 Let's solve a couple of questions on slit width in a single slit diffraction. Now for the first one, we have a blue-green light of wavelength 4500 angstrom, which falls on a diffraction slit and creates a central maximum of angular width delta theta, 10 to the power minus 3 radians. The question is to figure out the slit width and we can assume that the central maximum casts a tiny angle on the slit. Alright, as always pause the video and first try this question on your own. Okay, hopefully you have given us a shot. Now we need to figure out the slit width and we know the wavelength of the light and we know that the central maximum is casting an angle of 10 to the power minus 3 radians. Now we can start by drawing a single slit and a screen and how the pattern looks like on the screen. So this is how it can look like. These rays that are going to meet at the first minimum, they are making an angle of theta. You can see they are making an angle of theta with the horizontal and this part difference right here, this part difference, this right here. So this would be, this would be a sin theta. But the thing is that the central maximum is casting a tiny angle on the slit. So we can, we can assume, we can assume theta in place of sin theta. We can take theta in place of sin theta. We know the condition for minimas. It is a sin theta, which is equal to equal to n lambda. Now we can, we can use theta, the small angle approximation. We can use a theta and for the first minima we can use n equals to 1. So this is, this is equals to lambda. Now this means that a, the slit width, this is equal to lambda divided by theta. All right. And we also know that the central maximum, the central maximum is casting an angle of 10 to the power minus 3 radians. So if we look at this diagram, this is the total angle that the central maximum is casting and you can see that. And this angle is not just theta, it is theta plus theta. This is true theta. So 2 theta, this is equal to 10 to the power minus 3 radians. So theta would be, theta would be 5 into 10 to the power minus 4 radians. All right. Now we place theta over here and in place of lambda we place 4500 angstroms. So we can first change, change this to, change this to meters. And when we do that, this would be 4500 into 10 to the power minus 10 meters because 1 angstrom is 10 to the power minus 10 meters. So after placing these values and let's, let's do that a, this is equal to 4500 into 10 to the power minus 10 divided by 5 into 10 to the power minus 4. And this comes out to be equal to 9 into 10 to the power minus 4 meters. This will be, this will be 9 into 10 to the power minus 4 meters. And this really is 0.9 millimeters. All right. Let's look at one more question. Here we have Thomas who engineers a very small diffraction slit that creates a wide central maximum of angle of it 120 degrees. When light of wavelength 2000 into root three angstrom is shined on it. What is the width of this slit? So this question looks very similar to the previous one, but here there is no small angle approximation as such. The angle that the central maximum is casting, it's actually 120 degrees. The angle of it is 120. It's not a small angle. Okay. Again pause the video and first try this question on your own. Okay. Hopefully you have given this a shot. Now just like in the previous one, we will start off by drawing what the situation is. So we have a slit of width A and the rays that are coming out from the two ends, they are meeting at the first minima point. Let's say that B, this is the first minima point. So this is the first minima point. And if that is true, then that means that the part difference between A and B, if we can draw that, this part difference, this part difference right here, this would be, this would be A sin theta and that will be equal to lambda. Because for the first minima point, we will, generally we write N lambda, but for the first minima, N is equal to one. So it turns out to be A sin theta equals to lambda. We need to figure out the width of the slit. So A really is lambda divided by sin theta. Now we know what lambda is. It is 2000 into root 3 angstrom. So we can write 10 to the power minus 10 meters divided by sin theta. Now what is theta over here? Well, if we look at the question, we see that the central maximum is of an angle of it 120 degrees. Now, since we know that the central maxima lies between the first two minima points, it kind of looks like this right. So the central maxima, it lies between these two, these two first minima points. And if this angle is 120 degrees, if this angle right here, if this is 120 degrees, then half of this would be 60 degrees. So this angle theta, this angle really is 60 degrees. Because the rays that are coming from this end A, they are really going to this point B and they're making an angle of 60 degrees, half of 120 with the horizontal. So this is sin 60 degrees and sin 60 degrees is root 3 by 2. So root 3 really gets cancelled off. 2 remains, but let's just write everything. So we have, this is root 3 by 2. Root 3 gets cancelled off. This 2 goes to the top, becomes 4000. Then 4000 into 10 to the power minus 10. That is 4 into 10 to the power minus 7 meters. And we can write this as 0.4 micrometers. Because 1 micrometer, that is equal to 10 to the power minus 6 meters. So this is option C. All right, you can try more questions from this exercise in the lesson. And if you're watching on YouTube, do check out the exercise link, which is added in the description.