 Let's solve a couple of questions on intensity variation in a double slit interference experiment. Here is the first question. Now it says that in a double slit experiment, the two light waves arrive with different phase differences at various points on the screen. Which is true right? Because we get bright fringes, you also get dark fringes. And it says that the intensity of the central bright fringe is k units. What is the intensity at a point where the phase difference is pi by 4? And it's given that the intensity of light coming from the two slits is equal. Alright, before I get into this, why don't you pause the video and give this one a try. Alright, hopefully we have given this a try. Now in this one, the question says that we need to figure out the intensity at a point where the phase difference is pi by 4. And we know the intensity of the central bright fringe which is k units. So first of all, let's write the relation using which we calculate the intensity and that is given by the intensity at a point that is equal to 4 I0 which is the intensity of the light coming in into cos square pi which is the phase difference divided by 2. This is pi by 2. Now we know that the intensity of the central bright fringes k units. And we also know that the phase difference at all the bright fringes is in multiples of, it is in multiples of 2 pi's. And at the central bright fringe, it is just 0 degrees. So if we put in this value, if we put in this value over here, let's see what do we get. So we get the intensity at the central bright fringe that is k units. This is equal to 4 I0 into cos square cos square 0 divided by 2 which is just 0. And we know that cos 0 is 1. So cos square 0 degrees also 1. So it turns out k is equal to 4, 4 I0. Let's call this our first equation. This might be useful to us later because we can see the options are in terms of k. So we have somehow established a relation between the intensity of light coming in and k itself. Now the question is to figure out the intensity at a point where the phase difference is pi by 4. Now delta pi or pi, this is pi by 4. So let's plug in that in this relation and see what do we get. We have the intensity at the point where the phase difference is pi by 4. This is equal to 4 I0 into cos square pi by 4 divided by 2. So that becomes pi by 8. In place of I0, let's substitute k by 4 using the first relation. I0 is equal to k by 4. So when we do that, this is 4 into k by 4 into cos square cos square pi by 8. Now cos square pi by 8 is a little difficult to figure out. So if you have a calculator if you're using one, then right way you can see that this will be 0.85. But without using the calculator, what we can do is we know that 2 cos square pi, so one identity, one trigonometric identity is 2 cos square pi by 2. This is equal to 1 plus cos pi. So what we can do is we can take 1, 2 from this 4 and write this as 2 cos square pi by 8. So 2 cos square pi by 8, this will be equal to 1 plus cos pi by 4. And 1 plus cos pi by 4 will be 1 plus 1 by root 2 because that is what cos pi by 4 is. And so 1 plus 1 by root 2, 1 by root 2 will be 0.707, so this is 1.7. So 2 cos square pi by 8, this comes out to be equal to 1.7. And now what we can do is we can plug that back over here. This is 2 into k by 4. We took 1, 2 from this 4 and just wrote this before cos square so that we get this identity. And in place of 2 cos square pi by 8, we have 1.7. So this is equal to 3.4 k divided by 4. When we remove this decimal, this becomes 34k divided by 40. So this is 17k divided by 20. And that makes this the right answer as option C. Alright, let's move on to the next question now. So the second one says, consider the interference experiment below. The displacement graphs of the light waves at the slits S1 and S2 look like the following. The intensity of a maximum is W units. And the question is to figure out the intensity at the point O. And this is the point O, right at the center, point O on the screen. There is a note which says O lies on the perpendicular bisector of S1, S2, which basically means that it is right at the center of these two slits. And it also says that the intensity of light coming from the two slits are equal. Okay, again, before I get into this, maybe try attempt this one on your own first. Alright, hopefully we have given this a shot. One thing that we can clearly see when we look at these two displacement graphs is that the two waves, they are not in sync at the source slits. Or it seems that they have a phase difference. They have some phase difference between them. So this point O right here, it might not be a maximum because light is traveling the same amount of distance from S1 and S2 till O. But they are out of phase from each other. Let's calculate how precisely are they out of phase. If we try to draw this displacement graph at S1 on top of S2, it will look like this. And using this, we can try and figure out how much are they out of phase. So if we draw a line like this, this right here would be pi by 2 and this right here could be pi. So we can see that they are pi by 2. The phase difference between them is pi by 2. That is delta pi. Or they are 1 by 4 cycle apart from each other. Now we know the intensity of a maximum that is W units. And the phase difference always at a maximum, the phase difference always at a maximum is 0 degrees. Or it could be 2 pi, 4 pi, 6 pi. So intensity at the maximum that is W, this is equal to 4 I0 into cos square cos square pi by 2. And cos square pi by 2 at a maximum, this will always be 1. So I0, this is equal to W by 4. Now we need to figure out the intensity at the point O. So let's do that. Intensity at O, this is equal to, this is equal to 4 I0, which is the intensity at both of these slits, into cos square cos square pi by 2. Since the light from the two source slits is travelling the same distance till the point O, the phase difference between them still remains as pi by 2. None of the light travels an extra distance. So that could lead to some extra part difference and finally some change in phase difference. So if we write in case of pi, if we write pi by 2, this becomes cos square, cos square pi by 4, because there is a pi 2 already present. And cos pi by 4 is 1 by root 2. So cos square pi by 4 would be, that would be 1 by 2. So this is 4 I0 into 1 by 2. And this becomes 2 I0. And we already know that I0 is given by W by 4. So this is, this is equal to 2 into W by 4. And the final answer, the intensity at point O, this comes out to be equal to W by 2. Alright, you can try more questions from this exercise in this lesson. And if you're watching on YouTube, do check out the exercise link which is added in the description.