 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 some where 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 paddle 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 formed 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 represents 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 It's 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 1 and 2 they are out of question now. So it is between 3 and 4. All right Now you can play smart here little bit and say that anyways Since the lens is displaced vertically. So vertical displacement of the image will happen Why coordinate will not remain 0 so you can mark option 3 and move ahead. Okay, that that is like Playing smart here and not 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 you is F1 minus D So V by U becomes F2 divided by F1 plus F2 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 F2 delta divided by F1 plus F2 minus D Okay, now this is the height from the principal axis of the second lens Okay, but why coordinate is what why 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-coordinate 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 axis 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 If it if something like this comes it has to be from J advanced, okay? All right. So let us take up other 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 2 lambda d by a Okay, so this will be written as 2 into lambda which is 6 into 10 is power minus 7 capital D is 2 Divided by a now a is 10 is power minus 3 Okay, so you will get 2.4 mm So question number 12 option 4 is correct Why are you 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's on either side is To lambda d by a Okay, others. What are you getting question number 13? Poor week retracted its message his message Poor week 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 our 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 sign of 30 which is half Okay, so this is point 72 alright, 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 be 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 point seven two Minus two times Well, let's keep two common minus 30 degrees Okay, so that is our 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 Okay, all of you got 14th a diminished image of an object is to be obtained on a screen One meter from it. This can be achieved by placing Convex lens of focal length less than point two five meter concave lens now first of all this is Getting formed on the screen right so since it is getting formed on the screen it has to be a real object Okay, so plain mirror creates a virtual object. Sorry virtual image So since it is formed on the screen it has to be real image So plain mirror is not possible even convex mirror forms virtual image Even this cannot be put on the screen right and then concave lens also Will create a virtual image so it has to be option three only Okay, now let's talk about the 15th one and I Specialist prescribes practicals having combinations of convex lens of focal and 40 in contact with the concave The power of the Combination so you can see that similar concept was tested in J earlier also So simply you need to write down the combined power as one by F1 should be in meters plus One by F2 now you have to take care of the science also it is a concave lens So minus of point two five you have to write Okay, so I think you'll get What minus one point five? Okay, this is equal to four and this is ten by four so Yeah Okay, so let's quickly move to the next few questions So you might be seeing that majority of the questions are straight forward from this chapter but Sometimes questions get twisted. So there is a high chance that an easy question will come from this Particular chapter. I mean these two chapters. So make sure you Do good amount of practice because high chance that? You know if you do some practice, you will get the Answer in the exam This is not Okay, all of you getting question 16th answer as pi As an option number three It's a single slit experiment of Okay, let me solve question number 16 those who have got 16 can move to question number 17 parallel monochromatic beam of light incident on a narrow slit So this is the narrow slit it incident here diffraction pattern is formed Here in this screen, okay At the first minimum of the diffraction pattern the phase difference Between the rays coming from the two edges of this slit is what? Now first minima location You should know Suppose this is distance y. Okay. This ends y should be equal to lambda d by a where a is the Width of the single slit, okay Similar expression comes for the maxima in the double slit experiment, but for single slit experiment This is what the case is okay minima happens to be at a distance of lambda d by a Okay, now we need to find the phase difference between the rays coming from the two edges of this slit, okay now Let's connect these two rays. So one is this one and Other one is this so we need to find the phase difference between these two Okay, now we know the standard procedure of finding the phase difference, right? You drop a perpendicular on All right, so we are live. All right. So there is a standard way to find the You know phase difference or path difference. You just drop a perpendicular like this and This is the path difference. This is delta x okay now In order to find delta x what we typically do is we connect this line like that Okay, and then, you know, right this this is the angle theta Then even this angle is theta. So delta x is This distance which is a sign of theta, okay? And if theta is very less then delta x I can also approximate it as a tan of Theta fine. So this is delta x now corresponding to this delta x The phase difference is 2 pi by lambda into delta x so 2 pi by lambda Into delta x which is a tan theta now tan theta is what y divided by this distance, which is d Okay, y is what lambda d by a Divided by d which is tan theta. Okay. Now you see that Lambda get cancelled a get cancelled and d get cancelled. Okay. So phase difference is 2 pi Okay, so 16th option number 4 is correct Many of you have said 3 right Okay, so I guess you have tried question number 17. Also, let's see how we can go about it Now this kind of question where it there is no definitive Option as in it doesn't talk about Distance being 5.4 centimeter or 20 centimeter like that. It doesn't talk about a definitive statement, but it just tries to You know find out the estimate of it where exactly it will be between which and which point This can be done best by using ray diagram. Okay Now, let us see how we can utilize ray diagram to answer this question We have concave mirror, which is placed on the horizontal table This is the horizontal table. So we have a concave mirror like this this is the concave mirror and axis is Vertically upwards. Okay. So this is the axis. Oh is the pole and C is the center of curvature. So let's say C is here Okay, a point Object is placed at C only. So there is an object here. Okay, so the image will be where right now Image will be at C only Okay, because whatever ray that comes out of C will hit the mirror normally. So the ray will retrace its path Okay, now we are filling The mirror with water. Let us say this is the water Okay, so if I fill this with water What will happen here is that the ray that was normal to the mirror Suppose this ray if it keeps on going Then it will hit the mirror Normally and the ray will just reflect back But what will happen now because of the refraction the ray will bend towards the normal, right? So if I zoom this I will get a situation like this suppose Ray is coming like this this is the normal to the Water-air interface so light will just bend Towards a normal like this. Okay, and this is the Path for the normal this thing. This is the path for the As in This is the path where it hits the mirror normally. Okay, now it will hit like this. Okay, so for the mirror the object is Object appears to be somewhere there at the top beyond C Okay, then only light will you know appear to bend towards a normal So the image is slightly away from the C. Okay, so that is why when image is slightly away from C The object will come Closer to the as in between O and C it will go Right, so it will be a real image located between point C and O all right, so this actually Will be solved best if you are able to Analyze the ray diagram. Okay, because there is no definitive statement over here. You just have to just you know Evaluate the options. Okay, all of you clear about this any doubt type in yes or no All right. See anyways, this is getting recorded. So you can always refer this any moment So what is the answer for 18th again? It is an theoretical question, right? It's a theoretical one All right, should we solve now? Fine question number 18 first of all yellow light will be like of the order of 500 Nanometers right and This is a wavelength of yellow light and wavelength of X ray is like Think one two hundred Armstrong Okay, or whatever it may be you know that the wavelength of X ray is very very less than The wavelength of yellow light or frequency of X ray is very high compared to the frequency of yellow light. Okay now in the Young sorry in the single-state experiment or In fact in the double-state experiment also both of the cases the assumption is that the the size of slit is Comparable to the wavelength Okay, now if you use X ray the wavelength is so less that for X ray It is not a small hole It's not a small hole where you can see it say that it's a single-state experiment Right, it's not a narrow opening for X ray. It is like, you know a huge opening So there will be no fringes that will be observed So that is my option number four is correct over here Okay, because light will bend From the edges or diffraction will happen only when wavelength is comparable to the size of the hole Okay Now to the 19th one 19th. What do you think is the answer? Okay, now this is a thin slice of the cylinder Thin slice that cut out of the cylinder. So this cylinder actually it's a part of cylinder that comes out of your screen Okay, so When you look at this ray, okay And if you look at all the rays suppose are hitting along the line Which line is coming out of your screen then all of these line Will be similar right they will encounter same amount of path difference They will have any you know refraction over here Like like this it will go and then it will Simply bend away from the normal and then reflection will happen So all the lights which are in this line the line which comes out of your screen will encounter this scenario Fine, so there is a line symmetry over here Fine, so there is a line symmetry does not a circular symmetry. So that is why it will be straight the fringes will be straight Okay so these are I think These are not numerical. So you just have to know actually some concepts to solve these theoretical ones But it is very easy to go wrong in these theoretical questions because You know you can think in any direction. So it becomes tricky at times Whereas numerical you'll get a definitive answer. Oh You're saying option D fringe is spacing is spacing Increases as we go outwards see outwards means you are Coming out of the screen. Okay, if it is straight Then there is no question of spacing when you are coming out of the screen because along the same screen same Along the one straight line. There is only one fringe. Okay In last option, it was not clearly written whether it is a straight or a circular Fringe they are talking about when they talk about spacing Okay, good that you have answered the second one Initially in a compound microscope the intermediate image. We know intermediate image is inverted real and magnified So this is just knowledge-based question. If you have studied Your NCRD properly Then you'll get it immediately Question number 20 Are you able to read it properly because the font size became very less a Should I upload it again? So 20th question is on the lens maker formula So we have this formula 1 by f is equal to mu 2 by mu 1 minus 1 multiplied by 1 by r1 minus 1 by r2 Right, so yes. Okay. Okay fine Legible Now it is fine Anyways Fine, so we have a concave lens which has effective index 1.5 So basically mu 2 is refractive index of the material which is 1.5. Okay, so same radius of curvatures were there All right in the concave lens So it was like this This is the concave lens, right? So So basically for this Rates of curvature if you take minus r and for that side you have to take it as plus r Okay, so r1 is minus r and r2 is plus r Okay, so When you immerse in a medium of refractive index 1.75 then what will happen? Let's say 1.75 minus 1 1 divided by minus r minus of 1 divided by r Okay, so this will become equal to 1.5 minus 1.75 to minus 2 by r Okay, so this is 1 by f Fine, so you are getting focal length as positive. So f is Greater than zero so focal length is positive means what it means that when parallel rays come and hit this lens It will get converged forward then only you'll say that focal length is positive, right? So it becomes a converging lens Okay So concave usually is diverging but when the mediums refractive index becomes more than The material refractive index the same Diverging lens becomes or it will act as a converging lens. Okay, so it becomes a converging lens So option 3 and 4 they are out of question Okay All of you answered third for this Focal length is 1.75 divided by 0.25 Into r divided by 2. Okay, so this is This is 4 into point eight seven five times r Which is 3.5 times r Okay, so question 20 option 1 is correct How do you know r1 is minus r? See r is geometrical property if it is a concave lens, right? So this side should be like this Okay, so central curvature will be on the left hand side Okay, so if central curvature on the left hand side your amazing distance from here You are moving against the incident rate, right? So for concave A Surface radius center curvature is this side, right? But that is a reason why radius of curvature of this side is negative You we are using sine convention All right