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Predict/Calculate An obiect is located to the left of a convexlens whose focal length is 34 $\mathrm{cm}$ . The magnification produced by the lensis $m=3.0 .$ (a) To increase the magnification to 4.0 , should the objectbe moved closerto the lens or farther away? Explain. (b) Calculate thedistance through which the object should be moved.

(a) see work(b) see work

Physics 103

Chapter 26

Geometrical Optics

Wave Optics

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Simon Fraser University

McMaster University

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All right. So for this question, we have the Comdex lens of focal length of 34 centimeters on object distance that is unknown. That's creating an image of unknown distance. Except we do know that the magnification is initially three we're trying to do is raise it to four. And what we want to know is that if we're raising magnification from 3 to 4, do we have to move the object closer to or further from the lens? And we want to find that out initially, just analytically, without any calculation. And then we're gonna find that, actually, by how much we actually have to move the the object. So in order, find it analytically were to write down our lens equation. One over f is gonna be equal to one over Dio plus one over D I I'm gonna write down a magnification equation Magnification. It's gonna equal three initially, which is equal to negative. D I over dio So what every one of these equations initially tell us? Well, the first thing you figure out is that the eye is virtual. It's not really That's because our magnification is positive. If our magnification were negative, then we have a real image, but it's positive and we have a convex lens. So we have a virtual image. Next you want to know. Is it because we have a virtual image and we have a convex lens? The distance of the object must be within the focal length. That's the only way that we can create a virtual image of the convex lens. So there two things we can find out just by looking at our equations. The next thing we could find out is we have to analyze. We realize we're trying to raise the magnification from 3 to 4. We're trying like this number here bigger. So in order to do that, we must obviously have to make the numerator of this fraction bigger. We're trying to increase d. I essentially make d I a greater and negative number. So how we're gonna do that? We're Look at our lens equation. We're in. Rearrange it actually were to say one over f minus the object distance one over the object distance. Excuse me as an equal to one over the image distance, so this should show us exactly what we have to do in order to make the image distance a greater negative number. All right, so in order to make the image isn't a greater negative number, we need to make its inverse a smaller negative number. Try and follow my logic here. The M assistance has to be greater. So it's inverse. Must be smaller because this is a difference here in order to make the smallest number possible or a smaller number as faras. This inverse goes the universe of D I that we need to make the inverse of F and the inverse of dio as close together as possible so that their difference is very small. Does that make sense? We need to make the inverse of F in the universe of Dio very close together numerically so the difference is small to create a small inverse for one over d I. So if we're doing that, we're trying to make this smaller than it already is. We must get dio closer to deep tow f than it already is. We know Dios on the near side of F come when compared to the lens, so you must be moving away from the lens. So the answer for a is away from once, However, what we want to know next is by how much are we moving away from the lens? And so we need to do some math in order to figure that out. First math that we have to do is you have to set up our lens equation. Except now we're going to plug in the numbers of me now. So I would say one over half is just gonna be won over 34 centimeters. That's going to be equal to one over Dio plus one over D I. We can use our relationship of the magnification being equal to negative D I over Dio. In order to create a new relationship between d i. India, we can multiply my deal on both sides and say that three times dio people to negative d i If we plug that in to our lens equation, we can actually solve for a value of Dios. We can say 1/34. It's gonna be able to one over Dio minus 1/3 dio which should give you 2/3 dio Now if we do this correctly will say 34 it's going to be equal to three D over two imploding and or a calculator. We should get an initial object distance of 22 0.67 centimeters. Now that's where object is initially placed. In order, Teoh create a magnification of three. We could find out where our images using this number and our lens equation, but we're not going to cause the images not of importance in this question, were only trying to figure out why, how much we change the distance of the object from the lens. So that's our initial distance of the object from the lens. Let's find our new distance when our magnification isn't three anymore. But it's now four that's gonna change up our math just a little bit. We're still going to say 1/34 is equal to one over Dio, except now it's gonna be subtracted by 1/4 Dio and that will give us 3/4 Dio you then say 34 is equal to four dio over three and I'm gonna make this a dio Prime causes are new dio so dio prime should be equal to 25.5 centimeters and this is the new distance of our objects in order to raise the magnification from 3 to 4. Now, this isn't our answer or answers the difference between these two. So we still need to find that out. Where I say Dio prime minus dio is gonna be equal to 2.8 centimeters approximately. And we already answered part A. Which means that we know that we're moving 2.8 centimeters away from the lens and this is our answer.

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