 So today I will be taking up previously a JEE questions, which are of mains level. So effectively I'm 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. So you'll see that apart from mechanics, these chapters, a 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'll just put a 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 one, right? Because glass effective index is more than airs effective 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 one should be equal to V2 by lambda two, okay? So lambda two by lambda one 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 two by lambda one. So that is why since velocity is decreasing, wavelength will also decrease, okay? Second one, 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 two, okay? And the distance between the slits and the screen is double. So capital D is two times the earlier D and D is becoming D naught by two, okay? So you have fringe width four 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're 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 four, 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 four is easier and you have taken it first, okay? So a 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 one by F1 plus one 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 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 incident 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 reflection. So basically you're saying this theta can be any angle beyond 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 effective index 4 by 3. So I can say that 1.5, which is 3 by 2, sine of I should be equal to 4 by 3 sine of 90 degrees, okay? So I will get sine of I to be equal to 8 by 9, okay? So basically I'm getting I is equal to sine inverse 8 by 9. So my angle should be more than sine inverse 8 by 9, and angle of incidence cannot be more than 90 degrees, okay? So if angle is more than sine inverse 8 by 9, I can as well say sine of that angle is more than 8 by 9, okay? Because sine is increasing function between zero and 90 degrees. So that is the reason why option one is correct over here. Both option one is correct, okay? Any doubt on these two questions? Is it clear? Type in yes or no. I'll move to next one. I have 70 questions. I don't know how many we'll 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 resultant intensity will be maximum when cos of phi is equal to one, okay? If cos of phi is equal to one, this will be equal to root of I1 plus root of I2 whole square, okay? This is max, and minimum will be when cos of phi will be equal to minus one, which will be root I1 minus root of I2 whole square, okay? So this will come out to be nine 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 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 one and this is mu two. So mu one sine of I is equal to mu two 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 two by mu one, 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 two by mu one. 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 one sine of critical angle to be equal to mu two sine of 90 degrees, fine? So sine of critical angle will become equal to mu two by mu one, okay? And mu two by mu one is what? Okay, are you able to hear me? All right, so that is the reason why option one is correct because I is also equal to R, fine? Any doubt, guys? 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 seven unless you have 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 eighth question, first one is correct. Okay, so many of you are getting option four to be correct for question number seven. 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 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'll 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, one by V plus one by U is equal to one by F, okay? Now if I differentiate this, okay? I'll get one by V square dV minus one by U square dU is equal to zero, okay? Focal length is a constant. Now why I'm doing like this is because dU represent a very small deviation in object distance. So dU can be equal to B, fine? 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 one 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 U square F square divided by F minus U whole square into U square into B. The U square will get canceled, 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 four is correct, okay? Fine, now please attempt question number eight. See when this light hit here, it will have angle of incidence of 45 degrees after differentiation. See, once you differentiate, you'll 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 eight. 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 one, 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 one into sine of 90 degrees, right? So mu has to be greater than one by sine 45 which is root two. So if mu is greater than, 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. You can refer to your notes. You can refer to NCRT book, okay? So keep that with you for any reference. This is not a test which we are doing. This is just a problem practice session. So feel free to refer to any equation or any formula which you tend to forget. Okay, ninth, option one, Lalitha is saying, you remember, magnification of an astronomical telescope when object and image both are at infinity is FO by FE, okay? And you can quickly draw a diagram also, representing the telescope. Now this is objective and this is eyepiece, okay? Now it is astronomical telescope, so of course it will look at the far away object, so objects are at infinity. So the image will get formed at the focus of the objective, so this is let's say intermediate image, I1. So this distance has to be equal to FO, okay? And the final image is getting formed at infinity. If final image is getting formed at infinity, this has to be focus of eyepiece also. So this length should be equal to FE, fine? So separation between objective and eyepiece is 36 centimeter. So FO plus FE is given as 36 and magnification is given as five, all right? So from here, you'll get the answer. So FO is five times FE, so FE will become six, okay? So FE is six, so I'll just mark option four and move ahead. I don't need to even find out FO because FE has a unique value in all four options and I'm getting FE as six. Anyways, FO plus FE has to be 36, so FO becomes 30. What about 10th? Question number 10th, guys, okay. I'll do question number 10th. See here, it talks about division from the prism and it is written that it is a thin prism, fine? So what does it mean? We know that refractive index is given as sine of A plus delta M by two divided by sine of A by two, okay? So if the prism is thin, okay? Then angle A has to be very, very less. So thin prism will look like this. This is a thin prism, okay? So angle A of the prism is very less, okay? That is the reason why sine of the angle, I can approximate it to the angle itself. So mu will become equal to A plus delta M by A, okay? Now this is something which, you know, if you're doing it for first time, then yes, it is not a straightforward one, but then this is being tested again and again where they talk about dispersion or deviation from the thin prism. So when it talks about thin prism, that's how you approximate the formula, okay? So delta becomes mu minus one into A, all right? So basically deviation depends on the refractive index as well as angle of prism like this, okay? Now if it is producing no deviation, okay? If it is producing no deviation, then whatever deviation this prism is creating, okay? Should be countered by the other prism. So that is why it has to be kept, you know, inverted so that this ray becomes parallel to that ray and there is no deviation, fine? And in case there is no deviation, I can say that both of them have rotated the light by equal amount. So like for example, the first prism has rotated the light by let's say mu one minus one into A one. So this is a deviation from the first prism. This should be equal and opposite of the deviation from the second prism. So like this you have to write, all right? Now angle of second prism you can get by writing this equation, mu one minus one divided by mu two minus one into A one, okay? So mu one minus one is 0.54, then 0.72. So 54 divided by 72 times A one. A one is four degree, four degrees and this is 18 thriza and this is 18 forza. So you'll get three degrees. So that is an option three is correct over here, okay? So other than those who have answered, any doubt? Any doubts on these two questions? Please type in yes or no.