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A concave lens has a focal length of $-39 \mathrm{cm} .$ Find the image distance and magnification that result when an object is placed 31 $\mathrm{cm}$in front of the lens.

Thus, the image distance is $[-17 \mathrm{cm}]$Thus, the magnification produced by the lens is $[0.55]$

Physics 103

Chapter 26

Geometrical Optics

Wave Optics

Miguel V.

November 24, 2021

It would be helpful to explain how to derive the inverse to get the distance image of -17cm

University of Michigan - Ann Arbor

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All right. So for this problem, you have an object that's sitting 31 centimeters in front of a conclave lens as a focal length of 39 centimeters. We want to know exactly what the image distance is and what the image magnification is. And in order to find these two things, we're gonna have to use the image, uh, the lens equation. Excuse me and magnification equation. So let's set up our lens equation. First, we're going to say that one over f is gonna be equal to one over the image distance plus one over the object distance and a compliant or known values. We know the focal length to be negative 39 centimeters. We know the object distance to be 31 centimeters. We don't have to use meters thes equations because they're just ratio equations. So you centimeters and we'll say that one over the image distance. It's gonna be equal to one over negative 39 minus 1/31. And if we take this difference here or the summation rather and then find its inverse, we can say that d I is gonna be equal to negative 17 centimeters. Now we find the magnification will use the notification equation, saying that M it's gonna be equal to negative distance of the image over distance of the object you can plug in are known values. So instead, modifications would be equal to negative negative 17 over the distance of the object. We're just gonna be 31 and they will give us a magnification of 0.55 and these are two answers.

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