 Let's explore lens formula, magnification formula, sign conventions, and you solve a numerical from NCRT. So pause the video and just read this numerical first. All right, we are asked to draw a ray diagram. So let's do that first. So we are given that we have a converging lens of focal length, 10 centimeters, and an object is kept 25 centimeters away from it, and the height of the object is given as 5 centimeters. So we've seen how to draw a ray diagrams. We draw one ray of light that's parallel, it goes through the focus, and another ray of light going through the optic center goes undeviated. And that gives us the image over here. And our goal now is to find out what the image distance is, what the size of the image and the nature of the image is, how do you do that? So this is where we use the lens and the magnification formula. F stands for focal length, V is the image distance, and U is the object distance. And for magnification, which we define as height of the image by height of the object, for lenses happens to be image distance divided by the object distance. Now this might seem very familiar, very similar to the mirror formula, but there is a difference. In the mirror formula, you use a plus sign, here you use a minus sign. And over here for the mirror formula, you get a negative, but here you get a positive. So be very careful about this when you are applying it over here. Okay. Before we plug in the numbers, the next important thing is to draw sign conventions. What that means is that these, this formula only works when you use appropriate signs for all these values. And how do we decide the science? Well, the sign convention says you draw the object always to the left of the lens or the mirror. And then you choose the right side of the lens as positive, left side of the lens as negative. You choose above as positive and below as negative. This is for heights and this is for the distances. Okay. With that in mind, we can now write down what is given to us. Our object distance, which is you is given as 25 centimeters, but it's on the left side of the lens. So it is negative. The height of the object is given to us as five centimeters. But this is above the height for the height you have to look above or below. It's above the principle axis, so that's positive. And we are given the focal length is 10 centimeters. Now, which focus do you use? This or this? Well, from the red diagram, it's clear that this is the focus. And so that focus is on the right side, so it is positive. And now we can plug in and we can find our V. So feel free to do that yourself first. Okay. Let's plug in. So I get one over 10 equals one over V, which I don't know, minus one over U, which is minus 25. And if you simplify it, and I did it all, you can pause the video just to check the simplification. But if you simplify it, eventually we get V to be 16.67 centimeters. Now what does that mean? First of all, you have a positive sign. That means it should be on the right side of the lens, which is exactly what we got from the red diagram. Then we also see that it is 16.67. That means from the lens, this distance is 16.67 centimeters. So we found the position of the image. The next we need to find the size for that we will use magnification formula, which is height of the image, which we need to find that's the size, thereby the height of the object, which is given to be five centimeters. That equals V, which is we just found out 50 divided by three. And I like to use fractions because there is a division here. There's a division here. And so things can get canceled or divided nicely divided by U, which is minus 25. And again, if you simplify, this goes five times and this goes 10 times. So we get height of the image to be minus 10 divided by three. That gives you minus 3.33 centimeters. Now, what does the minus sign say? The minus sign says that it has to be below the principal axis. Exactly what we got. And the height, the size is 3.33 centimeters, which is diminished compared to the object. That's exactly what we got over here. And because it is negative, the height is negative, it's below the principal axis. We know that this is inverted compared to the object. And inverted images are always real images. So we've also found the nature. It's real image. Let's try the next problem. We're given a concave lens of focal length 15 centimeter forms an image, 10 centimeter from the lens. How far is the object placed from the lens? Draw the ray diagram. So why don't you pause the video and first see if you can draw the ray diagram yourself? All right, here we go. So here is the lens. We have 15 centimeters. We know the image is formed over here. I get that from the ray diagram. And I see that the object distance is what we need to calculate. And again, look at the ray diagram. A parallel ray of light goes deviate. And again, if you look at the ray diagram, we first draw a ray of light parallel to the principal axis that diverges away, appearing to come from the principal focus. And then another ray of light through the optic center goes undeviated. So the two rays appear to come from here. And that's where my image is. And now again, before I substitute over here, I have to choose the right side as positive. The left side is negative. The upwards is positive. And the downwards as negative for the height. OK, so let's see what's given. U is not given to us. We need to figure that out. We are given the focal length is 15 centimeters. Which focal length do we use? This one or this one? Well, from the ray diagram, you can see it is this one. Because this is where it appears to go from. And that is on the left side. So this is negative 15 centimeters. And then we are given v this time. What is v? It's 10 centimeters from the lens. Positive or negative? It's on the negative side, you can see. So that's minus 10 centimeters. And there we go. We have everything. We can now just plug in and calculate. So 1 over minus 15 equals 1 over minus 10 minus 1 over U. And again, if we simplify, you get U to be minus 30 centimeters. Again, feel free to pause and check the simplification to make sure that you're also getting the same answer. Anyways, now what does it mean that U is minus 30 centimeters? Well, the minus sign tells me that it is on the left side. And that's exactly what the ray diagram is telling me as well. So that's nice. And it tells me that it is 30 centimeters from the lens. So this distance must be 30 centimeters. And that again looks very good from the lens diagram. It is farther away from the focus. So everything looks great. Now in this question, we are not asked what the size and the nature of the image is. But if it were asked, we could just use the magnification formula, substitute and get the answer.