow. We did not discuss this point. We made holes, we did tell that the gap between both holes is very small, but we did not talk about whether the holes should be small or big. Child, if the size of those slits that we made is large, then will there be multiple sources within that one slit? Sir. And our next topic is diffraction. And if there are multiple sources, then a single source will start doing one more thing on its own. What is its name? Diffraction, which we do not want right now. Right now, we only want what? Interference. You do not know diffraction yet. I am telling you this much, if you did not take the sources narrow, you took the sources close, but you took very big sources, you took such big sources, then, brother, within this source itself, there will be multiple atoms, multiple particles, multiple sources, which will start doing what among themselves? Diffraction, which is the brother of interference, which we do not want to happen right now. That is why we have to take sources as narrow as possible. Next, the interfering wave must travel nearly in the same direction. This is obvious that they should go in the same direction. Common sense. If S1, the source, is throwing light this way, and S2 is throwing it this way, then it will become very difficult for interference to form. We will have to make it with reflection. Yesterday, a child also asked after the class, that if I take two sources of light and make them reflect, what if I make them reflect? Yes, it is possible, as long as, as long as the phase difference between them remains constant, then you can make them reflect, you can make them refract, it does not make any difference, right? So, here we are saying, we are not saying that if you do not do all this, then interference will not happen. We are saying, if you want to do it well, then do this, take the light in the same direction. The sources should be monochromatic. This point has been discussed, I think. So, let's move this forward. The interfering wave should be in the same state of polarization. This has been deleted from our syllabus. Do you want to study it for 5-10 minutes? There is polarization. Come on, polarization should be the same, this is something. Let's study it today for 5-10 minutes. To have sufficient fringe width. If this is the most important, and we know, if the fringe width should be sufficient, then small d should be small, and large D should be large, and wavelength should be as much as possible. What is written? The distance between the two coherent sources should be small, and the distance between the two sources and the screen should be large. Is that correct? Here comes the displacement of interference fringes, which means that if the entire YDSE setup is put into a denser medium. If YDSE setup is immersed in denser medium, for now, understand water, because mostly water comes in exams. So, in that case, your previous fringe width was lambda d / d, but in denser medium, it will become beta dash, let's call it lambda dash d / d. Also, the refractive index, which we just taught, will be lambda upon lambda dash. So, from here, lambda dash will come as lambda upon refractive index. According to this, your new fringe width will come, if I put lambda upon n in place of lambda dash, then what will remain in multiplication for us? d / d. Now, lambda d / d is our old beta. So, our new beta will be old beta divided by refractive index. And you all know that if it is a denser medium, then the value of refractive index will always be greater than one. And if it is greater than one, then the new fringe width will always be smaller than the old one. And if it is smaller, then you will not enjoy it. Understood? Okay. Now, how much shift will come in the fringe? We are going to find fringe shift. Yes, of course, child, if we have to find fringe shift, then we will subtract. Fringe shift means delta beta. So, what will come? We will subtract beta dash from beta. Let's do something like this. Have we done something like this before? Yes, in which we had found normal shift, that is, apparent depth minus real depth. Ray optics, something like that is happening here. So, how much shift will come, brother? It should have been lambda d / d, but in water, it became lambda d / d, how many times? Up times. So, from here, lambda d / d is taken common. (1 - 1/n) came. So, from here, our fringe shift is beta (n - 1) / n. And if you remember, the formula for that also came the same. The same formula came, remember the formula for shift? Real depth multiplied by (refractive index - 1). Now, its formula came the same. So, this will come. Now, if, child, now if we talk about a new topic today, which is also our last topic of this chapter, but again, it will run today and in the next class. And this is diffraction. This is diffraction. What is diffraction?

This means it has two definitions, son. First, the bending of light at the edges of obstacles. Bending of light at the edges of obstacles, or it has another definition: when light enters the shadow area due to bending at the edges. That is, has any one of you ever made a shadow? When light... in childhood. Now, perhaps, light... in your childhood, in our time, when we used to light candles, etc., now there's the process of internet and all. But earlier, what used to happen? What would you light in the evening? A candle. Now, perhaps, you don't light candles. You'll turn on the flash, whatever you'll do. So, when a candle was lit, we used to make strange designs with our hands on the wall. Or, if you didn't make it, suppose you've seen your own shadow. If you've ever noticed your shadow, it's never sharp. A shadow is never sharp. Why does this happen, son? Why does this happen? Understand this in two ways, son. Son, I know many children who memorize the derivations of diffraction, but they never know what diffraction is or why it's happening. Please. Here we have a screen, like a wall. Here we have some obstacle. This is an obstacle. This is an opaque object, through which light cannot pass. And somewhere here is a source. Somewhere here we have a source of light, which will release light rays. Now, you all know that the light that is going to be released from here and from here, no light ray will be able to pass between them. That is, these light rays will stop. This light ray will also stop. This light ray will also stop. All the night light rays in between will be blocked. But would it be correct to say that the light rays beyond this will go there? All the light rays beyond this will go there, and the light rays below will also go there, and all the others, okay? Tell me, is it correct to say that in this area, approximately in this area, a shadow will form? This is a shadow, because light rays did not reach this area. This is the principle of shadow formation. Now, what happens is, when light falls on the edge, on the edge of the obstacle. Now, it's two-dimensional. If you look from the front, our obstacle is a circle, so the shadow will also be a large circle. So, when light falls on the edge, what happens is that upon falling on the edge, the light bends. What happens? It bends. Bending means towards the normal or away from the normal? No. Light has photons, they are going. Understand it like this. I am explaining this to you in a very macroscopic way. Imagine light falling. Imagine light. Imagine very small balls of light that are going straight, with nothing to stop them. They are going straight. Nothing is stopping them. They are going. But here they stop, they hit and come back. That can also happen. But when they fell on the edge, they deflected from their angle. Now, after deflecting, they can come here, and they can go there. Oh, meaning, this ball, suppose it deflects, it can come here, and it can go there. Tell me, if light enters the light area, you won't even know which area this is. This is the light area where light will fall. This area is also where light will fall. And this is a shadow. So, as soon as the light fell here, it bent. Bending means some light rays bent and came here, and entered which area? The shadow area. Now, say yes, sir. And some light rays are in which area? The light area. Similarly, some light rays entered which area? The shadow. And some light rays entered which area? The light. Now, the light rays that entered the light area, we don't have any problem with them for now. Although only they are useful to us, because when light falls on light, there will be interference. So, only they are useful to us. But right now, I am telling you why the shadow is blurry at the edges. If this bending of light didn't happen, would the shadow be sharp? But due to bending, some light, not all, some light entered which area? The shadow area. And some light entered which area? The area that blurred the boundary of the shadow. So, you cannot define where the actual shadow is. For this reason, shadows are blurry at the edges. This is not our objective. Our objective is for light to enter the light area. But right now, I am explaining why the shadow is blurry. Understood? Now, can the opposite also happen? Meaning, its opposite? This time, we have changed the form of the obstacle. Exactly how? This time, we have placed a complete obstacle here, but we have made it like this from the middle. Last time, light passed from the sides, and was blocked from the middle. This time, it's allowed to pass from the middle, and blocked from the sides. And here lies our source of light. Here it is. Okay? What will happen? This light will stop. Will it go? It will stop. This light will stop. Okay, this light will stop. And the other light will also stop. But the next light will fall on its edge. What will happen to it? Let's see. Right now, all other light rays will stop, brother. And a shadow should form there. Okay? And all the light rays in between will go. And in a similar way, we will also take light from one edge, so that we can define where our shadow is and where our light is. Okay? So, this area, this area, is what this time? Light. Say, die. This area is light. This area is light. And this area is what this time? Shadow part. Shadow. And this area is also shadow part. Shadow part. Son, a child is asking, why do colors disappear from shadows? What, son? What, son? Give the answer to this. What are you saying? Color means light. There is no light. See, child. In the world, or what? If everything remained like this, if it remained like this, then the shadow would be sharp. Shadow. But when light fell on the edges, did the light bend? If this light that was going bent, then some light went here, and some light also went here. What we are concerned with right now is that light which entered the shadow. What we are concerned with right now is the light which entered the shadow region, and made our shadow blurry. Understood? Say. But now, what am I going to ask you? Our main objective is not to see the shadow blur. That is the cause of diffraction. Our objective is not that. Does this represent the wave nature of light? Oh, light fell on the edge and bent and went here and there. So, it's showing particle nature. In a way, we... What is the name of the chapter we are studying? We are concerned with wave nature. What nature is polarization showing? Polarization is showing which nature? Wave. Particles are oscillating like this, like this, and like this. Oscillation means wave. What nature did YDSE represent? It has to show wave nature. Now, tell me, intelligent children. The question. I called you intelligent. You lowered your heads. The question. Do you see YDSE here? Raise your hand if you see it. I say, you need not just human eyes, but the eyes of the mind. Do you see YDSE here? No, no. On this board. Brother, you are absolutely... On this board, do you see YDSE anywhere? I see it. You also saw it. Oh, very good. And you also saw it. Oh, very nice. And say it again. What do you see? How? Tell me quickly. Okay, okay, okay. Let's go. If you didn't see it, you saw it. Don't show it to others, man. Understand. What light came out from here? We only showed you yellow light twice. But actually, when light fell here, was it going like this, like this, like this, like this? And its wavefront will be like this. Coming. Coming. Feel it. Like this, like this, like this, like this. And here, say. Here also, light is like this, like this, like this, like this. Oh, people from Pune might come. Like this. This dark. Look here, brother. And from here, this wavefront? Yes, sir. Did we see something like this a little while ago? Brother, here it looks like this is a source. This is a source. Both sources are emitting their... What are they creating? Interference. Sir, this is double slit. Meaning, the same thing will happen here that happened in double slit. The only difference is, there we did it with a double slit. There we did it with a double slit. Here we are doing it with a single slit. Those slits were very small. There was a distance between them. Here, its width will be D. Understand diffraction first. What is it? When light falls on the edge. Where does it fall? Where? And after falling on the edge, what happens? It bends. What happens? It bends. This is called diffraction, due to which shadows appear blurry. But our objective is not to study this. Our study is: when light goes to the edge and bends, it actually interferes there. This light, if it enters our light area, what will it do? Interfere. What will it do? What will it do? Correct? Now listen. The biggest task. It will take me 15 minutes to explain this to you. If you understand, then in the next class, the derivation will be very easy. Very easy. Okay, I am telling you, what I am going to explain to you will not be found in any book. Not in our syllabus. But the sleep of an intelligent child is disturbed if it's not explained. Look, what did we study? That in YSE, the path difference for bright fringes will be... Say, yes, sir. And for dark fringes, the path difference was... There are two options. First, memorize it. But the second option is to understand why this is so. In 16 years of teaching, more than half the class won't understand even after explaining, because this is not something to be understood in class. But because there is such a big difference, it will trouble you mentally. What happened here that the position of bright and dark fringes interchanged? And such a big difference will occur. In the next class, the positions of YDSE and YSDE will switch. Why did this happen? To understand this, we will do it today. If you get it, fine. If not, no problem. This was our double slit experiment. But in single slit, there will be only one slit. Okay? First of all, we will have a screen here. Light will go from here and reach here. Does something remind you? Light will go from here and reach here. Does something remind you? You should get the feeling of YDSE here. Okay? Interference will form there. The distance from the center will be... We will find... The derivation will change, but the objective process will be almost the same. But why did the position change? Listen. How many players are needed to play a cricket match? Meaning, India is playing against itself, son. 22 players are needed, right, son? 22 players are needed, right? 22 players are needed. What are you saying? Extra players? Then umpires? Then cheerleaders? What are you saying? Okay, listen. How many players are needed? 22. But in the ground, there are 11 from one team, and 11 from the other. So, not all players of the batting team are being used at that time. Say, son. Yes, sir. Okay, I'm asking a difficult question. I need to capture a wavelength. You know what an antenna does? It captures waves. It captures waves. So, suppose I need to capture the entire wave. This is the wave, and its wavelength is 2 meters. So, to capture the entire wave, how big an antenna do I need? More than two? Why? Two meters. But if I tell you, I'll make an antenna of one meter and capture half the wave, and replicate it, because the wave will be the same. You understand? In the old days, people made very large antennas because they thought that to capture the entire wave, you have to capture the entire lambda. But then it was understood that it's not like that. If I make an antenna of half wavelength and capture half the wave, then I'll invert the next one and replicate it. Then another great person came and said, it's not necessary to make half. You can even make half of half. Meaning, you can make an antenna of 0.5 and capture this one, and mirror image it. These things are being discussed here. And replicate it and invert it. See, you've captured the entire wave. Did you not leave anything? It's not possible. No, no, no. Because this wave is completely individual. How can you replicate it by half? Son, suppose, suppose I explain it to you in computer language. You want to make the design of a wave. So, make this arc, copy it. It's done. Now copy this entire thing, invert it, attach it. See. So, did you need an antenna as big as the entire wave to capture the entire wave? Meaning, length of antenna needed is lambda by 4. How much is needed? How much is needed? This was also in our syllabus. The name of the chapter was Communication Systems. That's done. In that, we have to study the height of antennas, etc. But you are listening. It's starting from here to here. This is a single slit. In double slit, how many particles were there? Double. Here, it's a single slit. Let's assume there are 100 particles from here to here. We have to give a number. How many particles from here to here? 100. So, to make the first bright fringe, lambda, lambda, lambda, lambda was used, because it was n lambda. Say, yes, sir. Now listen. If we assume that one came out from here. See this. Okay. Let's assume this is a way to explain. So, out of 100 particles, 50 will be here, and 50 will be below. This much math comes. Half of 100 is 50. So, 50 here, 50 there. Now, if I want to make it dark, what do I want to make? You know it will be dark. But for that, a full lambda is needed. How? If this particle is oscillating upwards, then this particle is oscillating downwards. Let's work with numbers. If this particle is 1, then 51 is. So, will particle 1 cancel particle 51? Particle 1 says, go up. Particle 51 says, go down. Cancelled. Dark fringe is formed. What is formed? It says go up. Particle 52 says, go down. Same strength. Oh, it's a replica. Yes, sir. Cancelled. 7, 57. Oh, I skipped in between. Surprised. 49, 99. 50, 100. Meaning, to make it dark here, the overlap that is going to happen. To make this dark, these 50 particles will have to overlap with these 50 particles. Meaning, basically, to make it dark, lambda by 2 is not being used. Lambda is being used. What is being used? 50 particles are being made, but they cannot make it alone. The other 50 must also be there. Say. Meaning, when these 50 overlap with these 50, then only dark is formed. This means, lambda by 2 to lambda by 2 particles overlap, then only dark is formed. This means, lambda is used. Meaning, overall 50 particles are there, but you need 100 for it. Here we have an obstacle. And in this obstacle, there is a slit here. Will this slit be small or big? Small. Very small. And here, as always, is the screen. The distance between these two is again, as always, capital D. Somewhere here is the source. Just the arrangement will be different. How to place the source. That's all. And the length of this slit is what, brother? Yes, son. Find it later, please. Now look here. What will happen? Remember, a lot of light will come out from here. I have explained that to you in the last class. Light will bend on the edge. So, some light, not some, but a lot of light will bend from here and fall on the entire screen. We have only shown one light ray here. And light rays will also come out from here. They will go to the entire screen. Again, we have shown only one light ray. We have chosen a point here. And that point, like last time, we will call it A1, A2 if we give it, then you will think that there are two sources here. So, instead of calling it A1 and A2, we will call it A and B. Okay? Can also give it L1, L2. It's not like that. And from here, we have made a reference line, which is what? What is it? What is it doing down? Brother, what are you doing? Why are you making it? Why now? Okay, son. Understand. If you remember, in YDSE, to reach the path difference, we used two slits. We used math. First, in this triangle, we applied Pythagoras. Then in this one. Then we subtracted both. Then we made some assumptions and finally got the path difference. Here, we will find the path difference in a slightly different way. Again, the requirement is. Listen. First of all, I am drawing another reference line from here. Oh, oh, oh, oh. This is a reference line. This is not any light. This light ray has made an angle of theta with our zero angle, meaning the midpoint. How much angle has it made? This is something new we have made, which was not there last time. And the distance of this point from the center is what? It's x. This was also there last time. Now, the thing to listen to is, this distance x, we are going to call it the linear distance of the fringe. What did we call it? We didn't give it any name like this before. We just called it distance. But this time, we are giving it what name? Linear distance of fringe. Because this time, we are going to find two types of distances. One is linear, and one is angular. One is linear, one is angular. Meaning, the linear distance of point P from the center is x. And the angular distance of point P from the center is theta. See, this line, how much angle has it traveled to reach point P? It has reached by traveling theta angle. Say, okay. Let's consider. Let's consider a slit of width small d. Let's consider a slit of width small d and a screen at distance big D. A screen at distance big D from the slit. If P is any point on the screen, where P is any point on the screen where light rays are performing interference to form fringes, then small x is the distance from the center. Small x is the distance from the center. And theta is the angular distance from the center. Theta is the angular distance from the center. Now, if this is the case, listen carefully. See, its beginning is with what? With what? Path difference. There, the line came to two slits. Path difference. How am I going to find it here? Listen. I am drawing a line from here, which is perpendicular. Let's change its color. This perpendicular fell on this. It fell, and let's name this point capital X. So, I am saying the path difference is BX. What is it? BX. Let's understand how. Are A and B parallel? Yes. Did you forget? Is this slit very narrow? Is this screen very far? Yes, sir. So, are A and B approximately equal? And approximately parallel too? And when two lines are parallel to each other, to find the path difference between them, you have to find this distance. I have asked you this question before. When we were doing electric potential at any point of a dipole, I asked you, how would you find the difference in length between these two fingers? What is the first method? Measure its length, measure its length, and subtract. Which we did in YDSE. Here, we are adopting another method. We are saying, let's place them parallel to each other. Now, the extra length that has come is the difference. Understood, sir? If I tell you to tell me the difference between my height and his height, and we are very far away, how will you do it? You will measure my height, measure his height, and subtract. But there is another way: I tell you to come and stand parallel to me. You will automatically understand how much difference there is. But to compare like this, both have to stand parallel to each other. Are these two light rays standing parallel? Yes. Even if they don't look parallel in the diagram, we know they are approximately parallel. And if this is the case, then this line that came from A to X is their common line. So, this difference that has come, this difference that has come, is what? It is BX. So, BX is our path difference. But how to find this BX? Come. If this angle is theta. Okay. There is also a line here. Imaginary. Do you know its length? Do you know its length? Listen very carefully. If this angle is theta, then will the reference line going from here also have an angle theta? Yes, sir. Because all these lines are parallel to each other. Do we have to make them parallel? What does it mean? Do we have to make them? If you don't understand, then this is another reference line. And this is another reference line. We... Now, if this angle is theta, then this is also what? Theta. Because all three are parallel to each other. Say, brother. Yes, sir. And if this angle is theta, then this angle is 90, and this angle is also 90. So, this is common. So, pay attention to the red color. This angle is also theta. So, this is also what? Theta. Brother, the diagram is full of math. This is also what? Understand. So, if this angle is theta, then math says, the one opposite to theta is sine. What is it? So, this is BX, and this is theta. So, this BX is sin theta. What is it? What is it? Meaning, using proper math, we have proved the path difference between two lines to be equal to b sin theta. Where we proved the path difference to be equal to XD/D with effort. This is the effort. But on the pages, it's just two lines. No. Say, brother. Okay. Now, the path difference has come. Is theta big or small? It's very small. So, can we write the path difference as theta instead of sin theta? Because theta tends to zero. So, from here, the path difference has come. For dark fringes. For dark fringes, what will the path difference be? N lambda. It's already done in the last class. For dark fringes, what will the path difference be? N lambda. No one is asking you the reason. N lambda. If this is the case, then is d theta equal to n lambda? So, theta is equal to n lambda by D. Is this linear distance or angular distance? Is this linear distance or angular distance? Angular. This is linear distance or angular distance? Angular. This is linear or angular? Angular. Okay. Also, the formula for theta is arc upon radius. What is it? Arc upon radius. So, if this is theta, then the arc is x, and the radius is D. Do you need math to study physics? Who says that? See, math will defeat math. Again, the formula for theta is arc upon radius. So, this is the arc, and this hand of mine is the radius, which is equal to D. So, can I write arc upon radius instead of theta? Yes, sir. So, from here, what will be the formula for x? D theta. What will it be? D theta. So, if I write x by D instead of theta, you will get something that you already know. And that is n lambda D by D. Have you heard this name somewhere? Yes, sir. Does this formula sound familiar? Yes, sir. Oh, this formula came for bright fringes in YDSE. This formula came for bright fringes in YDSE. Now, this formula has come for dark fringes in single slit. There, the value of n was 2, 3, 4. Here also, the value of n is 2, 3, 4. The only difference is, there the center was bright. Here also, the center will be bright. Meaning, you cannot put 0 in place of n. You cannot put 0 in place of n. Let's find the position of dark fringes. We have to find both positions. Angular and linear. Both angular and linear. Dark. First dark. For the first dark, what will you put for n? You will put 1. So, what will theta be? Lambda by D. And what will x be? Lambda D by D. You are putting 1 in place of n, right? And you know that 1 means plus minus. This is the first dark. Which dark are we finding? Second dark. For the second dark, what will you put for n? You will put 2. So, what will be the angular? Say, 2 lambda by D. And what will be the linear? 2 lambda D by D. It will come here and there. So, plus minus. Which dark? Third dark. The third dark will come when the value of n is 3. So, what will theta be? Say, 3 lambda by D. And what will x be? 3 lambda D by D. Say, plus minus. Plus minus. So, this means, the position of dark fringes is overall. If we talk about theta, it is lambda by D, 2 lambda by D, 3 lambda by D, and so on. And similarly, if we talk about linear, it is lambda D by D, 2 lambda D by D, 3 lambda D by D. Done. Whose position has come? Now, use your brain. What will come between dark fringes? If you want to make fringes, what will come between dark fringes? What are you saying? Behind Tyagi, behind the child. Definitely bright fringes. Why did I feel like bright fringes? Definitely bright fringes. Okay. Understood, son. Okay. One question. If bright fringes come between dark fringes, then tell me the formula for bright fringes. Keep in mind, you have to start the value of n from 1. I have asked you a very difficult question. You know what the answer will be. It will be 2n-1/2 or n+1/2. You just have to choose between minus or plus. This is the difficult task. I am telling you, the answer is on the board. The answer is on the board. Think about minus or plus. Have you thought? How many children are saying minus? Write it. The rest also write it. How many children are saying plus? Okay. You said something. Nothing will come. Neither plus nor minus. Why? You chose minus. Can you tell me after thinking a little? What do you feel? It's okay. You chose plus. Why? Because it has to be positive. Okay. You chose what, brother? Yes. Why? Bright fringe is in between. Wow, son. You are saying anything. You are saying anything. You are saying anything. See, the answer will tell you. Yes, sir. Because if we start from zero, then plus. If we consider the center as bright, then plus. Okay. I will tell you whether plus or minus will come. One child is saying, both plus and minus can come. Can this also happen? Let's check. Okay. First, tell me, can two bright fringes have another bright fringe in between them? If there is one bright fringe between two bright fringes, then it's just one bright fringe. Now, listen. If the first dark fringe is at lambda D by D, and the first dark fringe from the other side is at minus lambda D by D, then at the center, there will be a bright fringe. Yes. You have used this much intelligence. Okay. This means, if I use the formula 2n-1/2, then put n=1. Put n=1. What will come? 1/2. How can 1/2 come? 1/2 is here. Oh, it comes between 0 and 1. Yes. So, how can the minus formula come? Okay, sir. Let's put plus. Put plus. What will come? Lambda by 2. You chose minus. Why? Because you thought it was okay. You chose plus. Why? Because it has to be positive. Okay. You chose what, brother? Yes. Why? Bright fringe is in between. Wow, son. You are saying anything. You are saying anything. You are saying anything. See, the answer will tell you. Yes, sir. Because if we start from zero, then plus. If we consider the center as bright, then plus. Okay. I will tell you whether plus or minus will come. One child is saying, both plus and minus can come. Can this also happen? Let's check. Okay. First, tell me, can two bright fringes have another bright fringe in between them? If there is one bright fringe between two bright fringes, then it's just one bright fringe. Now, listen. If the first dark fringe is at lambda D by D, and the first dark fringe from the other side is at minus lambda D by D, then at the center, there will be a bright fringe. Yes. You have used this much intelligence. Okay. This means, if I use the formula 2n-1/2, then put n=1. Put n=1. What will come? 1/2. How can 1/2 come? 1/2 is here. Oh, it comes between 0 and 1. Yes. So, how can the minus formula come? Okay, sir. Let's put plus. Put plus. What will come? Lambda by 2.

Put 3 by 2. 3 by 2 means one and a half. And after one comes one and a half, then two, then two and a half. Then some might still not have understood, some might have. Yes, some have understood. From some, I understood, but some still haven't. And especially those who didn't raise their hands for anything, they will only understand when I do the full derivation. In 10 minutes, the answer is what will come. Plus, it will come, minus cannot come. If minus comes, then you will have to start from the position of 'a', which is meaningless for the event, it will be wrong. So, if you have to start from 'a', then where do you have to go? Let's go to bright. We will fill this whole board, then we will make the intensity curve. What will we make? You will understand. Now, for bright, the path difference has to be kept as 2n plus 1 by 2 lambda. And the path difference is also equal to what? It is also equal to what? Path difference is d theta. Hey, what is the path difference also equal to, son? So, from here, d theta becomes 2n plus 1 by 2 lambda by d. And what did we just talk about? Okay, this is the formula for angular. Now, if we want to find the linear one, what will I put in place of theta? Take d to the other side, so x becomes 2n plus 1 by 2 lambda d by d. Have you seen this formula somewhere? Yes, this formula came in dark. Even if there is a difference of plus or minus, this formula came in dark. In 'd', which is for bright now. Let's go to the center. So, what is going to be formed? Bright. That is done, brother. So, we don't need to worry about the central bright. For the central bright, x will be zero and theta will be zero. Now, we are going to the first bright. For the first bright, what will we put in place of n? What will we put in place of n for the first bright? We will put one. Now, if we put one in place of n, then what will x1 become? Tell me. Put one. It becomes 3 by 2 lambda d by d. See, the problem is in the first one itself. If the first one is 3 by 2 lambda d by d, then the next one will be 5 by 2. And if the first one is half, then the next one will be 3 by 2. So, I have to explain the first one to you. The rest will come. And if I go to find theta, it will be 3 by 2 lambda by d, according to this formula. Then let's go to the second bright. So, x will come because you have to put 2 in place of n. So, x will become 5 by 2 lambda by d. And theta will become 5 by 2 lambda by d. Putting it in, what did it become, brother? 7 by 2 lambda d by d. And theta became 7 by 2 lambda by d. So, if we have to give the positions of the bright fringes, then what will they be? Tell me. They will be angular. Okay, plus minus will be applied. They will be angular. 3 by 2 lambda by d, 5 by 2, 7 by 2. 3 by 2 lambda d by d, 5 by 2. Only capital D comes extra on top. Now, here also, our aim is to find the fringe width. The only difference is that there we found the fringe width, and here we will find two fringe widths. We will find the central one separately and the secondary ones separately. Let me explain. First of all, let's make the intensity distribution curve here, in which all your answers are hidden. Intensity distribution curve. We had also made this for YDSE. So, here we have the center. What is going to form at the center? Bright. And after bright, what will come? So, dark will come. Where is the first dark? Tell me, brother. The first dark is at lambda d by d, if we are talking about linear distance. And here at minus lambda d by d. Now, I am asking you, if you had taken the formula, first of all, this is our central bright. Now, I am telling you, if you had taken 2n minus 1 by 2, then the formula would be 2n minus 1 by 2 lambda d by d. Now, if you put n equals 1 in this, then you will get the answer 1 by 2 lambda d by d. Now, tell me, this is zero, this is lambda d by d. So, 1 by 2 lambda d by d is coming here. How can it be here? A bright fringe when there is already a bright fringe there? Understood? If there is already a bright fringe at zero, and there is dark, then how can a bright fringe come between a bright fringe and a dark fringe? So, this means that we should not take the formula minus 1. We should take the formula plus 1. We should not take the formula minus 1. We should take the formula plus 1. Understood? Say, brother. Absolutely sure. Now, tell me, where is the next one? Okay, one more big difference. I will also give you the reason for this. The reason is, although there is so much common sense that if YDSE was happening, then in YDSE, there were two fresh sources of light, and they were taking the light to the screen. But in diffraction, light bends from the corners. Now, if light bends from the corners, then as we move away from the center, will there be a loss in intensity? And if there is a loss, then our intensity will keep decreasing. So, the first difference here is this. There, if you remember, we had made this intensity the same. But it is not like that here. Here, the intensity will fall gradually. It will fall gradually. This is 22. Now, if this is the case, then something like this will be formed here. Now, you understand that this is what? This whole thing is what, brother? Bright. This is bright. This is bright. This is bright. So, here, the first dark comes at lambda d by d. Where is the second dark? Tell me. 2 lambda d by d. And where is the third dark? 3 lambda d by d. And where will the next one go? 4 lambda d by d. So, yes, then where will the first bright fringe come? 3 by 2. What is this distance? 3 by 2 lambda d by d. And this is 5 by 2. And this is 7 by 2. Similarly, the second dark fringe here will be 2 lambda d by d. Here, 3 lambda d by d. And again, this is our first bright fringe. From this side, where will it be found? Where will it be found? Between these two, minus 3 by 2. And this will be found at minus 5 by 2. And this will be found. So, here, one thing that I have shown strangely in the entire diagram is that the intensity is falling. Say, the intensity is falling. Now, I am asking you the next reason. What am I asking? I will give you the next difference. So, the first difference is that the position of dark and bright fringes has changed compared to the previous graph, if you have made it. The position has changed. But do you see one more very big, very major difference here? If you see it, then think about it. Online students, write it down. One student is asking what is bright. Then your future is not there, son. At least, son, the distance we have shown here is linear. It is angular. We can also show it angular, but here, since we are talking about distance, we have shown it linearly. If it is angular, then capital D will not come. The rest is the same. Pointed out the second difference. Yes, very good. One student named Praj has told it. Others also need to see. You tell me now. Now you will tell me. You were not even looking at it four times. Come on, tell me. Tell me, it is your duty to tell me, because you are not looking at it at all. Look with your heart once. Think. Someone's post is on Instagram. It has come. Haven't seen it yet? Oh, man. The difference is visible. A very big difference is that here, the fringe width of the fringe is double than the others. If you haven't seen it, then I will show you. Look here. Who is this? The first bright fringe. Who is this? The second bright fringe. What is the distance between these two? Obviously, lambda d by d. Okay, if you are getting confused with bright fringes, then come to dark fringes. Fringe width is the difference between any two consecutive bright and dark fringes. So, what is the difference between this dark fringe and this dark fringe? It is lambda d by d. What is between this dark fringe and this dark fringe? And what is between this dark fringe and this dark fringe? So, here, the fringe width is d. Here also, the fringe width is lambda d by d. But the difference between this dark fringe and this dark fringe is lambda d by d. Meaning, the central one is double the width of the others. We are solving it now, but the most major difference comes here. So, here, the fringe width that we get, we have two. One is the central fringe width, which we are calling beta naught. Now, okay, when we were finding the fringe width, I said any two consecutive bright and any two consecutive dark. But here, if we are finding the central one, then we have to subtract this from this, because we are finding the central one. So, what will happen? Tell me. It will be lambda d by d minus lambda d by d. So, its width will be 2 lambda d by d. Correct. But if I find the secondary fringe width, then the secondary fringe width, meaning this. In this, you take any two consecutive bright and dark fringes, because this time the formula for dark is good, the 'n' lambda one. So, let's take that. Okay. So, from n lambda d by d, what will be subtracted? n minus 1 lambda. Last time also we did this. Just remember, these were bright fringes. Now, these are dark fringes. So, here it will be lambda d by d. So, can we write overall that beta naught is twice of what? Beta. Say, brother. So, the difference from here to here is beta naught. And this is beta. So, beta naught is double of beta. It used to take about two classes to read. Today, we will not work too hard on this. What is polarization? Let's see, brother. First of all, you have to understand that there are two types of light in the world. One is polarized light, and the other is unpolarized light. First, you have to understand these two types of light, only then you will be able to understand what polarization is. So, what is polarized light? Light in which particles oscillate only in a single plane. Here, you have to remember ninth class. There are two types of waves, broadly longitudinal and transverse. In longitudinal, the particles, in the direction the wave is going, are moving back and forth, back and forth, back and forth. Everyone knows this? Yes, sir. But in transverse, they are perpendicular to the direction the wave is going. If the wave is going this way, then the particles can move like this, like this, like this, anywhere. But they are oscillating perpendicularly to it. What is such a wave called? Transverse. And what type of wave is light? Transverse. We have studied this. Light is a transverse wave. So, light in which particles oscillate only in a single plane is called polarized light. Which means, assuming that our light ray is going in this direction, if the light ray is going here, then the particles are vibrating up and down, up and down, up and down, like this. But it is also possible that they are not vibrating up and down, and they are vibrating in and out, meaning, if we talk with respect to this board, then if the light is going like this, then the particles are vibrating in and out like this, in this plane, in this plane. So, we show in and out with a dot like this. Like this dot. Understand? Dot means it is not the vector dot. Dot means particles are oscillating in this plane. And this means oscillation. Think, imagine. If, if the light is going this way, then particles can oscillate like this, and they can also oscillate like this. Meaning, and they can also oscillate like this. But can they oscillate in between? Meaning, suppose the light is coming towards you. Then, in what I have drawn, the particles are oscillating up and down. And in what I have drawn, they are oscillating in and out. Because oscillation. So, either the particles are oscillating up and down, meaning the red face, or they are oscillating like this. But this is also a face. And this is also perpendicular to it. Yes, sir. And this is also like this. How many planes are there? Infinitely many. [Music] Oscillate in more than one plane. Particles are oscillating in more than one plane. Like, if our light is going this way, then in this, particles are oscillating up and down, in and out, up and down, in and out, in different planes, brother. So, what type of light is this? Unpolarized. What is it? Unpolarized. Meaning, in this, some particles are oscillating like this, and some like this. But in this, there were either like this or like this. But in this, particles are oscillating in different planes. So, what is this called? Unpolarized light. What is it called, son? Unpolarized. Now, can you tell by looking at it whether a light is polarized or unpolarized? Absolutely not. Okay, according to you, the light coming from the tube light is which one? You don't know. Don't talk nonsense. We don't know. We don't know what light is coming out. But a hint can be given that most of the light rays around us are unpolarized. How can we know that the light we have is polarized or unpolarized? For that, you have to learn a word, which is what? Polarization. The process in which unpolarized light is polarized. Such a process in which we have unpolarized light, we will convert it into polarized light. What is this called? Polarization. What is it called? Polarization. Come, brother, what do we study? How do we do polarization? Listen. Have you ever seen an entry like this inside a park? Like this? Have you ever thought why there is such an entry in the park? What are you saying? Polarization using a polarizer. What are you saying? Very good. Which animals? Dogs. Big animals like cows and bulls. And oh, yes, absolutely correct. Dogs. What are you saying, son? What is he saying? We have such an aunt in our neighborhood. Absolutely. If someone is going, she says something like this, turn around. She says, nothing. So that no drug addict enters, no one enters on a bike. Well, crocodile, snake, scorpion. Son, do you know what a scorpion is? It's not a car. Scorpion. Listen, son. That gate is made in such a way that big animals, which mainly eat grass and damage plants, will not be able to go. I swear, I am telling the truth. I have seen a cow going like this. I have seen it. Have you ever seen cow dung nearby? So, where is it coming from? Someone is bringing it and keeping it. So, cows come at night. They are just fooling you, making you think that they will stop like this. But the flexible cows, son. Okay, honestly, I have never understood that revolving gate in hotels. I have never understood its sense. I have tried to think a lot about why it is there, that we have to go by rotating like this. That door keeps rotating. And in some hotels, it keeps rotating. In some, we have to rotate it. I always imagine, what if there is a fire inside? How difficult will it be for everyone to escape? Because it takes me four to five seconds to come into resonance. I wait for two to four rotations first, thinking, "Brother, because son, you are going to a five-star hotel. People are there. You are wearing a coat and pants. And suddenly, you fall. And minus. That's why." Oh, polarizer. We are doing polarization with a principle similar to how we are stopping animals but letting humans go. With this exact principle, we are going to do polarization. What are we going to do? Polarization. Let's understand a word first. What do we need to learn? Polarity. What do we need to learn? Polarity. Oh, ho. Listen, son. Suppose this is a wall, and in this wall, there is a gate here. There is a gate here. This gate is like this. Okay. This gate is like this. And let's imagine. We have to imagine. It is not like this. Let's imagine. There are two types of people in the world. One is normal, like us, who walk like this. Okay. And suppose there are also some people who walk like this. Understood. Give me the pen. I said it doesn't happen like this. You have to assume that brother, some humans are walking like this, meaning on their legs, and some are lying down and moving with their navel or somehow. Okay. Now, listen. And this door, will it let both types of people pass? 50 people came like this. 50 people came like this. And 50 people came like this. You have to imagine these people. Oh, man, they are coming, they are coming, they are coming. Will this door let these 50 people pass? Yes, sir. But the 50 people who are coming by rolling like this, suppose they made a vow, "We will go to this hotel crawling." So, suppose 50 people are coming by crawling like this, like this, like this, rolling like this, rotating like this. Will they get stuck here? Will they get stuck here? So, which people will be able to pass? Those who are walking like this. And which people will stop? Those who are lying down. Now, you have to understand that just like this door, we also have polarizers. Who are they? Polarizers. These are crystals made of polymers, which are so fine that you cannot see them with the naked eye, because they have to stop the particles of light, photons. Now, if unpolarized light comes, imagine that in the polarizers, there is a door like this. There is a door like this. This door has many such doors, very small doors, microscopic doors. I will remove the door. Imagine the door. Okay. Now, these doors. Which light is coming through these? Look here. What is this light? Unpolarized. How is it unpolarized? Because particles are oscillating up and down, and in and out. So, if this light comes to the door that was there, which we are calling a polarizer, will it let these particles pass, but block these? And the light will become polarized. What will it become? What will it become? Polarized. And the process of polarization in this way is called polarization using a polarizer. And these polarizers are also called polarizers. They have different names. Like, here we have a polarizer, or you can call it a polarizer. Okay. Imagine, brother, that if the door was like this, then what would happen? Then, if I make this door like this, then the black particles would be able to pass, and these would be blocked. You can understand this much. So, if unpolarized light is coming from here, and this is a polarizer, which we can also call a polarizer, and if this light passes through it, then what will happen? This light was passing. So, will it let these particles pass, and block these? So, will the light that goes forward, the light that passes through it, look something like this to you? And if I had kept it like this, then the light that would pass through it would look something like this to you. I am telling the truth. But what should be said is that the light has become polarized. It is polarized light. Okay. Now, how will we know? Is the light still polarized or not? You don't know. How will you know? You don't know. But for now, let's say it has become polarized. Now, if the intensity of light was I here, and the intensity of light is I naught, then will this I naught be I naught by 2? Because you blocked half the light. 100 people left home. 50 people could pass through the door, son. So, half the intensity is absorbed there. Not destroyed, but the intensity that could pass was only I by 2, I naught by 2. Now, tell me, I naught by 2. Meaning, the intensity has halved. Does this mean that the light was polarized? Not necessarily. Let's understand how. Because, if I tell you, if I tell you, if we look through this bottle. If I tell you, son, is the letter 'A' visible on my hand right now? Tell me, son. Visible? Yes. We will also write 'R', son. You focus on yourself. Is the letter 'A' visible to you, son? Yes, sir. Is the light coming out of this 'A' polarized? It is unpolarized, son. Because light is hitting it and coming to you. So, it is unpolarized. If I keep this bottle in front of it, then obviously, the letter 'A' is still visible to you, son. Meaning, at least those in front can see it. But its intensity has decreased. So, is this a polarizer? Are you crazy? It's a water bottle. You tell me. We also halved the intensity like this. How did we do it? Because in ray optics, what is the formula? So, is the light polarized there? No, son. Halving the intensity does not mean polarization. So, how to find out? How to find out that the light has become polarized? The answer is, put another polarizer there. But its name will change. It is the same thing, but its function is different. So, the name is different. The function is different. So, its name is Analyzer. What is the name? Because now this is going to analyze whether the light is polarized or not. How is it going to do it, son? Let's go back to that hotel. Do you remember the hotel door? Do you remember the hotel door? Or why are you taking the pen back? Do you remember the hotel door? So, the 50 people who came straight are here, and here are the 50 people who were coming by lying down. So, there was a door here. So, the door blocked these and let them pass. Now, how will I know that the people who passed through the door now are the straight ones? Just. Hey, we are visible to you. Humans are visible, but photons are not visible. Photons are there. So, the door in front, let's lay it down. Put two doors. The door in front, let's lay it down. And after laying it down, this is the first door. Okay. These are the remaining people who are oscillating like this. Now, the door in front, it was a door like this. Now, keep it like this. What will happen? How many people will pass through the second door? Zero. Meaning, the ones who were lying down were blocked by the first door. Son. Meaning, this is the straight one. Meaning, the light was polarized. Okay. If I keep it like this, then what is happening? How many people were there? 50. How many passed through the doors? This also means that the light has become polarized. Meaning, if the light had come like this, and if it was still unpolarized, then whether you keep the door like this or like this, half the light particles would pass. But they are not passing. This means that the light has become polarized. So, what do we call these gentlemen? Analyzer. It is the same door. It is the same polarizer. But it is called an analyzer. Yes, absolutely correct. Okay. One student is saying that you just said that light is not necessarily in these two planes. It can be like this. Absolutely. But the light we have taken now is oscillating in two planes. What will happen if it is like this? We haven't given you any formula yet. Yes. So, if this happens, then here, let's assume we have made it perpendicular. Okay. If it is placed in this situation, then what will be the intensity coming out from here? It will be the same. And if it is kept exactly perpendicular, then what will happen? Hey, what will happen? It will not come. You are telling the truth. It will not come. So, what have I done now? I have kept it at a certain angle. What is its name? Which is a polarizer. Now, what will happen? Tell me. If I keep it at a certain angle, then the answer is, will it be zero? It will not be zero. Because actually, some particles of light will be oscillating like this. Meaning, some intensity will come. But the intensity that will come here, I will call it I double dash, and its formula will be I naught cos squared theta. What will it be? Now, look, if you want to know where this formula came from, then again, I have to go into a little more detail. But this law is called Malus's Law. What is it called? Malus's Law. Yes, for now, I will explain it very briefly, because we are wasting time. I will explain it briefly, that where this formula actually came from. The formula came from. The formula came from. If the particles of light are coming like this, and my polarizer is also like this, meaning the angle between them is zero, then the intensity will be maximum. But if the light is coming like this, but the polarizer is like this, meaning the angle is 90, then no light will pass. Now, tell me, in trigonometry, what is such a function that is zero at zero and maximum at zero? Cosine. But how did the square come here? Because the value of cosine also goes into the negative. But intensity can never be negative. So, cos squared came. I have explained it to you in a very bad way, but I am explaining it now, that brother, the intensity that will go after the analyzer, the intensity you will get will be what? It will be the intensity before the analyzer, times its cos squared theta. Times its cos squared theta. Understand? Not? Yes, son. I told you, Ridhi, that this formula did not come just like that, that Mr. Malus was sitting and he thought, "Let's put cosine, and let's put zero." Oh, sorry, "Let's put square." Yes, son, it must have had a derivation. Just like the formula for the heating effect came. We said, "Brother, it should depend on I, it should depend on I, and it should depend on T." So, for understanding, it should have been a formula. But when it was derived or observed, it was depending on the square. What was it? Square. So, it is a proper derivation. So, I hope you enjoyed this entire one-shot with the offline students. And I did this intentionally so that you don't have fear of this chapter. Otherwise, this chapter seems very difficult, very strange to children. So, to remove this fear, this one-shot was put like this. Please tell me how you liked this one-shot. Otherwise, we have started live sessions here. YouTube. This is 11. I have combined all the PPTs. Have all your wave optics doubts been cleared or not? We are bringing part two question answers very soon in live. We will bring one more session of part one. We will also bring question answers for part two. So, subscribe. Ring the bell. If you miss notifications, then otherwise, we will meet in more such lectures. Until then, thank you so much.