 Let's use the magnification formula for mirrors for a couple of questions. For the first one, we have a 10 cm long pencil which is placed in front of a convex mirror, a 2 cm long virtual image forms above the principal axis and 8 cm behind the mirror. Find the object distance U of the pencil. Let's write the answer in the Cartesian sign convention. Okay, as always pause the video and first draw everything, draw what you can and try the question on your own. Alright, hopefully you have given this a shot. Now for any question in ray optics, the best approach to begin with is always to try to draw the mirror and the axis, the object and everything. So we have a convex mirror. Let's try and draw a convex mirror to begin with and okay, good enough. Alright let's now make a principal axis. So principal axis can look, it can look somewhat like this. Okay, so this is the convex mirror. So this outer surface, that is where the light would be incident and we have a 10 cm long pencil. So let's make that, let's say that this right here is the 10 cm long pencil and we can write 10 cm. Now a 2 cm long virtual image is formed above the principal axis and 8 cm behind the mirror. Let's say that this much is 8 cm and a 2 cm long image is formed. It is above the principal axis and it is at a distance of 8 cm and this is 2 cm. Okay, we need to figure out the object distance of the pencil. Object distance is this distance right here and we need to figure this out. Now first let's use the sign convention and write down these values. So to do that, the height of the pencil, let's use a different color, let's say okay, the height of the pencil, so this is the height of the object, this is plus 10 cm. I'm writing plus because vertical distances are positive above the principal axis and similarly the height of the image is also plus 2 cm. Now we also know the image distance V and this is actually also plus 8 cm. And let's go back to the sign convention. So light is really incident, light is incident, light is incident on this side of the mirror. This is where the light is incident. So that will be towards the right, so direction would be towards the right. Light is incident on the mirror to the right. So horizontal distances will be positive on the right hand side and negative on the left hand side. So therefore we see that V, this distance is to the right, to the right hand side in the direction of incident light, this is plus 8 cm. And U, something that we need to figure out, object distance, we do not know what that is. Okay, now let's use the magnification formula. So that is magnification, let's write that over here. That is height of image divided by height of object, this is equal to minus V by U. Now height of image is 2, 2 divided by 10 that is the height of object, this is equal to minus 8 divided by U. And when you work this out, U comes out to be equal to minus 40 cm. So this is minus 40. And the minus sign makes sense because U is on the left hand side to the direction of the incident light and according to the Cartesian sign convention, this distance would be negative because it is opposite to the direction of incident light, which is to the right. So this right here, this is minus 40, minus 40 cm. Okay, let's look at one more question. Here we have Meghana who places an object 15 cm in front of a concave mirror. She observes that a 4 cm high inverted image is formed, 12 cm in front of the mirror. Find the height of the object. Okay, just like in the previous question, we will start off by drawing what the question is, what the question is telling us. So let's make a concave mirror and okay, good enough. Now let's make a principal axis. So there it is. Let's show that this is a concave mirror. So this will do the work. All right. Now she places an object 15 cm in front of a concave mirror. We don't know the height of the object. We don't really know that. But still let's just, let's just say that this is the height. This is the height, but we do know the distance. This is 15 cm from the mirror. And we know that an image is formed 4 cm high, which is inverted and 12 cm in front of the mirror. So in the same side, and it is 4 cm high. Now images 4 cm high inverted. So it will look like this. Let's say this is, this is 4 cm and the distance from the mirror is 12. So this right here is 12 cm and this the height, this is 4. Okay. Now let's write down, write down with the sign convention. So U, U will this be plus or minus. For you to pause the video and think about this. U is the Cartesian sign convention. Will U be plus 15 or minus 15? So turns out U is minus 15 cm because the direction of incident light is to the right. So the horizontal distance is to the right of the mirror would be positive. That is in the direction of the incident light. But everything on the left hand side opposite to the direction of the incident light, that would be, that would be negative. So U is negative, V will also be negative because again, that is, it is opposite to the direction of the incident light. This is minus 12 cm. And we know the height of the image. Now this will again be negative because vertical distances below the principal axis are taken as negative. And we need to figure out the height of the object. So let's again use the magnification formula. This is H i divided by H o. This is equals to minus V by U. Now H i is minus 4 and H o, we do not know what that is. V is minus 12. So this becomes minus of minus 12 divided by minus 15. That is U minus this minus gets cancelled off. And even these, even these minuses, they get cancelled off. So when we work out H o, this comes out to be equal to 5 cm. This height right here, this is 5 cm. All right. You can try more questions from this exercise in the lesson. And if you're watching on YouTube, do check out the exercise link which is added in the description.