 Let's solve a couple of questions on coherence and interference in Young's double slit experiment. For the first one, we have two coherent sources S1 and S2 and a point P which lies on the perpendicular bisector of the segment S1 and S2. So it lies right at the center of S1 and S2. What is or are definitely true about the two light waves? And we have five options out of which more than one can be correct. Why don't you pause the video and try to attempt this question on your own first? All right, let's get into this. Now we have two coherent sources of light. So these two sources are coherent. What could two coherent sources of light really mean? So two coherent sources, it means that the two sources, they have the same frequency. They have the same frequency, which means that they have the same wavelength because all of them, they are sources of light, they are traveling with the speed of light. So if they have the same frequency, they must have the same wavelength. And that means that the color, the color of the light must be the same. Color must be the same. So if we read the options, option A is definitely right. They are of the same color. Option B says they have the same phase at the sources. Now it is entirely possible for these two sources to be out of phase or to be in phase or to have a constant phase difference, maybe of pi by three or two pi by three, pi by six. And it is not at all necessary that they need to have the same phase at the sources. There can be a phase difference between these two coherent sources of light. So option B is not right. Option B is wrong. Option C says phase difference at sources equals the phase difference at P. Now, if there is some phase difference at the sources, let's say pi. The option is saying that even at P, it has the same phase difference. If you look at the part traveled by the light from the source S1, that would be this path right here. We can see that this is the same as the part traveled by the light from the source S2. There is no extra path traveled by any of these two sources. There is no extra phase difference added between the two sources of light. Whatever is the phase difference at the source, it is the same as that at point P. This can only be true if they are traveling the same amount of path. So option C is right. Now, option D says they have the same intensity. Again, for two sources of light, which are coherent, they need to have the same frequency and also they need to have a constant, constant phase difference. It doesn't matter whether they have the same intensity or different intensity. If they have different intensity, we won't get entirely dark fringes because they will never be the case when I1 minus I2 is equal to zero because they are not equal to each other. There will be some brightness between the bright fringes as well. You will never get a completely dark fringe, but that does not mean that the two sources of light are not coherent. They are still coherent because you are getting an interference pattern on the screen. So option D here is wrong. And the last option says they have the same phase at P. Now, this can be true, this can be true for some cases, but it is not a necessary condition for two sources to be coherent. The only two necessary conditions are that it should have the same frequency and a constant phase difference. Okay, so for this one, we have A and C as the right answers. Let's move on to the next question now. Here we have two coherent sources of light, S1 and S2, which emit light of wavelength 600 nanometers. The light waves, they have a phase difference of pi by three at the sources. And they interfere constructively at a point P. R1 and R2 are the position vectors from S1 and S2 respectively to P. And the question is, which of the following is most likely to be true? And we have five options out of which one is correct. Again, as always pause the video and give this question a try. Okay, let's get into this. Now the question says that there is a phase difference of pi by three at the source. There is a phase difference of pi by three at the source. What would this mean? We can try and think about this in terms of wavelength, because then we will be able to plot the two waves from these two sources. So as a general rule, we know that if the phase difference between two sources is two pi, then they could be zero or lambda distance apart from each other. And if the phase difference is pi by three, then we can say that they could be lambda divided by two pi into pi by three apart. So this is lambda by six. This is lambda by six. So now if we draw the two sources of light, they could look somewhat like this. Let's say that this wave is from S1 and this wave is from S2. These two sources of light are out of sync. They are not in phase. They have a phase difference of pi by three. So that corresponds to a path difference of lambda by six. Now that means that the distance between these two waves, that would be lambda by six. So the distance between the peaks right here, that is, this is lambda by six. Even this distance, this distance between the peaks is also, this is also lambda by six. Similarly, even, even this distance will also be lambda by six. Now let's say if we had a screen at this position, at this position right here and our point P, our point P is this point. This is our point P and we have a screen at that point. We can see that the wave from S1 is at its peak, right? But the wave from S2 is not. This is where the wave at S2 is. It is not at its peak. So we will not be getting a constructive interference if these two waves travel the same distance, same amount of path from the source. And when this is the case, these two are traveling the same amount of distance. But we are getting a constructive interference at point P because the peaks are not matching with the peaks. They're not on top of each other. So maybe one wave has to travel more or the other wave has to travel less so that, so that these two peaks, they are on top of each other and you get a constructive interference. Let's look into this some more. Let me make some space. So now here we have a wave from S1 and this is the wave from S2. And a screen was, a screen was at this position, this position right here. We can see that the peaks are matching with the peak so there won't be any constructive interference. But let's say if the wave from S2, that is a red wave, if it looked like this, then we can see that the peaks are matching. So now there will be a constructive interference at this point, the point P. But what happened so that the constructive interference is happening? We can see that now the wave from S2, it is moving an extra path. It is moving some extra distance so that there is some constructive interference. And what is this extra distance? This is exactly equal to, this is exactly equal to lambda by 6. Because we saw that a distance between two adjacent peaks or two adjacent points, that would be lambda by 6. Part difference of pi by 3 corresponded to a distance of lambda by 6 between the two waves. So even this distance is lambda by 6. And that is the extra amount of distance that the wave from S2 has to travel so that the peaks coincide and you get a constructive interference. But it's not just lambda by 6. Even if this point coincided with this point, then again you will get a constructive interference. So for this point to be able to coincide with this point, the red wave will have to travel lambda by 6. It is already traveling lambda by 6 more, plus lambda more. Plus lambda more, one complete wavelength. That is the distance between the two crests, right? So it has to travel lambda by 6 plus lambda. And similarly, if you extend this red wave further, there can be one more crest, there can be one more peak, which can coincide with this peak, and you can get a constructive interference. In that case, the wave will have to travel 2 lambda more. Apart from traveling lambda by 6 more, it has to travel 2 lambda more than that. Now let's come back to our question. Let me bring the question back. So we see that the part difference has to be lambda by 6, plus 1 lambda, 2 lambda, 3 lambda. So we can write this as n lambda. And we know that the wavelength is 600, so this would be 600 divided by 6, that is 100, plus 600 into n. Now let's look at the options. Now let's look at the options. The first one says that mod of r2 minus r1 is equal to 100 nanometers. Here we need to be a little careful, because we know that r1, r2, these are position vectors. And r2 minus r1 will actually be this distance right here. This is r2 minus r1. This is according to the rules of vector subtraction in a triangle. This is your r2 minus r1. And this is the distance between the slits. It has nothing to do with the part difference of the light from s1 and s2. So we can write away ignore option A, and with a similar line of reasoning, we can ignore option C as well. That leaves us with B, D, and E. Let's look at option B. Option B says mod of r2 minus mod of r1, that is equal to 700 nanometers. This kind of makes sense, because now if in place of n, in place of n, if we place 1, then this would be 100 plus 600. And that will give us 700 nanometers. The part difference has 700 nanometers. So option B makes sense. This is right. Option B says this is equal to 1, 2, 5, 0 nanometers. This will never be really possible, because no matter what value of n we place here, we will always get the final part difference as an integral multiple of 100. We can get it as 100, 700, or when we place n as 2, we will get it as 1300, and then 1900, so on and so forth. So it will always be a multiple of 100. Option D is wrong. And again, with similar line of reasoning, option E is also wrong. So for this one, the right option is option B.