 Alright friends, so here is a question on wave optics, let's 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's 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, you know, 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's 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's 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 what, S3 is a location of center maximum, isn't it? So at S3, the intensity will be equal to, let's call it as I3, that will be equal to 4 times I0 because that is a location of center 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's 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's 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 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.