 Let's solve a couple of questions on single slit diffraction where the slit width is changed. For the first one we have a diffraction experiment where the slit is made smaller by a fifth. And this causes a change in the diffraction pattern on the screen. The question is to figure out what happens to the central maximum. There is a note which says that assume that the central maximum casts a tiny angle on the slit. And we have to choose two answers out of these four. As always, why don't you pause the video and first give this a try. Alright, hopefully we have given this a shot. Now the question is asking us to think about what happens to the central maximum if the slit width is changed. So before we go into the options, before we see what angle of width or size means, let's try and draw the setup. So here we have the screen and here we have the single slit. We can see that the distance between them is capital D and the slit width is A. And now the diffraction pattern on the screen because of the light waves interfering at different points, it looks like this. You have a central maximum with a great intensity and the intensity decreases as you go further away from the center. The question is asking us to think about central maximum. So let's just focus our attention on this region of the interference pattern. And we need to think about angle of width and size. We can think about size first. Size could be just the width of this central maximum. It could be the width. So it could be this distance right here, the entire width of the central maximum. And the angle of width is the angle that the central maximum makes on the slit. So wherever the central maximum is ending, whatever angle it makes on the slit, that would be the angle of width. And that could look somewhat like this. This complete angle, this complete angle would be the angle of width of the central maximum. And I've divided this into two as half of the central maximum is over here and half is over here. So these two angles are the same. But this total angle is the angle of width of the central maximum. Since the angles are the same, we can call both of them as theta. Now in single slits, we know that the part difference A sin theta, A sin theta, the part difference between the waves coming out from the single slit. This was equal to, this was equal to M lambda for destructive interferences. So wherever we have a minima, the part difference is equal to M lambda, an integral multiple of lambda. So this condition is wherever a minima is formed. And we can see that the first minima, that would be the first order minima or M equal to 1, that would be A sin theta equal to equal to lambda. The first minima is formed where the central maximum ends and theta over here is small because a note says that the central maximum is casting a tiny angle on the slit. So this entire angle is tiny, so half of it will be tiny as well. We can approximate sin theta as theta. We can write this as A theta that is equal to lambda. And particularly for this point where the central maximum ends, where the first minima forms, this is the first order minima. So we can call this as theta 1, this is also theta 1. And over here theta 1, theta 1 becomes equal to lambda by A. Now the angle of width of the central maximum will be double of this. It will be 2 theta 1. So when we write that, when we write 2 theta 1, this becomes equal to 2 lambda by A. This is the angle of width of the central maximum. And now we can see what is happening to the slit. The slit is being made smaller by a fifth. So this is A old, we can say this is the original A. The new A, the new A becomes equal to A by 5. And we see what can that do to the angle of width. This 5 will go up and it will be 10 lambda by A. So what happens is your angle of width increases by 5 times. It increases by 5 times. We can now think about the size, this entire width of the central maximum. For this, we can take help of the triangles that are being formed over here. We can focus on this top right angle. So the angle is theta 1, this angle is 90. And this side of the triangle is half of the total width. So we can call that W. Later we can write 2W to figure out the total size. We can write this as tan theta 1 equal to W, W could be the side, this side length. W divided by capital D. Capital D is the base of this right angle triangle. And that theta 1 is sin theta 1 divided by cos theta 1. When theta 1 is small, sin theta 1 can be approximated as theta 1. And cos theta 1 is just 1. Because when theta 1 becomes really small, tending towards 0, let's say, cos 0 is 1. So we approximate cos theta 1 as 1 for smaller angles. Now let me write this over here. This becomes theta 1 equal to W by D. And W is just 1 half. So we can write 2W, that is the size of the central maximum. So this is 2D theta 1. And in place of theta 1, we can write lambda by A. So this becomes 2 lambda D by A. And now again, when the slit is made smaller by a fifth, this A, the new A, the new A becomes A by 5. A by 5. And again, this 5 goes to the top, gets multiplied by 2. And the width just increases by five times. You have this 5 going to the numerator. So the size, the size increases by five times. All right, now let's go to the next question. Here we have the width of the slit being tripled in size. And we can see how the size changes. What happens to the intensity of the central maximum? We need to choose one answer out of these five. Again, pause the video and give this a try. All right, so over here, we need to think about intensity of the central maximum. And when we try to recall the intensity of a wave, we can say that intensity was always proportional to the square of the amplitude. And we can think of the slit. We can think of this slit as being made up of, let's say N, N secondary, N number of secondary sources. So you have the primary source, which is sending all these light waves. And then when it passes through the slit, you have all these secondary sources which act as individual sources, creating their own waves. And that results in an interference and you see the pattern on the screen. This is what Huygens' principle said. All of these points act as secondary sources, creating the pattern. So if there are N number of such sources, let's say there are N number of such sources, and each source creates a wave with an amplitude of, we can say the amplitude is A with an amplitude of A. Then the amplitude at the central maximum, the amplitude at the central maximum will be the sum of all of the amplitudes produced by each of these sources. That is what central maximum is. This is a constructive interference. All of the sources are interfering constructively. So total amplitude at the central maximum, that would be N into A. So this is the amplitude. Intensity, we can say, is proportional to N A square, N A whole square. Now the width is being tripled. So now there are three N sources. And each source produces a wave of amplitude A. So the total amplitude of the central maximum, that would be three N A. And now the intensity, now the intensity is proportional to three N A whole square. When we take three outside, this becomes nine N A whole square. So we see that the intensity has increased nine times. And that makes this option B. All right, you can try more exercises from this lesson. And if you're watching on YouTube, do check out the exercise link in the description.