 Alright, so let's move to the next one this one should I do it see here you will have this angle as 30 degree because of reflection. So like that you will get this angle to be 60 degrees and similarly you can get this angle to be 60 and then this also become 60 so you will get equilateral triangle. So every time you get equilateral triangle only fine every time you will get equilateral triangle. So you can see that each equilateral triangle is occupying this much length ok which is how much which is equal to the side length of the equilateral triangle which how will you find I have this length this length I have to be equal to 0.2. So in this right angle triangle I can get the side of the equilateral triangle which is this to be equal to 0.2 divided by cos of 30 this is the side length ok. So a will come out to be 0.2 divided by root 3 by 2 so this is 0.4 divided by root 3 is this in clear ok. Now let's calculate number of equilateral triangles. So in 2 root 3 meters number of equilateral triangles will be equal to 2 root 3 divided by 0.4 divided by root 3 ok this will be 6 divided by 0.4 ok so 5 and this is 3. So you will have 15 equilateral triangles fine you will have 15 equilateral triangles and every equilateral triangle you will see that every edge of it represents one reflection ok. So if there are 15 equilateral triangle then you may tend to think that 15 into 3 as in 45 reflections are there but then you will see that this reflection this one this one all these reflections are counted twice ok. So if you just count 2 reflections as in suppose this equilateral triangle you count only 1 and 2 then this equilateral triangle you just count 3 and 4. So per equilateral triangle if you count just 2 reflections then you will be covering all the reflections ok. So hence number of reflections will be equal to 15 into 2 which is 30 ok that is an option number 2 is correct over here alright. Now you may see that towards the end this will be the situation where the last equilateral triangle this one will be one of the possible reflection ok where it will get reflected and go away alright. So answer is actually 30 or 31 but then 31 is not in the option so that is the option 2 is correct. Any doubt guys anything this is more of I think geometry question or more of mathematics question than the physics one. In case you have any doubts please type in we will move to the next one start solving this question I will upload another one these are the last 2 questions for today going to be 7.30 ok are you getting 32nd option 3 all of you see angular position is defined as suppose this is the point for which you need to find angular position let us say this is the point this points angular position is this angle fine that is how you define the angular position of a point on the screen ok. Now if theta is the angular position do you know that the path difference between these 2 is nothing but d sin theta ok which can be approximated to d tan theta as well fine. So let us say that I mean the path difference between these 2 is d sin theta ok now intensity has to be 1 fourth so intensity formula is what I is equal to 4 I naught cos square phi by 2 ok 1 fourth of the maximum intensity now maximum intensity is 4 I naught so 1 fourth of that is I naught so you can find out the value 5 for which I becomes equal to I naught by using this formula ok so you will get cos of phi by 2 to be equal to plus minus 1 by 2 now plus minus 1 by 2 actually represent one point above and the other similar point below so we will just focus on the point above so I will just take the positive value over there. So pi by 2 should be equal to what angle phi by 2 will be equal to pi by 3 ok so the phase difference should be equal to 2 pi by 3 fine if phase difference is 2 pi by 3 what will the path difference part difference should be equal to lambda by 2 pi into phase difference that is how you convert between phase difference and path difference right so this will be equal to lambda by 3 fine now d sin theta if you equate that to lambda by 3 you will get theta is equal to sin inverse of lambda by 3d that is why option 3 is correct over here fine now try out question number 33 some of you might have class to start at 730 I guess those guys who are from Kormangala NPS Kormangala they can leave this and join that mathematics class ok this is anyway getting recorded and this just problem solving so please join that alright anyone 33 no one should I solve it ok let me solve it now so I will just briefly discuss this question probably then I will leave it after that so that you can also do it your own ok so first of all PC is a wave front the P and C are the points on the same wave ok hence P and C are in phase phase of P is equal to phase of C ok now what happens this point C moves forward get reflected and reaches P ok so between this ray and this ray ok the path difference is CO plus OP and because of hard reflection lambda by 2 this is the path difference ok so always take care of the reflection also when there is a hard reflection lambda by 2 you should add alright now CO and OP you can find out OP is what this angle is theta ok so this D is given and here it is 90 degree so OP you can write it as D divided by cos of theta is OP ok and once you get OP you have a 90 degree angle here as well you can get CO as well CO is OP cos of 2 theta ok which is D by cos theta into cos of 2 theta cos of 2 theta is cos square theta minus sin square theta ok so you can use all that and then just substitute the value of OP and CO in this particular equation and you need to have a conceptive interference over here right so for conceptive interference delta X should be at least equal to lambda so you can just put the value over here as lambda and then just simplify it and you will get the answer I will tell you the correct answer over here 30th the correct answer is this 33rd ok fine so that's it for today from my side so we have we have tried to cover many different aspects of optics today so I hope you will now be able to revise optics in a more efficient way and if some question come in competition you should be able to get it but then again you know at times it may happen that a very difficult question can be put forward in the exam so it's okay that you have studied optics you know very nicely and you're not able to get the answer in the exam so don't fight with the question paper just leave it if you're not able to solve in couple of minutes so there are you know total 30 questions in physics so you can attempt others but then having said that there is a big I mean there's a good probability that easier questions will be asked from optics okay fine so for today you can like this session and then leave it bye bye