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Two lenses, with $f_{1}=+20.0 \mathrm{cm}$ and $f_{2}=+30.0 \mathrm{cm},$ are placed on the $x$ axis, as shown in FilsuRE $27-28 .$ An object is fixed50.0 $\mathrm{cm}$ to the left of lens $1,$ and lens 2 is a variable distance $x$ to the right of lens $1 .$ Find the lateral magnification and location of the final image relative to lens 2 for the following cases:(a) $x=115 \mathrm{cm} ;$ (b) $x=30.0 \mathrm{cm} ;$ (c) $x=0 .$ (d) Show that your result for part (c) agrees with the relation for the effective focallength of two lenses in contact, $1 / f_{\text { eff }}=1 / f_{1}+1 / f_{2}$ .

a) $=15.8 \mathrm{~cm}$b) $=-0.316$c) $=12 \mathrm{~cm}$d) $f=12 \mathrm{~cm}$

Physics 103

Chapter 27

Optical lnstruments

Wave Optics

Cornell University

Rutgers, The State University of New Jersey

Simon Fraser University

Hope College

Lectures

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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.

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