 Let's solve a couple of questions on fringe width in double slit experiment. For this one, the two slits are illuminated by a monochromatic light of wavelength 500 nanometers, and the slits are 0.12 millimeters apart, and the screen is kept 40 centimeters away from the plane of the slits. The question is to figure out the distance, the distance between the central bright fringe and the third bright fringe. Alright, before I get into this, pause the video and give this a try. Alright, now let's see what the question has given us. First of all, let us draw the screen and the two slits. There you go. And it says there are two slits illuminated by a monochromatic light of wavelength 500 nanometers. So we can have a light source giving out a monochromatic light of wavelength. The wavelength of this light is 500, 500 nanometers. The slits are 0.12 millimeters apart. So that is the distance, as you can see from the cursor, this distance right here, this is 0.12 millimeters. And also the screen is kept 40 centimeters away from the plane of the slits. So that distance is right here, this is 40 centimeters. And now we need to figure out the distance between the central bright fringe and the third bright fringe. Now as we have seen, when these light waves pass through the slits, they undergo diffraction and the diffracted waves from both of these slits will interfere with each other and that will result in a pattern on the screen. So the light waves after diffraction can look somewhat like this and when they are incident on the screen, you will get a pattern which looks like this. You will have places where the waves are interfering constructively, so you have a bright fringe at those places and at the other places where the light waves are interfering destructively, you get a dark fringe. Now the fringe width, the fringe width is the distance between the center from one bright fringe to the adjacent bright fringe. So one fringe width would be this distance right here. We can denote this by the symbol beta. This is the distance between the central bright fringe and the first bright fringe. The second bright fringe is right over here and this distance is also, this distance is also beta. Similarly, this distance right here, this will also be beta. This is the distance between the second bright fringe and the third bright fringe. Let us give numbers to the fringes also. This is the center fringe. This is the first bright fringe. This right here is the second and this is the third. You can see that the intensity of the bright fringes, it decreases as you go further away from the center. The question is asking us the distance between the central bright fringe, that is this, and the third bright fringe. So that would be adding fringe widths, that is beta, three times. So we have the fringe width beta and this is equal to lambda, that is the wavelength, into capital D, the distance between the plane of slits and the screen, divided by the distance between the slits, that is small d. We need to find a distance between the central bright fringe and the third bright fringe. So this would just become, this would just become, we are interested in three times of fringe width. So we multiply both the sides with three and there you go. We need to calculate three lambda d divided by small d. Let us put in the values and see what do we get. This becomes equal to three into lambda, that is 500 nanometers, so 500 into 10 to the power minus nine. We need to be very careful about units in such questions. Then capital D, which is 40 centimeters, so that is 40 into 10 to the power minus two, divided by, divided by small d, that is 0.12 into 10 to the power minus three meters. And when you work this out, you will get the distance between the central and the third fringe to be as 0.005 meters. But we need to express the answer in millimeters. So this becomes, this becomes five millimeters. So the answer over here is five. Let's move on to our second question now. For the second one, we have a monochromatic light of wavelength 620 norm, which shines on two slits separated by 0.5 millimeters in an interface experiment. And the question is to figure out the distance of the screen, how far should the screen be placed from the plane of the slits to get a fringe width of 1.2 millimeters. Again, before I get into this, pause the video and give this a try. Hopefully you have tried this on your own. This is very similar to the question that we did previously, but we need to find something else in this one. Okay, now let's see what all information is provided in the question. So firstly, the wavelength is 620, this is wavelength. We have the distance between the slits, that is 0.5. So this is small d. And we need to figure out capital D, how far should the screen be placed? So let me put d and a question mark next to it. And we also know the fringe width beta, that is 1.2, this is 1.2 millimeters. Okay, now let's bring the setup in. So we have the screen, the source and the two slits. And you have light waves incident on the slits which undergo diffraction. And then the waves from both of these slits, they can constructively and destructively interfere. And as a result, you get a fringe pattern on the screen, which looks like this. Now we know the distance between the slits, that is 0.5. We need to figure out this distance, capital D and we also know the wavelength of this light, that is 620 nanometers. And we also know the fringe width, that is the distance between the centers of two bright fringes. So that is this distance right here, this beta, this is given to be as 1.2 millimeters. All right, let's again use the expression which relates the fringe width with a wavelength and the distance between the slits, also the distance between the plane of slits and the screen. So that expression is given by beta, this is equal to, this is equal to lambda into D, capital D divided by small D. In this case, we need to figure out capital D. So let's keep capital D on one side and take everything else on the other side. When we do that, we can get capital D, this is equal to beta into small D divided by the wavelength. Okay, now let's try and solve this. This would be 1.2 into 10 to the power minus three, because this is in millimeters, into 0.5 into 10 to the power minus three divided by the wavelength and that is 620 into 10 to the power minus nine. And when we work this out, we will get the answer as 0.9 meters. So this answer, the distance between the plane of the slits and the screen, this is 0.9 meters. You can try more questions from this exercise in this lesson and if you're watching on YouTube, do check out the exercise link in the description.