 All right, friends, so here is a question on Veyoptic 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 this is a little bit different from the routine questions of Young's Diversite 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'm 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 phi, the phase difference. So for a particular position, let's 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. Thank you.