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Part € Complete previous part(s)Now remove the two lenses at € +20.0 cm and -20.0 cm and replace them with a single lens of focal length f3 at € 0...

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Part € Complete previous part(s)Now remove the two lenses at € +20.0 cm and -20.0 cm and replace them with a single lens of focal length f3 at € 0. We want to choose this new lens so that it produces an image at the same location as before.Part DWhat is the focal length of the new lens at the origin? Express your answer in centimeters, to three significant figures or as a fraction View Available Hint(s)TempEes Symbolslado redo' resbt keyboard shortcuts helpf3CmSubmit

Part € Complete previous part(s) Now remove the two lenses at € +20.0 cm and -20.0 cm and replace them with a single lens of focal length f3 at € 0. We want to choose this new lens so that it produces an image at the same location as before. Part D What is the focal length of the new lens at the origin? Express your answer in centimeters, to three significant figures or as a fraction View Available Hint(s) TempEes Symbols lado redo' resbt keyboard shortcuts help f3 Cm Submit



Answers

Two lenses having focal lengths $f_{1}=+9.0 \mathrm{~cm}$ and $f_{2}=-18.0 \mathrm{~cm}$ are placed $3.0 \mathrm{~cm}$ apart. If an object $2.50 \mathrm{~cm}$ high is located $20.0 \mathrm{~cm}$ in front of the first lens, calculate (a) the position and $(b)$ the size of the final image. (c) Check your answer graphically.

So in this question, as you can see from here There are two lenses which are At a separation of five cm from each other. The local and Afghan is five cm After his 10 centimeter and an object of fight. An object of height 2.5 cm is placed in front of the lenses. The distance of the object from the first glance is 15 cm. You can see this this distance is definitely 15th century to now. What we need to do is we need to find the final image and the size of the final image, the position of the time limit and the size of the final image. And let us help you get that. So that can be done For the first lines. I'm applying one by B -1 by US Secretary, one by F. So one by I'm calling. This has given this is the image hawk formed by the wastelands. So one by women -1 by I'm taking this you As -15 -15 Is equipped with one upon focal length is five cm. So from here we see one by women is equal to one x 5 And then mine is one x 15. Okay, uh, that's in place. Uh we want is coming out to be So everyone is coming out 3 7.5. Send me down They left. Everyone is 7.5 cm. What does it mean? That means this lens? We'll try to make an image somewhere here which is definitely 2.5 cm away from these lands. And From here to here it is 7.5. All right, So now this image will serve as an object. This will behave as an object for the second lens. So applying lens formula and serene So we can write one by Image from my second lens. I'm calling widow -1 by you. The value of you in this case for the second length. Will we Plus 2.5 because it is lying in this direction so it is positive. So I'll write it 2.5 is equal to one by F. So f. We can see here it is tense and either it is 10. So if you do this calculation a little bit of population you can the value of we do this is the position of the second hand This is coming out to be two cm was in place. Final limits the final image. This is the final limit would be at a distance of at a distance to syndicated from the second lens or from the first planes. The distance of the final limit is seven centimetre. Okay, so that's pretty simple. We have found the final position. Now what we can do is we can find the magnification is also so this was the first land salt we can get developed magnification in the first place. So everyone is it clear to we upon you. So everyone is a good way upon you which means 7.5. This is the value of the image divided by object to us at a distance minus 15. So everyone is coming out to you. Magnification is coming out the week -0.5. Mhm Right, so like ways we can get the magnification for second case. Also it is equal do once again we upon you. So what was the family? Which finally which was two cm. Okay, so we is too And you was how much? Okay, you was 2.5. 2.5. Okay, so we can see the magnification that we get here. He's coming out to me M two is coming out to be 212 point vibrate, it is zero point eight. So this is the magnification and and we know okay, we know very important concept. The final image divided by final image size divided by final sunny he midsize debated whether object size. Her combination of lenses for net combination is always a great product of magnification. All right, so the product of magnification is 0.5 -0.5- 0.8. This is M1 and this is an too. And at I the final limit size would be equal to object size into uh this, I am wanting to emigrate someone into him to value. We already happier we can secure the values here, Object size was uh 2.5, 2.5 into anyone is -0.5. M two was 0.8. So if you calculate this image slices coming out to me minus one centimeter, so this is not seven. This is one. Okay, so this is our answer.

Okay, so this is quite complicated problem. But let's simplify little the things we know that we have object that's represent with this. Aargh. We have, uh, converging lens. That's for today. Here, every have another krone verging lands in here. We know that the distance between the object in the first lens is just 50. Same team enters. We know that the distance between the two lenses his ex and we also know that the focal lens off these lenses is that's put in here F one if one is 20 saying team enters in f too sturdy ST Emitters. This is one, and this is too. Okay, so we know that the objectives place here and the two off the two lenses are converging type of Flynn's. And we want to discover the position off the image and the magnification off this system. 43 different excess we want to discover for X one equals 115 for X two equals two. Let's see. 30 same team enters. I think Tim enters here also in the final X tree. It's going to be for zero. Saint team enters. Okay, so it's a long problem to solve So first of all, I think about the problem itself. We know this is a two land systems, the boat. Both the lenses are converging type of lens. And we know that since the separation between the lens, a different from zero, we can work with one lands at a time, discover in the position of the image to the first lens, then discovered the position of the object of the second lands and finally the positions of the image to the second lands, which is the position off the image of the whole system. So let's do this. So let's begin calculating the position off the image off the first lens we know using the Finland's equation that this is just going to be one divided by F one minus one, divided by the 01 And the 01 is just position off the object related to the first lens. We know this is 50 centimeters. Okay, How these two, the power off minus one. So then we have the focal lens off. The 1st 20 isn't emitters, so it's just one divided by 20 minus one divided by 50 Oh, the power of minus one. This is just 33.3 70 meters. Okay, Now we have to get a little attention, because what is the position off the object related to the second lens? This is just going to be X minus 33.3, because accident separation between the two lenses. So now we can calculate the position off the image related to the second lands. And this is going to be one divided by F too miners one divided by he 02 Oh, to the power of minus one. So we have for two one divided by dirty miners. The position of the object, we calculate it. That is X minus 33. So this is going to be ex milers 33.3 oh, to the hour off minus one. And we know that this is the general equation that will solve the location of the image for each one off this access in here. And what is the magnification? While the magnification, it's just the multiplication between magnification that's put in black. The magnification is just magnification of the first lands will supply by magnification. The second mince and by definition, is going to be d. I weren't Times D i, too divided by D zero one in the zero to So with his values, we have dirty dream going Tree E I true, divided by 15 15 times X minus 33 point tree And this is the general equation that will solve or magnification part off the problem. So now we can calculate both the position and the magnification for each one of the exes. So for the Frank's first, X X one equals one 15. Think teenagers. We have to stew equations in here. Let's put in the x x one. So we're going to have did the I won it close. Let's see one divided by dirty miners 115 miners 33.3 or the bride or dividing one the power of minus one. So this is just 47. Same team, mater's. That's the first answer position. Now the magnification magnification Just going to be using the 2nd 1 Hmm. Let's see. Early three point tree times 47 that we just discover if I did buy If T. Times 115 miners 33.3 This is just judo point 38 and this is the answer to the first I 10 of the problem right in a now the second item. Let's see, we have to calculate it when the ex is equals to Turkey. Think team enters So a Scot Cledus position d i to just going to be, Let's see one divided by dirty one divided by 30 minus one, divided by already minus 33.3 Was your power of my nurse one. So this is just three same team enters Okay In the magnification, there's going to be 33.3 times three all these divided by 50 times dirty minus 33.3 and this is equals two minus 0.61. And that's the answer to the second I turn of the problem turned itin of the problem we have when x it goes zero when X equals zero, we haven't here. Let's see the distance off. The image is going to be one divided by 30 again miners, one divided by minus 33.3 because access zero all to the power of minus one. This is 15 point eight same team occurs, and the magnification it's going to be 33 43 times 15 0.8 divided by 50 times minus territory or three. Cut this. I'm really going to have miners zero point 316 That's the magnification when the distance between the two lances zero. Now, for the final part of this problem, in the final part of this problem, we have to demonstrate that the answer. Let's put in here. The answer. So did I can see is going to be equivalent when we use their relation between the focal lens that we learned, which is defective. Focal lens off one divided by f effective equals. This is a F equals one divided by F one waas, one divided by F too. So the only way to the most rapists is first calculating. He's effective focal ends, and the second way is using the results that we have for the distance of the image and the position of the object. Using the Finland's equation off one divided by zero plus one divided by deep I. So let's prove this. We have in here one divided by 20 plus one divided by 30 which is the two focal lenses, so the effective if the effective focal length when the distance between the two lenses. Zero going to be to ALF. Same team actors. Okay, And what is the focal and off this system? When we calculate each part separately, we're going to have a distance of the object. Distance off the object is just one divided by 50 plus one, divided by the distance off the image, which is 15 0.8 and then we have to a focal end it close 12 centimeters. Oh, so so using the both matter, that's we're going to get this.

Okay, So in this problem, we have to calculate a lot of things we know that we have object with, Ah, height off two point 05 some teenagers. And we know that this object is in front, off converging lens on a distance. Let's go. The 01 off, dirty same team injures. We know that the focal lent off this converging lens is 20.5 same team majors. Okay, so we know that a second land lens, it's put our distance d off dirty. Same team, mater's from the first lens. And what else do you know? I think we only know this r K. And the problem wants to discover what is the final height off the image after we put the second lens. So our key let's work with one lens at a time. So the first thing we need to calculate is the distance off the image produce by the first lens distance off the image of the first lense. So this distance using the feelings equation just going to be one divided by F minus, minus one divided. Bye D zero one. Okay, All this to the power off minus one. So this is just going to be one divided by 20.5 minus one, divided by dirty all to the power of minus one. So the distance off the image produced by the first Lands is equal to 64 point seven. Same team enters second. We need to calculate it. Ah, the distance that we're going to call the distance of the object off the first off the first of the second lens. Actually, so that's called the D 02 This is just going to be charity. That is the difference between the second to the first lens minus 64 point seven equals minus 2034.7. Same team enters. Okay, so now that we have the distance off object off the second lens, the distance of the image of the second lands just going to be different lens a creation all again with this time we have this second focal length, and we know that the second focal lens is 40 minus 42 points, five minus one divided by D 02 All of this what if our minus one so we can calculate this? The problem gives us the second for Colin off one divided by minors, 42.5 minus one, divided by minors, 34 0.7 all just to the power of miners. One. So this distance, the distance of the image it's just 190. Think team enters. Okay, so now that we have all the distance, that's with a different color in all the distance, we can calculate it. The final height off this image. Because we know from Chapter 26 the magnification is a way to calculate it. That's what's in here. Let's call the age all right that we want to calculate its equal to magnification times 80 So the magnification is going to be the magnification produced by the first Lent times. The magnification produced by the second lens times the initial height. And we know also from chapter 26. Did this magnification has values off miners? D'oh! I won. Divided by the 01 The 2nd 1 is also miners. D I too divided by D 02 times 80 And we know all this data here. So let's institute. Yes, we're going to have 64.7 times 190 times. Two point you know. Five. All these divided by 30 times minus 24.7. 34. 37 Here, 34 0.7. This is going to be equal finally, to minus 24 0.2 ST Emitters. So we know that the final image Israel and inverted because of the minor sign. And that's the final answer to this problem. Thanks for watching.

The force says we have Iran. It was one of what if one minus what you want in life. So that it was one over negative four minus 1/6 in the words, it was negative. 2.4 sending me so we to come out to be the total distance which is a T minus six centimeter minus. If you want hooches, Negative viewpoint. Full centimeter. It was boarded foreign. Full centimeter. No, we tried. Defied the stool which is one of our three to just one working too endless off that it was one of one for the important fool. Thus you do. Is that total distributed the immigrant? The object descended. We don't minus 18 centimeters. It was a fat. It was six point 55 sending me good. And because if the response give this is Ah, I want hard. You dance now. Don't have magnification equals. He wants you over. If you want to do equals number steam. We get the magnification to be negative. Zero for 33 So I'd of the image will become an indication that side of the object It was negative. Zittel, born 33 times were sending ago equals mega before


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