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An object with a height of 2.54 $\mathrm{cm}$ is placed 36.3 $\mathrm{mm}$ to the left of a lens with a focal length of 35.0 $\mathrm{mm}$ . (a) Where is the imagelocated? (b) What is the height of the image?

(a) 0.98 $\mathrm{m}$(b) $-68 \mathrm{cm}$

Physics 103

Chapter 26

Geometrical Optics

Wave Optics

Jennifer G.

December 18, 2020

Why did you use meters for magnification and then multiply that by the height in centimeters? Don't they need to be the same units?

Cornell University

Hope College

University of Winnipeg

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all right. So for this problem you have an object that's placed 36.3 millimeters in front of the 35 millimeter focal length lens, and the height of the object is 2.54 centimeters. And there's two things we want to know. You want to know the distance of the image will be from the lens, and I don't know what height the image will be. So in order to find the distance of the M attributions lens, we use our lens equation. When I say 10 graph, it's gonna equal to one over D I plus one over Dio. So if we plug in our knowns, we know F is 35 millimeters should be equal to one over d I, plus the object distance, which is 36.3 millimeters so we can say one over d. I is equal to 1/35 minus one over 36.3. Now, if we plug this into our calculator and then take the inverse of it, we should get a D. I off 977 millimeters, which is equal to 0.9 eight meters approximately all right now in order, find the height of the image we need to find the magnification That's good. That's being applied to the image. In order from the magnification, we can use the object and image distances and then apply those theory journal height of the object. So we're gonna say magnification M is gonna be equal to negative distance of the image over the distance of the object Plug in our knowns. Modifications will be equal to negative distance. The image we just calculated his negative 0.98 completed in your sign and provide that by the distance of the object, which is 0.0 363 and meters sure units match here, and that should give us an answer of negative 27. And that's our modification. So same magnification is also equal to the height of the image Over the height of the object. We can say that the height of the image is equal to magnification times the height of the object. We know both of these suing say the height of the image is just evil to negative 27 times the height of the object, which is 2.54 centimeters, and that will give us an answer of approximately negative 7 68 centimeters Now what does it mean for us to get a negative height? Well, just means our image has been inverted, and that's our final answer.

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