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Consider three lenses with focal lengths of $25.0 \mathrm{cm},-15.0 \mathrm{cm},$and 11.0 $\mathrm{cm}$ positioned on the $x$ axis at $x=0, x=0.400 \mathrm{m},$ and$x=0.500 \mathrm{m},$ respectively. An object is at $x=-122 \mathrm{cm} .$ Find (a) the

a) 0.382 $\mathrm{m}$b) $-0.4$

Physics 103

Chapter 27

Optical lnstruments

Wave Optics

Cornell University

Rutgers, The State University of New Jersey

University of Washington

Hope College

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Okay, so in this problem we have three lances and we have on objects. So let's try to draw this. We have an object here at position, except those miners were 122 centimeters. We have the first lens. It is a concave Linse because the focal length it's positive at position zero. Let's call this F one equals zero point 25 meters. We have this second lens. This lance is not Kovacs. It's concave. So let's see. This is a concave lance opposition 0.4 meters. This is actually a meters is not saying teenagers and the focal lens of F too is zero point 15 meters in their final land. Lentz is a cove excellence at position. Zero point five meters with a focal lens off 0.11 meters. Okay, and what the problem wants for us. We have to calculate the position and the magnification off the image. So let's move on. We know that the position off, let's do disease and under a bridge, the position off the image produce by the first lens using the Finland's A creation is just one divided by F one miners, one divided by D 010 so the power of minus one. So we have the focal lens, which is 55 centimeters. So you have one divided by 25 miners, one divided by 122 ST T Mater's all to the power of minus one. I have to put this intimate trees year. I wasn't wrong. And then we have to calculate this. We're gonna get let's see starting a one 0.4 state team enters. Okay, So 31.4 sing team enters is the position of the image produced by the first Let's no less discover. Let's discover the position off the object related to the second lands is going to be the separation between the two lances, which we can see here. That is 0.4. So we have 0.4 miners the position off the first image, which is little 0.314 which gives us, you know, points, you know, 86 meters. Okay, so this is the positions of the object. Now let's calculate the position off the image related to the second lands. It's going to be one divided by f too minus one, divided by D 02 Oh, what did your father of minus one? Which gives us one divided by minus 15 miners, one divided by eight point 56. 56. Let's see. No, actually 0.6. So let's write this 8.6 when six owe to the power of miners. One. This is closer to minors. Five point 45 Same team enters. Okay, so now we can calculate the position off the object related to the turtle ends. So this is just the separation between this deterred in the second lens. She's, you know, 0.5 miners, you know, 0.4. Is it a point once minus the image, which is five miners minus careful with the unit is his minus zero point 0545 meters. So calculating audience, we're gonna get zero point 155 meters now Finally, let's calculate to the position of the human Ege off this whole system. It's going to get one divided by. I have three miners, one divided by D zero three, the power of minus one. So we get one divided by 11 miners, one divided by 15.4 five. Minar swung. This is equal to 38 saying team enters So the location, the position of the turd image is just Let's see in here uh, position of the finally image is just the position off the turd lands, which is 0.5 below is the distance off the image to the lens, which you're going to get, you know, point 38. So this gives us the precisely location of zero point 80 or 88 meters. Now let's think about the magnification. We know that magnification is just the multiplication off all the magnification Sze and we know this is just miners d i wa he i to thee I treat divided by d 01 d 02 India zero tree. We have all this now, so let's institute. We're gonna get miners 31.4 times minus 5.45 times 38 oh, divided by it's fink 1 to 2 times 8.56 times 15.5. So multiplying all this we have a magnification finally off zero point four. So we have a positive magnification and the images upright. But the image

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