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A telescope has lenses with focal lengths $f_{1}=+30.0 \mathrm{cm}$ and$f_{2}=+5.0 \mathrm{cm} .$ (a) What distance between the two lenses willallow the telescope to focus on an infinitely distant object andproduce an infinitely distant image? (b) What distance between the lenses will allow the telescope to focus on an object that is 5.0 $\mathrm{m}$away and to produce an infinitely distant image?

a) 35.0 $\mathrm{cm}$b) 36.9 $\mathrm{cm}$

Physics 103

Chapter 27

Optical lnstruments

Wave Optics

Cornell University

University of Michigan - Ann Arbor

Hope College

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So in this problem, we have a telescope with two focal lances. F one equals to Turkey ST Emitters, and half two equals two five. Same team enters. Okay. Ah, The first question is, let's see what distance between the two lenses. We will allow the telescope to focus on infinitely distance object. Okay, So to answer this question, the first item we just need to some therefore Collins is because this is addition. Off balance is so it's just f one Wassef, too. So this is just 35. Same team enters. That's the first answer. Okay, what next? We need to discover the location, actually. Let's see the second question. We need to discover what distance between the two lances. We were allowed to telescope to focus on object that is five meters away and produce an infinitely distance image. Okay, so this time we need to discover the distance of the image first. So that's what we're going to discover. Let's see, we have in here the distance off the image. It close one divided by F one minus. It should be a minus one divided by the zero where the zero is the distance of the object. All this to the power off. Minus one. Okay, so we have here. One divided by dirty miners weren't divided by the distance off. The object is five meters or 500. Same team enters minus one. So after calculate this, we know that the distance of the image is going to be in 31 point nine. Same team enters. Okay, Now we can calculate the distance between the lances. The distance between the lances is just going to be distance off the image. Unless the focal lens off the I is. So this is just five. This is just 31.9 plus five. So it's just 36.9 thanked emitters. And that's the final answer. This problem, thanks.

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