Professor Anderson

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And this is the mathematical tools for figuring out exactly where that image is going to lie. So let's draw our picture again. And this is our mirror. Had a focus about there. And when we have an object We know that it's going to form some image somewhere. We're not exactly sure where yet. But this thing is our object. This is our image. There are some distances that we can measure here. One of them is this. Distance to the image from the mirror, from the apex of the mirror, to the image what is that distance? Likewise we can measure the distance to the object. Okay, from the apex of the mirror to the object what is that distance? And then finally we can, of course, measure the focal length. Which is from there to there. So the mirror equation allows us to determine where that image is going to be located. And the mirror equation is the following. 1 over do plus one over di equals 1over f. And remember that F was equal to R over 2. Okay, this thing is going to help you determine where di is. So let's take a look at the last clicker question that we just looked at and verify that this equation is in agreement with that. In that equation, we had the following. We had that the focal length was 1 meter. But we also had that do was 2 meters. Can we solve this now for di? Oh yeah, that doesn't look too bad, right? Let's take this equation and let's solve it for di. All right so we've got one over do plus 1 over di equals 1 over f. So let's move the do to the other side. I got 1 over di equals 1 over f minus 1 over do. And now I know all those numbers, so we can just plug them in. 1 over di equals 1 over f, which is 1 over 1 minus 1 over do which is 1 over 2. All right 1 minus 1/2 is 1/2. So 1 over di is equal to 1/2 and if I flip it back over I get di is equal to 2. And that's exactly what we found on that picture a second ago. If the object was 2 meters from the spherical mirror then the image was also 2 meters from the spherical mirror. In this mirror equation, all these numbers are positive if they are to the left of the mirror. So do measured to the left is a positive number. di measured to the left is a positive number. F measured to the left is a positive number. If you solve the equations and you end up with a negative sign for di that means the image is on the other side of the mirror, it's on the right side of the mirror, okay? And that'll happen when you start to get very close to the mirror. If we have an optic axis here, and we have a plane mirror then we know that an object will have a length do. But its image is on the right side. And it's to the right, and therefore di is negative do. So now if you go back to your equation, right, we had 1 over do plus 1 over di equals 1 over F we're going to have 1 over do minus 1 over do equals 1 over f and if 0 equals 1 over f the only option is f is equal to infinity. Remember that focal length is related to radius of curvature and so you're really asking the question, "what is the radius of curvature of a flat mirror?" In other words, what saw would cut out a flat mirror? A saw that had a radius that was infinity.

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And this is the mathematical tools for figuring out exactly where that image is going to lie. So let's draw our picture again. And this is our mirror. Had a focus about there. And when we have an object We know that it's going to form some image somewhere. We're not exactly sure where yet. But this thing is our object. This is our image. There are some distances that we can measure here. One of them is this. Distance to the image from the mirror, from the apex of the mirror, to the image what is that distance? Likewise we can measure the distance to the object. Okay, from the apex of the mirror to the object what is that distance? And then finally we can, of course, measure the focal length. Which is from there to there. So the mirror equation allows us to determine where that image is going to be located. And the mirror equation is the following. 1 over do plus one over di equals 1over f. And remember that F was equal to R over 2. Okay, this thing is going to help you determine where di is. So let's take a look at the last clicker question that we just looked at and verify that this equation is in agreement with that. In that equation, we had the following. We had that the focal length was 1 meter. But we also had that do was 2 meters. Can we solve this now for di? Oh yeah, that doesn't look too bad, right? Let's take this equation and let's solve it for di. All right so we've got one over do plus 1 over di equals 1 over f. So let's move the do to the other side. I got 1 over di equals 1 over f minus 1 over do. And now I know all those numbers, so we can just plug them in. 1 over di equals 1 over f, which is 1 over 1 minus 1 over do which is 1 over 2. All right 1 minus 1/2 is 1/2. So 1 over di is equal to 1/2 and if I flip it back over I get di is equal to 2. And that's exactly what we found on that picture a second ago. If the object was 2 meters from the spherical mirror then the image was also 2 meters from the spherical mirror. In this mirror equation, all these numbers are positive if they are to the left of the mirror. So do measured to the left is a positive number. di measured to the left is a positive number. F measured to the left is a positive number. If you solve the equations and you end up with a negative sign for di that means the image is on the other side of the mirror, it's on the right side of the mirror, okay? And that'll happen when you start to get very close to the mirror. If we have an optic axis here, and we have a plane mirror then we know that an object will have a length do. But its image is on the right side. And it's to the right, and therefore di is negative do. So now if you go back to your equation, right, we had 1 over do plus 1 over di equals 1 over F we're going to have 1 over do minus 1 over do equals 1 over f and if 0 equals 1 over f the only option is f is equal to infinity. Remember that focal length is related to radius of curvature and so you're really asking the question, "what is the radius of curvature of a flat mirror?" In other words, what saw would cut out a flat mirror? A saw that had a radius that was infinity.