 Let's recall everything about lenses and mirrors to help you in your board exams. On the right, you can see the index. So if you want to jump to any particular topic, feel free to do that by going to that time. All right, so let's start with the basics, the naming and the properties of the mirrors and lenses. So can you quickly pause and recall? What are the names of these mirrors and lenses? Go ahead, give it a try. All right, so the names are dependent on the shapes. So if you look at the first one, when you shine a ray of rays of light onto this on the reflecting side, notice that it hits the cave part of the mirror, and that's why it's called a concave mirror. On the other hand, over here when you shine light on the reflecting side, you will see it hits the bulged part of the mirror, and that's why it's called convex mirror. Vex means bulge. What about the lenses? Well, here you can shine light from any direction. So if you shine light, say, from the right again, notice it again hits the bulged part of the lens, and so this is called the convex lens. Over here, when you shine light on it, notice it hits the caved part of the lens, regardless of which direction you hit it, and that's why it's called concave lens. Okay, now let's think about what happens to these rays of light once they hit the mirror and the lens. So again, can you pause and think about this? Remember, recalling is the best way to learn for exams. So pause and think about what happens to the... All right, so for a concave mirror, after reflection, we'll find the rays of light will be directed towards the center. They'll be focused at a point called the principal focus. So this is a converging mirror. On the other hand, if you look at a convex mirror, since it's bulged out, you will see it will diverge the rays of light. Again, these rays will appear to now come from a point. Not really doing that, but they appear to come from a point. Again, that point is what we'll call the focus. What about the convex lens? Well here, refraction happens, not reflection. The rays of light will just go through and they will bend and over here, in the case of convex, we will again find it will converge the beam of light to a particular point, the focus. So this is a converging lens. And if you look at this one, we'll find that this will diverge the rays of light, again, appearing to come from a single point. And so you can immediately see concave mirrors and convex lens are similar in the sense they're both converging devices. And similarly, if you look at convex mirror, it is similar to concave lens, such that they're both diverging optical devices. And therefore, when you look at ray diagrams, there'll be no wonder that these will be very similar to each other, similar cases when it comes to images, image formation. And these two will have similar cases when it comes to image formation. Now of course, there's one big difference, which was a very common mistake I should do, is that when it comes to mirrors, you'll find that the focus lies right in between, exactly in between the center of curvature and the pole. Center of curvature is the center of this imaginary sphere, right? So focal point lies right in between the two. And the same thing will be for convex mirror as well. But that's not true for lenses. If I were to take the center of curvature, there is no direct relationship between the center of curvature and the focus. And so that's not true for lenses, mainly because over here, refraction is happening and how much the rays of light bend, depend, not only on the curvature, but also on the material. So long story short, here, we will not use center of curvature because there's no relationship with focus, but over here, we will. With that in mind, let's now jump to ray diagrams. Let's start with concave mirrors. Imagine we have an object kept beyond the center of curvature. Where would its image be formed? We can figure that out by drawing a couple of rays. So one ray of light, I like to shoot parallel to the principal axis because I know that ray, after hitting the mirror, goes through the focus. We just saw that all parallel rays of light should go through the focus. Another ray of light, what I could do is shoot it through focus because I know that after hitting the mirror, it's gonna go parallel to the principal axis. Now wherever these two lines, these two rays intersect, we know that's where the image needs to be formed, but we can also have a third ray just in case. So for example, I can draw another ray of light passing through the center of curvature because I know that this is where the normal lies and so any ray of light along the normal will just bounce back and so all the three rays will now intersect at this point giving you an image over there. But we don't need three rays, so I'm just gonna erase that. And we get an image over here. This is a real image because the rays of light are being focused. It can be captured on a screen. It's real, it's inverted. It is diminished. So when the object is outside or beyond the center of curvature, we get a diminished real inverted image between F and C. All right, can you try using ray diagrams to figure out where the image will be in these cases? Go ahead, pause the video and give it a try. All right, for the second case, we'll have similar ray diagrams, one ray parallel to the principal axis, another ray passing through the focus, and you will see they'll meet somewhere over here giving us an image which is beyond C and now enlarged compared to the object. So you can kind of see as the object comes closer to the focus, the image ends up going farther away, but most importantly, the image starts becoming larger and larger. Okay, what about over here? Here, again, I can draw one ray of light passing, parallel to the principal axis, passes through the focus after reflection, but I can't draw another ray of light through focus because it won't hit the mirror. This is where I'm gonna draw the second ray along that normal. So this is the ray along the normal, which is basically passing through the center of curvature. I know now that ray is gonna just bounce back and it's gonna go over here. Now notice that these two rays after reflection are not focusing at any point. They're just diverging away. They're not converging, they're diverging away. And so they appear to be diverging from a point behind the mirror, which means now I get a virtual image. This image is virtual because you can't capture it on a screen. All virtual images are erect and notice the virtual image is enlarged. And so with concave mirrors, you can get both real and virtual images diminished and enlarged images as well. All right, let's now jump to a convex mirror. Again, can you try drawing ray diagram to see where the image of this is formed? Go ahead, give it a try. Okay, let's see. We need to be a little bit careful over here. Now this ray of light will not go through the focus because it has to reflect. Remember, it's a mirror. So the reflected light will reflect in such a way it appears to come from the focus. Remember, it's a diverging mirror. So it appears to come from the focus. It doesn't go through the focus. And another ray of light, what we can do is we can pass it along the normal, basically along this line joining the center of curvature. We know it's gonna hit the mirror and just bounce back. And again, we get two diverging rays, which appear to be coming from here, giving us a virtual erect image. Notice that over here, we'll always get virtual erect image, regardless of where you keep that object. You can always test that. You can try that yourself. We'll always get an image between P, I forgot to mention the P over here, between P and F. It'll always be virtual and diminished. So what big difference you see between concave and convex? Well, we see that concave can do both real and virtual. Convex can only always give virtual images. What's the difference between the virtual images of concave and convex mirrors? Well, concave is always enlarged. As you can see, we don't have to remember this. Ray diagrams help you. But for convex mirrors, it's always diminished. Okay, now with that in mind, let's jump to lenses. What happens if we have an object which is kept outside or beyond 2F? Notice, remember, for lenses, we don't have center of curvature. We don't use that. We use twice the focal length. Where will the image of that object be formed? Now, you may be wondering, why do we have two values of F over here? That's because you can keep objects on either side. If the object is on this side, the rays of light will be coming this way and they will be converged over here. So this will become the focus. If the object was this side, then this will become the focus. And so I put this as green. So this is the focus. And so one ray of light, we can pass parallel to the principal axis. We know after refraction, this will refract, it will pass through this focus, okay? A second ray of light, what we can do for lenses, well, something very similar to the mirrors, we can pass it through the focus. And we know after refraction, it will go parallel to the principal axis. But you know what? Instead of doing that, what I like, I like to pass a ray of light straight through the center of the lens because the ray of light through the lens goes undeviated, okay? And so now I know that these rays are intersecting at this point, this point over here. And so that's my image. And so notice we end up with a real inverted but diminished image between F and 2F. Okay, your turn. Can you draw ray diagrams for these two cases of convex lenses? All right, hopefully you've tried. So for the second one, it's gonna be very similar. We'll end up with a real inverted image. This time it will be enlarged. It will be a little bit farther away from the lens but it'll be enlarged. What happens over here? Well, if you draw the two rays, this time you'll find they are no longer converging at any point, so we don't get a real image anymore. Instead, they're diverging and appear to come from a particular point. And so we end up with a virtual image. Why is this virtual? Because you can't capture it on a screen, it's erect. All virtual images are erect. Now you can take a moment and just compare these two. You see they're very similar. Concave mirrors are similar to convex lenses. No surprise because you already saw they're both converging devices. So lastly, this brings us to concave lens. What happens over here? Again, pause and try. All right, we'll start with the usual one ray parallel to the principal axis. Where does it go? Does it go through this focus? No, because over here this is the focus. A concave is a diverging lens and that's why this is the focus. And so that ray of light will diverge appearing, appearing to come from this point. All right, so need to be very careful when dealing with concave lenses. So that's one ray. Another ray we can just shoot straight through the optic center just like before you go on deviated and you can now see these two rays of light again diverging and they appear to come from this point. And as a result, we end up with a tiny, tiny virtual image. And regardless of where you keep this object, you'll find you'll always get a virtual diminished image. So we saw all the cases. Let's put it all together. In the converging case, we find that there are three possibilities. One, we can get real image. One, we can get virtual image. You can also get no image at all. If the object is kept at F, we will get no image because the rays of light will be parallel. But for diverging devices, we'll always get a virtual image, only one case. And this actually is a bonus question for you to think deeply about. Why is it that for converging we have three cases but for diverging we have only one case? I won't answer that, but something for you to think about. Anyways, since there are three cases for converging mirrors and lenses, real, virtual and no image, we need to know when each of them happen. So when do we get real images? We get real images when the object is outside the focus, outside the focus, outside the focus, real images. When your object goes inside the focus, you get virtual images. So the focus is the dividing point. What about the size? Well, again, if the object is closer to the focus, you'll end up with bigger image. Notice, closer you go to the focus, bigger the image. Closer to the focus, bigger the image. Same case even here. If you go closer to the focus, you get big image. Farther from the focus, small image. Farther from the focus, small image. And of course, at C, the object and the image will be having exactly the same size. I've not drawn that over here. Here at 2F, object and the image will end up having exactly the same size. And remember, the difference between these two is that the converging mirrors and lenses, they can form virtual images, but they'll be always enlarged. These will always form virtual images which are diminished. Now that we have put all of this together, let's see if we can recollect by asking some questions. So here are four questions for you. We have given the nature of the image and you need to find out which mirror can do that, which mirror and which lens are capable of doing that. So again, great idea to pause and see if you can recollect what we just learned. Don't refer back, but please recollect. Please see if you can answer these questions. All right, let's start with the first one. Real diminished image. Which mirror can do that? Well, I look at real. I know that real images can only be formed by converging devices. So this is a converging mirror which is our converging mirror, concave mirror. Real enlarged image. Again, I just look at real. I know real can only be formed by converging devices. You need to focus it at a point. Only converging devices can do that. So I need a converging lens, which is a converging lens. Convex lens is a converging lens. The third one, virtual enlarged image. Well, virtual images can be formed by both, but virtual enlarged, virtual enlarged can only be formed again by a converging device. Okay, that's what we saw. And so which is our converging lens? Again, convex lens. Finally, virtual diminished image. This is formed by a diverging object, diverging mirror. Which is our diverging mirror? Well, convex mirror is the diverging mirror. In fact, it's easy for me to remember this because I've seen convex mirrors on parking lots and everything. And I always see virtual, erect, diminished, small sized images. So I just remember this. Everything else is converging devices. Okay, ready for a challenge? Now comes follow up questions. For each of these cases, where should you keep the object? For example, if you want to get a real diminished image, where exactly should you keep an object in front of concave mirror? Similarly over here, where should you keep the object? Again, can you try and remember? All right, let's see. For the first one, I want a diminished image for a converging mirror. We just saw that the farther I go from the focus, the smaller the image becomes, right? So I need to be outside F for real. And for diminished, I need to be far away. So basically beyond C. Can you imagine that? Can you visualize that? We need to go beyond C. For the second one, real enlarged image for convex lens. Where do I get enlarged image? Well, again, for real, I need to be outside F, but for enlarge, I need to be close to F. Remember, closer you have to F, more bigger the image. So outside F, but close to F. Yeah, that's between F and 2F, not C, because of lens. So between F and 2F, between. All right, third one. I need a virtual enlarged image, virtual image for convex lens, for converging. When do I get virtual images for converging lenses? Or we saw we need to be inside F, right? So clearly it should be between F and the lens, or we can say between F and the center of the lens, which is usually represented as O. For the last one, I want virtual diminished image in a convex mirror. Well, for convex mirror, like a diverging mirror, we only have one case. So regardless of where you keep the object, you will always end up with this, and so you can keep this anywhere you want. Now, if you are troubles recollecting all of this, no worries, you can always jump back to the previous section, recollect, and come back over here, and see if you can answer them on your own. If all is well, we can now move on to the next part where we'll start talking about the mirror and the lens formula. These formulae connect the object distance, usually denoted by U, the image distance, V, and the focal length. Same here as well, object distance, image distance, and the focal length. And there's something called magnification formula, which connects the heights of the objects and the image with their distances. Again, the heights of the object and the image with their distances. So, can you recall what the mirror formula and the lens formula are? And what the magnification formula for the mirror and the magnification formula for the lens are? Again, grehedia to pause and see if you can remember. All right, here we go. For the mirror, the mirror formula is one over V plus one over U equals one over F. And the magnification formula for the mirror, where M, which is magnification, defined as the height of the image divided by height of the object, is given as minus V over U. Okay, what is it for the lens? It's pretty similar, but important for lens. The lens formula becomes one over V minus one over U equals one over F. So, the negative comes in the lens formula. But for the magnification, we see no negatives at all. Now, clearly, when we have to remember the signs for all the formulae, it's very easy to get confused. So, here's what I do. I only remember the mirror formula. And I remember that the mirror formula is all positive, right? All the values are positive. Then I know magnification formula becomes negative. Okay, once I know this, I know for lens, it's opposite. I know for lens, this becomes negative. And for magnification, this becomes positive. Hopefully, this will help you remember and get less confused with the signs when you're applying them to numericals. But before we apply them to numericals, we need to understand sign conventions because all these values can take both positive and negative values. So, what are the signs? When do they become positive? When do they become negative? Again, can you recall that? So, here's how it goes. We first consider the center of the mirror or the center of the lens as the origin. Either the pole is the origin or the optic center is the origin. Once we do that, one side of the origin is positive, another side is negative. One side is positive, another side is negative. How do we remember which is positive, which is negative? Here's how I like to remember. Always, incident direction is chosen as positive. Incident direction. So, over here, incident direction or incident light comes this way. So, the incident direction of the pole over here to the right becomes positive values, positive positions. Opposite direction becomes negative positions. Same thing over here. The incident direction is to the right. So, all the positions to the right side of the origin or the optic center becomes positive. All the positions on the left side, in this case, becomes negative. So, these are for the positions, right? We also have signs for the heights. That's a little bit more straightforward. Anything above principal axis, we call that as a positive height. Anything below the principal axis, we call that as the negative height. So, just to give an example, in this particular drawing, the object distance is negative. Notice on the negative side, image is also on the negative side. So, image distance is also negative. Focus is also on the negative side. So, focal length is also negative. Object height is positive because it's erect. Image height is negative because it's inverted below the principal axis. What about over here? Well, object is on the negative side. So, object distance is negative. Image is on the positive side. Image distance is positive. Focal length is on the positive side because focus on the positive side. Focal length is positive. object height is above the principal axis positive, image height is below the principal axis negative. With this, we can now solve numerical. What kind of numericals do we get? Well, most of the numericals, one of the two will be given to us and will be asked to find the third and the fourth one. For example, we might be given u and f and we might be asked what is v and then we might be asked what is the magnification. Or sometimes we might be given v and u and we might be asked what is m and we might also be asked what is f. Let's take some examples. So here's the first one. Usually when it comes to lenses and mirrors, I like to first draw a diagram, then see if I can write the data from that and then see if I can solve the problem. So great idea for you to pause and read this problem and see if you can try this yourself first. All right, let's see. It's given the image of the candle flame kept 30 centimeter from the mirror, forms on the screen, 60 centimeters from it. So this means we have a candle flame which is kept 30 centimeters from the mirror and the image of it forms on a screen, 60 centimeters from it. So can we immediately tell which mirror it is? We can, right? Because it's given that it forms on the screen. That's the clue. It's a real image. And which mirror gives us a real image? There's only one mirror which can do that. So it's concave, converging mirror. Concave mirror can do that. That's something that we already saw in our previous section. So that's the first answer. We're done with that. All right, now to calculate focal length and magnification, next let's immediately draw a diagram. So here's our diagram. Here's a concave mirror, 30 centimeters in front of it is the object, 60 centimeters behind it is the real inverted image. Remember, real images are always inverted. So from this we can now immediately write down the data. So we know U is 30 centimeters, object distance. V, the image distance is given to be 60 centimeters. And remember to use sign conventions. Remember signs that everything on the incident direction of the pole is positive. So everything to the right over here is positive. Everything to the left is negative. So this would be negative. This would be negative. And now we have to calculate what the focal length is. What formula do we use for focal length? We use the mirror formula. And what's the mirror formula? What about the signs? I remember that mirror formula has everything positive. So one over V plus one over U, use U one over F. And if we substitute in this, we will now find F turns out to be negative 20 centimeters. The negative sign is basically saying that the focal length is also, the focus is also on the negative side, which makes perfect sense. For a concave mirror, focal length is always negative. So it again confirms it's a concave mirror. If we had gotten positive, that would be wrong. So something would be wrong. And the focal length is 20 centimeters. So we got that. What about the magnification? How do we calculate magnification? Well, magnification formula. Remember the magnification formula? Since the mirror formula has positive, magnification formula has a negative in it. So M is going to be negative V by U. And again, if we substitute the values of V and U, we will get M to be equal to negative two. What does that mean? It means that the image is twice as big as the object, but it's real and inverted. That's why the negative sign. As the magnification is negative two. That's our answer. We have solved our problem. All right, let's go to second question. Again, pause and see if you can try and solve this yourself. All right, a two centimeter tall pin is 10 centimeters away from a concave lens of focal length, 10 centimeters. So immediately we can draw a diagram over here. We are given there is a concave lens of focal length for 10 centimeters, and the object is also kept 10 centimeters away from its height is given two centimeters. We need to find it's where the image is, position, nature, and the height of the image. So quickly we can go ahead and draw a ray diagram, what we have seen before, and we'll see we'll end up with a diminished image. But now let's go ahead and calculate that. So let's go ahead and write the data. We have U to be given as 10 centimeters, and again, if you look at the sign convention, everything on the positive side, on incident direction is positive, on the opposite direction is negative. So you get negative 10. Focal length is given to be negative 10 as well. And we are given the height of the object, which is positive. Why is that positive? That's because the height is above the principal axis. Anything above the principal axis is positive height. We have to find the image distance first. To do that, we're gonna use the lens formula. Remember the lens formula? For mirror formula has everything positive, lens formula has a negative signage. So I remember one over V minus one over U equals one over F. And if we substitute, we will now find V to be equal to negative five centimeters. And if you look at the diagram, you will immediately see the negative sign makes sense because the object is, the image is on the negative side. So we found the image distance or image position. Now we need to find the nature and the height. Well, we immediately know from the diagram it is virtual. Concave lens always gives you also virtual, but let's calculate it. To calculate, we go back to the magnification formula, which does not have a negative sign. Remember, lens formula has negative. Magnification formula for lens has no negative. If you now substitute in this, you will get magnification to be half. This means that the image is having half the size of the object. It's positive because image is erect compared to the object. That's what it's saying and that makes sense. So immediately we know it's a virtual image. It's height is half. Half of two centimeters is one centimeter. So immediately we understand the height of the image is one centimeter. But again, you can use the formula for magnification, which is H-I by H-O. That's the definition of magnification. That is half. And if you substitute, you'll get H-I as one centimeter. And there we have it. That is the height. It is virtual and it is diminished. We have found everything. Okay, since you are loving it so much, here's one last question before we wind up. Give it a try. It's a little different, but do give it a try. Okay, this is given a lens produces an erect five times enlarged image of a pin kept 20 centimeter in front of it. What's the lens? Which lens focal length and image distance? Again, let's look at the lens. It's producing an erect five times enlarged image. So it's a virtual image. Erect means virtual and enlarged image. Which lens produces erect enlarged image? That's a convex lens, something we saw before. And in such case, for lenses, we will find that the virtual image will be on the same side as the object. So immediately we can kind of draw this and say, look, this is the kind of drawing that we have. And so we can now write the data. What does the data say? Well, we know a pin is kept 20 centimeters front of it. So U is negative 20 centimeters. And by the way, we already know it is a convex lens. That's done. This is convex. U is negative 20, why is it negative? Well, the incident direction, if you go, that's positive. This is negative. We know now magnification is given five times enlarged. So in this question, magnification is given to be plus five. Why is it plus? Because it's erect. Because the image is of the same orientation as the object. So that's plus five. And we need to find the focal length and image distance. How do we do that? Well, since magnification is given, this time we can go directly for the magnification formula of the lens. Magnification formula is M equals V over U. And if you substitute, you will get V equals negative 100 centimeters. So there we have it. That's your image distance. So this distance is negative 100, and that makes sense. It's on the negative side. So image distance is 100 centimeters. And now we need to calculate the focal length. To calculate the focal length, you can go back to lens formula. So if you use the lens formula, and if you plug in value of U and value of V, we will now get F to be plus 25 centimeters. And that makes sense because for the convex lens, focal length is on the positive side. So this turns out to be 25 centimeters. And that means again, we found the answers to all of them. Now, if you're struggling to answer these questions, don't worry, I struggled as well when I was learning this. The key is to practice and practice and practice. And so that's the whole idea behind this video. To recall and give you more practice so that you would be more ready for your exams. All the best.