 Hi, I'm Zor. Welcome to Unia Zor Education. We continue talking about light. It's a very interesting and, well, very difficult actually, kind of science, type of science, because light is actually a very, very complicated subject. Well, we all know it's electromagnetic oscillations, well, oscillations of electromagnetic field, I would rather say, or electromagnetic waves. And it's transverse waves, not like sound, which is longitudinal. They are going like this. They are making more dense as they go, and they basically oscillate the molecules of air oscillate along the direction of sound in light. I cannot really say it mechanically because it's electromagnetic field, but the oscillations of electrical and magnetic components are across. So the light goes this way with electromagnetic field is oscillating perpendicularly to direction of propagation of light. Okay, fine. Just a few words. We have discussed in the last lecture the interference of two rays of light which go from two independent points, like two slits in some kind of object, whatever, and you have a screen and the ray from here and the ray from here towards any point here will interfere with each other. Why? Because the difference in these two lengths. Now, since there is a difference if the light comes to these two points as, let's say, flat wave front, which has exactly the same phase in these two points, then according to Huygens principle, these two points are sources of new oscillation of electromagnetic field. And into this particular point, S, let's say, since A S is longer than B S, the number of waves here and here will be different. So if they are in phase here, they might not be in phase here. And that's why depending on the position of the S we had, dark and light spots. Now we will talk about a little bit more complicated case. Now in these cases, these are two points, let's put it this way, which are sources of the secondary electromagnetic waves. Now we will talk not about these two very narrow slits, which are basically like points in the section. We're talking about section obviously. Now we will talk about a bigger opening. Well, it can be the lens, for instance, photo equipment. It can be our pupil of our eye. But in any case, it's something substantial, greater than the point. So I assume that the flat waves of light go, this is the front, and the whole thing actually is opening here. So every point gets the light, and from every point rays go into this particular point S. So my question is, so let's say this is the middle line, middle between A and P, and A, B is equal to G, and this is L, the difference, the distance to screen from this opening. And we obviously assume, as physicists obviously do always, that this is a very small one relative to this, because it's needed for approximation and simplification of calculations, you will see. Alright, so with these assumptions, so at points, at every point, at A to B, we have opening. So the light comes in, and all these points have exactly the same phase, because we're talking about some flat wave front coming into the whole thing. Now, my question is, what will be on the screen? Well, first of all, you will obviously have a bright spot, which goes basically parallel from A to a screen, and from B to a screen going down. But further outside of this bright spot, we will see again, light and dark areas. Well, if this is a slit, then we will have these lines on the screen. If this is a round hole, we will have concentric light and dark circles around the center spot. Okay, but no matter what we're talking about, two-dimensional section of the whole picture, so let's assume that we have this particular situation. And my question is, what will be at point S, depending on, well, location of the point S? Now, if this distance is, let's say, X, so my question is, what will be at point X? Okay, now let's think about it. First of all, the situation is more complex than in case of interference, because interference, we had only two sources, points. Now we have basically all the points from A to B. Well, mathematically, I could have said infinite number of points. From the physical standpoint, I'm always kind of hesitant to say infinite number of sources of light, because, you know, physics is different. It's a real life. So, photons might have certain dimensions, I don't know, but in any case, whatever it is, we assume this mathematical model, which means every point, which can be anywhere from A to B, and S can be again everywhere on the screen, will have some light from all the points from A to B. Okay, great. So, we know that if two rays of light come in phase, they will interfere constructively, they will enhance each other. Let's say this is my one line, and this is my another line, and if I will add them together, you see, the maximum goes to maximum, so I will have greater maximum, and minimum goes to minimum, I have a greater minimum, so I will have something like this, okay. Now, if, however, my waves are in anti-phase, which means maximum goes to minimum, minimum goes to maximum, something like this. So, this is the middle line, and this is the middle line, and if I will summarize them together, what would be, this is, let's say, plus maximum, this is minus, the same maximum, so I will have zero, and I will have zero everywhere, so I will have a straight line, I will have no light at all. So, only in case my two waves are in anti-phase, and what it means is, what is the shift one from another? This is the lambda, which is a wavelength, and right? From crest to crest. Now, here from crest to crest is this, so we are shifted by how much? By lambda over 2, right? So, this graph is shifted to the right, or to the left, whatever you want to say, by lambda over 2 from this graph. So, these two are in anti-phase, not just outer phase. Outer phase can be anything which is not exactly in phase, but anti-phase means exactly by lambda over 2. It means they will just cancel each other. So, let's just think about one particular case. The case is A s minus B s equals lambda, delta. So, delta is the difference between the longest and the shortest. Now, obviously, if you will compare the lengths of the ray from any point from A to B to S, as the point will move, let's say, from B to A, the lengths would increase, but if S is on this side of the middle line, right? So, every line from here would be, from here would be further than line from here, right? So, as we move from B to S to A, it will increase. Now, we are assuming we are above, so this is this bright spot. This is the bright spot, which, when the rays of light go through this aperture, it will just go straight without any kind of interruption and it will light this particular area. So, I should probably put it a little bit further down to the right. It will be more obvious. So, this is my straight line. S is somewhere here. So, obviously, as we move from B to A, the lengths of this is increasing monotonously. So, I assume that A s is by lambda greater than B s. Now, more precisely, if this is middle of this aperture, and this is x, then the lengths of A s would be equal to A s would be what? L square, and this is x plus half of this, the opening is D, so it's D over 2. So, L square plus x plus D over 2 square and square root. That's the Pythagorean here, right? Now, B s would be square root of L square plus x minus D over 2 square. If this is x, then this is x minus D over 2. And the difference between them is difference between these two radicals, right? Now, in the previous lecture, I actually simplified it a little bit because A s minus B s is equal to A s square minus B s square divided by A s plus B s, right? P minus Q times P plus Q equal P square minus Q square. Everybody knows this little algebraic formula. Now, the difference between these two squares, let's see what will be the difference. L square would cancel with this L square, because this, we're talking about square, which means we get rid of the radical and minus sign. So, L square goes, x square goes, D over 2 square goes, and the only thing which will remain would be 2 times x times D over 2, which is x, D. And this is minus x, D, but with a minus sign, so it will be plus x, D. So, it will be 2x, D. Now, on the bottom, I will have some of these two radicals, and I was saying that if L is great and x and D are small, then some of the, then each one of these, this is a little bit greater than square of L square plus x square. This one is a little less than L square plus x square. But if you add them together, you might consider that's approximately equal to this. With large L, that's basically true. So, this is approximately equals to this. And what is x divided by, x divided by the square root of L square? Now, this is the line from the middle to x. That's basically a sign of this, right? We are dividing this, x, divided by square root of L square, L square plus x square, which is hypotenuse. So, it's a sign of this, which is the same as sign of this, which is incident angle. So, it's approximately this, which is D times sine theta, where theta is incident angle. Now, again, we are considering that the Ds relatively small, and these two are more or less exactly the same angle as this one. So, again, it's a physicist level of precision. It works if the screen is at a substantial distance from the opening. But in any case, the precise answer is basically this. Now, let's forget about this little formula. It doesn't really matter. I will write it here. The delta is approximately D sine theta. Now, I'm totally thinking about one specific case when this is longer than this by lambda. That's very important now. That's the logic of this. Here is the... Now, if my difference is lambda, so as I'm going from B to A, my distance is gradually increasing until it reaches the lambda at the very end, difference between these two, between some kind of a point M. So, the difference between M, S minus B, S would be less than lambda, but greater than zero, right? As M, S is moving from B to A. Now, since my length is monotonously increasing, but not the length, the difference between M, S and B, S is monotonously increasing, there is one point where the difference is equal to M to lambda divided by 2. Okay, so, it's not necessarily... I mean, it's definitely not in the middle of this AB, but it's somewhere, because it grows maybe a little bit slower in the beginning and faster at the end. It doesn't really matter. So, there is some point M here, some point M here, where the difference in length between M, S and B, S is exactly lambda over 2. Okay, so, let's just consider two rays, B, S and M, S. Now, the difference in the length is lambda over 2, which means there is a difference in phase. That's exactly the situation which I was just talking about like this. The difference in length is lambda over 2. Well, that means that since this ray should cover... Now, they come here into these points M and B in phase. So, they are together at the same time. Maximum goes to maximum crest to crest, trough to trough. But since this is longer by lambda over 2, it will come a little later and it will be a trough from the M where the crest will be from the B, or vice versa. It will be a crest from the B, where trough is from the M, but trough from the B. And what happens? They will cancel each other. Great. So, we have one ray and another ray, and they cancel each other. Excellent. Now, let's move a little bit from the B to... Just doesn't really matter by which point. It will be slightly... Now, it will be B prime, S will be slightly bigger than S, plus something like delta. Okay. Now, delta is very small. It's definitely less than lambda. Definitely less than lambda over 2. So, we did not really reach point M yet. It's somewhere between M and B. Now, since my length is monotonously increasing, I will step from the M by some sink where to M prime, where M prime S would be equal to M S plus the same delta. Since this is increased by delta, I will start moving from M to the left, and I will also find something which is greater by lengths by delta. Now, what happens between M prime S and B prime S? Well, the difference between them is equal to exactly the same as difference between B S and M S, which is what? Lambda over 2. B S minus M S, or M S minus B S is lambda over 2. Okay. Which means what? Which means these two will also cancel each other. And now, we obviously understand that for each point from B to M, I can find corresponding point from M to A with which this ray cancels that ray. And vice versa. I mean, if I will start from the M moving to the left by some kind of delta, I can find another point between M and B, which is exactly by the same delta longer than B S. And again, the difference between this and this would be always the same. So, I can find from each point, I can find corresponding here with the ray cancelling this one. And from each from this, I can find this, which ray would cancel this. So, what it looks like that all the rays from M to B would be canceled by all the rays from A to M. It's a one to one correspondence between two segments, if you wish. And again, even if it's a continuum mathematically, like a real segment in terms of infinite number of points on it, or it's a physical kind of thing. If it's a physical, then with certain very reasonable approximation it's still true. So, for every point which is a source of secondary light according to the Huygens principle, I can find another point on this area from A to M, which will cancel this one and vice versa. Which means that this point S would be dark, all the rays would cancel each other. So, if this is true, the difference between distances is lambda, then it would be a dark spot. Now, if we deviate from this point left or right, obviously the amount of light will increase, because then I will have certain rays which will find, now the difference will be either greater or less than lambda, which means I will find something which is canceling each other, but I will find something which does not have a cancellation pair. So, this pairing would not work as easily. You see, this is lambda, so I divide it with another lambda two difference. Now, if it's greater than lambda or less than lambda, I will not be able to do it easily. So, let's just think about when is the next a really true dark spot. Well, that's very easy to say, that's true lambda. Why? Well, because if this is true lambda, I divide it by lambda and lambda, now all the rays from here would have difference lambda over 2 and lambda over 2 and here as well. So, it's two dark spots, two dark spots will result in a dark spot. And actually, anything of this type, n times lambda where n is integer number, would result in a dark spot. Now, if delta is approximately d times sine of theta, that gives me basically all the different theta where we will have dark spots. Obviously, we just resolve this for sine of theta, d times sine theta is equal to n lambda. So, for h n, we will have certain value of theta because d we know and lambda we also know, we are talking about some light, some monochromatic light. So, we have the length of the wavelength. So, we know the theta. In a more precise calculation, if you are interested to do it more precisely with all these radicals, then it would be much more difficult to resolve it for x. If you want to resolve it for x precisely, then it would be like, I would say one radical minus another radical equals two to this and you have to resolve it for x, which is kind of, I would probably say, you will have an equation of the fourth degree. If you will square it once, then square another, that's kind of difficult. So, physicists decided to do it easily and for large l, that's sufficient precision. By the way, if anybody wants to resolve this for x precisely based on exact Pythagorean theorem, how this equation can be presented in terms of x, not in terms of theta, be my guest, send it to me, I will publish it on my website. So, these are the dark spots. Now, as I was saying, the light spot would be in between, but where will be the brightest light spot? Well, it's somewhere in between the dark spots, right? Now, light will not be like exactly some kind of a bright, and then it will be a dark, black, basically spot, and then again, light and then dark. No, the boundaries will be gradual. So, from the brightest light, it will go down, down, down to the darkest black, basically, and then again. But we can still think that something like this would be the brightest spot for every end. Well, not forever end, up to certain number and we will talk about this. Now, why? Because let's say this distance is, well, in a simple case, let's say n is equal to zero. So, let's say, not zero, one, let's say, okay? So, n is equal to one. So, it will be three second lambda, okay? Now, what happens here is the following. I can divide this area into three parts. Now, the difference between this and this is lambda over two, difference between this and this is lambda two, and difference between this is lambda over two. So, I will have difference between these lengths, three lambda over two. Now, these will cancel each other, because the difference between this and this is exactly lambda, and this will be completely uncancelled. There will be no pair which will be completely out of phase, because to be out of phase, you need to be lambda two, right? But this is, the whole thing is lambda two, and this is already canceled. So, there is no other raise. To the left, it's a border, and on the right, all rays will cancel each other. So, there are no more rays, rays to cancel these rays. So, all of these would actually light up this point. Or if you wish you can basically just logically do it slightly differently. Now, these is two which can cancel each other, but then the light will be from this one. It doesn't really matter how you count, but one third of the total number of points here would actually light up the point s. So, this is for the lights, and without 0.5 would be for the darks. Now, again, just think about this as a simple exercise in, I don't know, geometry basically, no more than that. And obviously, we are approaching it very, very approximately here. Now, let me just finish it up with one very, very small observation. Are these dark and light spots going definitely here? Well, obviously not, because if you will use this formula, g times sine of theta equals to, let's say, n times lambda. And sine equals to 0.5 times lambda divided by d. Now, sine cannot be greater than 1, obviously. So, these rays will not go more than with incident angle greater than 90 degrees. So, from this, n times lambda divided by d should be less than 1, and n should be less than d divided by lambda. So, d is the distance between, well, the size of the opening and lambda is the wavelengths. So, if you divide one by another, that's basically number of lines or rings, whatever you will have around this main brightest spot, which is right across this opening. So, there is a certain number. So, if d, let's say, is a couple of micrometers, then you will have, and lambda is 0.5 times 700 nanometers or whatever. You will have basically like 5, 5, 4, depending on the wavelengths, obviously, and on the opening, size of the opening. And then another consideration, obviously the intensity of the light, even in the light spots, I mean, is decreasing, because we are going at a greater and greater incident angle, which means these lines would be a little bit wider and less bright, because the energy would be spread a little bit. Now, what's interesting is, if you use this formula, which does not depend on l, now what does it mean? If my l is further here, for instance, we will have the same angle, we will have the same dark spot. It will be here and it will be here. So, if we will move our screen further, the number of spots will be light and dark spots will be, or lines or concentric rings will be the same, but they will be wider, each one of them will be wider, obviously. Okay, so that's it. At conclusion, I have to recommend you, as usually, to read notes for this lecture. So, this lecture is presented on Unisor.com. Course is called Physics for Teens. You go to Waves. That's the big part of the course. And next, under the waves, you will have Phenomenal Light. And the diffraction of light is one of these lectures dedicated to Phenomenal Light. So, I do recommend you to read. There are some nice pictures over there, much better than I do, obviously. And some formulas, obviously. But it's always useful to basically read the textbook with the same exactly material as you have heard during my lecture. Okay, by the way, I didn't mention the site is completely free. There are no advertisements on the site. So, just open for everybody. Free for all, all for free. Okay, with this, thank you very much and good luck.