 Hi, I'm Zor, welcome to Inizor Education. Today we will talk about a practical problem, which basically all the people know about, they all see this type of phenomenon if you wish, and it's about the water depth problem. Now, what's interesting about this particular problem? When I was in high school, that was the problem in one of my exams. So in any case, it's about light diffraction, it's about light refraction. And using whatever the knowledge we have already obtained about refraction and its laws, we will try to solve this particular problem. We will try, we will not exactly solve it, but we will do approximate solution. Okay, this lecture is part of the Unizor.com's course called Physics 14. So if you go to the website, you will see this course and Maths 14, the prerequisite course as well. And just using the menu, you can find the particular part of this course, which is waves, and inside the waves, the topic is Phenomenon of Light. This is one of the lectures in that particular topic. Now, I do recommend you to use the website rather than, for example, you can find this lecture on YouTube using the search engine, but again, you have to understand this is part of the course. So I'm always using the knowledge which have been presented before in any subsequent lecture. So there is a very logical connection between the lectures. And that's why I do recommend you to use the website directly using the menus. You can find the prerequisite lectures, etc. Plus, the website has exams. By the way, all is free. There are no advertisements. There is no financial strings attached or anything like that. You don't even have to sign in if you don't want to. Information is not going anywhere anyway. Okay, so let's talk about this particular problem. Now, I believe everybody knows whenever you look from the top, from the air on the top of the water, inside the water, you will see whatever is on the bottom or your own legs if you are in the sea or something in the river. You will see everything closer to you. And basically, the question is how to explain it and how to quantify that particular phenomenon. So that's what this lecture is about. It's a rather simple thing. And we will get to some approximate solution. And, well, it will see anyway. So let me just explain first about how we see things. Now, when I see a flower or a house or whatever else, how do I see these particular objects and how can I make an opinion about the distance to these objects? Okay, here is how. First of all, my eyes are not emitting the light. They are absorbing the light, right? Now, if I see something, it means that something actually is the source of light. Well, it can be source of light itself or it can be reflected light. So, but whatever it is, reflected or the source, well, if it's a tree, obviously, if I see the tree, I see a light reflected from the tree, from its bark, from its leaves, from its branches. And if I see some object, let's say a stone, which is on the bottom of the water and I'm looking from the top, how do I see this stone? Why do I see this stone? It's because the light goes inside from sun, let's say, and it's reflected by this stone and it goes from this stone into both of my eyes. Now, let's just forget about water for a second. Everything is in the air. If I see something, how can I judge the distance to this particular object? Well, I have two eyes. If I have only one eye, then I cannot reasonably judge the distance unless I kind of know from experience this particular distance. But in honesty, it's two eyes which allow me to make an opinion about how far the object is. Why? Because the closer object is to my eye, the closer my eye should actually be turned to focus on this particular object. So the muscles which are actually moving my eyeballs are more tense if I'm looking at the closer object. If I'm looking at something in infinity or very far, my eyes practically going parallel to each other. But if I look at something close, my eyes should be closer, which means my muscle will be tense. Also, the closer the object, the lens inside the pupil should be more curved to focus on the retina, to focus the image of the object on the retina. And then in the retina there are nerves which are transferring this information to the brain. And the brain now makes a judgment based on muscle contraction of the eyeballs and the muscle contraction around our lenses in the pupil. Based on this, we make an opinion about how far the object is. Okay, so it's all about two eyes and the angle these eyes have to really be focusing to make a judgment. The closer the object is, the wider this angle is because this is when we go straight to the far object, it's almost zero angle between these two directions. But if I'm of closer, it becomes greater and greater. Not to 180 degree, but probably 120 at least. Okay, so let's just now introduce the water inside that picture. So let's say these are my two eyes. I make a much wider eyes than really is. Now this is the water. And let's say the object, this is the bottom of the water. Okay, so this is water and this is air. Okay, now let's say we have somewhere in the middle an object here. Now, we know that the light from the object, let's say it's reflected light from the sun, goes to my eye, not along a straight line, but along this line. Okay, so these are the lines of the light from this object, which go to the air. Now, this is an incident angle. Let's call it alpha. And this is a refracted angle, which we will call beta. Alpha is smaller than beta because the speed of light inside the water is smaller than in the air. We know that, right? So again, the light goes from this to the eyes, to both eyes. Now, let's see how the distance to the particular object is actually judged by our brain. Well, the brain sees the light which is coming from this. So from the brain's standpoint, my object is here. So this is the real location of the object. And this is the object how I judge the distance to this from my face. This is my face, right? Obviously closer. So that's the reason why we see objects closer when we look at them from the air through the water. Okay, so we have explained this phenomenon from the physical standpoint. Now it would be nice to evaluate the whole thing mathematically. How closer it is. All right, let's just do it. Okay, let's consider that this distance is h above the water. The water has distance, the depth of the water is d. And this particular, so this is d. And this particular distance is x to the point as we see this particular object. So what we will do right now, we will use the laws of refraction, as we know them, to evaluate x basically relative to d. How closer the object appears to be relatively to its real depth. Okay, so let's do it. We need some space obviously. Now, we do need this distance between the eyes. Let's put it s. It's very important for solving all these triangles, right? So if this is my left and this is my right eye. This is location of the object in real life. This is location as we see it by brain. And let's put this point c, because we need it for some triangles I guess. Okay, now we need this point d. And this is s. Okay. Now, let's think about the distance between the eyes and a couple of triangles which we can just derive from here. The distance I'm talking about od, which is s divided by 2. s is a given thing, the distance between the eyes, so we know that. We also know h and d. H is our distance from the face above the water and d is the depth of the water where the object is located in the bottom of this. Okay, so this s2, this od is equal. Okay, on one hand, this distance can be calculated as this is also angle beta, right? So from this triangle, this triangle. Now, lb is a straight line of this, right? Because we are kind of making the continuation of the line to find the virtual object which we think we see. So from this triangle, and this is also angle beta. Yeah, even from this triangle it's probably even better. Ldb. So this od, the s divided by 2, can be calculated based on this height, which is h plus x, multiplied by tangent of angle beta, right? If we divide ld by bd, that would be tangent of this angle. So od which is equal to s divided by 2 is h plus x, h plus x times tangent beta, right? Okay, from another point, we can always consider the same lengths as sum of two lengths. This one from this triangle, now it's also beta, we know this is h, so this piece is equal to h times tangent beta, plus this piece which is equal to this piece, and from this triangle it's d which is depth times tangent of this angle, which is also alpha, right? This is alpha and this is alpha, because the lines are parallel. So plus ac, which is d times tangent alpha, d times tangent alpha, and this is exactly the same as divided by 2. Well, for a simple solution, which is not actually a solution, I'll tell you why. What we can do is we can subtract these things, one from another, now h will cancel out, h times tangent beta, and what will remain is x times tangent beta is equal to d times tangent alpha, which means x divided by d, which is kind of ratio between the perceived depths of the water and the real depths of the water. It's equal to tangent alpha divided by tangent beta. Now, in most of the textbooks, whenever we are analyzing this, they're saying, you know what guys, tangent and tangent are very close to sine when the angles are small. So we don't know alpha and beta, but we can define their ratio tangent alpha divided by tangent beta as approximately equal to sine alpha divided by sine beta. And this particular ratio we know from the refraction index to be equal to n air divided by n water, where n is refractive index of refraction. So this is definitely the right equality. This one is approximation, and this approximation is good only if angles alpha and beta are very small. So that's why I told you that this is not really a solution. It's an approximate solution. And it's also good if, let's say, you are in the sea, let's say, so you are stanging and the water is, let's say, at the level of your knees and then there is something over there. Then it's really relatively far, relatively to what? Relatively to distance between your eyes. And that's why the angles are really small. And then you can do this type of calculations. And by the way, what's approximately? It's approximately equal to 0.75. So it's like trick waters. So our, this picture is not really proportional because if this is the real water and this is real air, the point is supposed to be somewhere here. So the ratio between x and g is 0.75 trick waters. Now I draw this picture just to make it a little bit more visible because if these angles are very small, it's kind of difficult to draw. But again, the difference between this distance between eyes and this is supposed to be really large comparing to. So this is much smaller than this. Only then these particular calculations are correct. If not, if not, the problem becomes much more complex mathematically because what we have done is this is our first equation, this is our second equation. And this, the law of refraction is our third equation. Now using these three equations, we have three variables, alpha, beta, and x, which we don't know. So we can actually solve it. However, well, I tried myself to solve these equations. I mean, it goes to very complicated calculations because tangent is not really a sign. You can express tangent through sign, but then it becomes with square roots of whatever, sine divided by square root of 1 minus sine square. It becomes much more complicated equations. I think I've got it as a polynomial equation of the fourth degree, which is not really a nice thing to solve. I mean, maybe it can be solved in certain cases analogically. In most cases, probably it will be difficult. Let's put it this way. Anyone who wants to try, welcome. And if you will come up with a good analytical solution to these three equations, these three variables, you're welcome to do it. Send me your write up. I'll put it on my website and with reference to the author, et cetera. So in any case, this is the story with this particular problem. So as you see, to get an approximate solution, it's a very simple thing. But in any case, the approximation of like 0.75 should give you a nice picture of what exactly is happening when you are looking at the water. And it's really kind of far distance from your eyes to the bottom of the sea level, let's say. Now, if it's something much smaller on the scale, let's say you look at the water in the pot. So your distance to the pot is very small and the pot is not very deep. So the angle relatively to distance between your eyes will not be small. So this approximation is not really good. So if you will put a coin, let's say, on the bottom of the pot, the pot will be half filled with water, let's say. And then you see that particular coin from the top, from the air. It will be closer definitely. But the calculations will not be exactly the same. In any case, probably something like 0.75 is, again, it's an approximation. So it can be 0.7 or 0.8. But it will be in that range. So it will definitely look closer to you than it really is, but not by much. About three quarters, whatever. So that's it. That's the problem which I have been given on my exams in high school. And I wish you like it. I think it's very kind of practical. And it goes exactly into the material which we have learned just recently. So maybe I will include some similar problems into exams for this particular topic of phenomenon of light. Okay, that's it. Thank you very much and good luck.