by Patrick Ford

Hey, guys, let's do an example of five centimeter tall. Object is placed 10 centimeters in front of a convex mirror. If the radius of curvature of the mirror is two centimeters, where is the image located? Is the image real or virtual? Is the image upright or inverted? And what is the height of the image? Okay, now the first three questions are all given to us by the image distance. The answer to those questions is given to us by the image distance right, based on the image distance, we know where the image is located. Obviously, we know whether it's real or virtual based on the sign, and we know whether it's upright or inverted, based on whether it's real or virtual. So. Really, those first three questions are answered by the same piece of information. All right, in order to find that image distance, we need to use our mirror equation. That one over the object distance plus one over the image distance is one over the focal length. Question is, now, what is the focal length? Well, the focal length is gonna be our over to in magnitude, but since this is a convex mirror we know by convention, the focal length has to be negative. So I'm gonna put a little negative sign in here so that I don't forget. And I don't mess up the problem because I used the wrong sign. The radius of curvature is two centimeters, so it's negative to over to which is negative. One centimeter. Now, I can use the mirror equation, so let me rewrite it to solve or sorry. Toe isolate for the image distance. This is one of us minus one of us. Not. This is gonna be one over. Negative one minus 1/10. Okay. And if you want, you can simplify this to use the least common denominator. This is negative. 10/10 minus 1/10 which is negative. 11/10. And then that makes finding the image distance as simple as just reciprocating the answer. Okay. This is where a lot of students make mistakes. You're finding one over s. I You are not finding, s I. This is not the final answer. The reciprocal of it is the final answer. And that is negative. 091 centimeters. Kate. That's the image distance now. Is this image real or virtual Well, it's negative. So it is virtual. Is it upright or inverted? Well, since its virtual, it has to be upright. So this is virtual and upright. Okay, those two have to go together and they always go with the negative image distance. The only thing left is defined the height of the image which is given to us by the magnification. So the magnification and I'm just gonna drop the sign because the sign is, frankly, a waste of time. We already know that it's upright, so we don't care about the sign of magnification. The image distances 0.91 the object distances 10. So this is gonna be 009 That's the magnification. That means that the image heights is 0.9 times the object type, and the object is five centimeters tall. So this is 0. centimeters. So if we want to sum up all the information about this image, it is located 0.91 centimeters from the mirror, technically behind the mirror, it is a virtual image which is upright and has ah, height of 0.45 centimeters. Alright, guys, that wraps up this problem. Thanks for watching

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