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REFERRING TO EXAMPLE $26-5$ Suppose the radius of curvature of the mirror is 5.0 $\mathrm{cm}$ . (a) Find the object distance that gives an uprightimage with a magnification of $1.5 .$ (b) Find the object distancethat gives an inverted image with a magnification of $-1.5 .$

(a) 0.83 $\mathrm{cm}$(b) 4.2 $\mathrm{cm}$

Physics 103

Chapter 26

Geometrical Optics

Wave Optics

Simon Fraser University

Hope College

University of Sheffield

McMaster University

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this question. We need to apply the mirror equation 22 different situations. Okay, so we have a mirror readers off curvature 5 p.m. Okay. And this is our on cape mirror because the radios off culture is positive. Yeah. So our if five cm. Okay, in part A. We want to find the object distance that gives an upright image with M equals to 1.5. Yeah, que So em is equal to minus the eye over the old, which is 1.5. This means that b I is equal to negative 1.5 d o. And then, um, they also given that the regis off culture, it's 5 p.m. So the focal line iss have the regis off curvature. But this is 2.5. Yeah, okay. Be using one over the oh, plus one over the I equals to one over f. So, uh, we just want to find the oh, okay. And then Oh, so, uh ah. I wonder where the oh plus one of minus never the 1.5 d O thesis is going to be won over three d o. And then this is equal to 1/2 0.5. Hey, this is a wonderful f. Okay. And so, uh, p O s 2.5 divided by three is equal to 0.83 Yeah. Okay, so this is the answer for a And then be, um we are given that, uh, is equal to negative 1.5, and then this is equal to negative The i d o. This means that the I is equal to 1.5 g o. Then we repeat the same step using one over the O. That's one of the I equals to one of the F. In this case, you also want to find object distance. Okay, so we have one over the old yes, 1/1 0.50 And then this is, uh, five over tree CEO. And this is a go through 1/2 0.5 PM and so b o is equal to 5/3 times, 2.5 a. M. And you get 4.2 c m as the object distance. Okay, so that's all for this question.

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