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Galileo devised a simple terrestrial telescope that produces an upright image. It consists of a converging objective lens and a diverging eyepiece at opposite ends of the telescope tube. For distant objects, the tube length is the objective focal length less the absolute value of the eyepiece focal length. (a) Does the user of the telescope see a real or virtual image? (b) Where is the final image? (c) If a telescope is to be constructed with a tube of length 10.0 cm and a magnification of 3.00, what are the focal lengths of the objective and eyepiece?

a. \text { See work for solution. }

b. \text { See work for solution. }

c. \text { See work for solution. }

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{'transcript': "Okay, So in this problem we have I got Lillian telescope. And in the first itin off the problem, we want to calculate the focal eyepiece lens over the stethoscope, knowing that he has a magnification off for off four and a focal objective. Lend off one point 25 meters. Okay, So, first of all, we know that the total magnification is just the f objective divided by the the focal lens off the eyepiece. But since the leading telescope is constrict with are converging lens and a convex lens, we have a minor sign in front of the magnification. So to calculate the focal, I've piece that we want to discover it's just going to be minus for times one point actually divided. We have the miners f objective, divided by magnification. So this is just going to be minus one point 25 divided by four. This is equals two minus 31. See question miners 31 point three. State team enters. Okay, so the second item off the problem, we went to discover Ah, the barrel lens off the telescope. And we know by definition that the barrel length is just the objective. Plus the eyepiece. Focal ends some. So this is just one point 25 minus zero point 31. This is just, let's see one point 56 meters and that's your final answer. So this problem. Thanks for watching."}

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