Hey, guys. Let's do an example. A 5 centimeter tall object is placed 10 centimeters in front of a convex mirror. If the radius of curvature of the mirror is 2 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. Alright?
In order to find that image distance, we need to use our mirror equation:
1 s + 1 s' = 1 fQuestion is now, what is the focal length? Well the focal length is going to be r over 2 in magnitude. But since this is a convex mirror, we know by convention the focal length has to be negative. So I'm going to 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 2 centimeters, so this is negative 2 over 2, which is negative 1 centimeter.
Now I can use the mirror equation. So let me rewrite it to isolate for the image distance:
1 s - 1 s' = 1 ( - 1 ) - 1 10And if you want, you can simplify this to use the least common denominator. This is negative 10 over 10 minus 1 over 10, which is negative 11 over 10. And then that makes finding the image distance as simple as just reciprocating the answer.
K? This is where a lot of students make mistakes. You are finding one over s'. You are not finding s'. This is not the final answer. The reciprocal of it is the final answer, and that is negative 0.91 centimeters. K? That's the image distance.
Now is this image real or virtual? Virtual? Well, it's negative, so it is virtual. Is it upright or inverted? Well, since it's virtual, it has to be upright. So this is virtual and upright. K? Those 2 have to go together and they always go with a negative image distance.
The only thing left is to find the height of the image which is given to us by the magnification:
m ; = -s' sSo the magnification and I'm just going to 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 the magnification. The image distance is 0.91. The object distance is 10. So this is going to be 0.09. That's the magnification. That means that the image height is 0.09 times the object height. And the object is 5 centimeters tall. So this is 0.45 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 a height of 0.45 centimeters. Alright guys, that wraps up this problem. Thanks for watching.