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Current time:0:00Total duration:8:53

Video transcript

- [Instructor] Let's say we have a concave mirror of focal length two centimeters, and in front of it, we're gonna keep an object, and let's say that the object is kept six centimeters in front of that mirror. Where will its image be? In this video, we're gonna find out exactly where that image would be. All right, now, in some of our previous videos, we actually discussed about this. We've seen that there are a couple of ways you can solve this. One way would be to actually, you know, draw this on a piece of paper and then use your ruler to figure out exactly, by drawing real diagrams, figure out exactly where the image is going to be. But if we don't wanna do all those drawings, then another method would be to build a master formula, a master formula which connects the object distance, the focal length, and the image distance, and the one which will work for all cases, for concave and convex mirrors, and we can just substitute in that and get our answer. So in this video, we're not going to build the master formula. If you are interested in figuring out how to derive that master formula, there a bonus video, and you can just go there and watch that. But we will see what that master formula is, we'll understand it, and then we'll use it to solve the problem, all right? So that one formula, which is going to help us figure out solve problems like this, that formula looks like this, and it's called the mirror formula, for obvious reason, because it works for mirrors. And let's see what these things are. There are three things you can see. F stands for the focal length, so that's something we already know. V is the image, image distance, so this is image distance, and that's something we don't know in this problem. And u is the object distance, these are the letters that we usually use for mirrors, object distance. And one super important thing to remember is that this is a sign-sensitive formula, which means when we're substituting the numbers into it, we need to take care of their signs, they can positive values and negative values, and to quickly recall why we do that, that's because we didn't differentiate between measurements done in front of the mirror and measurements done behind the mirror. We assign one of those measurements as positive, the other measurement is a negative. And do to quickly recall how we figure that out, which is positive and which is negative, we always choose the incident direction as positive. So notice that here is the object and here is the mirror, and the incident ray would be this way. So incident direction, right side is positive. This means that all the positions to the right side of the mirror will be positive positions. Anything on the right side of the mirror will be positive, and anything to the left of the mirror will be negative positions. And if this thing is not super clear to you, then it would be great to pause this video and go back to previous videos, where we've discussed this in great detail. Watch that and then come back over here. All right, now that we have all our signs ready, let's look at what we have. We need to find the image distance, that's what we need to figure out, so this is what we need. What do we know? We know the object distance, that is six centimeters, object is six centimeters from the mirror, and so u is six centimeters, but remember, we need to take care of the signs. Notice that the object lies on the negative side because all the positions here are negative. So u will be minus six centimeters. Then, what else do we need, we need focal length. We can see the focal length, the principal focus also lies on the negative side, and therefore, this focal length will be negative two centimeters. Hope that is clear. Anything that lies on the negative side will get negative distances. All right, okay, so we have two things, f and u, and we need to figure out what v is. And so now, then, whatever follows is just going to be substitution and algebra. So, great idea to pause the video and see if you can do the algebra yourself. Now, before you pause, a quick thing. I want to emphasize on what these negative signs mean in this particular example. So once we have these sign conventions, when we substitute the negative six over here, we're telling the formula that the object lies on the negative side, which means in front of the mirror. If we had substituted plus six by mistake, then the formula will assume that the object lies on this side, which is behind the mirror, and I know that sounds silly, but that's what the formula would assume, and then we'll probably get the wrong answer. Similarly, when we substitute the negative two centimeters over, we're telling the formula that we're dealing with a concave mirror. How, because only if it is a concave mirror, the focal point will lie on the negative side, as is in this example. And if we had substituted plus two centimeters over here, then it would assume that the focal point is on this side, meaning it would assume it's a convex mirror, again, it would give us a wrong answer. Just telling you how important it is to include proper signs in our formula. All right, pause and try to do the algebra yourself now. All right, let's do this. So if we substitute, we'll get one over f, which is negative two, that will equal one over v plus one over u, which is negative six. Now, I'm not gonna substitute the centimeters because everything is in centimeters, so my image distance will also be in centimeters. I like to put the negative side, negative sign on the numerator, so call this minus 1/2 equals one over v, negative times positive is negative, so minus one over six, excellent. Now, I need to isolate one over v because that's what I want to figure out, so I'm gonna try to get rid of this negative one over six, by adding one over six on both sides. So if we add one over six on both sides, on the left-hand side, we're adding one over six, we will also add one over six on the right-hand side, and when we do that, that'll cancel with this one or negative one over six. And so, we then operate one over v. Okay, so let's solve this. We have to add two fractions. We're gonna make the common denominator, two and six, our cm is six. And so I'm gonna multiply this by three, multiply this by three, it'll get negative three plus one, plus one, that's gonna be negative two, so we can just write that down somewhere. Okay, minus two over six, and that's one over v. And it's getting a little crowded, but you know, it's almost done, minus, we can just cancel this, two goes one times, two goes three times. and so, we have one over v equals negative one over three, so we want v, so we'll do the reciprocal, v will be negative reciprocal of this one, one over three reciprocal is three, as we get negative three centimeters. And there we have it, we have found what v is. So what this means is that the image is three centimeters away from the port, from the center of this mirror. Remember, whenever we are using this formula, all the distances are from the port. So, three centimeters, but which side? Well, the sign tells me it's on the negative side, and so, I know it's in front of the mirror. So it would be somewhere over here because this is two, it's a given, so this would be somewhere over here, three centimeters. And there we have it, we have solved our problem. So, to quickly summarize, we learned what the mirror formula is, which connects focal length, object distance, and image distance, and to use this formula, we have to use our sign conventions, and the sign convention that we use is we consider the incident direction is positive and then the other direction becomes negative. Now, before we wind up, one thing I need to mention is something that we should always remember is that a sign convention that we're using is just a standard. If you ever forget that, don't worry, you will still get the right answer, so what I mean is if, by mistake, when you're solving this problem, if you chose the left side as positive, and the right side as negative, but you substituted your values accordingly, then you would still get the right answer. You would eventually get that the image is three centimeters in front of the mirror. But usually, we like to choose the incident direction as positive, and we like to stick to this sign convention.