 All right friends, so here is a question on wave optics chapter. So you can see on your screen, let's read the question first. Consider an arrangement shown in the figure by some mechanism the separation between the slit S3 and S4 can be changed. So you can now make out that you know this is it's a little bit different from the routine questions of Young's Divergent Experiment. Here we have four slits, two at one end and two at a distance D from it. And then there is a point P on the common perpendicular bisector, we need to find the intensity measured at point P. Let's assume that intensity to start with here it is given as I0. So intensity at S1 and intensity at S2 is given let us say I0. And we need to find what is the maximum possible intensity at point P. Now if you visualize it one by one, you can see that point P is a location of central maxima for S3 and S4 slits. So for S3 and S4, P is a central maxima. So if at S3 intensity is I1, due to symmetry intensity at S4 will also be I1. So intensity at point P should be equal to four times of I1. Now if I change the value of Z, intensity I1 will change. Why it is like that? Because if you see here my dear friends that at slit S3, both the waves from S1 and S2 will reach and before coming out from S3 they will have an interference themselves. So depending on what kind of interference it is constructive, destructive, partially constructive, the intensity I1 will vary. Now in order to get the maximum possible intensity at point P, I need to have maximum possible intensity at I1. At S3 I need to have maximum possible intensity. So more is the value of I1, more is the value of IP finally. So we know that intensity at any location is given as four times I0 cos square 5 by 2. So I am talking about intensity in the plane of S3 S4, 5 by 2. So let us say that this intensity I am trying to find out at location of slit S3. So this will be equal to I1 only. Now if I change the value of Z, I can change the value of 5, the phase difference. So for a particular position, let us say for a particular value of Z, there might be constructive interference going on at I1. So intensity I1 will become equal to 4 I0. Wherever there is a constructive interference going on, the value of cos square 5 by 2 will become equal to 1. So I1 will become equal to 4 I0. So intensity at point P will become what? Intensity at point P will become 4 into 4 times I0. So at point P I can essentially get 16 times the intensity with which I started with. So like this you have to do this particular question. All right friends, so here is a question on wave optics. Let us see what is this about. So it refers to a figure. You can see here what is going on. The two slits S1, S2 plays symmetrically around the center line. I are illuminated by monochromatic light of wavelength lambda. The separation between the slits is small d. So the light transmitted by the slit falls on the screen, plays at a distance capital d from the slits. The slit S3 is at a center line and slit S4 is at a distance of Z from S3. So another screen is placed at a further distance capital D from sigma 1. So this is let us say sigma 1 and this is sigma 2. Find the ratio of maximum to minimum intensity observed on sigma 2 if Z is equal to this, this and that. So as a function of Z, basically I need to find what is the intensity on sigma 2. So let us try to see, first of all you need to understand that if you just look at S3, S4 and this screen Z2 or sigma 2, you will see that it is like a Young's Diversite experiment. But before going to screen 2, there will be an interference that will happen on sigma 1. So even sigma 1 if you look at with respect to S1 and S2 is a Young's Diversite experiment. So there will be let us say one wave coming from S1 that will interfere with S2 at location of S4. Similarly, over here there will be an interference happening. Now S3 is a location of central maximum, isn't it? So at S3, the intensity will be equal to let us call it as I3 that will be equal to 4 times I0 because that is a location of central maximum. Now at S4, the path difference between the two waves which I am talking about this wave and that wave that will be equal to Z small d by capital D. So this is the path difference. So the phase difference will be what? Phase difference if I write phi will be equal to 2 pi by lambda times delta x that will be equal to Z d by capital D. So now using the formula, I can get intensity where the slit 4 is equal to 4 times I0 cos square phi. So depending on different values of Z, you will get different values of phi. So this A, B, C if you put the values one by one, you can get the intensity at 4. So I have intensity at 3. Let us call it as equation 1 and this as equation number 2. And there might be the case that at 4, destructive interference happens. So if at S4, destructive interference happens, then only one wave comes out from S3. So there is no scenario of interference. So whatever intensity of S3 will be the intensity everywhere on screen sigma 2. But if suppose it is not 0, then I have to use this formula. So this will be equal to I3 plus I4 plus 2 times under root I3 I4 cos square, let us say further phase difference will be theta. So theta by 2. Now I3 and I4 you can get from here. And since maximum and minimum values are only r's, so I can say that I max is when cos square theta by 2 will be equal to plus 1. So I max will be equal to root of I3 plus root of I4. And I minimum will be equal to sorry root of I3 plus root of I4 the whole square. And here I minimum will be root I3 minus root I4 the whole square. You can treat I3 as root I3 whole square. So when this becomes minus 1, it becomes root I3 whole square plus root I4 whole square minus 2 I3 I4, which is nothing but root I3 minus root I4 whole square. So like this you can get the minimum and maximum values and just substitute the values of z, you will get the exact values. So that's it with respect to this particular question. I hope you have learned something today. And in case you have any doubts, feel free to get in touch with us. Thank you.