 So, the first topic of ray optics should be what? Reflection. Write down. Reflection. Fine. Write down. First law. Are you studying optics of first time? You have studied it in class 10th. And you have also studied mechanics in class 9th. You saw the difference? In 11th, what you studied in 9th, there was a bit difference. The difference is not much when you are in 12th and what you have learned in 10th. The difference is only about broader application. The depth remains more or less similar. Fine. So, like the way I always say that the first two chapters are the most difficult chapters. Fine. Electrostatics. So, do the electrostatics very well. Other chapters will be automatically, you know, they are very easy. Fine. So, first law, you remember first law? Write this down. Angle of incidence. The angle of incidence and angle of reflection are equal or angle i is equal to angle r. Fine. Now, what is angle i? Whenever you say angle, angle has to be between two lines. Isn't it? What is the first line? Incident ray. Second line? The normal. Fine. So, incident angle is angle between normal and incident ray. And reflected angle is reflected ray and normal. Right? Straight forward. Right? So, these two angles should be always equal. Fine. So, can you draw a diagram showing all these? So, you have this plane and this is normal and light is coming like this. Can you show incident angle and reflected angle quickly? Also, you have a spherical surface. This is the center. Light comes like this. Show these angles. And then, you have this surface. You understand, right? This hash, it means that side is silver. This is the center. Now, light should always come from the other side of the silver. Now, show the angle of incidence and reflection in all these three scenarios. Very good. So, this angle is angle of incidence. And if this reflects off, then this angle is reflection. So, i is equal to r. In this case, first you need to draw what? Normal. How do you draw normal? Through the center. It should pass through the center. Right? So, this is normal. Incident ray and there will be this reflected ray. So, this angle will be equal to that angle. This is i and this will be r. Fine? Now, here this is normal. So, like this. This is i and this is r. Fine? Suppose, there is some irregular surface. There is a surface like this and light comes. Then, what will you do? How will you draw normal and this thing? Normal, not tangent. There is a perpendicular tangent like this. So, this will be the angle of incidence and this is angle of reflection. So, you have good idea about it now. So, when you draw ray diagram, these things will be very useful. So, this is the first law. Second, what is the second law? Incident light, reflected light and the normal. They all lie in the same plane. Incident ray, normal and reflected. What does it mean? Suppose, these are the three lines. For example, this. One of them is this is incident ray. This one is normal and this one is reflected. They all should be in the same plane. If you look like this, it should look like only one line. It should not be like this. Now, they are not in the same plane. Fine? They lie in a different plane. So, now, like this also they are lying in a different plane. But if they are in a single plane, it will look like this. So, that is what this law is about. Now, how will you write this condition mathematically? Suppose, vector A is incident. This vector is incident ray. Vector B is normal and vector C is reflected. They all should lie in the same plane. So, how will you write this condition? Will A cross B be perpendicular to C or not? Yes or no? A cross B will be perpendicular to both A and B. It should be perpendicular to plane containing both A and B and in the same plane C is also lying. So, if they are perpendicular, so I can say that A cross B dot with C is equal to 0. This is the condition. Fine? There can be several other ways also you can write the same thing, but this is one of the proper way of writing this condition. So, these are the two laws of reflection. Fine? So, now we will apply these laws in different kinds of scenarios and the easiest of the scenario is what? A plane mirror. So, let us try to apply that. So, this is suppose a plane mirror. This line is normal to mirror or not? It is normal. Now, suppose here is this object. Now, you have to find out the image of this object. Now, in order to find the image of the object, you should know what? Object and where is the observer? Where is the observer? Is it that side? Behind the mirror, observer will be this side. Suppose here is the observer. Observer is normal and not allowed to look at it directly. Light ray will directly reach the eyes of the observer from the object, but this is not observer is doing. Observer is looking at the mirror. So, observer will only see what comes through the mirror. Getting it? When you are standing in front of mirror, you can directly look at yourself like this, but you do not do it. You look at the mirror. What comes from the mirror? So, you have to find out how the light ray reach the observer. Assume that entire of this side is observer only. Wherever light ray goes, observer can see it. Draw the ray diagram to find the location of image. One ray. See, what is the easiest ray to draw here? A ray that goes like this straight. What will happen to that ray? What is the angle of incidence for this ray? This one. So, angle of reflection should be 0. So, this should come back. So, any light ray that is along the normal will just come back after reflection along the same path. In a circular, if there is a ray that is along the radius, it will come back because there also angle of incidence is 0. So, this is one ray that goes. So, observer will feel that image is along this line somewhere. But observer do not know where it is. If you draw only one ray, but if you draw another ray, then you should be able to locate the image. Now which ray you should draw now? So, you can draw any ray in fact, but if you draw this, what will happen to this? This angle is what? Incident angle. They will be reflected. How it will get reflected? It will get reflected like this. This angle will be what? This is also I. Now, will these two ray ever meet? These two ray, they will not meet, but they will appear to meet where? At the other side. They will appear to meet where? At the other side. They will appear to meet somewhere here. Now in order to find the image, the light ray does not need to actually meet. They can appear to be meeting at a point. So, this guy will feel as if light rays are coming from this point. So, this is the image of this point A. So, this will be suppose A dash. Now what will be the image of this point where it will be? Will it be along this line? Will it be along this line? The image of this point will be along this straight line. So, if you draw the ray diagram, you will see that image will be here only. So, if you track down all the points, you will get the image like this here. This is the image. Now, if this is distance D, what is this distance? Prove that. This angle is I. What about this angle? I only alternate opposite angles. I. This side is common. This is 90 degree. This is also 90 degree. So, this angle is also same. This side is equal to that side. So, it is a congruent triangle. If it is a congruent triangle, then this distance will be equal to that distance. So, write them characteristic of the image. Characteristic of the image due to a plane mirror. The image will be at the same distance as the object is, same size, virtual, erect and laterally inverted. So, when you stand in front of mirror, when you see the image, your right becomes left, your left becomes right. That is why in the ambulance, the ambulance is written in other way. So, that when somebody sees from the rear view, then we can see that there is ambulance coming in and anyway, you can give the ambulance path to pass. So, observer will feel that the image is behind the mirror. But can you put a screen behind the mirror and locate the image? You cannot there are no actual light rays there. It is just appear to meet. So, virtual image you cannot put on the screen. So, whenever in the question, it comes that there is a screen on to which image is falling. It automatically means it is a real image. Otherwise, you cannot put it on the screen.