 Hi, I'm Zor. Welcome to Unisore Education. I would like to spend some time to explain some basic principles on which holography is basically created. Holography is something which allows us to see some image in three-dimensional space. So, I'll talk about how basically can be done. Only very basic principles, without much technical details, obviously. I think it's very important to understand the principle. And that's basically the whole education up until some college level probably is supposed to be geared to. You have to know the principles, you have to understand what's behind some event or characteristic or property or something. Now, this lecture is part of the course called Physics for Teens, presented on Unisore.com. I suggest you to watch this lecture from the website. If you found it on YouTube, for example, it's only the lecture. The website contains the course basically, which means all the lectures logically related to each other with menus properly referenced to the videos, and also every video contains every video has a supplement, a textual part, which basically like a textbook. So, the same information which I'm trying to convey orally right now is written right besides the video lecture on Unisore.com. And the site is completely free, by the way. There are no advertisements, etc. Also, there is a prerequisite course called Mass for Teens on the same website, which I would suggest you to familiarize with. The physics without mass actually is not a subject to learn by itself. Now, to create image of three-dimensional image of some object well, it's like two different stages. First, we have to record that image on some media, and then recreate it. It's like photography. Photography has some kind of an object. You make a picture. This picture is somehow recorded on some device, whether it's a memory chip or some photographic chemical-based composite or whatever. And then you can view it separately as a photograph without the object from which it was taken. Same thing here. So, we have to be able to record the image on some media, and then when the object is no longer there we have to be able to recreate it in some place in a three-dimensional way, so we'll look at this object as if it was at that time. Before going into the complex object, today we will concentrate only on an object which contains only one point. So, let's consider something which is called point light. Point light is a point from which the light is emitted. So, if you consider, for example, sun as a point of infinitely small dimensions, the rays are going into all the different directions. So, this is our object which we would like to record the image of it, and then to be able to recreate that object in some other place using the media on which it was recorded. Now, I'm suggesting right now something which, well, some famous physicist came up with as idea how to record and then later on recreate the image of a point light. Okay, so here is what was suggested by some smart guy. I forgot his name. In the textual part of this lecture I do put his name. It's some Hungarian name Gabbard or something. I don't remember. Gabor, I think. Okay, so here is what he has suggested. Let's consider we have a point light. Now, let's put here some kind of a, let's call it a screen. I think it makes sense right now to consider this screen to be a glass. Now, what's very important and that's the fundamental for holography is that this light, this point light emits monochromatic light. Now, it's very, very important. Again, it's fundamental for holography. So this point light emits holographic, emits monochromatic light. So the light has exactly some kind of wavelengths, lambda. Now, as it emits, all the rays are obviously in phase with each other. Which means on the same distance from this source of light, rays are coming in phase. But they're not coming in phase onto this screen, obviously, because the difference of the length between this and this is different. So if it's a spherical surface around the point light, then on that spherical surface, all the rays will be in phase. But on a flat surface, they will not. So we don't need this spherical anymore. We will record these lights somehow on this surface. Now, question is how can we record it. But right now, without anything else, light is actually shining on all surface of the screen evenly without any kind of difference. Everything will be maybe a little faster. The light faster will reach this particular point because it's closer than that. But after it's reached, light is light. Also, you might actually consider that the amount of light the light will be more dense here than further down. And the further you go, the intensity of the light will probably be per unit of square centimeter or whatever of this screen will be less. But again, that has nothing to do with recording an image of this point. We cannot, just using this screen, we cannot recreate the position of this point. Okay, so what can we do? Here is a suggestion. Let's have another source of light. But in this case, it will be with a flat wave front. So all the rays are parallel. Here all the rays are radial. But these are all parallel. Also consider it's exactly the same monochromatic light of the same wavelengths. That's very important. I have something like this picture in the text part of this lecture on Unisor.com. It's probably a little bit better, but in any case. So we have these rays of light and these rays of the same monochromatic light, the same wavelengths. Okay, now let's assume, and it doesn't really jeopardize the generality of our presentation, let's assume that the light which goes here to a middle point. So this is S, this is A. So A is the closest. The S is perpendicular to a screen. So that's the closest distance between points of light and the screen. So one of the parallel lights goes along this line S8. Let's assume that this light which goes parallel rays and the light which goes from the S, from the point source to the same point. Let's consider they are in sync. They are in phase. Now, if they are not, I'll tell a little bit later what happens. If they are not, it has absolutely no difference for the further explanation. Okay, so these are the same, which means if they are the same, point A would be the point of constructive interference between this ray which comes from one of the parallel rays and the ray which comes from the point source S. So point A would be light, bright. Okay, because these two rays reinforce each other. They are in phase. Now, let's consider that the point B is such that Sb minus Sa is exactly lambda. What happens then? Well, let's consider the point B and let's consider one of the parallel rays which goes to the same point B. Now, will this parallel ray and Sb ray will be in phase? Well, let's consider all parallel rays are in phase with each other. Which means that this particular ray and this will be in phase. Now, but this one of the parallel and Sa radio are in phase as well. We have assumed this, right? Which means that the difference between the phases at point B, between this and this would be lambda. These are in phase. This one is lambda greater than this. Which means that the difference in phase between this and this would be lambda, which means they are in phase again. So, whenever the difference in phase is exactly the wavelength, we will have they are in phase. So, if this is true, point B will be in phase. Now, if Sb minus Sa is equal to 2 lambda, it will be somewhere here. Let's say it's B prime. Then again B prime would be bright spot because this ray and this ray will have a difference in phase equal to 2 lambda, which means they will be again in phase. So, all the points which are characterized by this equation all the points will be bright. And middle points between them will be dark. This will be dark, this will be dark, this will be dark. So, we will have bright and dark spots if this is a screen. Well, actually, if it's a screen in a three-dimensional world, it's like this, right? Then we will have a circle of bright spot. That would be a circle. That would be a circle, right? So, we will have bright and dark circles around the midpoint A on the flat screen, which is perpendicular to this board. Correct? Because if it's a circle let me put it in three-dimensional way the difference is exactly the same between Sa and Sb, wherever the point B is on this circle, right? It will be like a column. All these rays which are following on the circle around the point A will form a column and it will be exactly the same distance between the top of the column and any of the point on its base. So, we will have some kind of bright and dark circles, all concentric around A. So, the picture if we will view it from this point would be this bright, dark, bright, dark, bright, etc. Now, what happens if this central ray of the one of the parallel rays is not exactly in face with the ray which comes from point S? Well, then in the middle it will be instead of the bright spot it will be dark spot and white will be around it. So, again, it doesn't really change principally the whole picture. The most important is that we will have dark and bright concentric circles around the midpoint on the screen. Okay. Now, this is called interference picture. It's interference between all the rays which are coming from the point source and all the rays which are coming from all the parallel rays. Now, my statement right now is that this interference picture is very specific for location of the point S. It's also specific for its brightness and intensity of the light, but that's a secondary question. I'm talking about location. Because if you change location of point S either closer to the screen or up or down, the whole picture will change, obviously. Because the difference in length will be difference, etc. There are still concentric circles of bright and dark light but it will be different. So, it's very specific. Interference picture on this screen is basically somehow representing the location of the point S. That's what's very important. Now, I would like to consider this interference picture as a recording of the point S. What's with my voice? Now, how can I do it? Well, if this is a glass and on the glass I see dark and bright circles, well, let me do it this way. I will just paint the dark circles in some black paint. The bright ones will remain transparent. The glass is transparent. So now, I have this plate, if you wish, glass plate with dark circles at some point, concentric circles painted with paint. I'm stating that this is actually a picture of a three-dimensional picture of this point S. So, now question is how can I represent it? Alright. First of all, let's just talk about is it really like feasible to do this type of thing? Let's think about this particular equation. Let's consider this length as X. So, SB will be now, and consider the distance from S to A would be D. Then D squared plus X squared would be SB squared. Now, SA is G. So, what I can say is that this particular criteria for the bright spot D squared plus X squared square root minus G is equal to N times lambda. From which we can find X. For each N, we can find X. So, XN is equal to what would be equal. This would be plus so D plus N lambda squared minus D squared square root. So, for every N I have the location of this of a bright spot. How? It's actually a radius of a bright circle on the screen, right? For N is equal to 0, I will have D squared minus D squared. It will be 0, which means location will be right here. That's the first bright spot. Assuming that the parallel one of the parallel lights which goes into point A and the radial are in the face. Now, if not, it will be different. It doesn't really matter. The next one, so X0 is equal to 0. X1 is equal to, if N is equal to 1 it will be square root of D plus lambda squared minus D squared. X2 would be square root of D squared plus 2 lambda squared minus D squared, etc. Now, if you will consider this obviously the greater N is, the greater XN is. If N goes to infinity XN goes to infinity. So the reduces are greater and greater and greater of the bright spots. All the bright concentric circles will be of greater and greater radius to infinity. Obviously it's not infinity because our screen is finite, etc. They are talking about mass right now. What's actually another interesting thing is to find XN plus 1 minus XN. So how close to each other are these concentric circles become as N is increasing? Now, it's very easy to show that in the beginning the distance will be relatively large between, let's say, between 1 and X0 the difference will be this. Between X2 and X1 difference will be between this and this. It will be smaller and it can be shown that as N is increasing to infinity the whole thing actually is going to lambda. Why? Because if N is really great then this D and this D don't really make much difference if N is really great. D is constant. So when N goes to infinity this addition and this subtraction don't really mean much. So the whole thing would actually be N lambda square root. This is small relative to this which is equal to N lambda which means that the difference between N plus 1 and N's radiuses would be actually lambda, closer and closer to lambda. So the circles become closer and closer to each other and the difference between them would be closer and closer to lambda. Lambda is very small actually. We are talking about light. So it's impossible to paint our screen with a black paint let's say with a distance between them equal to 500 nanometers which is half a micron practically impossible. So we are not talking about technicality here. This is idea. So let's assume that I was able to paint in black the dark circles and leave the light circles completely transparent. So in depth what I am staging is sufficient to recreate the image relative to our screen. So let's say there is no more this point, this light which comes from point S. We took it away. We still have the screen. The screen has dark circles and transparent circles in between. So let me do it this way. What if I will put my lights from this direction exactly the same monochromatic light exactly the same wavelengths. What happens here? Well they will reach light points and dark points. Dark points obviously would be not transparent but the circles which are light will let the light through. So the light goes through. Let's consider every spot which is light, which is transparent and this light goes out. Now you remember the Huygens principle which actually says that each point where the light is reaching is becoming a source of waves in all directions and these waves are interfering each other. And because of interference of neighboring rays the light which goes straight will go straight and the rest one we don't really see much. So we can consider as a new point source every light spot which is here. It goes to all direction. Again whenever light reaches the point this point becomes the source of new points. It's a new point source from which the light goes in all directions. But it's waves, right? So the waves whenever the wave comes to a particular point it spreads the waves around it in all directions. And the water especially is very easy to see. If you say for instance the surface of water and you drop a stone here you will have circular waves but when it reaches some kind of an obstacle here with a small opening from this opening it will be. If you have two openings it will be like this. And these will interfere with each other. So that's how waves are spreading around. Alright fine. So every point every transparent point is a new source of light. What happens next? Well all these lights will interfere with each other. Whenever they are interfering with each other if they are at certain point in phase this would be a constructive interference. If not it will be destructive. So what happens next? Well let's consider these only these rays from the spots where the light goes through to some point S. If this point is on the same distance as original point source we know that all these rays have the difference between SB and minus SA would be some kind of a multiple of lambda. So if A is where my screen used to be and B is one of the points where the light goes through and S is a point where the point source used to be before this point will be a point where all these rays will come in phase and will enhance each other. So it will be a bright spot in space. All other points yes some rays would actually reach any other points but how many of them will be in phase? Well significantly less. I mean it's just an accident if they will be in phase. Here we have designed the whole system in such a way that as many rays as many transparent spots or transparent circles actually on the screen where. So it's a lot of intensity in this particular because all these transparent lights will have particular rays which are coming here into every other point we will have some rays and God only knows whether we'll be in phase or not in phase it's all kind of much less probable that they will come into significantly bright spot. So if experiments actually were made and obviously this point was the bright point as a result of the illuminating this screen from the opposite side and that's actually the most important principle. So whenever you have recording of the point source with these dark circles dark and bright circles and you have the light which goes through the transparent bright, which used to be bright spots they will lit this particular point where the original point source used to be very brightly and this is the most important part in holography. Everything else is kind of technicality. So instead of having the glass for instance and painting the dark circles with paint, which is basically impossible the practical implementation involves kind of a plate with I think silver based compound, which under regional light the bright spot actually becomes like pieces of silver and the dark spots become some silver compound so basically you have where the bright spots where you will have like mirrors you will have circular mirrors and whenever the light goes instead of this we have the light from here again original light, original parallel light so there is no point source anymore but there are silver circles on this plate and we direct exactly the same parallel light of the same wavelengths we will have the bright spot exactly where it used to be so this is the principle and it's one of the to implement this principle like in practice is definitely possible it's kind of beyond the scope of this lecture my point right now was to explain the main idea behind the holography that you have the monochromatic light which comes from the point source which would like to actually record the point source now recording is done on this plate and if you will take this plate somewhere else parallel lights of the same wavelengths you will have an image of this point relatively to the screen on the same distance basically where this point source used to be originally well that's the principle again, I suggest you to read the text for this particular lecture, it has a little bit better picture and my goal was to explain what's behind this technology, technology is developing definitely and there are many different little things which are making the whole thing better, but the most important idea is this, you have two different lights, one is the point light which you would like to record location relatively to the screen and the parallel lights which help basically to create an interference picture which becomes an image of this point coded in some way interference pictures some kind of a code in which we have recorded the location of this particular point and this code looks like in this particular case for a point light, it looks like concentric circles of different range and they represent the location of this point, now to recreate the location we need again the same parallel lights, the same it's called reference beam and in the absence of this point light we will have a bright spot exactly where it used to be ok, that's it, thank you very much again, read the text for this particular lecture, it's very important and there are some good pictures so you will understand how the holograms are made thank you very much and good luck