diff --git "a/transcriptions_1021_1024.json" "b/transcriptions_1021_1024.json" new file mode 100644--- /dev/null +++ "b/transcriptions_1021_1024.json" @@ -0,0 +1,957 @@ +[ + { + "video_id": 22672, + "transcription": " Hi friends, we are going to discuss some important questions in the chapter Wave Optics. We know that wave optics is not a chapter where we study many things. Because we have to study three phenomena. One is the Huygens principle, then the interference, diffraction and polarization. It is a very small chapter. The questions asked will be simple. We will discuss some questions related to polarization in this video. You can do at least questions and study. Practice more questions. You should focus more to learn theory is to learn questions. So, learn questions and then reverse track it. Learn questions and then reverse track it. Try this strategy. Especially in physics. Learn theory through questions. We don't need to learn theory like that. Understand it like that. Let's see this question. So, this is the story of the story of the story of the story of the storyaroids placed one after the other in the path of the light. Taking the intensity of incident light as 100%, the intensity of out coming light that can be varied in the range of incident light as 100% each, the intensity of the light that can be varied in the range is what we are asking. So, we know that this is a Polaroid. This is Polaroid 1. The light passes through this through 100% intensity So we know that the light coming out of this is 100 by 2, that is equal to 50%. We don't have to look at anything, we know that it is 50%. Now we have to put this light through another polaroid. So this is polaroid 2. So what is the special passed through another Polaroid So this is Polaroid 2 So what is the feature of this Polaroid? When light passes through this we can apply a low here If we rotate this Polaroid from theta angle the light intensity coming out from this If this is I1 this is I1 and I0 So the light intensity from second polaroid is I2 That is equal to I1 into cos2 There is a law called asus low is the intensity of light coming from the second polaroid or analyzer is equal to I1 cos square theta. So it is called as Malus low. So I2 is equal to I1 cos square theta. theta. So here theta can be varied from 0 degree to 90 degree. From 0 degree to 90 degree. Because, if this Polaroid is like this, then 50% of light will pass through it. If we keep it perpendicular, then 50% of light will not pass through it. 0% will pass through it. So, if theta is equal to 0 degree cos 0 is equal to 1 that is, the whole light will pass through i1. See, I have substituted in the equation. i2 is equal to i1 cos square 0 that is, i2 is equal to i1 i1 is equal to 50% So, 50% light will pass through it. Now, another% is passed through this. Now, what is the other possibility of this? What is the possibility of eta? It is possible to get 90. So, I2 is equal to I1 cos square 90. How much is cos 90? cos 90 is 0. So, it is possible to get 0%. So, I2 is equal to 0 so what all possibilities are there 0% is possible 50% is also possible so our variation from here to here will be 0% to Malus law. We can use this as a simple question to understand Polaroid. Let's see the next question. The polarizing angle for a transparent medium is theta. V is the speed of light. Then the relation between theta and V. C is the velocity of light in air. We have learnt a law. It is polarization by reflection. See, an light ray is coming and is incident. Imagine that light is going to another medium from air. So, if light is incident at a particular angle, called polarizing angle. Light will have reflection and light will have refraction. This refracted ray and reflected ray will be perpendicular. And reflected ray is polarized. This is called as polarization polarization by reflection an unpolarized light ray is incident at a particular angle called polarizing angle or Brewster's angle the reflected ray will be completely polarized this is the concept of low so here we have learnt a concept if the refractive index of this medium is n, then we can write a great equation called Brewster's law. What does Brewster's law say? It says that the refractive index of this medium is equal to tan of this thing. So we can write a great equation called n is equal to tan theta. So, if this is another medium of refractive index n1, if this is another medium of refractive index n2, how do we actually find this? I will just explain. This is our incident angle called theta. What happens to light? Reflection happens and light refracts. This is 90 degree. This is angle of refraction r. If this is theta, this is theta. Angle of incidence is equal to angle of reflection. So, this r is 180. So, r is equal to, 180 is equal to, we already know that I is theta. applying Snell's law we are going to apply Snell's law here the refractive index of this media is n1 and n2 so what will happen when we apply Snell's law when we apply Snell's law we are going to apply Snell's law here the refractive index of this media is n1 refractive index of this media is n2 so what will happen when applying Snell's law? when applying Snell's law, look n1 sine theta is equal to the refractive index of this media into the angle of incidence of this media n1 sine theta is equal to the refractive index of this media into the angle of incidence of this media is sin r. r is 90 minus theta. sin of 90 minus theta is cos theta. So, n1 sin theta is equal to n2 into cos theta. So, sin theta by cos theta is tan theta. tan theta is equal to n2 divided by n1. tan theta is equal to n2 by n1. So, we can reach this kind of that n2 by n1 is substituted by tan theta so we can write tan theta is equal to tan theta is equal to n2 by n1 is v1 divided by v2 because velocity is inversely proportional to refractive index we use this medium as air and this as a denser medium so the speed of air is c so it is c divided by v v2 is tan theta is equal to c by v so theta is equal to tan inverse c by v so in this option there is no c by v So we will put it back So cot theta is equal to v by c So theta is equal to cot inverse v by c So theta is equal to cot inverse v by c So we can get to the answer of cot inverse v by c So option B is our correct answer we have already learnt this we just need to apply it easy right? ready? next question two polaroids P1 and P2 are placed with their axes perpendicular to each other. So there are two Polaroids and their axes are perpendicular. So this is a Polaroid and this is another Polaroid. So their axes are perpendicular. Now, unpolarized light I0 is incident on P1. Okay, okay, okay. This is P1 and this is P2. So, a light with I0 intensity is incident on P1. Now, third polaroid P3 is kept in between P1 and P2 such that it makes an angle 45 degree with P1. So, in have another Polaroid to make 45 degree angle with P1. So, the angle we are making with P1 is 45 degree. So, the angle we are making with P1 is 45 degree. So, angle made by this P3 with P1 is 45 degree See this angle is 45 degree I hope you understood it correctly Now what is the intensity of transmitted light through P2? The light coming out through P2 is the intensity of the light We can see this very easily So first we will see comes out of P2 is what we are asked. So, first we have to find out the light intensity that comes out of P1. That is I1. After that, we have to find out I3. After that, we have to find out I2. So, we will find out each one of them. So, I0 is the light intensity We can find each one of them. So I0 is the incident. So I1 is the half of the incident light. That is I0 divided by 2. That is the most basic thing we know. Now what is I3? I3 is equal to I1 cos square theta. So, I will write here, I3 is equal to I1 cos square theta. So, I3 is equal to I1 is equal to I0 divided by 2 cos square 45. What is cos 45? 1 by root 2. How much is the square of 1 by root 2? 1 by 2. So, I3 is equal to I0 divided by 2 into 1 divided by 2. How much is I0 divided by 2 into 1 by 2? I0 divided by 4. So, I3 is equal to I0 divided by 4. So, what can we say about I3 is equal to I0 divided by 4 so what can we say about i3 is equal to i0 divided by 4 we can easily say that so that is the case now we have to find i2 by Mailloux's law i2 is equal to i3 cos square theta this is the angle we are making that is 45 this is the angle of third polaroid and the angle of second polaroid again 45 degrees if we draw this with angle we, we will get 45. So, we can substitute that. Therefore, I2 is equal to I3 cos square 45. We can substitute that. So, what will we get? This is equal to I3. I3 is I0 divided by 4 into cos square 45 is 1 by 2. So, this is equal to I0 divided by 8. So our answer is I0 divided by 4 into cos square 45 is 1 by 2 so this is equal to I 0 divided by 8 so our answer is I 0 divided by 8 so I 2 is equal to I 0 by 8 so option B is our correct answer very simple we can apply both sides in the model. I will repeat again this is what you should not mistake Don't mistake this axis I will draw it clearly This axis and this axis see in the picture. The angle of P2 and P3 is 45 degrees. And the half of the initial intensity will come out through the first polaroid. This is also very simple concept that we can understand clearly. So our answer is I0 divided by 8. Next question. Light is incident on a polarizer with intensity I0. A second prism called analyzer is kept at an angle of 15 degree from the first polarizer, then the intensity of final emergent light will be? This is a very simple question. There is a polarizer. The first polarizer is called as polarizer. And second polarizer is placed at an angle of 15 degree. So, how much is it placed at an angle of 15 degree so it is placed at 15 degree angle so this is analyzer so we know that the light incident to polarizer is I0 so the light incident from polarizer is I0. So the light intensity coming out of this is I0 divided by 2. So the light intensity coming out of this is Ia is equal to... This is the light intensity coming out of the polarizer. IP is equal to I0 divided by 2. So the light intensity coming out of the analyzer is Ia is equal Ip cos2 theta. That is equal to Ip is equal to I0 by 2. I0 by 2 cos2 15. To change cos2 15, I will give you a simple way. I will divide by 2 and do 2 into. So, I0 by 2 into 2 into 2 cos square 50. Now, notice this person. I am going to change this person. Let us know an identity. 2 cos square theta by 2 is 1 plus cos theta what is 1 plus cos theta? 2 cos square theta by 2 is 1 plus cos theta can be written as 2 cos square theta divided by 2 so 2 cos square 50 is I0 divided by 4 into 2 cos square 50 can be written as 1 plus cos 30 write it as double theta comes theta by 2 so 30 comes 15 so 2 cos square 15 can be written as 1 plus cos 30 cos 30 is root 3 by 2 so what is substituted is equal to I0 divided by 4 into 1 plus cos 30. cos 30 is root 3 divided by 2. Let's do something inside. I0 by 4 into 2 plus root 3 divided by 2. So, the answer is I0 divided by 8 into 2 plus root 3. So, our answer is i0 divided by 8 into 2 plus root 3. Option C is our correct answer. So, we have seen another question that is very simple and not applied in the malus. Ready? Very simple, isn't it? So, let's study and learn and write it. Next question. Unpolarized light is incident from air on a plane surface of a material of refractive index mu. At a particular angle of incidence i, it is found that the reflected and refracted rays are perpendicular to each other. So, we know where the unpolarized light is incident so this is unpolarized light having electric field vibrations in all directions this unpolarized light when incident on a particular interface light is reflected and light is refracted the reflected ray is completely polarized and the reflected ray is perpendicular to the plane of the incidence plane of incidence is perpendicular to the plane of incidence and the electric field vector is perpendicular to the plane of incidence so here electric field vibrations are always perpendicular to the plane of the incidence. You can see it on the screen. That is a property of this. Here, electric field vectors are... Here, this light is polarized. Let's learn that first. Reflected ray is polarized. And also, electric field vector is not parallel. It is perpendicular to the plane of incidence. even the property and a reflected light is polarized with its electric field vector is always perpendicular to the plane of incidence that is plane of incidence in a perpendicular right I think in a problem in the area electric field vectors were you gonna look I am a put up in the earth wake up here electric field vectors is always perpendicular to the plane of incidence. So the answer is very clear. It is option B. The rest is not clear. It is not parallel. It is not this. It is not this. It is inverse n. So the answer is option B. Ready? Next question. The Brewster's angle IB for an interface should be? We know that when Brewster's uncle should be I b for an interface. We know that when Brewster's angle is inserted here, I am giving theta as the angle. The light will reflect and the light will refract. There is an equation for Brewster's angle, n is equal to tan theta. This is called as Brewster's law and this theta is called as Brewster's angle or polarizing angle. So, we are familiar with this equation. Another thing we can know is that, the value of any material's refractive index will be between 1 and infinity. Any material's refractive index will be between 1 and infinity. We know tan\u03b8 is n. So tan\u03b8 is 1 less than n. I gave tan\u03b8 less than infinity. I gave tan\u03b8 tan 1 and infinity. So tan inverse 1 is less than theta or less than tan inverse infinity the logic of tan inverse is the same tan theta is equal to 1 then if tan theta is equal to 1 then theta is equal to tan inverse 1 we have applied that there. If tan theta is equal to \u03b8 is equal to tan inverse infinity. So, I just gave the tan to both sides as tan inverse. So, we get tan inverse 1. tan inverse 1 is 45 degree. And what is tan inverse infinity? tan inverse infinity is nothing but 90 degree. So, tan 90 is 90 degree degree. So, 45 degree less than theta less than 90 is the option. So, this is called as Brewster's law. Very simple law. Don't forget this. This is Brewster's law. So, this is Brewster's law. Let's move on to the next question. Light goes from water of refractive index 4 by 3 to glass of refractive index 3 by 2. The Brewster's angle is? We know that N2 divided by N1 is equal to tan theta. So, Ent the refractive index of the coming medium. The refractive index of glass is 3 by 2. N1 is 4 by 3. This is equal to tan theta. So, we have to write 9 divided by 3 by 2 into 3 by 4 is equal to tan theta. So, 9 divided by 8 is equal to tan theta. So, theta is equal to tan inverse 9 divided by 8. So this is the basic question we have to write 9 divided by 8. So, this is the basic question. So, remember this equation. Remember the equation of tan theta is equal to n. Remember the equation of n2 divided by n1. Next question. The critical angle of a medium is 45 degrees. Its angle of polarization is? Let's see the critical angle of a medium is 45 degree its angle of polarization is? We know that the equation connecting critical angle sin c and n is sin c is equal to 1 by n. So critical angle is 45 degree means sin 45 is equal to 1 divided by n. So sin 45 is 1 by root 2 that is equal to 1 divided by n implies n is equal to root 2. So, we can say that the value of n is root 2. So, if we get the value of n as root 2, we can say the polarization angle. Because tan theta is equal to n. So, tan theta is equal to root 2. Therefore, theta is equal to tan inverse root 2. So, this is a case where we get an answer of 57 degrees. So, remember that. Theta is equal to tan inverse root 2. We can reach a very simple equation. Let's move on to the next question. When light is incident from air to a medium at an angle 60 degree, it was found that reflected and refracted rays are mutually perpendicular. Then the refractive index of the medium is? Light is incident from a particular angle. That angle is 60 degree. So, the reflected ray and The reflected ray is usually perpendicular. What happens there? Polarization by reflection happens there. What angle can we call this angle? Polarization by reflection happens there. Reflected ray is completely polarized. What can we call this angle? Th the Brewster's angle. We can call theta as the Brewster's angle. That is 60 degree. So, we have asked about the refractive index. Refractive index n is equal to tan 60 degree That is equal to sin 60 divided by cos 60 Sin 60 is root 3 divided by 2 And cos 60 is 1 divided by 2 That is equal to root 3 So what is refractive index? It is equal to root 3 This is a very basic concept that Ready? Next question. Two polaroids A and B are kept crossed to each other. Yes. Ready? An ordinary light is of intensity 2I0. So see, two polaroid. If a third Polaroid C is placed between A and B at an angle theta with A, which makes an angle theta with A. So this is the Polaroid A, this is the Polaroid B and this is the. Now the question is, the intensity of light coming out of B is? If the light intensity is 2I0, half of it will come out. So, Ia is equal to 2I0 divided by 2. That is equal to I0. First understand that. So, Ia is equal to 2I0 divided by 2. That is equal to I0. Now, what is the intensity of light coming out of Ic? Ic is equal to Ia into cos square theta. So, theta is the angle here. So, Ia is I0. So,a into cos2 theta. So theta is the angle here. So Ia is I0. So what will we get from I0 cos2 theta? We will get the value of IC. So how much is the intensity of light coming out of this? IC is equal to I0 cos2 theta. Now let's see here. Yes. Ready? So this angle So, this angle is called theta. So, this angle is 90 minus theta. This angle is theta. So, what is that angle? That is 90 minus theta. So, the angle between C and B is 90 minus theta. See, I have drawn this concept very basically. I have drawn a line from here. This angle is theta. See, this angle is theta. So, this angle is theta. So, what is this angle? This angle is 90 minus theta. So, the angle between C and B is 90 minus theta. See, it is very simple. IB is equal to IC cos square 90 minus theta. The angle between them is 90 minus theta. Now, how much is cos 90 minus theta? That is sine theta. So the answer is Ic sin square theta Ic is I0 cos square theta so Ic is I0 cos square theta into sin square theta. I divided by 4 and multiplied by 4. So 4 into sin square theta into cos square theta. I wrote it down. I wrote I0 divided by 4 into 2 sin theta cos theta whole square. 2 sine theta cos theta whole square in the energy to sine theta cos theta in the array 2 sin theta cos theta whole square 2 sin theta cos theta is sin 2 theta So the final answer is I0 divided by 4 into sin square 2 theta Isn't it correct? I0 divided by 4 into sin square 2 theta is the answer So, this is our answer. Please understand that 2sin\u03b8 cos\u03b8 is sin2\u03b8. So, I will write sin2\u03b8 is equal to 2sin\u03b8 cos\u03b8. This is our answer. Ready? So now we will discuss questions about polarization. So see you again in the next section with more questions. Thank you.", + "timestamped_transcription": [ + { + "timestamp": [ + 0.0, + 48.0 + ], + "text": " Hi friends, we are going to discuss some important questions in the chapter Wave Optics. We know that wave optics is not a chapter where we study many things. Because we have to study three phenomena. One is the Huygens principle, then the interference, diffraction and polarization. It is a very small chapter. The questions asked will be simple. We will discuss some questions related to polarization in this video. You can do at least questions and study. Practice more questions. You should focus more to learn theory is to learn questions." + }, + { + "timestamp": [ + 48.0, + 52.0 + ], + "text": " So, learn questions and then reverse track it." + }, + { + "timestamp": [ + 52.0, + 56.0 + ], + "text": " Learn questions and then reverse track it." + }, + { + "timestamp": [ + 56.0, + 58.0 + ], + "text": " Try this strategy." + }, + { + "timestamp": [ + 58.0, + 60.0 + ], + "text": " Especially in physics." + }, + { + "timestamp": [ + 60.0, + 62.0 + ], + "text": " Learn theory through questions." + }, + { + "timestamp": [ + 62.0, + 90.0 + ], + "text": " We don't need to learn theory like that. Understand it like that. Let's see this question. So, this is the story of the story of the story of the story of the storyaroids placed one after the other in the path of the light. Taking the intensity of incident light as 100%, the intensity of out coming light that can be varied in the range of incident light as 100% each, the intensity of the light that can be varied in the range is what we are asking." + }, + { + "timestamp": [ + 90.0, + 85.76 + ], + "text": " So, we know that this is a Polaroid. This is Polaroid 1. The light passes through this through 100% intensity" + }, + { + "timestamp": [ + 85.76, + 406.0 + ], + "text": " So we know that the light coming out of this is 100 by 2, that is equal to 50%. We don't have to look at anything, we know that it is 50%. Now we have to put this light through another polaroid. So this is polaroid 2. So what is the special passed through another Polaroid So this is Polaroid 2 So what is the feature of this Polaroid? When light passes through this we can apply a low here If we rotate this Polaroid from theta angle the light intensity coming out from this If this is I1 this is I1 and I0 So the light intensity from second polaroid is I2 That is equal to I1 into cos2 There is a law called asus low is the intensity of light coming from the second polaroid or analyzer is equal to I1 cos square theta. So it is called as Malus low. So I2 is equal to I1 cos square theta. theta. So here theta can be varied from 0 degree to 90 degree. From 0 degree to 90 degree. Because, if this Polaroid is like this, then 50% of light will pass through it. If we keep it perpendicular, then 50% of light will not pass through it. 0% will pass through it. So, if theta is equal to 0 degree cos 0 is equal to 1 that is, the whole light will pass through i1. See, I have substituted in the equation. i2 is equal to i1 cos square 0 that is, i2 is equal to i1 i1 is equal to 50% So, 50% light will pass through it. Now, another% is passed through this. Now, what is the other possibility of this? What is the possibility of eta? It is possible to get 90. So, I2 is equal to I1 cos square 90. How much is cos 90? cos 90 is 0. So, it is possible to get 0%. So, I2 is equal to 0 so what all possibilities are there 0% is possible 50% is also possible so our variation from here to here will be 0% to Malus law. We can use this as a simple question to understand Polaroid. Let's see the next question. The polarizing angle for a transparent medium is theta. V is the speed of light. Then the relation between theta and V. C is the velocity of light in air. We have learnt a law. It is polarization by reflection. See, an light ray is coming and is incident. Imagine that light is going to another medium from air. So, if light is incident at a particular angle, called polarizing angle. Light will have reflection and light will have refraction. This refracted ray and reflected ray will be perpendicular. And reflected ray is polarized. This is called as polarization polarization by reflection an unpolarized light ray is incident at a particular angle called polarizing angle or Brewster's angle the reflected ray will be completely polarized this is the concept of low so here we have learnt a concept if the refractive index of this medium is n, then we can write a great equation called Brewster's law. What does Brewster's law say? It says that the refractive index of this medium is equal to tan of this thing. So we can write a great equation called n is equal to tan theta. So, if this is another medium of refractive index n1, if this is another medium of refractive index n2, how do we actually find this? I will just explain. This is our incident angle called theta. What happens to light? Reflection happens and light refracts. This is 90 degree. This is angle of refraction r. If this is theta, this is theta. Angle of incidence is equal to angle of reflection. So, this r is 180. So, r is equal to, 180 is equal to, we already know that I is theta. applying Snell's law we are going to apply Snell's law here the refractive index of this media is n1 and n2" + }, + { + "timestamp": [ + 406.0, + 408.0 + ], + "text": " so what will happen when we apply Snell's law" + }, + { + "timestamp": [ + 408.0, + 446.0 + ], + "text": " when we apply Snell's law we are going to apply Snell's law here the refractive index of this media is n1 refractive index of this media is n2 so what will happen when applying Snell's law? when applying Snell's law, look n1 sine theta is equal to the refractive index of this media into the angle of incidence of this media n1 sine theta is equal to the refractive index of this media into the angle of incidence of this media is sin r. r is 90 minus theta." + }, + { + "timestamp": [ + 446.0, + 450.0 + ], + "text": " sin of 90 minus theta is cos theta." + }, + { + "timestamp": [ + 450.0, + 455.0 + ], + "text": " So, n1 sin theta is equal to n2 into cos theta." + }, + { + "timestamp": [ + 455.0, + 457.0 + ], + "text": " So, sin theta by cos theta is tan theta." + }, + { + "timestamp": [ + 457.0, + 545.22 + ], + "text": " tan theta is equal to n2 divided by n1. tan theta is equal to n2 by n1. So, we can reach this kind of that n2 by n1 is substituted by tan theta so we can write tan theta is equal to tan theta is equal to n2 by n1 is v1 divided by v2 because velocity is inversely proportional to refractive index we use this medium as air and this as a denser medium so the speed of air is c so it is c divided by v v2 is tan theta is equal to c by v so theta is equal to tan inverse c by v so in this option there is no c by v So we will put it back So cot theta is equal to v by c So theta is equal to cot inverse v by c So theta is equal to cot inverse v by c So we can get to the answer of cot inverse v by c So option B is our correct answer we have already learnt this we just need to apply it easy right? ready? next question two polaroids P1 and P2 are placed with their axes perpendicular to each other. So there are two Polaroids and their axes are perpendicular. So this is a Polaroid and this is another Polaroid. So their axes are perpendicular. Now, unpolarized light I0 is incident on P1. Okay, okay, okay." + }, + { + "timestamp": [ + 545.46, + 766.96 + ], + "text": " This is P1 and this is P2. So, a light with I0 intensity is incident on P1. Now, third polaroid P3 is kept in between P1 and P2 such that it makes an angle 45 degree with P1. So, in have another Polaroid to make 45 degree angle with P1. So, the angle we are making with P1 is 45 degree. So, the angle we are making with P1 is 45 degree. So, angle made by this P3 with P1 is 45 degree See this angle is 45 degree I hope you understood it correctly Now what is the intensity of transmitted light through P2? The light coming out through P2 is the intensity of the light We can see this very easily So first we will see comes out of P2 is what we are asked. So, first we have to find out the light intensity that comes out of P1. That is I1. After that, we have to find out I3. After that, we have to find out I2. So, we will find out each one of them. So, I0 is the light intensity We can find each one of them. So I0 is the incident. So I1 is the half of the incident light. That is I0 divided by 2. That is the most basic thing we know. Now what is I3? I3 is equal to I1 cos square theta. So, I will write here, I3 is equal to I1 cos square theta. So, I3 is equal to I1 is equal to I0 divided by 2 cos square 45. What is cos 45? 1 by root 2. How much is the square of 1 by root 2? 1 by 2. So, I3 is equal to I0 divided by 2 into 1 divided by 2. How much is I0 divided by 2 into 1 by 2? I0 divided by 4. So, I3 is equal to I0 divided by 4. So, what can we say about I3 is equal to I0 divided by 4 so what can we say about i3 is equal to i0 divided by 4 we can easily say that so that is the case now we have to find i2 by Mailloux's law i2 is equal to i3 cos square theta this is the angle we are making that is 45 this is the angle of third polaroid and the angle of second polaroid again 45 degrees if we draw this with angle we, we will get 45. So, we can substitute that. Therefore, I2 is equal to I3 cos square 45. We can substitute that. So, what will we get? This is equal to I3. I3 is I0 divided by 4 into cos square 45 is 1 by 2. So, this is equal to I0 divided by 8. So our answer is I0 divided by 4 into cos square 45 is 1 by 2 so this is equal to I 0 divided by" + }, + { + "timestamp": [ + 766.96, + 762.0 + ], + "text": " 8 so our answer is I 0 divided by 8 so I 2 is equal to I 0 by 8 so option B is our correct answer very simple we can apply both sides in the model. I will repeat again this is what you should not mistake Don't mistake this axis" + }, + { + "timestamp": [ + 762.0, + 764.0 + ], + "text": " I will draw it clearly" + }, + { + "timestamp": [ + 764.0, + 785.2 + ], + "text": " This axis and this axis see in the picture. The angle of P2 and P3 is 45 degrees. And the half of the initial intensity will come out through the first polaroid." + }, + { + "timestamp": [ + 785.2, + 832.0 + ], + "text": " This is also very simple concept that we can understand clearly. So our answer is I0 divided by 8. Next question. Light is incident on a polarizer with intensity I0. A second prism called analyzer is kept at an angle of 15 degree from the first polarizer, then the intensity of final emergent light will be?" + }, + { + "timestamp": [ + 832.0, + 820.0 + ], + "text": "" + }, + { + "timestamp": [ + 823.0, + 828.0 + ], + "text": " This is a very simple question. There is a polarizer. The first polarizer is called as polarizer." + }, + { + "timestamp": [ + 828.0, + 835.0 + ], + "text": " And second polarizer is placed at an angle of 15 degree." + }, + { + "timestamp": [ + 835.0, + 947.38 + ], + "text": " So, how much is it placed at an angle of 15 degree so it is placed at 15 degree angle so this is analyzer so we know that the light incident to polarizer is I0 so the light incident from polarizer is I0. So the light intensity coming out of this is I0 divided by 2. So the light intensity coming out of this is Ia is equal to... This is the light intensity coming out of the polarizer. IP is equal to I0 divided by 2. So the light intensity coming out of the analyzer is Ia is equal Ip cos2 theta. That is equal to Ip is equal to I0 by 2. I0 by 2 cos2 15. To change cos2 15, I will give you a simple way. I will divide by 2 and do 2 into. So, I0 by 2 into 2 into 2 cos square 50. Now, notice this person. I am going to change this person. Let us know an identity. 2 cos square theta by 2 is 1 plus cos theta what is 1 plus cos theta? 2 cos square theta by 2 is 1 plus cos theta can be written as 2 cos square theta divided by 2 so 2 cos square 50 is I0 divided by 4 into 2 cos square 50 can be written as 1 plus cos 30" + }, + { + "timestamp": [ + 947.38, + 1125.5 + ], + "text": " write it as double theta comes theta by 2 so 30 comes 15 so 2 cos square 15 can be written as 1 plus cos 30 cos 30 is root 3 by 2 so what is substituted is equal to I0 divided by 4 into 1 plus cos 30. cos 30 is root 3 divided by 2. Let's do something inside. I0 by 4 into 2 plus root 3 divided by 2. So, the answer is I0 divided by 8 into 2 plus root 3. So, our answer is i0 divided by 8 into 2 plus root 3. Option C is our correct answer. So, we have seen another question that is very simple and not applied in the malus. Ready? Very simple, isn't it? So, let's study and learn and write it. Next question. Unpolarized light is incident from air on a plane surface of a material of refractive index mu. At a particular angle of incidence i, it is found that the reflected and refracted rays are perpendicular to each other. So, we know where the unpolarized light is incident so this is unpolarized light having electric field vibrations in all directions this unpolarized light when incident on a particular interface light is reflected and light is refracted the reflected ray is completely polarized and the reflected ray is perpendicular to the plane of the incidence plane of incidence is perpendicular to the plane of incidence and the electric field vector is perpendicular to the plane of incidence so here electric field vibrations are always perpendicular to the plane of the incidence. You can see it on the screen. That is a property of this. Here, electric field vectors are... Here, this light is polarized. Let's learn that first. Reflected ray is polarized. And also, electric field vector is not parallel. It is perpendicular to the plane of incidence. even the property and a reflected light is polarized with its electric field vector is always perpendicular to the plane of incidence that is plane of incidence in a perpendicular right I think in a problem in the area electric field vectors were you gonna look I am a put up in the earth wake up here electric field vectors is always perpendicular to the plane of incidence. So the answer is very clear. It is option B. The rest is not clear. It is not parallel. It is not this. It is not this. It is inverse n. So the answer is option B. Ready? Next question. The Brewster's angle IB for an interface should be? We know that when Brewster's uncle should be I b for an interface." + }, + { + "timestamp": [ + 1125.5, + 1130.0 + ], + "text": " We know that when Brewster's angle is inserted here," + }, + { + "timestamp": [ + 1130.0, + 1134.7 + ], + "text": " I am giving theta as the angle." + }, + { + "timestamp": [ + 1134.7, + 1447.74 + ], + "text": " The light will reflect and the light will refract. There is an equation for Brewster's angle, n is equal to tan theta. This is called as Brewster's law and this theta is called as Brewster's angle or polarizing angle. So, we are familiar with this equation. Another thing we can know is that, the value of any material's refractive index will be between 1 and infinity. Any material's refractive index will be between 1 and infinity. We know tan\u03b8 is n. So tan\u03b8 is 1 less than n. I gave tan\u03b8 less than infinity. I gave tan\u03b8 tan 1 and infinity. So tan inverse 1 is less than theta or less than tan inverse infinity the logic of tan inverse is the same tan theta is equal to 1 then if tan theta is equal to 1 then theta is equal to tan inverse 1 we have applied that there. If tan theta is equal to \u03b8 is equal to tan inverse infinity. So, I just gave the tan to both sides as tan inverse. So, we get tan inverse 1. tan inverse 1 is 45 degree. And what is tan inverse infinity? tan inverse infinity is nothing but 90 degree. So, tan 90 is 90 degree degree. So, 45 degree less than theta less than 90 is the option. So, this is called as Brewster's law. Very simple law. Don't forget this. This is Brewster's law. So, this is Brewster's law. Let's move on to the next question. Light goes from water of refractive index 4 by 3 to glass of refractive index 3 by 2. The Brewster's angle is? We know that N2 divided by N1 is equal to tan theta. So, Ent the refractive index of the coming medium. The refractive index of glass is 3 by 2. N1 is 4 by 3. This is equal to tan theta. So, we have to write 9 divided by 3 by 2 into 3 by 4 is equal to tan theta. So, 9 divided by 8 is equal to tan theta. So, theta is equal to tan inverse 9 divided by 8. So this is the basic question we have to write 9 divided by 8. So, this is the basic question. So, remember this equation. Remember the equation of tan theta is equal to n. Remember the equation of n2 divided by n1. Next question. The critical angle of a medium is 45 degrees. Its angle of polarization is? Let's see the critical angle of a medium is 45 degree its angle of polarization is? We know that the equation connecting critical angle sin c and n is sin c is equal to 1 by n. So critical angle is 45 degree means sin 45 is equal to 1 divided by n. So sin 45 is 1 by root 2 that is equal to 1 divided by n implies n is equal to root 2. So, we can say that the value of n is root 2. So, if we get the value of n as root 2, we can say the polarization angle. Because tan theta is equal to n. So, tan theta is equal to root 2. Therefore, theta is equal to tan inverse root 2. So, this is a case where we get an answer of 57 degrees. So, remember that. Theta is equal to tan inverse root 2. We can reach a very simple equation. Let's move on to the next question. When light is incident from air to a medium at an angle 60 degree, it was found that reflected and refracted rays are mutually perpendicular. Then the refractive index of the medium is? Light is incident from a particular angle. That angle is 60 degree. So, the reflected ray and The reflected ray is usually perpendicular. What happens there? Polarization by reflection happens there. What angle can we call this angle? Polarization by reflection happens there. Reflected ray is completely polarized. What can we call this angle? Th the Brewster's angle. We can call theta as the Brewster's angle. That is 60 degree. So, we have asked about the refractive index. Refractive index n is equal to tan 60 degree" + }, + { + "timestamp": [ + 1447.74, + 1451.42 + ], + "text": " That is equal to sin 60 divided by cos 60" + }, + { + "timestamp": [ + 1451.42, + 1455.4 + ], + "text": " Sin 60 is root 3 divided by 2" + }, + { + "timestamp": [ + 1455.4, + 1458.76 + ], + "text": " And cos 60 is 1 divided by 2" + }, + { + "timestamp": [ + 1458.76, + 1460.12 + ], + "text": " That is equal to root 3" + }, + { + "timestamp": [ + 1460.12, + 1461.64 + ], + "text": " So what is refractive index?" + }, + { + "timestamp": [ + 1461.64, + 1462.94 + ], + "text": " It is equal to root 3" + }, + { + "timestamp": [ + 1462.94, + 1525.18 + ], + "text": " This is a very basic concept that Ready? Next question. Two polaroids A and B are kept crossed to each other. Yes. Ready? An ordinary light is of intensity 2I0. So see, two polaroid. If a third Polaroid C is placed between A and B at an angle theta with A, which makes an angle theta with A. So this is the Polaroid A, this is the Polaroid B and this is the. Now the question is, the intensity of light coming out of B is? If the light intensity is 2I0, half of it will come out. So, Ia is equal to 2I0 divided by 2." + }, + { + "timestamp": [ + 1525.18, + 1666.76 + ], + "text": " That is equal to I0. First understand that. So, Ia is equal to 2I0 divided by 2. That is equal to I0. Now, what is the intensity of light coming out of Ic? Ic is equal to Ia into cos square theta. So, theta is the angle here. So, Ia is I0. So,a into cos2 theta. So theta is the angle here. So Ia is I0. So what will we get from I0 cos2 theta? We will get the value of IC. So how much is the intensity of light coming out of this? IC is equal to I0 cos2 theta. Now let's see here. Yes. Ready? So this angle So, this angle is called theta. So, this angle is 90 minus theta. This angle is theta. So, what is that angle? That is 90 minus theta. So, the angle between C and B is 90 minus theta. See, I have drawn this concept very basically. I have drawn a line from here. This angle is theta. See, this angle is theta. So, this angle is theta. So, what is this angle? This angle is 90 minus theta. So, the angle between C and B is 90 minus theta. See, it is very simple. IB is equal to IC cos square 90 minus theta. The angle between them is 90 minus theta. Now, how much is cos 90 minus theta? That is sine theta. So the answer is Ic sin square theta Ic is I0 cos square theta so Ic is I0 cos square theta into sin square theta. I divided by 4 and" + }, + { + "timestamp": [ + 1666.76, + 1666.16 + ], + "text": " multiplied by 4. So 4 into sin square theta into cos square theta. I wrote it down. I wrote I0 divided by 4 into 2 sin theta cos theta whole square. 2 sine theta cos theta whole square in the energy to sine theta cos theta in" + }, + { + "timestamp": [ + 1666.16, + 1725.82 + ], + "text": " the array 2 sin theta cos theta whole square 2 sin theta cos theta is sin 2 theta So the final answer is I0 divided by 4 into sin square 2 theta Isn't it correct? I0 divided by 4 into sin square 2 theta is the answer So, this is our answer. Please understand that 2sin\u03b8 cos\u03b8 is sin2\u03b8. So, I will write sin2\u03b8 is equal to 2sin\u03b8 cos\u03b8. This is our answer." + }, + { + "timestamp": [ + 1725.82, + 1729.06 + ], + "text": " Ready? So now we will discuss" + }, + { + "timestamp": [ + 1729.06, + 1731.1 + ], + "text": " questions about polarization." + }, + { + "timestamp": [ + 1731.1, + 1732.78 + ], + "text": " So see you again in the next section" + }, + { + "timestamp": [ + 1732.78, + 1737.28 + ], + "text": " with more questions. Thank you." + } + ] + }, + { + "video_id": 22671, + "transcription": " Hi friends, we are going to discuss some important questions in wave optics. We have already discussed a set of questions. Now we will discuss a set of questions related to diffraction and polarization. Without wasting any time, let us move on to the first question. On introducing a thin film in the path of one of the two interfering beam, the central fringe will shift by one fringe width. If refractive index is equal to 1.5, the thickness of the film is? If refractive index is 1.5, then we are asking out the thickness of the film. We can see that there are two slits S1 and S2. There we have a thin film of thickness T We can see that the new central maximum is formed here The shift happened here and now the new central maximum is formed here New central maximum will be formed at this point and there is an equation to find the distance y that is, the distance y. The path difference here is delta x is equal to t into n minus 1. This is an equation that everyone knows. delta x is equal to t into n minus 1. So, we can remember the is the path difference the path difference is equal to t into n minus 1 t is the thickness n is the refractive index so we know delta x is equal to yd divided by d. So, if we substitute that, t into n minus 1 is equal to y. Here, the y is one fringe width shift. So, we know the equation of fringe width. The equation of fringe width is beta is equal to lambda into capital D divided by small d. This is the equation of fringe width. So, this is the shift that happened. So, here y is equal to beta. Because the shift of fringe width is that much. So, it is beta into D divided by d. We can see that. item so T into n minus 1 is equal to beta into small d divided by capital D therefore T into n minus 1 is equal to fringe with this lambda capital D divided by small D into D divided by D on and the mechanic to me I'm at the bottom and I'm daddy by D on So, we get a equation like this Now we can write T into n-1 is equal to lambda That is T into 1.5-1 is equal to lambda the value of lambda is given here so lambda is given to you so t into 0.5 is equal to lambda 0.5 is equal to lambda t by 2 t by 2 is equal to lambda or rather t is equal to 2 lambda. This is a very easy answer to solve this question. This is a very simple answer to solve this question. This is a very basic question to solve this question. So this is the answer. t is equal to 2 lambda. We need to know many things to do this. First, we need to know the path difference equation. Second, we need to know the equation of y and thumb. Third, we need to know the equation of fringe width. What can we do only if we know all these things? This concept is very important. So let's see how to do it. First, we need to know the equation of y and thumb. Second, we need to know the equation of fringe width. Third, we need to know the equation of y and thumb. Fourth, we need to the equation of the fringe width. Only if we know all these things, we can apply this concept and get the answer. So, let's move on to the next question. This is a very simple question. We need to remember who is the person this. D is the distance between slit and screen. This is actually screen. And d is the distance between slits. d is the thickness of the film. The thickness of the film is T. CM is central maximum. These are the things we have to discuss here. Let's move on to the next question. Look at the next question. Young's double slit experiment is first performed in air and then in a medium other than air. Okay, it is found that 8th bright fringe in the medium lies were 5th dark fringe lies in air. It is found that eighth bright fringe in the medium lies where fifth dark fringe lies in air. Eighth bright fringe in medium means where is it? In medium. In a medium. Where was eighth bright fringe in medium?? there is the 5th dark fringe in air that is equal to 5th dark fringe in air so the hint given in this question is 8th bright fringe in medium is equal to 5th dark fringe in air So what is being said here is that when we place a media on screen, we get 8th bright fringe and same place we get 5th dark fringe in the same place in the air. So, what we can understand is, when we got this in the air, where did we get it? We got this in a medium. We got the 8th bright fringe at this point. Where did we get this? When we got this in the air, we got the 5th dark fringe at the same point. So, what is the question in air. The question is, the refractive index of the medium is? We need to know some things about this. What we need to know is, the wavelength of a medium is lambda dash, lambda air divided by n. We can remember this as important. The wavelength of the medium will change. How will it change? It is equal to lambda divided by n. This is one important thing to know. Second thing is 8th bright fringe in medium is equal to 5th dark fringe in air. So, I will write it again. 8th bright in medium is equal to 5th dark air. I will show you this figure again. So, what we can understand from this is that the distance to the 8th bright fringe is the same as the distance to the 5th dark fringe. So, look here. The distance to the 8th bright fringe is equal to the distance to the 5th dark fringe. Because we got both at the same point. So, that is what I am going to equate here. So, linear distance to 8th bright fringe is... We know the equation to find the distance to bright fringe. What is the equation to find the distance to bright fringe? y bright is equal to n lambda capital D divided by small d. Now, we know the equation to find the distance to the dark fringe The equation to find the distance to the dark fringe is 2n minus 1 lambda D divided by 2d These two equations are going to be applied In this, the medium of the 8th bright fringe is told So 8 into 8th bright fringe is n8 So, it is thexN8 So the median is lambda dash into D divided by small d. That should be equal to 5th dark fringe in air. So the equation is 2 into N5 2 into 5 minus 1 lambda air. So lambda is D. So, we can see that lambda is lambda. So, after solving this, we get 8 lambda dash. What is lambda dash? It is lambda air divided by refractive index so today I have substituted lambda divided by n that should be equal to 2 into 5 is 10, 10 minus 1 is 9 9 lambda divided by 2 is lambda lambda is cancelled n is equal to 8 into 2 is 16 divided by 9 16 divided by 9 is 1. 7 times 7 is 67 1.7 is our refractive index so if we know some basic things, we can easily solve this basic level question. So, remember this. The wavelength of lambda dash in a medium will change. And the equation to find the change in that wavelength is lambda dash is equal to lambda pi n. So, remember all these things. Then, remember the equation to find the distance to the bright fringe and the linear distance to the dark fringe. Next question. Assertion. Two point coherent sources of light S1 and S2 are placed on a line as shown. P and Q are two points on that line. If at point P maximum intensity is observed, then maximum intensity should also be observed at Q. P and Q are two points on that line. We could see the maximum density. Q is also the maximum density. The reason for that is, S1P minus S2P, that is the path difference, is equal to distance S2Q minus S1Q. So, if we want to see the maximum intensity here, which should be the interference that should be there? It should be constructive interference. So, which should be the interference that should be there? It should be constructive interference. So, If constructive interference is there, then delta x should be equal to n lambda. That is, the path difference between two waves should always be equal to n lambda. The question is already given. What is the maximum intensity in P? So, This is the path difference. S1p minus S2p. That is the path difference. Obviously we know that S1p is the distance between the two waves. That is the path difference between these two waves. So that is equal to n lambda. So P has the maximum intensity. Similarly, it is equal to S2Q minus S1Q. So S2Q minus S1Q is the distance between these two waves. That is also the maximum. If it is equal to n lambda, then the constructive interference will occur. And we will get the maximum intensity. So the reason is equal to n lambda, constructive interference will occur and we get the maximum intensity. So reason is correct. If at point P maximum intensity is observed, then maximum intensity should also be observed at Q. If the distance S1P minus S2P is equal to distance S2Q minus S1Q. Both assertion and reason are true and reason is the correct explanation of the assertion Because of this, the assertion came out correctly So is observed exactly in front of one slit. The distance between the two coherent sources is d and the distance between source and screen is D. The wavelength of light source used is? So, let's analyze this question very simply. These are the slits S1 and S2. If the second minimum is formed here, What is formed here? A minimum of 1 second is formed there Now, if a minimum of 1 second is formed there Let us know one thing Yes Look at this. This distance is d, the distance between slit and screen is d. This is the central maximum. So this distance would be d by 2. From central maximum to second minimum is y So y minimum Second minimum How we can find y minimum? Please understand this clearly This is our central maximum Because I took the correct central of the slit This is a Young's double slit experiment So if this is d, this will be d by 2 If this is d by 2, that will be d by 2 distance from central maximum to second minimum is given as y. How much is y? y is d by 2. Very simply, we can see that y is equal to d by 2. y is the distance from central maximum to second minimum the distance to dark band is y dark is equal to 2n-1 lambda divided by 2d n minus 1 lambda divided by 2d on another order substitute a medium second minimum so 2 into n-1 that is 2-1 lambda d divided by 2d that is equal to d divided by 2. So 2 into 2 is 4. 4-1 is 3. So, 3 lambda d divided by 2d is equal to d divided by 2. So, to cancel this, we have two. We are looking for the wavelength of light source here. So, how can we find the wavelength? If we stop the term of wavelength there, lambda is equal to d into d, that is d2 divided by 3 lambda. So, our answer is lambda is equal to d into d that is d2 divided by 3 by 3D. Option C is our correct answer. So, notice that the opposite point of one slit is the one where the second minimum is formed. So, the wave length equation we have asked is very direct. So, if we can do this question directly, if the central maximum is half of this, and if d by 2, then we can work out this question. Next question. The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern the ratio Imax minusImin divided by Imax plus Imin is equal to So, the intensity ratio is n. So, note that I1 divided by I2 is n. So, n is equal to n by 1. So, I1 is equal to n and I2 is equal to 1.1 is equal to n and I2 is equal to 1. I1 is equal to n and I2 is equal to 1. If you take it like that, it will be easy. Then what is Imax? We know the equation. There are two equations for Imax. We know the equation in terms of amplitude and in terms of intensity. The equation in terms of amplitude is a1 plus a2 the whole square. The equation in terms of intensity is root of i1 plus root of i2 the whole square. This is the equation in terms of intensity. So, this is the one I am going to apply this. This is equal to root of i1 is n here. So, root n plus root i2 is 1. So, root 1 is the same. So, root n plus 1 whole square is what we get when we substitute. So, what is root n plus 1 whole square and the root n whole square that is n plus 1 plus 2 root n in the book and the amitum I don't item item in an equation 1 in the an quarter to okay one in the angle to ease of the short oh what is the equation for I mean I mean the question the so the same I mean is equal to the look a amplitude in the So, this is the equation for this-min? I-min is equal to Let us know the equation of amplitude a1 minus a2 the whole square it is a1 minus a2 the whole square that is equal to root of a1 is i1 a2 is i2 so, root i1 minus root i2 the whole square what can we write? root i1i2 whole square So, on substituting these values, we get \u221an-\u221a1 whole square So, we can write it there. So, we can write a square plus b square minus 2ab So, we can write n plus 1 minus 2 root n So, we can write n plus 1 minus 2 root n n plus 1 minus 2 root n So, we got Imax and Imin We are asked Imax minus Imin divided by Imax plus Imin so let us find that out what is Imax minus Imin so Imax minus Imin divided by Imax plus Imin I max minus I mean divided by I max plus Imin is equal to Imax is n plus 1, I substitute n plus 1 plus 2 root n minus n plus 1 minus 2 root n so this will be divided by Imax plus Imin is, n plus 1 plus 2 root n plus n plus 1 plus 2 root n plus minus 2 root n So, this is how we can substitute Who cancels this? n minus n 1 minus 1 2 root n minus minus 2 root n that is equal to 2 root n plus 2 root n that is equal to 2 root n plus 2 root n that is equal to 4 root n in the mcindy amitum is on item at 2 root n minus 2 root n i'll cancel it away n plus n that is 2 n plus 1 plus 1 that is 2 2 common i to cut either way l get 2\u221an divided by n plus 1 So, our answer is 2\u221an divided by n plus 1 We can easily get to this answer So, we need to know this max a1 plus a2 whole square or root i1 plus root i2 whole square i mean a1 minus a2 whole square or root i1 plus root i2 whole square i-min is a1 minus a2 whole square or root i1 minus root i2 whole square If we know this equation, we can easily work out this question Let's see the next question The intensity at the maximum in a Young's double slit experiment is I0. The distance between the two slits is 5 lambda. D is equal to 5 lambda. Where lambda is the wavelength of light used in the experiment. the experiment what will be the intensity in front of one of the slits is d is equal to Phi lambda where lambda is a wavelength of light used in the experiment what will be the intensity in front of one of the slits on the screen placed at a distance d is equal to 10D okay it is yes the you would I on in the So, we know that the intensity of the interference fringe, at a distance d by 2, what is the intensity? We know that this distance is the distance between two slits d. the other one is d by 2 so here y is equal to d by 2 I am going to write delta x is equal to yd divided by d delta x is equal to y d by d Now, I am going to write some things in this equation delta x is equal to y d by 2 d is 5 lambda So, we will get the value of y by 5 lambda divided get y value. I am going to find out the people who are not able to substitute in this equation. So, I got y value as 5 lambda divided by 2. And I got d value as 5 lambda. Now, screen is placed at a distance 10d. So, d is equal to 10d. That is equal to 10 into, d is lambda, 5 lambda. So, D is equal to 10D. That is equal to 10 into 5 lambda. So, 10 into 5 lambda is equal to 50 lambda. So, I got D and D and D. This is the third and fourth equations. So, I am substituting 2, 3 and 4 in the first equation. So, what did I get? It is very simple. Delta x is equal to y is 5 lambda divided by 2 into d is 5 lambda. 5 lambda divided by d is 50 lambda. We can substitute it. Now, another thing substitute y is equal to d by 2. So, 5 lambda divided by 2. We have solved all the problems. So, on solving this, we get lambda lambda So, to lambda lambda cancel. So 5 times 50 is 10 times. 5 times 10 is 2 times. So our answer is lambda divided by 2 into 2 is 4. So what is delta x? Delta x is equal to lambda divided by 4. What is delta x? Delta x is the path difference. So we the path difference. It is lambda divided by 4. Now we are going to a complete difference. I am writing the values I got. I got the path difference. I got the value of delta x is equal to lambda divided by 4 in the value in a committee like lambda divided by 4 you know that face difference face difference a comma Delta Phi is equal to 2 pi by lambda into path difference on in the weekend the amateur parameter face difference Delta Phi is equal to 2 pi by lambda into path difference so phase difference delta phi is equal to 2 pi by lambda into path difference So, phase difference delta phi is equal to 2 pi by lambda into lambda divided by 4 So, lambda lambda cancelled 2 times The answer is pi by 2 that is equal to 90 degree So, our phase difference is 90 degree What we get as 90 degree? So, face difference is 90 degree So if we get 90 degree phase difference i is equal to i0 cos2 5 by 2 or 4i0 cos2 5 by 2 So i is equal to i0 into cos2. 5 is 90 divided by 2. 90 divided by 2 is 45. So, I is equal to I0 cos2, 45. cos45 is 1 by root 2. So, its square is 1 by 2. So, I0 divided by 2. So, what is the intensity? I0 divided by 2. Because cos45 is equal to 1 by root 2. When we substitute that, we can easily get the answer. Let's move on to the next question. The intensity ratio of maxima and minima in the interference pattern produced by two coherent sources of light is 9 is to 1 okay so here it is given as Imax divided by Imin is the ratio of maximum intensity to minimum intensity is 9 is to 1 so Imax by Imin is equal to 9 is equal to 1. This is the question. The intensity of light sources is the intensity. So we know Imax is a1 plus a2 the whole square. Divided by Imin is a1 minus a2 the whole square. This is equal to 9 divided by 1 We can write it as a square root of 9 So, if we take the square root of both sides a1 plus a2 is equal to a1 plus a2 divided by a1 minus a2 is equal to 3 is to 1 Because the square root of 9 is 3 and the square root of 1 is 1. So if we cross multiply, we can write a1 plus a2 is equal to 3 into a1 minus a2. So if we do it again, we can write a1 plus a2 is equal to 3a1 minus 3a2. So, let us rearrange it. a2 plus 3a2 is2 as 3a1 minus a1. So, a2 plus 3a2 is equal to 2a1. Therefore, a1 divided by a2 is equal to 4 divided by 2. That is equal to 2 by 1. so a1 by a2 is equal to 4 divided by 2 that is equal to 2 by 1 So A1 by A2 is equal to 4 by 2, 2 by 1 So what we have seen is not the ratio of amplitudes but the ratio of intensities Therefore I1 divided by I2 is equal to A1 by A2 the whole square because i is proportional to a square we all know that i is proportional to a2 so the answer we got here is a1 by a2 the whole square therefore i1 by i2 is equal to a1 by 2 by 1 whole square so the answer is 4 by 1 so what is the intensity? It is 4 by 1. So the thing we need to know is Imax by Imin is a1 plus a2 whole square divided by a1 minus a2 whole square. You just need to understand this concept. So we will discuss more questions in the next section.", + "timestamped_transcription": [ + { + "timestamp": [ + 0.0, + 65.64 + ], + "text": " Hi friends, we are going to discuss some important questions in wave optics. We have already discussed a set of questions. Now we will discuss a set of questions related to diffraction and polarization. Without wasting any time, let us move on to the first question. On introducing a thin film in the path of one of the two interfering beam, the central fringe will shift by one fringe width. If refractive index is equal to 1.5, the thickness of the film is? If refractive index is 1.5, then we are asking out the thickness of the film. We can see that there are two slits S1 and S2. There we have a thin film of thickness T" + }, + { + "timestamp": [ + 65.64, + 64.8 + ], + "text": " We can see that the new central maximum is formed here The shift happened here and now the new central maximum is formed here New central maximum will be formed at this point and there is an equation to find the distance y" + }, + { + "timestamp": [ + 64.8, + 306.0 + ], + "text": " that is, the distance y. The path difference here is delta x is equal to t into n minus 1. This is an equation that everyone knows. delta x is equal to t into n minus 1. So, we can remember the is the path difference the path difference is equal to t into n minus 1 t is the thickness n is the refractive index so we know delta x is equal to yd divided by d. So, if we substitute that, t into n minus 1 is equal to y. Here, the y is one fringe width shift. So, we know the equation of fringe width. The equation of fringe width is beta is equal to lambda into capital D divided by small d. This is the equation of fringe width. So, this is the shift that happened. So, here y is equal to beta. Because the shift of fringe width is that much. So, it is beta into D divided by d. We can see that. item so T into n minus 1 is equal to beta into small d divided by capital D therefore T into n minus 1 is equal to fringe with this lambda capital D divided by small D into D divided by D on and the mechanic to me I'm at the bottom and I'm daddy by D on So, we get a equation like this Now we can write T into n-1 is equal to lambda That is T into 1.5-1 is equal to lambda the value of lambda is given here so lambda is given to you so t into 0.5 is equal to lambda 0.5 is equal to lambda t by 2 t by 2 is equal to lambda or rather t is equal to 2 lambda. This is a very easy answer to solve this question. This is a very simple answer to solve this question. This is a very basic question to solve this question. So this is the answer. t is equal to 2 lambda. We need to know many things to do this. First, we need to know the path difference equation. Second, we need to know the equation of y and thumb. Third, we need to know the equation of fringe width. What can we do only if we know all these things? This concept is very important. So let's see how to do it. First, we need to know the equation of y and thumb. Second, we need to know the equation of fringe width. Third, we need to know the equation of y and thumb. Fourth, we need to the equation of the fringe width. Only if we know all these things, we can apply this concept and get the answer. So, let's move on to the next question. This is a very simple question. We need to remember who is the person this. D is the distance between slit and screen. This is actually screen. And d is the distance between slits. d is the thickness of the film. The thickness of the film is T. CM is central maximum." + }, + { + "timestamp": [ + 306.0, + 369.04 + ], + "text": " These are the things we have to discuss here. Let's move on to the next question. Look at the next question. Young's double slit experiment is first performed in air and then in a medium other than air. Okay, it is found that 8th bright fringe in the medium lies were 5th dark fringe lies in air. It is found that eighth bright fringe in the medium lies where fifth dark fringe lies in air. Eighth bright fringe in medium means where is it? In medium. In a medium. Where was eighth bright fringe in medium?? there is the 5th dark fringe in air that is equal to 5th dark fringe in air so the hint given in this question is 8th bright fringe in medium is equal to 5th dark fringe in air" + }, + { + "timestamp": [ + 369.04, + 360.0 + ], + "text": "" + }, + { + "timestamp": [ + 371.2, + 649.0 + ], + "text": " So what is being said here is that when we place a media on screen, we get 8th bright fringe and same place we get 5th dark fringe in the same place in the air. So, what we can understand is, when we got this in the air, where did we get it? We got this in a medium. We got the 8th bright fringe at this point. Where did we get this? When we got this in the air, we got the 5th dark fringe at the same point. So, what is the question in air. The question is, the refractive index of the medium is? We need to know some things about this. What we need to know is, the wavelength of a medium is lambda dash, lambda air divided by n. We can remember this as important. The wavelength of the medium will change. How will it change? It is equal to lambda divided by n. This is one important thing to know. Second thing is 8th bright fringe in medium is equal to 5th dark fringe in air. So, I will write it again. 8th bright in medium is equal to 5th dark air. I will show you this figure again. So, what we can understand from this is that the distance to the 8th bright fringe is the same as the distance to the 5th dark fringe. So, look here. The distance to the 8th bright fringe is equal to the distance to the 5th dark fringe. Because we got both at the same point. So, that is what I am going to equate here. So, linear distance to 8th bright fringe is... We know the equation to find the distance to bright fringe. What is the equation to find the distance to bright fringe? y bright is equal to n lambda capital D divided by small d. Now, we know the equation to find the distance to the dark fringe The equation to find the distance to the dark fringe is 2n minus 1 lambda D divided by 2d These two equations are going to be applied In this, the medium of the 8th bright fringe is told So 8 into 8th bright fringe is n8 So, it is thexN8 So the median is lambda dash into D divided by small d. That should be equal to 5th dark fringe in air. So the equation is 2 into N5 2 into 5 minus 1 lambda air. So lambda is D. So, we can see that lambda is lambda. So, after solving this, we get 8 lambda dash. What is lambda dash? It is lambda air divided by refractive index so today I have substituted lambda divided by n that should be equal to 2 into 5 is 10, 10 minus 1 is 9 9 lambda divided by 2 is lambda lambda is cancelled n is equal to 8 into 2 is 16 divided by 9 16 divided by 9 is 1. 7 times 7 is 67 1.7 is our refractive index so if we know some basic things, we can easily solve this basic level question. So, remember this. The wavelength of lambda dash in a medium will change. And the equation to find the change in that wavelength is lambda dash is equal to lambda pi n. So, remember all these things. Then, remember the equation to find the distance to the bright fringe and the linear distance to the dark fringe. Next question. Assertion. Two point coherent sources of light S1 and S2 are placed on a line as shown." + }, + { + "timestamp": [ + 649.0, + 645.48 + ], + "text": " P and Q are two points on that line. If at point P maximum intensity is observed, then maximum intensity should also be observed at Q. P and Q are two points on that line." + }, + { + "timestamp": [ + 645.48, + 767.0 + ], + "text": " We could see the maximum density. Q is also the maximum density. The reason for that is, S1P minus S2P, that is the path difference, is equal to distance S2Q minus S1Q. So, if we want to see the maximum intensity here, which should be the interference that should be there? It should be constructive interference. So, which should be the interference that should be there? It should be constructive interference. So, If constructive interference is there, then delta x should be equal to n lambda. That is, the path difference between two waves should always be equal to n lambda. The question is already given. What is the maximum intensity in P? So, This is the path difference. S1p minus S2p. That is the path difference. Obviously we know that S1p is the distance between the two waves. That is the path difference between these two waves. So that is equal to n lambda. So P has the maximum intensity. Similarly, it is equal to S2Q minus S1Q. So S2Q minus S1Q is the distance between these two waves. That is also the maximum. If it is equal to n lambda, then the constructive interference will occur. And we will get the maximum intensity. So the reason is equal to n lambda, constructive interference will occur and we get the maximum intensity." + }, + { + "timestamp": [ + 767.0, + 826.66 + ], + "text": " So reason is correct. If at point P maximum intensity is observed, then maximum intensity should also be observed at Q. If the distance S1P minus S2P is equal to distance S2Q minus S1Q. Both assertion and reason are true and reason is the correct explanation of the assertion Because of this, the assertion came out correctly So is observed exactly in front of one slit. The distance between the two coherent sources is d and the distance between source and screen is D." + }, + { + "timestamp": [ + 826.66, + 850.0 + ], + "text": " The wavelength of light source used is? So, let's analyze this question very simply. These are the slits S1 and S2. If the second minimum is formed here," + }, + { + "timestamp": [ + 850.0, + 840.0 + ], + "text": "" + }, + { + "timestamp": [ + 842.0, + 846.0 + ], + "text": " What is formed here? A minimum of 1 second is formed there" + }, + { + "timestamp": [ + 846.0, + 851.0 + ], + "text": " Now, if a minimum of 1 second is formed there" + }, + { + "timestamp": [ + 851.0, + 855.0 + ], + "text": " Let us know one thing" + }, + { + "timestamp": [ + 855.0, + 857.0 + ], + "text": " Yes" + }, + { + "timestamp": [ + 857.0, + 927.0 + ], + "text": " Look at this. This distance is d, the distance between slit and screen is d. This is the central maximum. So this distance would be d by 2. From central maximum to second minimum is y So y minimum Second minimum How we can find y minimum? Please understand this clearly This is our central maximum Because I took the correct central of the slit This is a Young's double slit experiment So if this is d, this will be d by 2 If this is d by 2, that will be d by 2 distance from central maximum to second minimum is given as y. How much is y? y is d by 2. Very simply, we can see that y is equal to d by 2. y is the distance from central maximum to second minimum" + }, + { + "timestamp": [ + 927.0, + 926.0 + ], + "text": " the distance to dark band is y dark is equal to 2n-1 lambda divided by 2d n minus 1 lambda divided by 2d on another order substitute a medium second" + }, + { + "timestamp": [ + 926.0, + 1005.0 + ], + "text": " minimum so 2 into n-1 that is 2-1 lambda d divided by 2d that is equal to d divided by 2. So 2 into 2 is 4. 4-1 is 3. So, 3 lambda d divided by 2d is equal to d divided by 2. So, to cancel this, we have two. We are looking for the wavelength of light source here. So, how can we find the wavelength? If we stop the term of wavelength there, lambda is equal to d into d, that is d2 divided by 3 lambda." + }, + { + "timestamp": [ + 1005.0, + 1227.0 + ], + "text": " So, our answer is lambda is equal to d into d that is d2 divided by 3 by 3D. Option C is our correct answer. So, notice that the opposite point of one slit is the one where the second minimum is formed. So, the wave length equation we have asked is very direct. So, if we can do this question directly, if the central maximum is half of this, and if d by 2, then we can work out this question. Next question. The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern the ratio Imax minusImin divided by Imax plus Imin is equal to So, the intensity ratio is n. So, note that I1 divided by I2 is n. So, n is equal to n by 1. So, I1 is equal to n and I2 is equal to 1.1 is equal to n and I2 is equal to 1. I1 is equal to n and I2 is equal to 1. If you take it like that, it will be easy. Then what is Imax? We know the equation. There are two equations for Imax. We know the equation in terms of amplitude and in terms of intensity. The equation in terms of amplitude is a1 plus a2 the whole square. The equation in terms of intensity is root of i1 plus root of i2 the whole square. This is the equation in terms of intensity. So, this is the one I am going to apply this. This is equal to root of i1 is n here. So, root n plus root i2 is 1. So, root 1 is the same. So, root n plus 1 whole square is what we get when we substitute. So, what is root n plus 1 whole square and the root n whole square that is n plus 1 plus 2 root n in the book and the amitum I don't item item in an equation 1 in the an quarter to okay one in the angle to ease of the short oh what is the equation for I mean I mean the question the so the same I mean is equal to the look a amplitude in the So, this is the equation for this-min? I-min is equal to Let us know the equation of amplitude a1 minus a2 the whole square it is a1 minus a2 the whole square that is equal to root of a1 is i1 a2 is i2 so, root i1 minus root i2 the whole square what can we write? root i1i2 whole square So, on substituting these values, we get \u221an-\u221a1 whole square So, we can write it there. So, we can write a square plus b square minus 2ab So, we can write n plus 1 minus 2 root n So, we can write n plus 1 minus 2 root n n plus 1 minus 2 root n So, we got Imax and Imin We are asked Imax minus Imin divided by Imax plus Imin" + }, + { + "timestamp": [ + 1227.0, + 1267.68 + ], + "text": " so let us find that out what is Imax minus Imin so Imax minus Imin divided by Imax plus Imin I max minus I mean divided by I max plus Imin is equal to Imax is n plus 1, I substitute n plus 1 plus 2 root n minus n plus 1 minus 2 root n so this will be divided by Imax plus Imin is, n plus 1 plus 2 root n plus n plus 1 plus 2 root n plus minus 2 root n So, this is how we can substitute Who cancels this? n minus n 1 minus 1 2 root n minus minus 2 root n that is equal to 2 root n plus" + }, + { + "timestamp": [ + 1267.68, + 1307.0 + ], + "text": " 2 root n that is equal to 2 root n plus 2 root n that is equal to 4 root n in the mcindy amitum is on item at 2 root n minus 2 root n i'll cancel it away n plus n that is 2 n plus 1 plus 1 that is 2 2 common i to cut either way l get 2\u221an divided by n plus 1" + }, + { + "timestamp": [ + 1307.0, + 1305.16 + ], + "text": " So, our answer is 2\u221an divided by n plus 1 We can easily get to this answer So, we need to know this max a1 plus a2 whole square or root i1 plus root i2 whole square" + }, + { + "timestamp": [ + 1305.16, + 1365.04 + ], + "text": " i mean a1 minus a2 whole square or root i1 plus root i2 whole square i-min is a1 minus a2 whole square or root i1 minus root i2 whole square If we know this equation, we can easily work out this question Let's see the next question The intensity at the maximum in a Young's double slit experiment is I0. The distance between the two slits is 5 lambda. D is equal to 5 lambda. Where lambda is the wavelength of light used in the experiment. the experiment what will be the intensity in front of one of the slits is d is equal to Phi lambda where lambda is a wavelength of light used in" + }, + { + "timestamp": [ + 1365.04, + 1368.22 + ], + "text": " the experiment what will be the intensity in front of one of the slits" + }, + { + "timestamp": [ + 1368.22, + 1725.0 + ], + "text": " on the screen placed at a distance d is equal to 10D okay it is yes the you would I on in the So, we know that the intensity of the interference fringe, at a distance d by 2, what is the intensity? We know that this distance is the distance between two slits d. the other one is d by 2 so here y is equal to d by 2 I am going to write delta x is equal to yd divided by d delta x is equal to y d by d Now, I am going to write some things in this equation delta x is equal to y d by 2 d is 5 lambda So, we will get the value of y by 5 lambda divided get y value. I am going to find out the people who are not able to substitute in this equation. So, I got y value as 5 lambda divided by 2. And I got d value as 5 lambda. Now, screen is placed at a distance 10d. So, d is equal to 10d. That is equal to 10 into, d is lambda, 5 lambda. So, D is equal to 10D. That is equal to 10 into 5 lambda. So, 10 into 5 lambda is equal to 50 lambda. So, I got D and D and D. This is the third and fourth equations. So, I am substituting 2, 3 and 4 in the first equation. So, what did I get? It is very simple. Delta x is equal to y is 5 lambda divided by 2 into d is 5 lambda. 5 lambda divided by d is 50 lambda. We can substitute it. Now, another thing substitute y is equal to d by 2. So, 5 lambda divided by 2. We have solved all the problems. So, on solving this, we get lambda lambda So, to lambda lambda cancel. So 5 times 50 is 10 times. 5 times 10 is 2 times. So our answer is lambda divided by 2 into 2 is 4. So what is delta x? Delta x is equal to lambda divided by 4. What is delta x? Delta x is the path difference. So we the path difference. It is lambda divided by 4. Now we are going to a complete difference. I am writing the values I got. I got the path difference. I got the value of delta x is equal to lambda divided by 4 in the value in a committee like lambda divided by 4 you know that face difference face difference a comma Delta Phi is equal to 2 pi by lambda into path difference on in the weekend the amateur parameter face difference Delta Phi is equal to 2 pi by lambda into path difference so phase difference delta phi is equal to 2 pi by lambda into path difference So, phase difference delta phi is equal to 2 pi by lambda into lambda divided by 4 So, lambda lambda cancelled 2 times The answer is pi by 2 that is equal to 90 degree So, our phase difference is 90 degree What we get as 90 degree? So, face difference is 90 degree So if we get 90 degree phase difference i is equal to i0 cos2 5 by 2 or 4i0 cos2 5 by 2 So i is equal to i0 into cos2. 5 is 90 divided by 2. 90 divided by 2 is 45. So, I is equal to I0 cos2, 45. cos45 is 1 by root 2. So, its square is 1 by 2. So, I0 divided by 2. So, what is the intensity? I0 divided by 2. Because cos45 is equal to 1 by root 2. When we substitute that, we can easily get the answer. Let's move on to the next question. The intensity ratio of maxima and minima in the interference pattern produced by two coherent sources of light is 9 is to 1 okay so here it is given as Imax divided by Imin is the ratio of maximum intensity to minimum intensity is 9 is to 1 so Imax by Imin is equal to 9 is equal to 1. This is the question. The intensity of light sources is the intensity. So we know Imax is a1 plus a2 the whole square. Divided by Imin is a1 minus a2 the whole square. This is equal to 9 divided by 1 We can write it as a square root of 9 So, if we take the square root of both sides a1 plus a2 is equal to a1 plus a2 divided by a1 minus a2 is equal to 3 is to 1 Because the square root of 9 is 3 and the square root of 1 is 1." + }, + { + "timestamp": [ + 1725.0, + 1722.0 + ], + "text": " So if we cross multiply, we can write a1 plus a2 is equal to 3 into a1 minus a2. So if we do it again, we can write a1 plus a2 is equal to 3a1 minus 3a2. So, let us rearrange it." + }, + { + "timestamp": [ + 1722.0, + 1765.06 + ], + "text": " a2 plus 3a2 is2 as 3a1 minus a1. So, a2 plus 3a2 is equal to 2a1. Therefore, a1 divided by a2 is equal to 4 divided by 2. That is equal to 2 by 1. so a1 by a2 is equal to 4 divided by 2 that is equal to 2 by 1" + }, + { + "timestamp": [ + 1765.06, + 1763.0 + ], + "text": " So A1 by A2 is equal to 4 by 2, 2 by 1 So what we have seen is not the ratio of amplitudes but the ratio of intensities Therefore I1 divided by I2 is equal to A1 by A2 the whole square because i is proportional to a square" + }, + { + "timestamp": [ + 1763.0, + 1806.0 + ], + "text": " we all know that i is proportional to a2 so the answer we got here is a1 by a2 the whole square therefore i1 by i2 is equal to a1 by 2 by 1 whole square so the answer is 4 by 1 so what is the intensity? It is 4 by 1." + }, + { + "timestamp": [ + 1806.0, + 1809.0 + ], + "text": " So the thing we need to know is Imax by Imin" + }, + { + "timestamp": [ + 1809.0, + 1812.0 + ], + "text": " is a1 plus a2 whole square divided by a1 minus a2" + }, + { + "timestamp": [ + 1812.0, + 1815.0 + ], + "text": " whole square." + }, + { + "timestamp": [ + 1815.0, + 1818.0 + ], + "text": " You just need to understand this concept." + }, + { + "timestamp": [ + 1818.0, + 1821.0 + ], + "text": " So we will discuss more questions in the next section." + } + ] + }, + { + "video_id": 22670, + "transcription": " Let's discuss some more questions. Let's get into the questions. If the intensity ratio of two waves is 1 is to 4, then find the ratio of their amplitudes. Intensity ratio i1 divided by i2 is equal to 1 divided by 4. We are asking about the ratio of the amplitudes of the by I 2 is equal to a 1 divided by a to the whole square neither could I so either the mocha 1 by 4 and then the intensity all ratio so square root of the way so random side is square root at all is 1 by 4, the intensity and ratio. So, if we take the square root, a1 by a2 is equal to 1 by 2. We get the answer very easily. We can see this as a very basic question. Next question. Two coherent monochromatic beams of intensities I and for I respectively are superimposed the maximum and minimum intensities in the resulting pattern is like I am for I am superimposed it upon the muck around I max in the parine the it is root I1 plus Root I2 the whole square. That is equal to root of I1 What is I1? That is 4i. I have taken 4i. I took I2 as the whole square. That is equal to root 4i is nothing but 2 root i. So it is 2 root i plus root i the whole square. We are going to write it in another way. 2 root i plus root i, that is 3 root i. We don't have to look at it. 2 root i plus root i, if we take root i as common, it is 3 root i. 3 root i the whole square is 9i. Ready? Because 3 square is 9 and root i is square i. Then what is min minimum intensity is very simple it is a direct equation there is no confusion in this root i1 minus root i2 the whole square is the equation of i min that is equal to root 4i minus root i the whole square root 4i minus root i the whole square okay root 4i minus root i the whole square we will get there. Now see this is equal to 2 root i minus root i the whole square Our final answer is 9i and i 9i and i are the maximum intensity to minimum intensity 9i and i are the maximum intensity to minimum intensity 9i and i are the maximum intensity to minimum intensity Next question In a single slit question in a single slit diffraction experiment using light of wavelength 500 nanometer the first minima falls at an angle 30 degree the width of the slit is? so we know that the width of the slit is the single slit diffraction pattern. A is the slit width. So, the angle here is the first minimum which is the central maximum. First minimum falls at an angle 30 degrees. So, this angle is 30 degree. So we know one thing in this case sin theta is equal to delta x divided by a sin theta is equal to delta x divided by a. So we have sin theta is equal to delta x by a this is the shadow sine theta is equal to delta x divided by a so we have sin theta is equal to delta x by a this is the sign theta is equal to delta x what is delta x other first minima first minima is n lambda divided by a first minima so it is lambda divided by a so we are asked a So, this is what we have been asked. can we find that? A is equal to lambda divided by sine theta. What is lambda? Lambda is 500 nanometers. So 500 nanometer divided by sine. Angle is 30 degrees. is 1 by 2 so 500 nanometer divided by sine angle is 30 degree sine 30 degree sine 30 is 1 by 2 so 500 nanometer divided by 1 by 2 1 by 2 it to only QLA thousand nanometer and the material nano is nothing but nano triana nano in the warning al 10 raise to minus 9 meter on a la so 10 raise to 3 into 10 raise to minus 9 meter. That is equal to 10 raised to minus 6 meter. That is equal to 1 micrometer. So, I think the answer is 1 micrometer. You can verify it. This is how we get the result. Provided, when we do this, there is a chance of error in some questions. You may have a chance of error and I may have a chance of error. We are doing all these like ready to go. We are doing all these like ready to cook. So, what is the chance of error in all these? There is a chance of error. Everyone will have an error. I will have an error and you too. So, understand that clearly. Try to do it again. Okay. Next question. In a single slit diffraction experiment using slit width 2 micrometer. Slit width A is equal to 2 micrometer. The second secondary maxima Secondary maxima. Second secondary maxima falls at the angle 30 degrees. Secondary maxima means secondary maxima. False set angle theta is equal to 30 degrees. The wavelength of light used is lambda. This is to verify. So, the second maxima is the case. So, it is 30 degrees. So, we know that sin theta is equal to delta x divided by a. So, we know that sin theta is equal to delta x divided by a. So, the maximum condition is 2n plus 1 lambda divided by 2a. So, a is equal to sin 30 degree is equal to 2 into 2n plus 1 lambda la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la So our final answer is 2 into 2 is 4. 4 plus 1 is 5. So sine 30 is 1 by 2. 1 by 2 is equal to 5 lambda divided by 4 into 10 raised to minus 6. So 2 and 4 are the answer. So lambda is equal to 2 into 10 raised to minus 6 divided by 5. to into 10 raised to minus 6 divided by 5. So 2 by 5 into 10 raised to minus 6 will be lambda or 2 by 5 micrometer. Let's see the next question.", + "timestamped_transcription": [ + { + "timestamp": [ + 0.0, + 4.0 + ], + "text": " Let's discuss some more questions." + }, + { + "timestamp": [ + 4.0, + 35.58 + ], + "text": " Let's get into the questions. If the intensity ratio of two waves is 1 is to 4, then find the ratio of their amplitudes. Intensity ratio i1 divided by i2 is equal to 1 divided by 4. We are asking about the ratio of the amplitudes of the" + }, + { + "timestamp": [ + 35.58, + 20.0 + ], + "text": "" + }, + { + "timestamp": [ + 27.24, + 31.7 + ], + "text": " by I 2 is equal to a 1 divided by a to the whole square neither could I so either the mocha 1 by 4 and then the intensity all ratio so square root of the way so" + }, + { + "timestamp": [ + 31.7, + 85.44 + ], + "text": " random side is square root at all is 1 by 4, the intensity and ratio. So, if we take the square root, a1 by a2 is equal to 1 by 2. We get the answer very easily. We can see this as a very basic question. Next question. Two coherent monochromatic beams of intensities I and for I respectively are superimposed the maximum and minimum intensities in the resulting pattern is like I am for I am superimposed it upon the muck around I max in the parine the it is root I1 plus" + }, + { + "timestamp": [ + 466.0, + 486.0 + ], + "text": " Root I2 the whole square. That is equal to root of I1 What is I1? That is 4i. I have taken 4i. I took I2 as the whole square. That is equal to root 4i is nothing but 2 root i. So it is 2 root i plus root i the whole square. We are going to write it in another way. 2 root i plus root i, that is 3 root i. We don't have to look at it. 2 root i plus root i, if we take root i as common, it is 3 root i. 3 root i the whole square is 9i. Ready? Because 3 square is 9 and root i is square i. Then what is min minimum intensity is very simple it is a direct equation there is no confusion in this root i1 minus root i2 the whole square is the equation of i min that is equal to root 4i minus root i the whole square root 4i minus root i the whole square okay root 4i minus root i the whole square we will get there. Now see this is equal to 2 root i minus root i the whole square Our final answer is 9i and i 9i and i are the maximum intensity to minimum intensity 9i and i are the maximum intensity to minimum intensity 9i and i are the maximum intensity to minimum intensity Next question In a single slit question in a single slit diffraction experiment using light of wavelength 500 nanometer the first minima falls at an angle 30 degree the width of the slit is? so we know that the width of the slit is the single slit diffraction pattern. A is the slit width. So, the angle here is the first minimum which is the central maximum. First minimum falls at an angle 30 degrees. So, this angle is 30 degree. So we know one thing in this case sin theta is equal to delta x divided by a sin theta is equal to delta x divided by a. So we have sin theta is equal to delta x by a this is the shadow sine theta is equal to delta x divided by a so we have sin theta is equal to delta x by a this is the sign theta is equal to delta x what is delta x other first minima first minima is n lambda divided by a first minima so it is lambda divided by a so we are asked a So, this is what we have been asked. can we find that? A is equal to lambda divided by sine theta. What is lambda? Lambda is 500 nanometers. So 500 nanometer divided by sine. Angle is 30 degrees. is 1 by 2 so 500 nanometer divided by sine angle is 30 degree sine 30 degree sine 30 is 1 by 2 so 500 nanometer divided by 1 by 2 1 by 2 it to only QLA thousand nanometer and the material nano is nothing but nano triana nano in the warning al 10 raise to minus 9 meter on a la so 10 raise to 3 into 10 raise to minus 9 meter. That is equal to 10 raised to minus 6 meter. That is equal to 1 micrometer. So, I think the answer is 1 micrometer. You can verify it. This is how we get the result. Provided, when we do this, there is a chance of error in some questions. You may have a chance of error and I may have a chance of error. We are doing all these like ready to go. We are doing all these like ready to cook. So, what is the chance of error in all these? There is a chance of error. Everyone will have an error. I will have an error and you too. So, understand that clearly. Try to do it again. Okay. Next question. In a single slit diffraction experiment using slit width 2 micrometer. Slit width A is equal to 2 micrometer. The second secondary maxima Secondary maxima. Second secondary maxima falls at the angle 30 degrees. Secondary maxima means secondary maxima. False set angle theta is equal to 30 degrees. The wavelength of light used is lambda. This is to verify. So, the second maxima is the case. So, it is 30 degrees. So, we know that sin theta is equal to delta x divided by a. So, we know that sin theta is equal to delta x divided by a. So, the maximum condition is 2n plus 1 lambda divided by 2a. So, a is equal to sin 30 degree is equal to 2 into 2n plus 1 lambda la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la So our final answer is 2 into 2 is 4. 4 plus 1 is 5. So sine 30 is 1 by 2. 1 by 2 is equal to 5 lambda divided by 4 into 10 raised to minus 6. So 2 and 4 are the answer. So lambda is equal to 2 into 10 raised to minus 6 divided by 5. to into 10 raised to minus 6 divided by 5." + }, + { + "timestamp": [ + 486.0, + 492.1 + ], + "text": " So 2 by 5 into 10 raised to minus 6 will be lambda or 2 by 5" + }, + { + "timestamp": [ + 492.1, + 498.6 + ], + "text": " micrometer." + }, + { + "timestamp": [ + 498.6, + 504.0 + ], + "text": " Let's see the next question." + } + ] + }, + { + "video_id": 22784, + "transcription": " Hi friends, we are going to discuss some important questions in the chapter called electromagnetic waves. This chapter is very simple and you will feel that even though the content is very less, the chapter that asks good questions is the electromagnetic wave. So never miss it. Ready? A capacitor of capacitance C is connected across a source of voltage V given by V is equal to V0 sine omega T. The displacement current between the plates of the capacitor would be given by... Displacement current is used. So, there is a capacitor like this. That capacitor is connected to a battery. The current Q is flowing through this. So, displacement current is used. So, we need such an equation. So, let need a equation like this. So, we know that I is equal to dQ divided by dt. So, this is equal to d by dt of Q is Cv. So, I is equal to d by dt of Cv. V is the voltage applied, that is, V0 sine omega t. So, d by dt of C into V0 sine omega t. d by dt of Cv0 sine omega t. So, I is equal to let's take Cv0 out. So I is equal to Cv0 into d by dt of sine Omega t. What is d by dt of sine Omega t? That is equal to omega cos omega t. Therefore, I is equal to Cv0 into omega cos omega t. Good equation, isn't it? We should learn this equation altogether. d by dt of Cv. Because, in some questions, we are asked to find I. When we are asked to find I, we can take C out of the equation. We get C into dV divided by dt. So, the rate of change in voltage, dV by dt, will be given in the question. Then we can find I. Then we have to remember this equation. So, we use your be used and this equation should be remembered and the last equation should be remembered. C V0 omega cos omega t. So, V0 omega C cos omega t. So, this is the equation for displacement current. It is a very simple equation to remember. Next question. A parallel plate capacitor of capacitance 20 microfarads is being charged by a voltage source whose potential is changing at the rate of 3 volt per second. So dv by dt is 3 volt per second. The hint given in the question is dv divided by dt is equal to 3 volt per second. Now, what is given in the question? The conduction current through the wire and displacement current. Both are same. Conduction current is same and displacement current is same. In this case, not in this case, but in all the cases, in all the cases, the conduction current is the displacement current. If so, we have given capacitance and this is given. So, which equation should be used? ID is equal to the equation we just learned C into dv divided by dt. So this is equal to 20 into, see the value of C is 20 microfarad. We need microfarad because we have micro in options. So 20 microfarad into, what is dv by dt? Rate of 3 volt per second. So, 3. So, answer is 20 into 3. That is 60 microampere. So, we got the value of displacement current here. That is equal to 60 microampere. So, what is displacement current? That itself is conduction current. What will come here? It will come equal. So, answer is 60 microampere. Let's move to the nextere. Next question. A plain electromagnetic wave of frequency 25 MHz travels in a free space along x direction. At a particular point in the space and time, E is equal to 0.3 J cap volt per meter. The magnetic field at this point is very simple coming in an equation C is equal to e divided by B that is equal to e is 0.3 LA 0.3 divided by what is C okay I'm gonna magnetic field a little in the so magnetic field B is equal to E divided by C so E is nothing but 0.3 and B C is 3 into 10 raised to 8 so our answer is 0.1 cut either way lay 0.1 into 10 to the power minus 8 Tesla in the cancer to LA 0.1 into 10 raised to minus 8 tesla. So, this is the question we need to apply. Ready? Next question. The magnetic field of a plain electromagnetic wave is given by this equation. This equation is similar to by is equal to b0 sine kx plus Omega t. So, comparing this with this equation, we are taking the propagation constant value k is equal to 2 pi by lambda, therefore lambda is equal to 2 pi divided by k. So, comparing these two equations, k is 0.5 into 10 raised to 3. So, 2 pi divided by K so comparing these two equations K is 0.5 into 10 raised to 3 so 2 pi divided by 0.5 into 10 to the power 3 in the middle it would have been okay 2 pi divided by 0.5 that is equal to 4 pi so 4 pi into 10 to the power minus 3 it is meter a meter so So, for pi into 10 raise to minus 3 meter. Next question the frequency of electromagnetic wave. So, omega is given as 1.5 into 10 raise to 11. So, omega is equal to we are taking the equation omega is equal to 2 pi f therefore, F is equal to omega divided by 2 pi so here omega is given by 1.5 into 10 to the power 11 divided by, comparing these two equations, divided by 2 pi. That is, this is our frequency, this much hertz. This will be our frequency. We just need to solve that. You just need to solve that. Then, the velocity of electromagnetic wave. Next, we need to find the velocity. How to find the velocity? Velocity V is equal to frequency into wavelength. Therefore frequency is 1.5 into 10 to the power 11 divided by 2 pi into wavelength is 4 pi into 10 to the power minus 3. So we will get free velocity a path of the solver the original velocity to next amplitude of electric field vector okay amplitude am langa and odika we have C is equal to e 0 divided by B 0 so what is e 0 that is amplitude of electric field vector which implies this is our equation. So E0 is equal to B0 into C. So B0 is here 2 into 10 to the power minus 7 into C is here 3 into 10 raised to 8. So our answer is 6 into 10 to the power 1 so 60 answer is 60 so E0 is equal to 60 E0 is 60 Newton per Coulomb E is equal to F by Q then what is direction of this electric field vector The way we wave is along x axis because the coefficient of k is kx. That means in which direction is the electromagnetic wave? It is on x axis. In which axis is the magnetic field? It is on y axis because b is given as y. So the electromagnetic wave is on x axis and the magnetic field is on y axis-axis. The electric field is on the z-axis. So, the electric field is along the z-axis because the direction of the electromagnetic wave is perpendicular to the magnetic field. If we try to understand the the direction of magnetic field vector, then this is the direction of electromagnetic wave. If the electromagnetic wave is on x axis, then the electric field will be along y axis and magnetic field will be along z axis. So this is a concept that we can understand. If you remember a movie like this, Yes, the electromagnetic baby cell. So, this is the equation or concept that we know. If you remember the picture, you can see the electromagnetic wave is along the x-axis. So, this is the case that we know. So, that is what we have asked here. And last is the expression for electric field vector. So, electric field vector can be written as e vector is equal to e 0 e 0 is 60 sine K X K is nothing but 0.5 into 10 raised to 3 X I didn't know Mary with the solo to plus Omega t 1.5 into 10 raised to 11 T and K cap LA in a you come here. Okay, so this is the expression for electric field vector. So we have done everything. Next question. In an electromagnetic wave, the amplitude of electric field vector is 20 V per meter. So E0 is equal to 20 V per meter. The average energy density of electromagnetic wave is average energy density of electromagnetic so total we have to find the total okay the average energy density of electromagnetic wave is average energy density of electromagnetic wave. So total we have to find the total. So let us know one thing. Let us find out the electric field energy. field energy is uE is equal to half epsilon zero into E square. E is equal to E0 by root 2. That is equal to half epsilon zero into E0 divided by root 2 whole square. This is E rms and this is the rms value. That is equal to E0 divided by root 2. Remember that the electric field is varying sinusoidally. Therefore, E rms is equal to E0 by root 2. This is the rms value. So, we will get half into epsilon 0 into E0 divided by 2 that is equal to 1 by 4 epsilon 0 E 0 square with a like that in that will be equal to 1 by 4 into epsilon 0 is 8.8 into a 10 raised to minus 12 into E 0 square what is E 0 square 20 square so 20 square is a soldier either you know grants it that is electric field energy density so energy density electric field other a league of similarly magnetic field energy density the molecular and ODI magnetic field energy and in city and Nandu DIA other a the question are you B is equal to half 1 by 2 mu 0 into B square LA so B square and the 1 by 2 mu 0 into b square. So b square is 1 by 2 mu 0 into b 0 divided by root 2 square. So this is equal to b 0 square divided by 4 mu 0. We know that after connecting, c is equal to E0 divided by B0 which implies B0 is equal to E0 divided by C which is equal to E0 is nothing but 20. 20 divided by 3 into 10 to the power 8. So we can substitute the value of B0 and UB and add both of them to get the total energy density. That is the aim here. If you complete it, the length of the video will increase. I have explained the logic. Ready? Next two steps. In an electromagnetic wave, the electric and magnetic fields are vibrating along the directions i plus j by root 2. is electric field I by root 2 plus J by root 2 this is the direction of electric field vector and what is the direction of magnetic field vector that is equal to I by root 2 minus J by root 2 the direction of propagation is the direction of propagation of an electromagnetic wave is given by E vector cross B vector that is always E cross B E cross B means we just have to cross this cross these two things then we can get the direction of propagation of E cross B. So, I by root 2 plus J by root 2 cross I by root 2 minus J by root 2. So, let's see the cross product of the coefficient. Cross product. You know cross product, right? If we see cross product, we will get the answer. So, here I cross. So, here i cross i is 0 and i cross j is 0. So, that term is 0. Minus i cross j is k. So, k cap divided by 2. We can do it by taking coefficient. Not only i and j. Yes, we can do it like this. So, the coefficient of this is 1 by root 2, coefficient is 1 by root 2. And here the coefficient is 1 k there. There is no coefficient. 0 0. So, this is i into this minus this. So, 0. Minus j into this minus this. 0. plus k into k into 1 by root 2 into minus 1 by root 2 that is minus 1 by 2 lh 1 by root 2 into minus 1 by root 2 it is minus 1 by 2 yes minus 1 by 2 minus 1 by root 2 into 1 by root 2 that is equal to 1 by 2 that is our answer is minus 1 by 2 minus 1 by 2 minus 2 in the room minus 1 by 2 minus 1 by 2 is minus 2 by 2 minus 1 so minus k cap so answer is minus Sarah minus Sarah minus Sarah minus are in the minus minus 1 k cap is our direction of propagation. Simple. These are connected questions. Remember them. So, cross product or direction of propagation of electromagnetic wave is along E cross B. You can try that. Next question. Light with an energy flux. Energy flux means intensity. Intensity is equal to 18 watt per centimeter square falls on a non-reflecting surface at the normal incidence. If the surface has an area of 20 centimeter square, find the average force exerted on the surface during first 30 minutes Yes so we have I is equal to E by 80 This is the equation of intensity. Intensity is equal to energy by area into time We can see this in JEE. So we have energy, let's change this. so this is equal to power by area. Because energy by time is power, right? So, power is equal to force into velocity divided by area. force into velocity divided by area. So we have this equation. Now what we have been asked is, what is the force at normal incidence. So we have I is equal to force into velocity divided by area. So force is equal to Ia divided by V. So we have I is 18 watt per centimeter square into area is 20 centimeter square. Don't change it again because there is already centimeter square here. So, if we take centimeter square here, it will be cancelled. Divided by velocity is 3 into 10 raised to 8. It is light. So, its speed is 36.18. 6 into 2 is 120. Into 10 to the power minus 8 will be that much Newton. So the speed is 36.18 into 2 is 120 into 10 to the power minus 8 is the Newton. So, the answer is very direct average force is given. Next question. In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency 2 into 10 raised to 10 hertz and amplitude 48 volt per meter then the amplitude of oscillating magnetic field is amplitude of oscillating magnetic field is what we have asked then magnetic field in the oscillation so Electric field amplitude is 48 volt per meter. What is amplitude of oscillation of magnetic field? We have asked B0. So we have C is equal to E0 divided by B0 that is equal to 48 divided by B0. So we have B0 is equal to E0 divided by C that is equal to 48 divided by 3 into 10 to the power 8 that is equal to 48. So 16 into 10 to the power minus 8 or 1.6 into 10 to the power minus 7 Tesla. Answer is 1.6 x 10 to the power minus 7 Tesla. Nothing. The only thing that is given in this is the answer. It is a question that can be done very simply. Next question. When light propagates through a material medium of relative permittivity epsilon r and relative permeability mu r the velocity of light V is given by okay C is equal to 1 by root of mu 0 epsilon 0 V is equal to 1 by root of mu epsilon that is equal to 1 by root of mu 0 mu r into epsilon 0 epsilon r clearly either in the The equation we know. c is equal to 1 by root mu0 epsilon 0. The velocity of the median is 1 by root mu epsilon. mu is mu0 mu r. epsilon is epsilon 0 epsilon r. Therefore, if we write this split, 1 by root mu0 epsilon 0 into root mu r epsilon r. Who is this guy? 1 by root mu0 epsilon 0. So it is C divided by root mu0 epsilon 0 into root mu r epsilon r. 1 by root mu0 epsilon 0 it is c divided by root of mu r epsilon r. Therefore, v is equal to c by root of mu r epsilon r. Option d is our correct answer. Easy question. Very simple question. Next question. For a plane electromagnetic wave propagating in X direction which one of the following combination gives the correct possible direction of electric field and magnetic field? So wave is propagating in X direction. So let us look at two things. Let us look at the option B. In this electric field and magnetic field direction. This is the electric field Let's see the option B. The direction of the magnetic field of the electric field is the same as the magnetic field of the electric field. Let's check one thing. What is the direction of the electromagnetic wave E cross B? Let's check whether it is coming from X. Let's check whether it is coming from X. Let's take the option B. It is J plus k and j plus k. Cross this. I have crossed this. So what will happen? j cross j is 0. j cross k is i. i and plus k cross j is minus i and k cross a is 0. So the answer is zero. So the answer is zero. Now let's take this option. This is b. Cross this. minus j plus k cross minus j minus k. Let's see. Minus j. Please understand this clearly. I will do it on a big screen. Minus j plus k cross minus j minus k, okay. Right? Minus j plus k cross minus j minus k. Look, minus j into minus j, or minus j cross minus j, it is 0. j cross j is 0. Plus minus j cross minus k. Look, minus j cross minus k. So, minus of, what will come? j cross k. j cross k is i. Minus i. Plus, no, minus minus is plus i. Minus j cross minus k. So, it is plus plus i. Ready? Now k cross minus j. What is k cross minus j? Minus k cross j. What is k cross j? It is minus i. K cross j is minus i. Plus k cross k. It is 0. So, we have to write minus and then k cross j. k cross j is minus i. k cross j is the opposite of arrow mark. So, i j k. k cross j is the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. That's what we see here is that if we do this, what will happen to us? A person who says i will come. If you do everything else, the rest of the answer will be zero. If we do that, we will get a term called i. Here, here, here, 2i is given, right? Whatever we do, the answer will be zero. So, answer is option c. Option c reckon number cover a correct answer I will next you see light with an average flux of 20 watt per centimeter square. So, I is equal to 20 watt per centimeter square falls on a normal reflecting surface incidence having surface area surface area is equal to 20 centimeter square yes the energy received by the surface during the time span of one minute so energy received during the surface time span so i is equal to e divided by a into t so e is equal to i into a into T so I 20 watt per centimeter square, A is 20 centimeter square and 1 minute is 60 seconds. So answer is very simple. That much joule will come. 6 times 12 is 12 and 12 times 12 is 24. 24 times 10 raised to 3 joule will be our answer. So you can find it out very easily. So very simple equation. Remember I is E by by a T I is called as energy flux I is called as energy flux or it is also called as intensity is energy is area T style okay next was the ratio of contributions made by electric field and magnetic field components to the intensity of electromagnetic wave is it is 1 is to 1 other electric field magnetic field contribution same I reckon it is 1 is to 1 when I can prove yeah number energy density is equal to half epsilon 0 e square on LA and UB UB is is equal to half into mu0 into B square. So, here B is equal to E by C is equal to E by B. Now, we will get b is equal to E by P. In that, B is equal to E by C. C is equal to 1 by root mu0 epsilon0. Using these two equations, we can prove that this is the same. Magnetic field energy density is equal to electric field energy density. We can prove proved the same. It can be proved practically like that, but we don't need to examine it. Here, electric field and magnetic field energy densities are equal. Electric field component and magnetic field component are of the same ratio. 1 is to 1. You can prove it. Using these two equations, you can see if this is connected to this. So, C is equal to E by B or E0 by B0 and C is equal to 1 by root of mu0 epsilon0. If we can get these two equations from this, it will be very easy. Ready? Next question. For a transparent medium, relative permeability and permittivity are given. The velocity of light in this medium is we have C V is equal to C by root of mu R epsilon R if another one to do this little therefore 3 into 10 raised to 8 divided by root of mu R is nothing but 1 point 0 into epsilon R is 1 point 4 4 so we have 1 point 4 4 414, it is root 2. So, after doing that, we will get a particular answer. That is the velocity of light. It is a simple question. So, never forget this equation. Connect it and remember it. Next question. The electric field of a plain electromagnetic wave of amplitude 2 volt per meter varies with the time and propagates along z axis the average energy density of the magnetic field magnetic field are just you can the car none joule per meter cube lana no go and the so average energy density of magnetic field on our decision so electric field in the amplitude 2 volt per meter. Therefore, E0 is equal to 2 V per meter along z axis. It is propagating along z axis. So, we have asked for the magnetic field energy density. u b is equal to 1 by 2 mu 0 b square. So it is equal to, we know b is equal to b 0 divided by root 2. So substitute this, 1 by 2 mu 0 into b 0 square divided by root 2 square, that is 2. So this is equal to 1 by 4 mu 0 b 0 square. So we know that C is equal to E 0 divided by B 0 therefore B 0 is equal to E 0 divided by C so after substituting 1 by 4 mu 0 into B 0 square it is E 0 square divided by C square so B 0 is equal to E to E0 by C. So, on substituting 1 by 4 mu0 into E0 given, E0 is 2 volt per meter, C given, all are substituted. 1 by answer. If there is any problem, please check. I think no mistake is there. But still, please check. There are more chances of making mistakes. Because there are many such things. The reason why I'm not doing all these is because there is no need to increase the length of the video. This is for you to practice. I have reached here to the full answer and I have not got any match. Neither do you you. Because these are all max. If you practice max, you will get something in the exam. If you do this, there is no point in doing it in the same way. If you do this, you will think, what to do next? What is the next step? Only then you will have a point. Next question. An electromagnetic wave is propagating in a medium with a velocity v is equal to v i cap. The instantaneous oscillating electric field of this electromagnetic field is along positive i axis. Electric field is along positive i axis. Then the direction of oscillating magnetic field of electromagnetic wave will be. The electromagnetic wave is propagating along x axis. Question is asked because, because it is a VI cap, its velocity vector is along x-axis. The electric field is along y-axis. Naturally, the magnetic field is on z-axis. It is along positive z direction. There is no doubt. We can easily get the answer. Next question. In an electromagnetic wave in free space space the root mean square value of electric field is 6 volt per meter the peak value of magnetic field is okay ERMS is equal to e 0 divided by root 2 therefore e 0 is equal to root 2 into ERMS that is equal to root 2 into 6 that is equal to 6 root 2. Therefore, we have C is equal to E0 divided by B0 therefore, B0 is equal to E0 divided by C. So, 6 root 2 divided by 3 into 10 to the power 8 that is equal to 2 root 2 into 10 raised to minus 8. So we are asked the peak value of magnetic field. 2 root 2 into 10 raised to minus 8. Root 2 is 1.414 so answer is 2.83 into 10 raised to minus 8. Simple. Try that. That's all. at the momentum and so like a time it okay next question out of the following options which one can be used to produce a propagating electromagnetic wave electromagnetic wave can be produced by a charge less particle no way a charge moving at a constant velocity no way accelerate and motion is needed a stationary charge in away an accelerating charge can only produce electromagnetic wave. That's a very direct theory question. Next question, when a charged particle moves in a circular path, it produces electromagnetic wave. Correct answer. Because charged particle in circular motion is an accelerated motion. So it has acceleration. Reason is also correct. Both assertion and reason are two and reason is the correct explanation of assertion. Simple question. Option A is correct. Next question. If E and B are the electric and magnetic field vectors of electromagnetic wave, then the direction of propagation of electromagnetic wave is along E cross B. Simple. Electric field and magnetic field are given. Direction of propagation of electromagnetic wave is along E cross B. Simple question. you seen uh light with the energy flux uh 25 into 10 raised to 4 watt per meter square falls on a perfectly reflecting surface e perfectly reflecting surface if the surface area is given the average force exerted so the same equation i is equal to e divided by 80. That is equal to P divided by A. That is equal to F into V divided by A. These are the key questions. F is equal to I A divided by C. That is equal to I. 25 into. 25 into 10 raised to 4 into A is given, it is area 15 centimeters square. So 15 into 10 raised to, see, here it is in centimeters square. Here it is in meters square, here it is in centimeters square. So it should definitely change. So 15 into 10 raised to minus 4 will come. It should change to centimeters meter. When we convert cm2 to m2, it becomes 10 raised to minus 4 will be divided by 3 into 10 raised to 8. This will be multiplied by 2 because it is a perfectly reflecting surface. When it is a perfectly reflecting surface, it should be multiplied by 2. would be force wear. You can calculate how much force wear it will cause. Next one. AM radio waves. We are asking about the wavelength of radio waves. The wavelength is very high. It is 10 square meter. Microwaves. You know the range. Radio waves, microwaves, infrared, visible light, ultraviolet, X-ray, Gamma rays. The most wavelength is radio waves. The lowest frequency is radio waves. The highest frequency is gamma rays. The lowest wavelength is gamma rays. Ready? This is something that we all know. So, the highest wavelength is 10 square meters. That is radio waves. Then, it is microwaves. So, the value of the microwaves is 10 raised to minus 2 meters. Then we have the infrared. The infrared is 10 raised to minus 4 meters. The X-ray is very less. It is 10 raised to minus 10 meters. Direct answer is A2, B3, C4 and D1. Option 4 is our correct answer. Even though we don't know the range of this, if we have the same idea, we can work out this question. It is very simple. Next question. The energy of electromagnetic wave is of the order of 15 keV. To which part of the spectrum does it belong? So energy is equal to hc by lambda therefore lambda is equal to hc by e. Just connect it as dual nature. hc's value is 1240 electron volt nanometer. Divided by energy is 15 kilo electron volt. 10 raised to 3 electron volt. So electron volt, electron volt cancelled. So this is 1.2i. So 1.2 divided by 15 nanometer. So this is in the range 0.083 nanometer. Now, I am going to show you how to make a beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, 1.2i. So 1.2 divided by 15 nanometer. So this is in the range 0.083 nanometer. When it comes in the nanometer range, the problem is that it is possible to be an X-ray. Because in the nanometer range, if we look at the chart, it is the-ray. So, we have completed all the questions in this. We can upload more questions later. All the best.", + "timestamped_transcription": [ + { + "timestamp": [ + 0.0, + 145.32 + ], + "text": " Hi friends, we are going to discuss some important questions in the chapter called electromagnetic waves. This chapter is very simple and you will feel that even though the content is very less, the chapter that asks good questions is the electromagnetic wave. So never miss it. Ready? A capacitor of capacitance C is connected across a source of voltage V given by V is equal to V0 sine omega T. The displacement current between the plates of the capacitor would be given by... Displacement current is used. So, there is a capacitor like this. That capacitor is connected to a battery. The current Q is flowing through this. So, displacement current is used. So, we need such an equation. So, let need a equation like this. So, we know that I is equal to dQ divided by dt. So, this is equal to d by dt of Q is Cv. So, I is equal to d by dt of Cv. V is the voltage applied, that is, V0 sine omega t. So, d by dt of C into V0 sine omega t. d by dt of Cv0 sine omega t. So, I is equal to let's take Cv0 out. So I is equal to Cv0 into d by dt of sine Omega t. What is d by dt of sine Omega t? That is equal to omega cos omega t. Therefore, I is equal to Cv0 into omega cos omega t. Good equation, isn't it? We should learn this equation altogether. d by dt of Cv. Because, in some questions, we are asked to find I. When we are asked to find I, we can take C out of the equation. We get C into dV divided by dt. So, the rate of change in voltage, dV by dt, will be given in the question. Then we can find I. Then we have to remember this equation. So, we use your be used and this equation should be remembered and the last equation should be remembered." + }, + { + "timestamp": [ + 145.32, + 185.8 + ], + "text": " C V0 omega cos omega t. So, V0 omega C cos omega t. So, this is the equation for displacement current. It is a very simple equation to remember. Next question. A parallel plate capacitor of capacitance 20 microfarads is being charged by a voltage source whose potential is changing at the rate of 3 volt per second. So dv by dt is 3 volt per second. The hint given in the question is dv divided by dt is equal to 3 volt per second. Now, what is given in the question?" + }, + { + "timestamp": [ + 185.8, + 183.0 + ], + "text": " The conduction current through the wire and displacement current. Both are same. Conduction current is same and displacement current is same. In this case, not in this case, but in all the cases, in all the cases, the conduction current is the displacement current. If so, we have given capacitance and this is given. So, which equation should be used? ID is equal to the equation we just learned" + }, + { + "timestamp": [ + 183.0, + 184.0 + ], + "text": " C into" + }, + { + "timestamp": [ + 184.32, + 367.18 + ], + "text": " dv divided by dt. So this is equal to 20 into, see the value of C is 20 microfarad. We need microfarad because we have micro in options. So 20 microfarad into, what is dv by dt? Rate of 3 volt per second. So, 3. So, answer is 20 into 3. That is 60 microampere. So, we got the value of displacement current here. That is equal to 60 microampere. So, what is displacement current? That itself is conduction current. What will come here? It will come equal. So, answer is 60 microampere. Let's move to the nextere. Next question. A plain electromagnetic wave of frequency 25 MHz travels in a free space along x direction. At a particular point in the space and time, E is equal to 0.3 J cap volt per meter. The magnetic field at this point is very simple coming in an equation C is equal to e divided by B that is equal to e is 0.3 LA 0.3 divided by what is C okay I'm gonna magnetic field a little in the so magnetic field B is equal to E divided by C so E is nothing but 0.3 and B C is 3 into 10 raised to 8 so our answer is 0.1 cut either way lay 0.1 into 10 to the power minus 8 Tesla in the cancer to LA 0.1 into 10 raised to minus 8 tesla. So, this is the question we need to apply. Ready? Next question. The magnetic field of a plain electromagnetic wave is given by this equation. This equation is similar to by is equal to b0 sine kx plus Omega t. So, comparing this with this equation, we are taking the propagation constant value k is equal to 2 pi by lambda, therefore lambda is equal to 2 pi divided by k. So, comparing these two equations, k is 0.5 into 10 raised to 3. So, 2 pi divided by K so comparing these two equations K is 0.5 into 10 raised to 3 so 2 pi divided by 0.5 into 10 to the power 3 in the middle it would have been okay 2 pi divided by 0.5 that is equal to 4 pi so 4 pi into 10 to the power minus 3 it is meter a meter so So, for pi into 10 raise to minus 3 meter." + }, + { + "timestamp": [ + 367.18, + 526.0 + ], + "text": " Next question the frequency of electromagnetic wave. So, omega is given as 1.5 into 10 raise to 11. So, omega is equal to we are taking the equation omega is equal to 2 pi f therefore, F is equal to omega divided by 2 pi so here omega is given by 1.5 into 10 to the power 11 divided by, comparing these two equations, divided by 2 pi. That is, this is our frequency, this much hertz. This will be our frequency. We just need to solve that. You just need to solve that. Then, the velocity of electromagnetic wave. Next, we need to find the velocity. How to find the velocity? Velocity V is equal to frequency into wavelength. Therefore frequency is 1.5 into 10 to the power 11 divided by 2 pi into wavelength is 4 pi into 10 to the power minus 3. So we will get free velocity a path of the solver the original velocity to next amplitude of electric field vector okay amplitude am langa and odika we have C is equal to e 0 divided by B 0 so what is e 0 that is amplitude of electric field vector which implies this is our equation. So E0 is equal to B0 into C. So B0 is here 2 into 10 to the power minus 7 into C is here 3 into 10 raised to 8. So our answer is 6 into 10 to the power 1 so 60 answer is 60 so E0 is equal to 60 E0 is 60 Newton per Coulomb E is equal to F by Q then what is direction of this electric field vector The way we wave is along x axis because the coefficient of k is kx. That means in which direction is the electromagnetic wave? It is on x axis. In which axis is the magnetic field? It is on y axis because b is given as y. So the electromagnetic wave is on x axis and the magnetic field is on y axis-axis. The electric field is on the z-axis. So, the electric field is along the z-axis because the direction of the electromagnetic wave is perpendicular to the magnetic field. If we try to understand the the direction of magnetic field vector, then this is the direction of electromagnetic wave. If the electromagnetic wave is on x axis, then the electric field will be along y axis and magnetic field will be along z axis. So this is a concept that we can understand. If you remember a movie like this," + }, + { + "timestamp": [ + 526.0, + 546.0 + ], + "text": " Yes, the electromagnetic baby cell. So, this is the equation or concept that we know." + }, + { + "timestamp": [ + 546.0, + 585.24 + ], + "text": " If you remember the picture, you can see the electromagnetic wave is along the x-axis. So, this is the case that we know. So, that is what we have asked here. And last is the expression for electric field vector. So, electric field vector can be written as e vector is equal to e 0 e 0 is 60 sine K X K is nothing but 0.5 into 10 raised to 3 X I didn't know Mary with the solo to plus Omega t 1.5 into 10 raised to 11 T and K cap LA in a you come here. Okay, so this is the expression for electric field vector." + }, + { + "timestamp": [ + 585.24, + 584.4 + ], + "text": " So we have done everything. Next question. In an electromagnetic wave, the amplitude of electric field vector is 20 V per meter. So E0 is equal to 20 V per meter. The average energy density of electromagnetic wave is average energy density of electromagnetic so total we have to find the total" + }, + { + "timestamp": [ + 584.4, + 605.34 + ], + "text": " okay the average energy density of electromagnetic wave is average energy" + }, + { + "timestamp": [ + 605.34, + 647.76 + ], + "text": " density of electromagnetic wave. So total we have to find the total. So let us know one thing. Let us find out the electric field energy. field energy is uE is equal to half epsilon zero into E square. E is equal to E0 by root 2. That is equal to half epsilon zero into E0 divided by root 2 whole square. This is E rms and this is the rms value. That is equal to E0 divided by" + }, + { + "timestamp": [ + 647.76, + 786.0 + ], + "text": " root 2. Remember that the electric field is varying sinusoidally. Therefore, E rms is equal to E0 by root 2. This is the rms value. So, we will get half into epsilon 0 into E0 divided by 2 that is equal to 1 by 4 epsilon 0 E 0 square with a like that in that will be equal to 1 by 4 into epsilon 0 is 8.8 into a 10 raised to minus 12 into E 0 square what is E 0 square 20 square so 20 square is a soldier either you know grants it that is electric field energy density so energy density electric field other a league of similarly magnetic field energy density the molecular and ODI magnetic field energy and in city and Nandu DIA other a the question are you B is equal to half 1 by 2 mu 0 into B square LA so B square and the 1 by 2 mu 0 into b square. So b square is 1 by 2 mu 0 into b 0 divided by root 2 square. So this is equal to b 0 square divided by 4 mu 0. We know that after connecting, c is equal to E0 divided by B0 which implies B0 is equal to E0 divided by C which is equal to E0 is nothing but 20. 20 divided by 3 into 10 to the power 8. So we can substitute the value of B0 and UB and add both of them to get the total energy density. That is the aim here. If you complete it, the length of the video will increase. I have explained the logic. Ready? Next two steps. In an electromagnetic wave, the electric and magnetic fields are vibrating along the directions i plus j by root 2. is electric field I by root 2 plus J by root 2 this is the direction of electric field vector and what is the direction of magnetic field vector that is equal to I by root 2 minus J by root 2 the direction of propagation is the direction of propagation of an electromagnetic wave is given by" + }, + { + "timestamp": [ + 786.0, + 966.5 + ], + "text": " E vector cross B vector that is always E cross B E cross B means we just have to cross this cross these two things then we can get the direction of propagation of E cross B. So, I by root 2 plus J by root 2 cross I by root 2 minus J by root 2. So, let's see the cross product of the coefficient. Cross product. You know cross product, right? If we see cross product, we will get the answer. So, here I cross. So, here i cross i is 0 and i cross j is 0. So, that term is 0. Minus i cross j is k. So, k cap divided by 2. We can do it by taking coefficient. Not only i and j. Yes, we can do it like this. So, the coefficient of this is 1 by root 2, coefficient is 1 by root 2. And here the coefficient is 1 k there. There is no coefficient. 0 0. So, this is i into this minus this. So, 0. Minus j into this minus this. 0. plus k into k into 1 by root 2 into minus 1 by root 2 that is minus 1 by 2 lh 1 by root 2 into minus 1 by root 2 it is minus 1 by 2 yes minus 1 by 2 minus 1 by root 2 into 1 by root 2 that is equal to 1 by 2 that is our answer is minus 1 by 2 minus 1 by 2 minus 2 in the room minus 1 by 2 minus 1 by 2 is minus 2 by 2 minus 1 so minus k cap so answer is minus Sarah minus Sarah minus Sarah minus are in the minus minus 1 k cap is our direction of propagation. Simple. These are connected questions. Remember them. So, cross product or direction of propagation of electromagnetic wave is along E cross B. You can try that. Next question. Light with an energy flux. Energy flux means intensity. Intensity is equal to 18 watt per centimeter square falls on a non-reflecting surface at the normal incidence. If the surface has an area of 20 centimeter square, find the average force exerted on the surface during first 30 minutes" + }, + { + "timestamp": [ + 966.5, + 994.46 + ], + "text": " Yes so we have I is equal to E by 80 This is the equation of intensity. Intensity is equal to energy by area into time We can see this in JEE. So we have energy, let's change this. so this is equal to power by area. Because energy by time is power, right? So, power is equal to force into velocity divided by area." + }, + { + "timestamp": [ + 985.76, + 1065.62 + ], + "text": " force into velocity divided by area. So we have this equation. Now what we have been asked is, what is the force at normal incidence. So we have I is equal to force into velocity divided by area. So force is equal to Ia divided by V. So we have I is 18 watt per centimeter square into area is 20 centimeter square. Don't change it again because there is already centimeter square here. So, if we take centimeter square here, it will be cancelled. Divided by velocity is 3 into 10 raised to 8. It is light. So, its speed is 36.18. 6 into 2 is 120. Into 10 to the power minus 8 will be that much Newton. So the speed is 36.18 into 2 is 120 into 10 to the power minus 8 is the Newton. So, the answer is very direct average force is given. Next question. In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency 2 into 10 raised to 10 hertz and amplitude 48 volt per meter then the amplitude of oscillating magnetic field is amplitude of oscillating magnetic field is what we have asked then magnetic field in the oscillation so Electric field amplitude is 48 volt per meter." + }, + { + "timestamp": [ + 1065.62, + 1305.36 + ], + "text": " What is amplitude of oscillation of magnetic field? We have asked B0. So we have C is equal to E0 divided by B0 that is equal to 48 divided by B0. So we have B0 is equal to E0 divided by C that is equal to 48 divided by 3 into 10 to the power 8 that is equal to 48. So 16 into 10 to the power minus 8 or 1.6 into 10 to the power minus 7 Tesla. Answer is 1.6 x 10 to the power minus 7 Tesla. Nothing. The only thing that is given in this is the answer. It is a question that can be done very simply. Next question. When light propagates through a material medium of relative permittivity epsilon r and relative permeability mu r the velocity of light V is given by okay C is equal to 1 by root of mu 0 epsilon 0 V is equal to 1 by root of mu epsilon that is equal to 1 by root of mu 0 mu r into epsilon 0 epsilon r clearly either in the The equation we know. c is equal to 1 by root mu0 epsilon 0. The velocity of the median is 1 by root mu epsilon. mu is mu0 mu r. epsilon is epsilon 0 epsilon r. Therefore, if we write this split, 1 by root mu0 epsilon 0 into root mu r epsilon r. Who is this guy? 1 by root mu0 epsilon 0. So it is C divided by root mu0 epsilon 0 into root mu r epsilon r. 1 by root mu0 epsilon 0 it is c divided by root of mu r epsilon r. Therefore, v is equal to c by root of mu r epsilon r. Option d is our correct answer. Easy question. Very simple question. Next question. For a plane electromagnetic wave propagating in X direction which one of the following combination gives the correct possible direction of electric field and magnetic field? So wave is propagating in X direction. So let us look at two things. Let us look at the option B. In this electric field and magnetic field direction. This is the electric field Let's see the option B. The direction of the magnetic field of the electric field is the same as the magnetic field of the electric field. Let's check one thing. What is the direction of the electromagnetic wave E cross B? Let's check whether it is coming from X. Let's check whether it is coming from X. Let's take the option B. It is J plus k and j plus k. Cross this. I have crossed this. So what will happen? j cross j is 0. j cross k is i. i and plus k cross j is minus i and k cross a is 0. So the answer is zero. So the answer is zero. Now let's take this option. This is b. Cross this. minus j plus k cross minus j minus k. Let's see. Minus j. Please understand this clearly. I will do it on a big screen. Minus j plus k cross minus j minus k, okay. Right? Minus j plus k cross minus j minus k. Look, minus j into minus j, or minus j cross minus j, it is 0. j cross j is 0. Plus minus j cross minus k." + }, + { + "timestamp": [ + 1305.5, + 1326.0 + ], + "text": " Look, minus j cross minus k. So, minus of, what will come? j cross k. j cross k is i. Minus i. Plus, no, minus minus is plus i. Minus j cross minus k. So, it is plus plus i. Ready? Now k cross minus j. What is k cross minus j? Minus k cross j. What is k cross j? It is minus i. K cross j is minus i. Plus k cross k. It is 0. So, we have to write minus and then k cross j. k cross j is minus i." + }, + { + "timestamp": [ + 1326.0, + 1329.0 + ], + "text": " k cross j is the opposite of arrow mark." + }, + { + "timestamp": [ + 1329.0, + 1332.0 + ], + "text": " So, i j k." + }, + { + "timestamp": [ + 1332.0, + 1605.22 + ], + "text": " k cross j is the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. So, we have to find out the answer to this question. That's what we see here is that if we do this, what will happen to us? A person who says i will come. If you do everything else, the rest of the answer will be zero. If we do that, we will get a term called i. Here, here, here, 2i is given, right? Whatever we do, the answer will be zero. So, answer is option c. Option c reckon number cover a correct answer I will next you see light with an average flux of 20 watt per centimeter square. So, I is equal to 20 watt per centimeter square falls on a normal reflecting surface incidence having surface area surface area is equal to 20 centimeter square yes the energy received by the surface during the time span of one minute so energy received during the surface time span so i is equal to e divided by a into t so e is equal to i into a into T so I 20 watt per centimeter square, A is 20 centimeter square and 1 minute is 60 seconds. So answer is very simple. That much joule will come. 6 times 12 is 12 and 12 times 12 is 24. 24 times 10 raised to 3 joule will be our answer. So you can find it out very easily. So very simple equation. Remember I is E by by a T I is called as energy flux I is called as energy flux or it is also called as intensity is energy is area T style okay next was the ratio of contributions made by electric field and magnetic field components to the intensity of electromagnetic wave is it is 1 is to 1 other electric field magnetic field contribution same I reckon it is 1 is to 1 when I can prove yeah number energy density is equal to half epsilon 0 e square on LA and UB UB is is equal to half into mu0 into B square. So, here B is equal to E by C is equal to E by B. Now, we will get b is equal to E by P. In that, B is equal to E by C. C is equal to 1 by root mu0 epsilon0. Using these two equations, we can prove that this is the same. Magnetic field energy density is equal to electric field energy density. We can prove proved the same. It can be proved practically like that, but we don't need to examine it. Here, electric field and magnetic field energy densities are equal. Electric field component and magnetic field component are of the same ratio. 1 is to 1. You can prove it. Using these two equations, you can see if this is connected to this. So, C is equal to E by B or E0 by B0 and C is equal to 1 by root of mu0 epsilon0. If we can get these two equations from this, it will be very easy. Ready? Next question. For a transparent medium, relative permeability and permittivity are given. The velocity of light in this medium is we have C V is equal to C by root of mu R epsilon R if another one to do this little therefore 3 into 10 raised to 8 divided by root of mu R is nothing but 1 point 0 into epsilon R is 1 point 4 4 so we have 1 point 4 4 414, it is root 2. So, after doing that, we will get a particular answer. That is the velocity of light. It is a simple question. So, never forget this equation. Connect it and remember it. Next question. The electric field of a plain electromagnetic wave of amplitude 2 volt per meter varies with" + }, + { + "timestamp": [ + 1605.22, + 1746.0 + ], + "text": " the time and propagates along z axis the average energy density of the magnetic field magnetic field are just you can the car none joule per meter cube lana no go and the so average energy density of magnetic field on our decision so electric field in the amplitude 2 volt per meter. Therefore, E0 is equal to 2 V per meter along z axis. It is propagating along z axis. So, we have asked for the magnetic field energy density. u b is equal to 1 by 2 mu 0 b square. So it is equal to, we know b is equal to b 0 divided by root 2. So substitute this, 1 by 2 mu 0 into b 0 square divided by root 2 square, that is 2. So this is equal to 1 by 4 mu 0 b 0 square. So we know that C is equal to E 0 divided by B 0 therefore B 0 is equal to E 0 divided by C so after substituting 1 by 4 mu 0 into B 0 square it is E 0 square divided by C square so B 0 is equal to E to E0 by C. So, on substituting 1 by 4 mu0 into E0 given, E0 is 2 volt per meter, C given, all are substituted. 1 by answer. If there is any problem, please check. I think no mistake is there. But still, please check. There are more chances of making mistakes. Because there are many such things. The reason why I'm not doing all these is because there is no need to increase the length of the video. This is for you to practice. I have reached here to the full answer and I have not got any match. Neither do you you. Because these are all max." + }, + { + "timestamp": [ + 1746.0, + 1749.0 + ], + "text": " If you practice max, you will get something in the exam." + }, + { + "timestamp": [ + 1749.0, + 1752.0 + ], + "text": " If you do this, there is no point in doing it in the same way." + }, + { + "timestamp": [ + 1752.0, + 1755.0 + ], + "text": " If you do this, you will think, what to do next?" + }, + { + "timestamp": [ + 1755.0, + 1758.0 + ], + "text": " What is the next step?" + }, + { + "timestamp": [ + 1758.0, + 1761.0 + ], + "text": " Only then you will have a point." + }, + { + "timestamp": [ + 1761.0, + 1906.2 + ], + "text": " Next question. An electromagnetic wave is propagating in a medium with a velocity v is equal to v i cap. The instantaneous oscillating electric field of this electromagnetic field is along positive i axis. Electric field is along positive i axis. Then the direction of oscillating magnetic field of electromagnetic wave will be. The electromagnetic wave is propagating along x axis. Question is asked because, because it is a VI cap, its velocity vector is along x-axis. The electric field is along y-axis. Naturally, the magnetic field is on z-axis. It is along positive z direction. There is no doubt. We can easily get the answer. Next question. In an electromagnetic wave in free space space the root mean square value of electric field is 6 volt per meter the peak value of magnetic field is okay ERMS is equal to e 0 divided by root 2 therefore e 0 is equal to root 2 into ERMS that is equal to root 2 into 6 that is equal to 6 root 2. Therefore, we have C is equal to E0 divided by B0 therefore, B0 is equal to E0 divided by C. So, 6 root 2 divided by 3 into 10 to the power 8 that is equal to 2 root 2 into 10 raised to minus 8. So we are asked the peak value of magnetic field. 2 root 2 into 10 raised to minus 8. Root 2 is 1.414 so answer is 2.83 into 10 raised to minus 8. Simple. Try that. That's all. at the momentum and so like a time it okay next question out of the following options which one can be used to produce a propagating electromagnetic wave electromagnetic wave can be produced by a charge less particle no way a charge moving at a constant velocity no way accelerate and motion is needed a stationary charge in away an accelerating charge can only produce electromagnetic wave. That's a very direct theory question. Next question, when a charged particle moves in a circular path, it produces electromagnetic wave. Correct answer. Because charged particle in circular motion is an accelerated motion. So it has acceleration. Reason is also correct. Both assertion and reason are two and reason is the correct explanation of assertion. Simple question. Option A is correct." + }, + { + "timestamp": [ + 1907.34, + 2085.08 + ], + "text": " Next question. If E and B are the electric and magnetic field vectors of electromagnetic wave, then the direction of propagation of electromagnetic wave is along E cross B. Simple. Electric field and magnetic field are given. Direction of propagation of electromagnetic wave is along E cross B. Simple question. you seen uh light with the energy flux uh 25 into 10 raised to 4 watt per meter square falls on a perfectly reflecting surface e perfectly reflecting surface if the surface area is given the average force exerted so the same equation i is equal to e divided by 80. That is equal to P divided by A. That is equal to F into V divided by A. These are the key questions. F is equal to I A divided by C. That is equal to I. 25 into. 25 into 10 raised to 4 into A is given, it is area 15 centimeters square. So 15 into 10 raised to, see, here it is in centimeters square. Here it is in meters square, here it is in centimeters square. So it should definitely change. So 15 into 10 raised to minus 4 will come. It should change to centimeters meter. When we convert cm2 to m2, it becomes 10 raised to minus 4 will be divided by 3 into 10 raised to 8. This will be multiplied by 2 because it is a perfectly reflecting surface. When it is a perfectly reflecting surface, it should be multiplied by 2. would be force wear. You can calculate how much force wear it will cause. Next one. AM radio waves. We are asking about the wavelength of radio waves. The wavelength is very high. It is 10 square meter. Microwaves. You know the range. Radio waves, microwaves, infrared, visible light, ultraviolet, X-ray, Gamma rays. The most wavelength is radio waves. The lowest frequency is radio waves. The highest frequency is gamma rays. The lowest wavelength is gamma rays. Ready? This is something that we all know. So, the highest wavelength is 10 square meters. That is radio waves. Then, it is microwaves. So, the value of the microwaves is 10 raised to minus 2 meters. Then we have the infrared. The infrared is 10 raised to minus 4 meters. The X-ray is very less. It is 10 raised to minus 10 meters. Direct answer is A2, B3, C4 and D1. Option 4 is our correct answer. Even though we don't know the range of this, if we have the same idea, we can work out this question. It is very simple. Next question. The energy of electromagnetic wave is of the order of 15 keV. To which part of the spectrum does it belong? So energy is equal to hc by lambda therefore lambda is equal to hc by e." + }, + { + "timestamp": [ + 2085.42, + 2086.38 + ], + "text": " Just connect it as dual nature." + }, + { + "timestamp": [ + 2090.12, + 2090.56 + ], + "text": " hc's value is 1240 electron volt nanometer." + }, + { + "timestamp": [ + 2093.72, + 2093.9 + ], + "text": " Divided by energy is 15 kilo electron volt." + }, + { + "timestamp": [ + 2095.18, + 2095.52 + ], + "text": " 10 raised to 3 electron volt." + }, + { + "timestamp": [ + 2097.04, + 2097.7 + ], + "text": " So electron volt, electron volt cancelled." + }, + { + "timestamp": [ + 2101.1, + 2101.56 + ], + "text": " So this is 1.2i." + }, + { + "timestamp": [ + 2105.08, + 2083.12 + ], + "text": " So 1.2 divided by 15 nanometer. So this is in the range 0.083 nanometer. Now, I am going to show you how to make a" + }, + { + "timestamp": [ + 2083.12, + 2085.28 + ], + "text": " beautiful, beautiful, beautiful, beautiful," + }, + { + "timestamp": [ + 2085.28, + 2087.68 + ], + "text": " beautiful, beautiful, beautiful, beautiful, beautiful," + }, + { + "timestamp": [ + 2087.68, + 2089.68 + ], + "text": " beautiful, beautiful, beautiful, beautiful, beautiful," + }, + { + "timestamp": [ + 2089.68, + 2091.68 + ], + "text": " beautiful, beautiful, beautiful, beautiful, beautiful," + }, + { + "timestamp": [ + 2091.68, + 2093.68 + ], + "text": " beautiful, beautiful, beautiful, beautiful," + }, + { + "timestamp": [ + 2093.68, + 2095.68 + ], + "text": " beautiful, beautiful, beautiful, beautiful," + }, + { + "timestamp": [ + 2095.68, + 2097.68 + ], + "text": " beautiful, beautiful, beautiful, beautiful," + }, + { + "timestamp": [ + 2097.68, + 2099.68 + ], + "text": " beautiful, beautiful, beautiful, beautiful," + }, + { + "timestamp": [ + 2099.68, + 2101.68 + ], + "text": " beautiful, beautiful, beautiful, beautiful," + }, + { + "timestamp": [ + 2101.68, + 2103.68 + ], + "text": " beautiful, beautiful, beautiful," + }, + { + "timestamp": [ + 2103.68, + 2128.0 + ], + "text": " beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, beautiful, 1.2i. So 1.2 divided by 15 nanometer. So this is in the range 0.083 nanometer. When it comes in the nanometer range, the problem is that it is possible to be an X-ray. Because in the nanometer range, if we look at the chart, it is the-ray. So, we have completed all the questions in this." + }, + { + "timestamp": [ + 2128.0, + 2132.0 + ], + "text": " We can upload more questions later." + }, + { + "timestamp": [ + 2132.0, + 2136.0 + ], + "text": " All the best." + } + ] + } +] \ No newline at end of file