 So today I will be taking up previous year JEE questions which are of mains level. So effectively I am taking up all those questions which have single option correct and they have come in JEE for like I think last 40 years JEE was there. So all those questions we are planning to take up today. Okay, so let us start. Others will join in. You will see that apart from mechanics these chapters lot of theoretical questions are also asked. Okay, so at times our complete focus is on numericals but then 30 or 40% of time a theoretical question comes. Okay, so I will just put few questions. You can start solving them. So this is the first one, this is the second one. Okay, Pulig is saying first one is the first option. See refractive index is defined as speed of light divided by speed of light in that medium. Okay, so here we know that refractive index is greater than 1, right? Because glass refractive index is more than air's refractive index, okay? So velocity has to be less than speed of light, okay? And frequency should be constant, okay? Frequency is what? Velocity divided by wavelength. So V1 by lambda 1 should be equal to V2 by lambda 2, okay? So lambda 2 by lambda 1 is equal to V2 by V1, okay? So basically if velocity is decreasing, wavelength will also decrease because their ratios are same, okay? So V2 by V1 is equal to lambda 2 by lambda 1. So that is why since velocity is decreasing, wavelength will also decrease, okay? Second one, we have fringe width defined as lambda d by small d, right? So when the separation between the slit is halved. So basically d is becoming d by 2, okay? And the distance between the slits and the screen is double. So capital D is 2 times the earlier d and d is becoming d0 by 2, okay? So you have fringe width 4 times, all right? So you can see that straightforward questions have come from this chapter, all right? So we'll quickly move to next one. In case you have any doubts, stop me and ask doubts, okay? Don't just sit quite. And I hope you are watching this on your laptop or desktop and not on your mobile phone because readability will be an issue if you watch it on a small screen. Do these two questions, okay? We have question number 4, answer, minus 1.5 d, that is correct. Third one, so like always you should attack the easiest question first. It doesn't matter in what sequence you are getting the questions. It is good that you have identified question number 4 is easier and you have taken it first, okay? So convex lens of focal length, 40 centimeters. So focal length is plus 40 because it is convex. And for concave, focal length is 25. So it is concave, so focal length is negative, fine? The power is 1 by F1 plus 1 by F2. But these F1 and F2 should be in meters. So this is 0.4 meter and this is minus of 0.25 meters. So when you substitute, you'll get minus 1.5 d. Third one, what is the answer? Okay, let us see how to solve the third question. So I will first draw this triangle, fine? So this is the triangle and then you have incident ray coming like this, okay? This is the incident ray and this ray has to reach side BC, okay? This ray will reach the side BC only when it gets totally internally reflected or if it graces the surface. So if it just graces the surface and travels like this, then also, you know, you can say that it has reached the surface BC, otherwise there is no other way it can reach, all right? So this is the limiting condition that should happen, at least this should happen, okay? So at least this much angle of incidence should be there, beyond this, your angle of incidence could be anything, okay? Now let us see what is the relation between angle of incidence and angle theta, fine? So we have this as normal, okay? This is the normal, so angle of incidence is what? Angle of incidence is this, okay? Let's call it I, angle of refraction is 90 degree, okay? So if this is theta, what is the relation between theta and I, angle theta is equal to angle I? Yes or no? This is 90 minus theta, here it is 90 degree, okay? So I is anyway 90 minus this angle which is 91 theta, so I becomes equal to theta, right? So since I is equal to angle theta and you can have any angle beyond angle I for it to have total internal refraction. So basically you are saying this theta can be any angle beyond, you know, whatever is critical angle over here, okay? So whatever is I is same as whatever is theta. So let's find out what is the angle I for this limiting condition which is also equal to critical angle, okay? So we have refractive index of prism 1.5 and the water has refractive index 4 by 3. So I can say that 1.5 which is 3 by 2 sin of I should be equal to 4 by 3 sin of 90 degrees, okay? So I will get sin of I to be equal to 8 by 9, okay? So basically I am getting I is equal to sin inverse 8 by 9. So my angle should be more than sin inverse 8 by 9, okay? And angle of incidence cannot be more than 90 degrees, okay? So the if angle is more than sin inverse 8 by 9, I can as well say sin of that angle is more than 8 by 9, okay? Because sin is increasing function between 0 and 90 degrees. So that is the reason why option 1 is correct over here, both option 1 is correct, okay? Any doubt on these two questions? Is it clear? Type in yes or no. I will move to next one. I have 70 questions. I don't know how many we will be able to solve today. Let's aim to solve all of them, okay? So you are solving the sixth one initially. So we know, right, that intensity of two interfering waves which have intensity I1 and I2 is given by this relation where phi is the phase difference, okay? So intensity, this resulted intensity will be maximum when cos of phi is equal to 1, okay? If cos of phi is equal to 1, this will be equal to root of phi 1 plus root of I2 whole square, okay? This is max and minimum will be when cos of phi will be equal to minus 1, which will be root I1 minus root of I2 whole square, okay? So this will come out to be 9 times I and this will come out to be I. Anyone got the fifth one? Okay, fifth, saying first one. Let's see question number five. If you've got question number five, you can move on to the sixth one in case you have not attempted that. Okay, question number five, let me solve here. First of all, angle I will be equal to angle R because that are, you know, incident and reflected rays. And if this is 90 degree, if this is 90 degree, then I have R plus R dash plus 90 degrees equals to 180 degrees, okay? So this gives me R plus R dash is equal to 90 degrees, okay? And R is equal to I, so basically I plus R dash is equal to 90 degrees, okay? So this is first equation. And then you can apply the Snell's law, okay? So let's say this is mu 1 and this is mu 2, so mu 1 sine of I is equal to mu 2 sine of R dash. Now R dash is 90 minus I, so I can write here 90 minus I, okay? So what I'm getting here is tan of I to be equal to mu 2 by mu 1, okay? So this is second equation. Sine of 90 minus I is equal to cos I, okay? So when you take cos I below sine I become tan I, so tan I is mu 2 by mu 1. Now at a critical angle, angle of refraction becomes 90 degrees, fine? So I can say that at critical angle of refraction you have mu 1 sine of critical angle to be equal to mu 2 sine of 90 degrees, fine? So sine of critical angle will become equal to mu 2 by mu 1, okay? And mu 2 by mu 1 is what? Okay? Are you able to hear me? All right, so that is the reason why option 1 is correct because I is also equal to R, fine? Any doubt, guys? All right, start solving these questions. Can a ray of light moving from denser to rarer get refracted? Yes, every time part of light will get reflected and part of light will get refracted, okay? That is always true. Doesn't matter what happens, you know? You may want to just study the reflection or you may just want to study the refraction of particular ray, okay? But in reality, both will happen. Only when it is total internal reflection, then yes, 100% reflection happens. Otherwise, partly will get refracted, partly will get reflected. All right, solve these questions. Now you will not be able to solve question number 7 unless you have, you know, done something like this before, okay? So I suggest in case this type of question come in exam, you just leave it and move to the next question, okay? So I made why 8th question, first one is correct. Okay, so many of you are getting option 4 to be correct for question number 7, all right? So now first of all, you have to pay attention to the wordings of the question, okay? So let me just draw the diagram first. So this is a concave mirror, all right? So this is a concave mirror and you have a shot. So here it is written short linear object. So the size of the object along the principal axis is negligible, all right? And it is given that it has very small length B, okay? It is at a distance U from the pole. Now what is given is that object is so small that it looks like a point object and its distance is given as U from the pole. So here is this object, okay? This object's distance is U, okay? Now at times you will get confused that whether the distance is given from the front of the object or from the back of the object, it doesn't matter, it's a short linear object. So you ignore its dimension when you calculate its distance from the pole, okay? Now we need to find the size of the image, okay? So we know that we have this mirror formula 1 by V plus 1 by U is equal to 1 by F, okay? Now if I differentiate this, okay? I will get 1 by V square dV minus 1 by U square dU is equal to 0, okay? Focal length is a constant. Now why I am doing like this? Because dU represents a very small deviation in object distance. So dU can be equal to B, right? So dV will represent the size of the image, okay? Which will be minus of V square by U square into dU, which is B, fine? Now the answer is not in terms of V. In the answer F and U are there. So simply you find out the V from here and substitute there, fine? So if you do that, you will get 1 by V or let me directly write V, V to be equal to UF minus F minus U, right? So dV is equal to minus of U square F square divided by F minus U whole square into U square into B. So U square will get cancelled, okay? So you have F square divided by F minus U whole square into B, okay? So seventh one, actually the option there is a correction which I have intentionally kept it like that. It should be square over here then only option 4 is correct, okay? Fine? Now please attempt question number 8. See when this light hit here, it will have angle of incidence of 45 degrees after differentiation. See once you differentiate, you will get dV in terms of dU, dV is V by U whole square into dU. dU is size of the object, which is B and then I substitute value of V from the mirror formula, okay? Now let's go back to question number 8. So you have angle of incidence equals to 45 degrees here, okay? So if angle of incidence is 45 degrees, then it may happen that some wavelength may get totally internally reflected, fine? So let's see whether that will happen here. So we have refractive index of the prism, let us say is mu and this is air which has refractive index as 1, okay? So mu into sine of 45 degrees, if this wavelength which has this refractive index is about to go under total internal reflection, then at least this should be, you know, if it is getting totally internally reflected, then mu sine 45 has to be greater than 1 into sine of 90 degrees, right? So mu has to be greater than 1 by sine 45 which is root 2. So if mu is greater than 1.414, then total internal reflection will happen, fine? So you can see that these two colors have refractive index more than 1.414. Hence these two colors won't come out, only red will come out which has refractive index of 1.39, okay? So at times, you know, we, in fact, almost every time we assume that refractive index of a medium is constant, but that is not the reality. Refractive index also depends on what is the wavelength of the light. Any doubt guys on these two questions? Anything? Great, Sanjana, can you share that solution with me after the class over WhatsApp? I'll circulate it to everyone. Okay. Any doubts? I'll move to the next.