 Let's learn mirror formula, magnification formula, sign conventions and solve this NCRT problem. Why don't you pause and quickly read this problem? Okay, we are given that there is an object of height 7.0 centimeters kept 27 meters in front of a concave mirror of focal length 18 centimeters. So when you draw it always makes it easier to understand at what distance from the mirror should a screen be placed so that a sharp focused image can be obtained, meaning we need to find the image distance from the mirror. Also we need to find the size and the nature. How do we do that? Well, we have to use mirror formula and magnification formula. Here it is. Mirror formula is one over F the focal length equals one over the image distance plus one over the object distance. This formula works for any mirrors for all cases. And magnification formula is defined as magnification is defined as height of the image divided by the height of the object. That's how we define magnification. And this for mirrors happens to be minus V divided by you minus of image distance divided by the object distance. Okay, with that, let's see what we know. And let's see if we can calculate the image distance. We know the height of the object. So let's write that 7.0 centimeters. We are given the object distance. 27 centimeters. We also know the focal length that is 18 centimeters. What we need to calculate is the image distance. And we need to calculate the size and the nature, which means I also need to calculate the height of the image. So how do I do that? Well, I have the mirror formula. I know F. I know you. So I can calculate V and so we can directly substitute, right? No. This formula you can only use by using sign conventions. Let's quickly remember what sign conventions is. So all these numbers have a sign. How do you know that? Well, when you draw, you should always draw in such a way that the object is kept to the left of the mirror like we have drawn over here. That's the first step. The second step is we choose all the distances to the right side of the mirror as positive and all the distances and positions to the left side of the mirror as negative. So in our case, notice the object is to the left side of the mirror. So that's a negative distance. Focus. Focus is on the left side of the mirror. So that is negative distance. What about the height of the object? Well, for height, we use another similar sign convention. Any height above the principal axis is positive. Any height below the principal axis is negative. So in our case, notice the height is positive. Okay, because it is above the principal axis. Okay. Now that we've used sign conventions, we can use these formulae. So let's quickly substitute in this. Now is a good time to pause the video and see if you can try on your own first. So you get one over minus 18 equals one over V, which I don't know, plus one over minus 27. And I have simplified it all. You can pause and see the simplification if you want. But after doing the algebra, this is what we get. One over V equals minus one over 54. This means from here, I get V equal to minus 54 centimeters. What does that mean? That means my image is 54 centimeters to the left because there's a negative sign. And 54 means somewhere over here. And therefore that's where I have to keep my screen in order to obtain a sharp image. So there you have it. I have solved the first problem. Next, I need to find the size and the nature of the image for size. I need to calculate the height of the image. I know the height of the object. I know V. I know you. So I can just plug in. So let's do that. So height of the image divided by height of the object that is given to be 7 centimeters, 7.0. That equals negative minus 54. V is minus 54 divided by U, which is minus 27. Now if we do the algebra, what will we get? This goes two times. And so height of the image is minus 14 centimeters. What does that mean? That means the negative sign is saying that the height is below. So this is going to be here 14 centimeters below the principal axis. And so that's the size. And I have also found the nature because it is inverted that tells me that this must be a real image because all real images will be inverted compared to the object orientation. Here's another problem from NCRT. Why don't you read and see if you can draw a diagram first yourself? Okay, we are given this time a convex mirror with an object of five centimeter height kept 20 centimeters in front of it. And this time we are given the radius of curvature, which means the distance from the center to the mirror that is 30 centimeters. We need to find the position of the image, its nature and size. Very similar to before, but this time I'm given radius. How do I find focal length? Well, remember, focal length is always half of the radius. So focal length or the focus will always be in between the center and the pole, which means I immediately know that this has to be half of this, which is 15 centimeters. And now if I use my sign convention, which is the same thing, keep the object to the left of the mirror, choose the right side of the mirror as positive, left side of the mirror is negative, above as positive, below as negative. Same thing. Let's see if we can write down our data. We know height of the object to be plus five centimeters. We know the object distance to be it's on the left side minus 20 centimeters. We know the focal length. What is that positive or negative? Focus is on the right side on the positive side this time. So plus 15 centimeters. We need to find what the image distance is, the position of the image, its nature and size, which means I need to also find the height of the image. So again, let's substitute and again, feel free to try it on your own first. So one over 15 equals one over V plus one over U, which if we simplify and I have again done it quickly, you can pause and check the simplification. You end up with V, the image distance being 8.6 centimeters. What does that mean? Well, it is positive, which means it's on the right side. It's inside the mirror. 8.6 means if this is 15, it will be somewhere over here. So this is where my image is going to be. So I found the position of the image. Now I need to find the height of the image and its nature. So I'm going to use this formula. So the height of the image divided by the height of the object, which I know to be five, that should equal negative V. I'm going to use fraction whenever there is division, fractions could help divided by U, which is minus 20. So the minus goes, this goes three times. And so if I simplify, I now get five into three divided by seven, that is 15 divided by seven, giving me 2.1 centimeter. I have rounded off to one decimal place. And so the height is positive, which means it's going to be above the principal axis, but it's going to be smaller. So it's going to be smaller. And so we found the height, which is the size, and we found the nature because it's above the principal axis, it has to be virtual. So this is how we use the mirror formula and the magnification formula along with the sign conventions to solve mirror problems.