 Any doubt guys on these two questions anything you want to ask How do you know R1 is minus R? See R is geometrical property. It 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, you're amazing distance from here You're moving against the incident rate, right? So for concave Surface rate is center of 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 Okay, let's move to next These two all right. You're not able to solve the first one 20 second See when the ray will pass through a glass slab The emergent ray will be parallel to the incident ray. So this ray Has to be parallel to that Okay, similarly here also This ray which will come out will be parallel Okay, this ray is parallel to that and this ray will be parallel to this, okay? And hence the angle remains alpha only that is our option 2 is correct Okay, what about 23rd? See in 23rd if you have equal widths Then intensity coming out from both the slits will be I naught and I naught Okay, if you increase the width of one of this slate to double the intensity of one of the slits becomes for I naught Okay, because area actually depends defines Area defines how much you be the intensity. So if you in if you double it Intensity becomes four times. So four I naught and I naught. So this is I one and this is I two All right. Now, you will see that the maximum possible intensity is root I one plus root I two The whole square which will come out to be nine I naught Okay, and minimum possible intensity is root I one minus root I two the whole square Okay, which is I naught. So earlier the minimum intensity was zero and maximum was for I naught now Minimum is I naught and maximum is nine I naught. So both Maxima and minima both of the intensities are increasing All right, any doubt on these two questions anything All right others What are you getting? so if If the person is able to see the object then the light from the object should reach the person Right and light will follow law of reflection. So it will get reflected from let us say edges From these two edges, it will get reflected then it will travel like this fine. So if This ray and that ray the person catches then he'll be able to see the object But if he goes beyond this point if it suppose is here Then he'll not be able to see the see this source or object because this is the maximum distance up to which The light from the source can reach on this vertical line Fine, so you just need to find out what is the distance between this point and that point? So that is the answer. Okay, so if I draw Let's say this line and that line. Okay So due to symmetry you'll see that this distance is D Okay, and then You'll see that You know, this is D. So this is D by 2 Okay, and this distance will also be D by 2 because these two triangles are congruent All right, and then you'll see that one two one two three and One four five. So triangle one two three is similar to triangle one four five, right and one two five Distance one to five is Two times the distance one to two. So if you use ratio of sides This length four to five length will be two times of two to three So if two to three is D by two, this has to be D and similarly this distance will also come out to be D Fine, so the total distance will be D plus D plus D which will be 3d That's why option number four is correct over here Fine. All right. So I think Alright, so break time is over Fine, so let me upload new question. Try solving this question. It refers to a figure So I'll just draw the figure over here Outside refractive index is N2 and inside refractive index is N1 Okay. Now try this So if ray comes out from CD then it has to undergo total internal reflection on ad Okay So that's the hint you can use So N2 is more than N1, right? So it will bend away from the normal So it has to at least graze the surface In order to reach CD So angle of refraction over here should be 90 degrees So this angle, let's say this is theta has to be critical angle Should I solve this? Okay, any other answer? Alright, so let's see This angle, let us say is R Okay, so N2 sine of Alpha should be equal to N1 sine of R Okay, so sine of R is equal to N2 by N1 sine of alpha Okay, or R is equal to sine inverse N2 minus N1 sine of alpha Fine. So if R is this The angle of incidence at the surface ad has to be 90 minus R because this is right angle triangle, right? So theta has to be equal to pi by 2 Minus R. Now R is sine inverse of that. So this will become cos inverse of N2 by N1 sine of alpha Okay, this is theta Okay Now we need to find the maximum value of alpha such that the ray comes out Okay, so if you increase the value of alpha Beyond alpha max then cos inverse of that will come out to be less Okay, if this bracket term increases cos inverse will decrease. So theta will decrease if you increase alpha Fine. So lesser the alpha better it is Okay. Now at an angle when it Just undergoes total internal reflection or there is this critical angle in that case you will have N1 sine of theta which is N1 sine of cos inverse N2 by N1 sine of alpha This should be equal to N2 Wait and N1 is more than N2. So this light will actually bend towards the normal So R is less than alpha max. So this ray diagram is wrong But the way we have to do this Whatever we have done is correct because we have not used the fact that R is less or more than alpha Okay, so this is N2 Sine of 90 degrees Fine, so we will get Sine of cos inverse N2 by N1 Sine of alpha Is equal to N2 by N1. All right, so Then you rearrange this term you will get alpha to be equal to sine inverse of this So that is why option number one is correct over here Okay, so lot of inverse technometric function usage is there in this particular question, but if you just Move ahead in a logical manner. You will arrive at the correct answer. Okay, so I'm moving to next question. Okay what Refracted ray will pass through CD without TIR See we are assuming that AD is very large the length AD is large Okay So if it has some angle with the horizontal it has to cross the AD So that's the assumption we are making. Okay Is it clear? Okay. Now solve 26 and 27 two beams of light having intensity is I and 4 I Interfered to produce a fringe pattern the phase difference is pi by 2 at A and pi at B The difference between resultant intensity is at A and B we have to find out So again, this is direct application of the formula resultant intensity is I1 plus I2 Plus 2 under root I1 I2 cause of phase difference Okay, so Initially when phase difference is pi by 2 intensity is I plus 4 I that is 5 I and When the phase difference is pi then It will be 5 I Plus 2 times under root of 4 I square cos of pi is minus 1 So this is I so difference in IA and IB is 4 I All right What about 27 27 there is a Young's double slit experiment 12 fringes are observed to be formed in a certain segment of the screen when wavelength of 6600 nanometer is used now when this change you need to find number of fringes observed in the same segment Now first of all you should know that all the fringes will have the same width in Young's double slit experiment right and the fringe width is given as lambda d by d D and the segment length will be same in both the cases whether you 600 or 400 the segment length is L only So number of fringes will be L divided by beta right, so let's say initially beta 1 is the fringe width and Number of fringes is n1. So n1 is equal to L by beta 1. Okay, so this will be L divided by Beta 1 is lambda 1 d by d. So a small d will go up Fine. This is n1 and n2 will be similarly L D divided by lambda 2 capital D, right? So this is equation 1 and that is equation 2. So if I divide these two, I'll get n1 by n2 To be equal to lambda 2 by lambda 1. Okay, so n2 will be equal to n1 times lambda 1 by lambda 2 Alright, so this will be 12 into lambda 1, which is 600 divided by 400 So we'll have 6 by 4 3 by 2 So it will be 18. So that is my option 2 is correct over here Any doubts on these two questions? anything Alright, so let us move to next try solving this Okay, so this is also a straightforward question if your concepts are proper So we have Ray of light that is passing through four transparent media their refractive indexes are different The surfaces are parallel if the emergent ray CD is parallel to AB then what should be the relation? So if AB is parallel to CD then You know if this is the normal then This angle should be equal to that angle Right, so this angle which is angle in refractive index mu 1 should be equal to that angle Which is angle in refractive index mu 4, right? So Snell's law says that refractive index into sine of the angle in that medium Which is this should be equal to Mu 2 into sine of angle here mu 3 into sine of angle there is equal to mu 4 Into sine of angle there which should be equal to this because these two rays are parallel So mu 1 will come out to be equal to mu 4, right? In case you have any doubts, please type in I'll move to the next one You'll be able to see an object only when the light from that point reaches your eye Okay, so use that Should I solve? Okay, let me solve it now beaker is filled with a liquid up to a height of 2h so Up to this height Let us say it is filled with liquid Okay, now He can see the lower end of the rod now. You'll be able to see the lower end of the rod only when light from the lower end Reaches the eye Okay, the light has to follow this path Then only this guy will be able to see it Right. We need to find refractive index of the liquid Now we know that this is the normal Okay, so if I apply Snell's law over here, let's say this is Angle I and then this has to be our Fine So we have refractive index over here is Mu so mu into sine of i Has to be equal to one which is refractive index of air into sine of r now you will notice here that The value of r is 45 degrees How because this length from here to here this length is H and Even that length is H and this is a 90 degree triangle this angle is 45 So even that is 45 because complete should be 90 degrees. Okay, so angle r is 45 degrees Okay, now if you look at angle i for that if you do do a small construction drop a perpendicular like this Okay, you will see that Are you guys able to hear me? Okay, let me share it again So tan i is 1 by 2 so from here you will get sine of i as 1 by root 5 Okay So you'll have mu into sine of i which is 1 by root 5 Equals to 1 into sine of 45 which is 1 by root 2. Okay, so you will get mu to be equal to under root of 5 by 2 So that is a option number 2 is correct over here How do you know that the light is meeting at the water level at its midpoint very good question Okay, so now Let us imagine All of you understood right how root 5 by 2 is coming now Amog is asking How do we know that light is meeting the water level at its midpoint? Okay, that is what the question is Now suppose it doesn't meet here it goes there then What will happen it will bend away from the normal okay, so This angle All right, this angle will be more than that angle as in What I'm trying to say this is the midpoint let us say okay So this is the midpoint is which is such a way that the light ray is able to reach the person's eye Okay, if assume that is a condition then if light ray is Going from right hand side of this surface. Okay, then what will happen here? Is that this angle of incidence since it is more than this one? So this is I1 and this is I2 since I2 is more than I1 it will bend further away Okay, so What is given here is that he can see the lower end of the rod when the liquid is being filled up to a height of 2h Okay, so this guy is just able to see it now If I talk about a point over here. All right, let us say I talk about point over there Then this angle of incidence is less. Okay now In order for it to You know, let's say the eye is over here. Okay, this is the eye which is seeing through this Slit, okay, so I The ray will has to travel From the eye the ray has to Go in this direction only because you can see that there is this opening, right? There is this kind of opening It is already fixed the direction is already fixed. The light has to enter this It's not just a hole like this It's a slit like that and the eye is seeing through the slit so When you say that light is going through and through of that slit then the red line is already fixed Isn't it and this red line is Passing through the center only because this With when you draw like this, this is angle 45 degrees Okay, this length is equal to that length now no matter up to what height you You move the water level up to The this this line the perpendicular will always hit the center line Are you able to understand what I'm trying to say here? Or you can understand like this also that the light has to travel this path All of you agree this light has to travel one to two if it has to reach the eye If light has to travel through one to two if it has to reach the eye, there is no other choice Okay, one has to be the midpoint because the angle is 45 degrees So I did not went into all that discussion because then unnecessary you will start thinking in Various other directions. So I wanted to keep it little simple, but then anyways it is good a very good question So Let's move to next question if intensity of the central maxima Remains unchanged, which is what I is equal to for I naught Cos square 5 by 2 right so in that intensity formula if you keep I1 is equal to I2 is equal to I naught You will arrive at this relation. Okay, so at maxima Intensity should be for I naught Okay, so whether it is central maxima or any other maxima in double slit experiment all have same intensity All the maximas, okay So cos of 5 by 2 should be equal to plus minus 1 or Phi should be equal to 2 n pi Okay, or you can say that the path difference should be Integer times lambda so all that you can consider here We need to find the minimum thickness of the glass plate over here now if glass plate has a thickness T At the location where central maxima is there. Okay, let's say This is the location where central maxima is there. Okay here the actual path difference is 0 Okay, the only path difference is coming because of the Slit, sorry because of the glass plate and because of the glass plate We know that path difference has to be equal to mu minus 1 into T Right, so when this is equal to n lambda then Consecting difference will happen or the same intensity as that of Central maxima was there that will be there. So T will be equal to n lambda by Mu minus 1 right so minimum thickness T minimum Will be lambda divided by Mu minus 1 okay lambda is what oh Lambda in terms of lambda we need to find out so lambda divided by mu is 1.5 So 1.5 minus 1 so you'll get two times of lambda as the answer so option 1 is correct over here Okay, you guys are making a lot of silly errors. Okay, it's not good You have to be careful. These are like sitters. You have to get it right You can't afford to be wrong in such easy questions