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Find the location and magnification of the image produced by the mirror in Problem 26 using the mirror and magnification equations.

Thus, the image distance of the concave mirror is $[1.2 \mathrm{m}]$The magnification of the image produced by the concave mirror is $[-0.6]$

Physics 103

Chapter 26

Geometrical Optics

Wave Optics

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question 26 says an object for the height of 33 centimeters is placed two meters in front of the con cave mere with the focal length of 20.75 meters. Ah, we're supposed to diagram this to determine the location and size of the image. And so what's do that quickly? I'm just going to sketch this. I'm not actually put it on graph paper, but would get reasonable with this. Ah, and then we can move on from there. So we have ah, focal length of 0.75 meters. So here's one meter. Here's two meters. Ah, this is our focal length. No, no. Our focal length of sorry. 0.75 So here's our focal length 0.75 meters. Um, our object is placed two meters away. So here's our object, and it's got a high of 0.33 meters. Um, 0.33 meters. Now, if we were to sketch this, we're gonna take a line straight off of the top of the object towards the mere, and that line will reflect through the focal point back here somewhere. And then we'll go off of the object through the focal point towards the mirror, and that's going to reflect back here. And so we see that our image is going to form right there. Um, this was one meter. This looks to be a little bit past it. And so our image distance is probably around 1.2 meters Will do the math to prove it in just a second. And then it says, What is the approximate size? Ah, well, if this is 0.33 meters and this is it's also about 0.2 meters high, but it's negative. It is an inverted thing. So let's go ahead and actually do the math on this and then ah, see what we get. So remember, one over Dio plus one over D I is one over f And so in this case, I have, um, won over two plus one over d. I is equal to one over 10.75 or one over D. I equals one over 10.75 minus 1/2. And so I'm going to invert every side of this to get my answer. And on my calculator, I'm just gonna push 0.75 inverse minus two inverse push, enter and then invert that and get 1.2. And so my image distances in fact, 1.2 meters. Now. Remember the ratio of the image distance to the object distance is also the ratio of their heights. So 1.2 divided by two is 20.6 and 0.6 times 0.33 is 0.1 night eight and so our height, eyes 0.198 and it is inverted.

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