 In this video, we will try to understand the concept of virtual object and also see how this concept is useful in figuring out the distance of image in convex or concave lenses. Let's start with some converging rays. These rays can meet at a point and form a real image, but if we introduce any optical element in between like a mirror or a lens, let's say we introduce a mirror, then these rays will get reflected and instead of meeting at this point, they will meet somewhere over here. So that could look somewhat like this. We have a real image forming at this point and no rays are passing through the mirror. But where is the object of this real image? Turns out if we extend these light rays, the point where the rays would have converged acts as a virtual object and we say that the mirror forms a real image of the virtual object over here. So if we extend these rays like this, the point where they appear to meet acts as the virtual object for this real image. Usually in mirrors, we always had a real object and we got a virtual image, right? And we can also get that over here using the idea of the principle of reversibility of light. If we reverse the direction of the travel of the light rays, if let's say we had a real object over here and then that object would be the source, the light rays would be diverging from that object sort of like this. Then this point from where the light rays are diverging, it becomes a real object and the rays are not converging at any point. So there is no real image. But if we extend these reflected rays, if we extend these reflected rays, they seem or appear to be coming out from this point. So we call this point as the virtual image. Now in place of the virtual object, we have the virtual image and in place of a real image, we have a real object. One thing that we can notice over here is that the objects, whether it is the real object or the virtual object, they are always connected to the incident ray. We can see that these two incident rays, they are diverging from this real object. For virtual object, we have these incident rays which appear to be meeting at this point. So objects are always connected to incident rays and similarly images are always connected with reflected and as we will see also refracted rays. Now here this real image, this real image is connected to this reflected ray and this reflected ray. Similarly the virtual image over here, that is connected to this reflected ray and this reflected ray. These rays are extended and they seem to be coming out from the virtual image point. So to quickly summarize what a virtual object is, whenever an optical element can be a mirror or a lens, whenever any optical element comes in the way of converging rays, then the point where the rays would have converged if there was no optical element, then that point is called the virtual object. That point acts as the virtual object. Let's explore more about virtual objects with lenses and refraction. Here we have a thick, a thick convex lens. It has these two refracting surfaces. Let's say there's a light ray from this point. We have an object at this point. This light ray will undergo refraction because the refractive witness on both the sides are different. This is made of glass and you have air on the left. So it undergoes refraction. Let's draw a normal to see where it could bend. Now it will bend towards the normal. So this light ray, this light ray will bend slightly. It will get slightly, somewhat like this, bends towards the normal. Now as the light ray moves through the glass, it again encounters a surface where it can undergo refraction because it encounters an interface between glass and air. It will undergo refraction again over here. So let's do that. Let's make it undergo refraction. We have a normal like this and then it undergoes refraction and forms, forms an image, forms an image at this point. So this is a light ray which comes out from the object and forms an image on the right side of, right side of the lens. Here we see that there are two interfaces. This right here is the first interface and this is the second one. If the second interface wasn't there, then the image, the image would have formed along this line. So if we try to extend this line, let me, I'll have to shift everything on the left. So let me do that. If you extend this light ray, this light ray like, like this, it's not entirely straight, but it's, it's really straight, almost straight, almost straight. Okay. So if this interface wasn't there, then this light ray would have just moved in this direction. Just assume it's a straight line. It's moving in a straight line and an image would have formed an image would have formed at this point right here. An image would have formed right here and you could imagine there are, there's not just one ray. There are many rays that are going like this. There's a ray which will, which will go from the bottom. It will get refracted, get refracted again, form a real image over here. But then if the second refracting surface wasn't there, it would have, it would have converged with this light ray, with this light ray over here. This, so this point, I dash, this point is really the virtual object. This is the virtual object for, for this real image. That's the virtual object for this real image. In the case of mirror, we saw that the rays that were about to converge, they were interrupted by the mirror and as a result of which the rays converged at some other point, but the point where they would have converged serves as the virtual object. Similarly over here, the point where the rays would have converged acts as the virtual object. So looking at these two refracting surface, we can try and draw a general principle out of this, and then we will see how to use that in numericals. So what we are really seeing over here is that there are these two steps. Let's think of them as two steps. There is this first interface, there is this second interface. The image from the first interface could never really be formed because the second interface came in between. So whenever this case happens, whenever the image of the first optical element is not formed and this could be a refracting surface, a thin lens as we will see or a mirror. Whenever the image of the first refracting surface is not formed, it acts as the virtual object for the next or the second or the coming refracting surface. In this case, the image of the first refracting surface never really formed because second came in between. So that point, this point started acting as a virtual object for the second interface. Let's write that and we can use variables like n to make it more general. We can write that if the image of the n-1 refracting surface or a reflecting surface, if that image is never formed, then that point acts as a virtual object for the n-th step or for the next refracting or reflecting surface. Here the image of the first refracting surface never really formed. So that point acted as a virtual object for the second refracting surface. And you might say that in the case of the mirror, in the case of the mirror we just had, we had these converging rays and they never really met. They met somewhere over here and the point where they appeared to meet that act as a virtual object. So we really only had one reflecting surface. There is an assumption that goes with it that if you have these two converging rays, they could be coming off from a really big mirror on this side or they could be a big converging lens on this side. So the rays could either be converging, they could be converging or they could be an object here, so the rays get reflected and they move in this direction. So what happens over here is even if these two rays, even if they are the product of some reflection of a mirror or a refraction from a lens, their image never really formed. The image of the first step never really formed. So then that point, because it was interrupted by the second optical element, in this case the mirror. So the point where the rays would have met acts as a virtual object for the second reflecting surface, in this case, in this case some mirror. Let's see how we can use this idea of virtual objects and also this principle in numericals. And in numericals we deal with thin lens. We don't deal with the convex thick lens that you see in front of you. We deal with thin lenses. Now the thing about thin lenses is that we don't really show rays undergoing refraction twice. We just show it undergoing refraction once and that from the center of the lens. We show it once. So a thin lens is just one refracting surface. We do not really assume that there are these two separated refracting surfaces in a thin lens. There is just one and the ray undergoes refraction once. That's the type of lens we work with in numericals. So let's see how we can use this idea in the case of thin lenses. Here we have these two thin lenses and let's say we have an object right over here. This is our object. The task is to figure out the distance of the image once the light rays undergo refraction from both the lenses, from both of these optical elements or from both of these refracting surfaces. Remember these are thin lenses, right? So the light ray will only undergo refraction once. Okay, usually in questions we are provided with focal lens. So let's say this is f1, this is f2 and we also are sometimes provided with the distance between the two optical elements. So that we can assume that is that is d. The last thing that we know, well not the last thing, we also are mostly let's say we know, we know the object distance and I'm calling this u1 because I called this as f1. This will act as the object for the first lens. One thing that we also know is a thin lens equation that is 1 by v minus 1 by u equal to 1 by f. This is the thin lens formula. All right, so rays will be diverging from the object and we will get two cases really. The first case, the first case would look, it could look somewhat, somewhat like this. You have the rays undergoing refraction and let's say they converge. They really converge at a point before encountering the second lens. They converge at this point. In this case, this point right here, it acts, it is really the image of the first lens, but this image then acts as an object for the second lens. So the light rates they keep on moving, they again undergo refraction. They again undergo refraction and then they finally converge. They finally converge at this point. So they meet, they meet at this point and this point acts as the object. So I'm writing o2, object for the second lens. They finally meet at this point, which is the final image position. And when we have a case like this, we really need to figure out this image distance. Let's call it v2. So we can use the thin lens formula to figure out v2. If we know this object distance as u2, then we can use this formula 1 by v2 minus 1 by u2. That is equal to 1 by f2. How do we know u2? We can again find that from the first, on the first lens. We already know this u1 and we already know f1. So we can find this, we can find this distance, that is v1. Once we do find v1, we can see that we are already provided with the value of d. So then we can figure out u2. But this is a case when the rays actually converge and meet at a point. Let's look at a different case. So now let's look at, let's look at this case when, when the rays after undergoing refraction, they are about to converge at a point, but before that they encounter a second optical element, a second lens, a second refracting surface and as a result of which they converge some more and they meet at this point. They meet at this point right here. This is the final image position. Now how do we figure out this distance? How do we figure out the final image position? Well, if we look at the thin lens formula to figure out v2, what we really need to know is u2 in this case because we already know what f2 is. We know that f2 is provided. That's the focal length of the second lens. So if you are able to figure out the object distance for this image, then we should be able to figure out the image distance. But where is the object of the second lens? We know that the rays were about to converge at a point, but they couldn't. The image of the first step was never formed. So that point where they would have converged, that point acts as the virtual object for the second refracting surface or the second step. So if we extend these lines, if we extend these lines, then the rays they would have met at this point. This point right here, this point acts as the virtual object for this image position, for this image right here. Let's make some arrows to see how the rays move. So if we label u2, the object distance, that would just be this, this much distance. You have a virtual object case for the second lens. But how do we figure out u2? Well, we know what u1 is. We know what f1 is. Can we use these two to figure out where the image of the first lens should be formed? Well, we can. We know that 1 by v1 minus 1 by u1, this is equal to 1 by f1. We know f1, we know u1. We can figure out what v1 is. And if we try to label v1 on this diagram, v1 would really be this long distance. It would be this, this long distance, this entire distance, that would be, that would be v1, right up till this point, right? Because the image would have formed at that point. Had there been no second lens with this condition of focal length and this condition of object distance, the image would really be formed at this point. But the second lens changed the story. So now we see that we are mostly provided with the distance between the lenses. We know this v1 and u2, u2 is really, u2 is, this is really v1 minus d. It's really v1, this whole distance minus this distance between the lens that gives us u2. Now, we know most of the values. We know u2, we know f2. So we can use them to figure out the final image distance. And that is how we can use the idea of virtual objects to figure out the distance of the final image whenever there are more than one thin lens. If the image from the first thin lens can never be really formed, then the point where the rays would have met acts as a virtual object for the second lens. Okay, you can try questions on this from the exercise in the lesson. And if you are watching on YouTube, the exercise link is added in the description.