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Two concave lenses, each with $f=-15 \mathrm{cm},$ are separated by7.5 $\mathrm{cm} .$ An object is placed 25 $\mathrm{cm}$ in front of one of the lenses. Find(a) the location and (b) the magnification of the final image produced by this lens combination.

a) 4.4 $\mathrm{mm}$b) 0.18

Physics 103

Chapter 27

Optical lnstruments

Wave Optics

University of Michigan - Ann Arbor

University of Washington

Hope College

University of Sheffield

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Okay, so in this problem, we have to conclave lances itch with a focal end. That's put F one and two off. How much? Minus 15 miners. 15. Same team Richter's. And we know that these two lances separated by a distance D off 7.5. Think team monitors and we need to calculated did final distance the location, actually, and the magnification off the final image produce by this lances. If we have object police at a distance off 25 centimeters off one of those lances. Okay, so this problem's quite competently complicated. So we're going to separate. Yeah, various steps. The first step. We need to calculate it. The image to the first lens. So where is going to be the image it's called the one I to the first lens is just a finance, a creation. So it's just going to be one divided by F minus one, divided by the initial position from the first lens to the hour of miners. One. So this is just going to be one divided by miners. 15 miners, one divided by 25 which is the initial position off the object. Ah, all discombobulated gives us minus nine point three seven five 70 metres. So this is the image produce by the first lens. Okay, the second step, This is the first step. The second step. We're going to subtract the first image distance from the separation distance. So the second lens initial position, it's just going to be this separation distance minus the distance from the image of the first lens. So this is going to be 7.5 minus come minors with minors plus nine point 37 five. So the initial position to the second lens, it's just going to be 16 points. 87 five Same team actors. Okay, the dirt stuff. Turn step. We're going to solve the Finland's equation to the second lens. So the image produced by the second Lentz image reduced by the second lense just going to be one divided by F minus one divided by in your shoe object of the second lands to the power of minus one. All right, so this is going to be one miners 15 minus one, divided by 16 16 points. 875 All this to the power of miners. One recall If we calculated this, we're going to have minor seven points, 94 same team mixtures is a four. Okay, so finally, the answer to the first item it's that is the location of the image. We just need to ad the separation distance to find a position off the image. So the image that we want to discover is just 7.5 plus miners. Actually, that's put minors in here seven point 94 which gives us mindless little point for for same team monitors. So a final image is located a 4.4 millimeters in front of the lens closest to the object. And what is the magnification out? My identification is just magnification, just a magnification of the first lens multiply by magnification of second lens, and we know from Chapter 26. The magnification is just a distance off the image, divided by the distance off the object before minors in front times distance off the image divided by the distance off the object. And we already have all this information. So the magnification, it's just going to be let's see minus nine nine Point tree 75 If I didn't buy 25 times my new seven point 94 if I did buy 16 points. Eight 75 So if we multiply all of this, we have a magnification of zero point 18.

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