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A converging lens of focal length 9.000 $\mathrm{cm}$ is 18.0 $\mathrm{cm}$ to theleft of a diverging lens of focal length $-6.00 \mathrm{cm} .$ A coin is placed12.0 $\mathrm{cm}$ to the left of the converging lens. Find (a) the locationand (b) the magnification of the coin's final image.

a) 9.0 $\mathrm{cm}$b) $-1.5$

Physics 103

Chapter 27

Optical lnstruments

Wave Optics

Cornell University

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

University of Washington

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Okay, so we know that we have in this problem coin. That was a place to the left off. Ah, cold verging lens. And this land's is left off a diverging lance. We know that the separation between the converging lens and the diverging lens iss Let's see, 18 centimeters. So this distance here is 18. Same team enters. Okay, what else do we know? We know that the coin displaced 12 centimeters from the converging lens. So this distance here is 12. Same team enters. We also know that the focal lens off the converging lens he's nine team enters, and we know that the focal and off the diverging lands is minor. Six send team enters. Okay? We have any other information? I don't think so. Okay. So what we want to calculate, we want to know what is the location and the magnification off this coin in this Toolan systems. We know that every time we have a tool and system which the two lands are separated by a distance, we have to work with one lens at a time and discovered the distance off the image off each one of those lenses. So let's begin the calculation. What do we have here? We know that the distance off the object to the first lens, the converging lens is just to al ve think team enters. We know that the focal lens is not. I don't need to rewrite this. We know that the focal lens off the converging lens is 1910 meters. So we can use the Finland's equation to discover the position off their image that the first lens reduce. So this is going to be one divided by F one miners, one divided by these. You know, one This is going to be to the power off minus one. So this is just one divided by nine miners, one divided by 12 Oh, to the power of minus one. So this is just 36 centimeters. Okay, Now the distance off the or project off the second lens is just the difference between this separation and the position off the image off the first lens. So this is just miners 18 70 metres. Okay, so we have the distance off the object of the succulence. We have the focal lens off the second lands. So using again the Finland's equation, we can discover the distance off the image off. The second lens is produced by the second lands. This is just one divided by I have two miners, one divided by the zero to oh R minus one. This is just one minor. Six miners, one divided by miners. 18 Oh, the power minus one. Finally, this gives us miners nine same team members. So the answer to the first itin what is the location of the image? The image is located at nine same team enters to the left off the diverging lens off die verging lens. Okay. And what is the magnification? The magnification off this system that's try to write the M in here we have that. The magnification is just the multiplication off the magnification of the first lens with the 2nd 1 And we know this is defined us miners. The one I divided by d 01 multiply by minors did, too. I divided by the zero to So we have all this and this is going to be that is six times miners ninth divided by 12 times minus 18. So the magnification he's positive 1.5. And that's the final answer to this problem. Thanks for watching

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