 Let's solve a couple of questions on overlapping fringes in double slit interference. For the first one, we have a beam of light which contains blue light of wavelength lambda 1 equals 6000 angstrom and orange light of wavelength 8000 angstrom. And this produces interference patterns in a setup of the double slit experiment. The question is to figure out which of the following bright fringes overlap and you have four options right over here. Okay, before I get into this, why don't you pause the video and first try this one on your own. Alright, hopefully you have given this a shot. Now let's try and break down this question. So first of all, let us bring in the double slit experiment setup. So you have the plane, the screen and the plane of the slits. And in this one, we have a beam of light which contains blue light and orange light also. So both the blue and the orange light, they get diffracted from the two slits and they produce an interference pattern on the screen. Now there will be places where there is a bright blue fringe and there will also be places where there's a bright orange fringe. And it's a possibility that some of those fringes might overlap. And that is what our goal is. The goal is to figure out which bright fringes overlap. So let's say that there is a bright fringe at this point due to both of these lights. There is a bright fringe due to the blue light, so you get a blue bright fringe. And also there is a bright fringe due to an orange light, so you get a bright orange fringe. Now one thing that is common in these overlapping fringes is their distance from the centre. They are at the same distance from the centre, right? So if we say that the N1 order of blue fringe, this could be any number, this could be any order, 16, 12, 11, 10, we don't know. Let's say the N1 order of the blue fringe that overlaps with the N2 order of the bright orange fringe. Again, this could be any number. And the distance from the centre is the same. Distance from the centre of these overlapping fringes is the same. So that means Y1 equals Y2 if we call this distance Y just in general. So for blue light it could be Y1 and for orange light it could be Y2. This distance is the same for both. And we can further express this mathematically. We know that Y1, the distance from the centre to the N1 order of bright fringe, that would be N1 into lambda 1 into capital D divided by small d. And this is equal to N2 into lambda 2 into capital D divided by small d. Here capital D is the distance between the two planes and small d is the distance between the slits. We can see that couple of quantities get cancelled. You have these two quantities getting cancelled. So at the end what remains is N1 lambda 1, this is equal to, this is equal to N2 lambda 2. And we can write this as N1 divided by N2 and that is equal to, that is equal to lambda 2 divided by lambda 1. Now lambda 2 is 8000 angstrom and lambda 1 is 6000 angstrom. So when you work out that fraction that comes out to be equal to 4 divided by 3. So we have 8000 divided by 6000 and that is 4 by 3. Now we have a relation between N1 and N2. N1 divided by N2 must follow the ratio of 4 by 3. And we can cross check this from the options that we have. So if we look at the first option we have N1 here is 16 and N2 is 9. And 16 divided by 9 is not 4 by 3. So the first option, the first option is wrong. The second option says N1 is 16 alright and N2 is 8. So N1 divided by N2 this becomes 2 right. So this is again wrong. Second option is wrong. Third one says 12th blue fringe. So 12 divided by 10 that is, that is 6 by 5. Again it's not 4 by 3. So even this is wrong. That leads us to the last option. This is N1 that is 12 divided by N2 9. So this is 4 by 3. So this is, this is the right option. The 12th blue fringe is overlapping with the 9th orange fringe. And it's not that just these two fringes are overlapping each other. Whichever fringe that is related in this manner N1 divided by N2 is equal to 4 by 3. So even the fourth, even the fourth blue fringe must be overlapping with the third, with the third orange fringe. And if you find equivalent fractions of 4 by 3, all of those fringes would be overlapping each other. Let's move on to our next question now. For the second one we have a coherent beam of light which contains two colors. Green, light or wavelength, lambda 1, 5700 angstrom and blue of wavelength 4500 angstrom. And this is used as a source light in a double slit experiment. We also know the ratio of capital D divided by small d which is 3000. We need to figure out the least distance from the central maximum where a green dark fringe overlaps with a violet dark fringe. Okay, now the dark fringes are overlapping and we need to figure out the distance. The index of variables over here is just basically specifying that the capital D is the distance between the screen and the plane on the slits and small d is the distance between the slits. Okay, again before I get into this, try this one on your own. Alright, hopefully you have given this a shot. Now, we need to figure out the least distance from the central maximum where a green dark fringe is overlapping with a violet dark fringe. So again, if these two fringes are overlapping, then their distance from the center is the same. That is the distance that we need to find out. So let's say for green light, for green light, we can call that distance as y1 and for violet light, let's call that distance as y2. And we know that y1, y1 is equal to y1 is equal to y2. Now, at this point, we can try and recall the distance at which a dark fringe is formed. Why don't you pause the video and try to recall it? Okay, so generally, generally a dark fringe, a dark fringe of nth order could be the first dark fringe, second dark fringe, third or fourth. That is formed at a distance of n plus 1 by 2 into lambda capital D divided by small d, right? And in this case, we don't know which orders of green dark fringe and violet dark fringe are overlapping. So we can just give them a general name. We can call, we can say that the n1 order of green dark fringe overlaps the n2 order of violet dark fringe. So that distance, in that case, that distance would be n1 plus 1 by 2 into lambda 1 into capital D divided by small d. And this is equal to, this is equal to n2 plus 1 by 2 into lambda 2 into capital D divided by small d. So right away, we can see that these two quantities, they get cancelled. And now again, we can try and figure out the ratio between n1 and n2 because we already know what lambda 1 and lambda 2 are. So let's try and work that out. This would be, this would be 2n1 plus 1 divided by 2 into lambda 1 equals 2n2 plus 1 divided by 2 into lambda 2. Over here, 2 gets cancelled and this is, this is good amount of math. This is 2n plus 1 divided by 2n2 plus 1. And that is equal to lambda 2 divided by lambda 1. Now zeros get cancelled and we can further simplify this fraction. Both of these numbers are divisible by 3. So this would be, this would be 15 divided by 19. Both of these are divisible by 3. And we can write 15 as, we can write 15 as 2 into 7 plus 1, 2 into 7 plus 1. 14 plus 1, that's 15 and 19 as 2 into 9, that is 18 plus 1. Now if we compare the left hand side with this, we can see that n1, n1 is equal to 7 and n2, n2 is equal to 9. So the overlap, the overlap closest to the center maximum, the least distance closest to the center maximum happens when the seventh green dark fringe overlaps with the ninth blue, sorry your violet dark fringe. Alright, but we're not entirely done with the question now. We need to figure out what that distance is. So what we can do is we can pick any, either of these two orders and put that into either of these two equations. So let's pick n1, 7 and place that, place that over here. Place and try to figure out what y1 is basically. Even if you figure out y2, that will give us the same answer. So y1, y1 this is equal to 7 plus 1 by 2, 7 plus 1 by 2 and that is multiplied by lambda into capital D divided by small d. Luckily we have the ratio of that, that is 3000 and lambda 1, this is 5700, this is multiplied by 3000. Alright, now we can see that the answer, the units of these values, they are in millimeters. But the wavelength is in angstrom and one angstrom, one angstrom is equal to 10 to the power minus 10 meters. So maybe we can write that, we can write y1 as 7 plus 1 by 2, that is 7.5, this is multiplied by 5700 into 3000 into 10 to the power minus 10. And we will get y1 in meters, then we can change it back to millimeters to pick the right option. So first let's calculate what y1 is. This comes out to be equal to 0.0128 meters and in millimeters you can multiply this with 1000. So this will be, this will be 12.8 millimeters and that is the last option, that is option D. Alright, you can try more questions from this exercise in this lesson and if you are watching on YouTube then do check out the exercise link which is added in the description.