 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 formulae which you tend to forget. Okay, ninth Option one Lalita 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 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 the 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 will get the answer So fo is five times fe to fe will become six Okay, so fe is six. So I'll just mark option four and move it I don't need to even find out fo because fe has a unique value in all four options and I'm getting fe has six Anyways fo plus fe has to be 36. So fo becomes 30. What about tenth? Question number tenth guys Okay, I'll do question number tenth 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 sign of a plus Delta m by 2 Divided by sign of a by 2 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 a sign 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 do it for first time then yes, it is not a straight forward 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 1 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 1 minus 1 into a 1. 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 1 minus 1 divide by mu 2 minus 1 Into a 1. Okay. So mu 1 minus 1 is point five four Then point seven two so fifty four divided by seventy two times a one a one is four degree Four degrees and this is 18 threesa and this is 18 forza. So you'll get three degrees So that is a option three is correct over here Okay So other than those who have answered any doubt And it out on these two questions, please type in yes or no Okay, let us move to next few questions this one, this is an Interesting question. So let us try to solve this take some time. No, hurry Whatever you do do it properly and try to get somewhere This is question number 11 That's one hint. I want to give if you Displace the lens vertically Horizontal distance will not change Okay, so let me solve now Now the first image Here we have focal length f1 for the First one and focal length f2 for the second one. Okay, small d is less than both f1 and f2 Okay, so we have the parallel beam of rays coming from the left-hand side So where it will coincide or where the image will try to get formed Image will be getting formed at the focal length right, so somewhere At a distance of f1. Let us say This is where the image is getting formed. So this distance Should be equal to f1 Right, so but then There is a second lens Okay, because there is a second lens actually may will not get from there This is this image, which is the image of the first lens Will behave as an object of the second lens. Okay, so in order to find the Final image distance. I need to use the lens formula for the second lens So 1 by v minus 1 by u is equal to 1 by f2. Okay now We represent the final image. Okay minus. What is you use this distance? Because I am applying lens formula for the second lens right and that distance will be equal to f1 Minus d Okay, because f1 is a positive quantity focal length of a convex lens This is equal to 1 by f2 Fine, so I will get v to be equal to f2 times f1 minus d Divided by f1 plus f2 minus d Okay, now This is the final image. So maybe final image gets formed over here somewhere and this distance Is given as this. Okay, but what is asked is the x coordinate? So x coordinate will be what this its distance from the second lens plus small d That is the x coordinate, right? So x coordinate will be equal to v plus D Okay, then you'll see that x coordinate will come out to be this Okay, which is in the option third and fourth both have x coordinates Okay, so one and two they are out of question now. So it is between three and four All right now you can play smart here a little bit and say that anyways Since the lens is displaced vertically. So vertical displacement of the image will happen Y coordinate will not remain zero. So you can mark option three and move ahead. Okay, so that that is like Playing smart here and you know taking some risk and moving ahead But then if you actually want to find the answer here, then you know, you can find out Like this first of all You have to understand that principal axis for the second lens Is here somewhere. Okay, and object Of the second lens is on the principal axis of the first lens Because image of the first lens is the object of second. So this is the object for the second lens Fine now this object I can treat it like an extended object for the second lens, okay, because What is the object's height objects height is its distance? Vertically from the principal axis Okay, so I can just track where this tip of the object goes After the refraction from the second lens and that is where The image will be Okay So magnification of the second lens Is basically v by u Okay. Now v is this and u is what u is f 1 minus d So v by u becomes f 2 divided by f 1 plus f 2 minus d This is the magnification. Okay, this magnification should be equal to size of the image divided by delta which is the size of the object So height becomes equal to f 2 delta divided by f 1 plus f 2 minus d Okay. Now this is the height from the principal axis of the second lens Okay, but y coordinate is what y coordinate is distance from the First principal axis because first principal axis that is a principal axis of the first lens is the x coordinate Sorry, is the y chord All right. So sorry for that This is the height from the second principal axis So distance from the first principal axis will be H minus delta because delta is a distance between two principal axes Okay, and that's how you get this as the correct option option number three Fine. So slightly difficult question. So if this kind of question comes in j mains I know I hope what you have to do, you know You have to leave this type of questions. It will just eat up your time. Okay. So this question Uh If it if something like this comes it has to be from j advanced. Okay. All right. So let us take up other questions these two questions So the 12th question is a single slit experiment. Okay and the distance between the first minima on either side Is nothing but the width of the central maxima And width of the central maxima I hope you remember is lambda d by two lambda d by a Okay, so this will be written as two into lambda which is six into 10 is power minus seven Capital d is two divided by a now a is 10 is power minus three Okay, so you will get 2.4 mm So question number 12 option four is correct. Why is like saying to oh guys be careful. Okay distance between The first minima on the either side it is saying. Okay The distance of the first minima is lambda d by a Distance between two minima on either side is Two lambda d by a question number 13 Okay, others. What are you getting question number 13? A poor weak retracted its message his message Poor weak said they will remain parallel and he said option one is correct See this ray these two rays they are hitting the first surface Along the normal right, so they'll just go straight And then hit the other surface fine You want me to do this now? Anyone got the answer? No one Okay, naman is telling one answer Others Okay, let me solve now. What is this angle of incidence? This is 60 degrees So this is also 60 degrees. So angle of incidence Is 30 degrees Okay, now it will bend away from the normal right So since prism is the denser medium, so it will bend Like this Okay It goes like that So this angle Is let us say r Okay, this is i So I can say that 1.44 Sign of i Is equal to 1 into sign of r Okay So sign of r is equal to sign of i is what sign of i is sign of 30 which is half Okay, so this is 0.72. All right. So r becomes equal to sign inverse 0.72 Okay Now the option is with respect to what angles these rays are making Okay, so if I get to know What is this angle then double of this angle will be the answer? Yes or no Because other ray will also come like this same angle. The other ray will also make Okay, so double of this angle will be the answer now. How much is this angle? If I know i and if I know r, what is that angle equal to b if I extend this line Like this what I'll get here is that this angle Is this angle Getting it now this angle is nothing but r minus i Okay, because total angle is r Okay, and This angle is i so this angle will become r minus i Okay, so two times of r minus i Is the angle between the two rays So that will be equal to two times sign inverse of 0.72 Minus two times Well, let's keep two common minus 30 degrees Okay So that is an option number three is correct over here fine Any doubt on this question guys type in yes or no quickly. I'll move to the next one