two-lenses-with-f_1200-mathrmcm-and-f_2300-mathrmcm-are-placed-on-the-x-axis-as-shown-in-filsure-27-

Question

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 fixed
50.0 $\mathrm{cm}$ 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 focal
length of two lenses in contact, $1 / f_{	ext { eff }}=1 / f_{1}+1 / f_{2}$ .

Nathan Silvano · Accepted Answer

Okay, so this is quite complicated problem. But let&#x27;s simplify little the things we know that we have object that&#x27;s represent with this. Aargh. We have, uh, converging lens. That&#x27;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&#x27;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&#x27;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&#x27;s see. 30 same team enters. I think Tim enters here also in the final X tree. It&#x27;s going to be for zero. Saint team enters. Okay, so it&#x27;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&#x27;s do this. So let&#x27;s begin calculating the position off the image off the first lens we know using the Finland&#x27;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&#x27;t emitters, so it&#x27;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&#x27;s just the multiplication between magnification that&#x27;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&#x27;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&#x27;s first, X X one equals one 15. Think teenagers. We have to stew equations in here. Let&#x27;s put in the x x one. So we&#x27;re going to have did the I won it close. Let&#x27;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&#x27;s. That&#x27;s the first answer position. Now the magnification magnification Just going to be using the 2nd 1 Hmm. Let&#x27;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&#x27;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&#x27;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&#x27;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&#x27;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&#x27;t here. Let&#x27;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&#x27;s going to be 33 43 times 15 0.8 divided by 50 times minus territory or three. Cut this. I&#x27;m really going to have miners zero point 316 That&#x27;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&#x27;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&#x27;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&#x27;s equation off one divided by zero plus one divided by deep I. So let&#x27;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&#x27;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&#x27;s we&#x27;re going to get this.

Zachary Warner · Answer

we&#x27;re told that we have two converging lenses with two different focal lengths. Lens one has a focal length F, one of 10 centimetres lanes lens to as a focal length of two of 20 centimeters, and they&#x27;re placed 50 centimetres apart. You can lick it, figure P 23.44 if you want to get a mint. A nice picture of what this looks like. So the final images is located between the two lenses, so it&#x27;s on the right side of lens bowling, but on the left side of lens, to which means for lens to it&#x27;s a virtual image and is looking at the position X is equal to 31 centimeters. So therefore, I confined the image distance from that lens, using the fact that, uh, lends to has located a 50 centimeters and then, uh, image. The image is 31 centimeters to the left of that or excuse me is that X equals 31 centimeters. So thank you to is equal the negative 50 minus 31 centimeters. The native signed because it&#x27;s a virtual image. It&#x27;s on the left side of the lens, so que tu is at 19 centimeters so using this information for party, it wants to figure out how far to the left of the first lens should the object be positioned in order for this situation to take place. Okay, so to do that, we&#x27;re gonna use, um, our equation one over Q in general, plus one over p in general were queue and queue is image distance and Pia&#x27;s object distance is equal to one over the focal length F. So, for what we&#x27;re gonna do is we&#x27;re gonna find the image, um, or the object for lens, too, because the object for Linz to is the image of lens one. And then we can use that to find the object of limbs one. So rearranging this equation to solve for P, we find pee too soon for dealing with the second lens is equal to que tu, which we know times f to which we also know divided by uh que tu minus f too. So, plucking those values into this expression, we find that this is equal to nine 0.74 centimeters. Okay, so object to p two is 9.74 centimeters to the left of the second lens. But remember, the second lens is located at 50 centimetres. So then to get Q one Q. One is I don&#x27;t think that you look a little nicer here, equal to 50 centimeters minus P, too, so Q one is equal to 40.3 centimeters, and they were gonna do the exact same thing as before. We had the equation for pee, too, but now it&#x27;s p one. So we&#x27;re gonna, uh, replace everything in that with Cuban and F one. This is Q one times F one over kyu won minus F one. So plugging those values in this expression, we find that this is equal to 13 0.3 centimeters making box. That in, is their solution. For part a part B wants us to find the total magnification from the two lenses so that total magnification is equal to the magnification from Lens one, which is minus Q one over P. One, multiplied by the magnification from lens to which would be minus Q two overpay to so we know all those values. Just go plug him into this expression and carry out that operation, and we find that the magnification is equal to minus 5.9 making box set in is their solution for part B, the magnification of the two lenses. That part C wants to know if the image that is produced is upright or inverted. To figure that out, we need two feet. We need to just simply look at the, uh, magnification from part B, and we see that it&#x27;s negative. So we can say that since M is negative, the image is inverted. If it was positive the image would be upright, we can box it in as their solution for part C in the final part of the

Zachary Warner · Answer

for a question. We&#x27;re told that we have a converging lens with the focal length. I call F sub C of four centimeters and a diverging lens, which has placed a distance D 10 centimeters away from that converging lens. Uh, with the focal length F sub d equals negative eight centimeters. It then says that you have an object that has placed a position X equals negative, too, and it wants to determine the X location of the final image that&#x27;s produced. So this object would be the object for the converging lens. And since it&#x27;s at X equals negative, too, and the converging lenses that X equals zero, that means it&#x27;s two centimeters to the left of the converging lens. Anything that&#x27;s to the left of a converging lenses considered positive, right? So piece of sea, which would be the object position for the converging lenses. Two centimeters and then we can use, uh, the relationship, which is one overpay distance. The object, plus one over Q distance to the image is equal to one over F, the focal length to solve first for the image that&#x27;s produced by the converging lens because that will be the object of the diverging lens. And then we can find the image produced by the diverging lens and then use that to find our exposition of that image. Okay, so first, let&#x27;s rearrange this to salt for Cube, so this will be Cousteau. See, this is the image produced by the converging lens. It&#x27;s equal to piece of sea times, the focal length of C, divided by piece of C minus the focal length of C. S. I just rearrange that equation above the sulfur cube. Plugging those values into this equation, we find that this is equal to negative four centimeters. Okay, so this negative sign means that it&#x27;s two to the left of the mirror or on the same side is the object. So now we need to find that position relative to the diverging linds. But if you&#x27;ll remember the diverging lenses 10 centimeters away from be converging lens, so then we can call this piece of D. The object location for the diverging lens is going to be that distance D 10 centimeters minus Ah Q. Some C. So then that means it&#x27;s 14 centimeters, so it&#x27;s 14 centimeters to the left of the diverging leads. That&#x27;s why it&#x27;s positive. Okay, so now we can use that to solve for Q sub d. So the exact same as this equation here separate? Placing all the seas with these. So we have Q sub d is equal to piece of D times, the focal length of the diverging lens. Remember, that&#x27;s negative eight centimeters divided by piece of D minus the focal length of the DIA bridging lens. Playing all those values into this relationship, we find that q sub d the image produced by the diverging lenses that negative 5.1 centimeters. Again, this negative sign means that is a virtual image or it is on the left side of the diverging leads. Okay, so if it&#x27;s 5.1 centimeters to the left of the diverging lens, we need to figure out its exposition. So the exposition of the diverging lenses 10 centimetres so 5.1 centimeters to the left of that we find that X is equal to D ah plus Q sub d. Right, so this is equal to So it&#x27;s 10 centimetres minus, uh minus that 5.1 centimeters or 4.9 centimeters. We can go ahead and lock set in as our solution to the question for part a Part B wants us to figure out the magnification of the two lenses. So that&#x27;s gonna be the magnification of the con Caitlin&#x27;s, which would be in minus que ce of C divided by p subsea multiplied by the magnification of the diverging lengths that remind Excuse of D divided by a piece of d So plugging all those values in this expression. Since we&#x27;ve already calculated them, we find that this is equal to zero 0.73 Linkenbach set in is their solution for party.

Mayukh Banik · Answer

at events one Fine. I&#x27;ll get distance off. Find emails. Distance, which is one of the F one minus one, is a one and once equals 112 by just 1.5 in verse. It was then sent it because so we would have. We do. It was s minus you want Waas? Yes. These 16.5 minus speak like five minus data. It was for something we do now. The second thence three is looking it at then speak is located at 19.8 centimeter. So we have you three was 39 for eight minus its location 19.8. It was Do nt sending me now Object distance at them Stree B one F three minus water Like you&#x27;d be in worse. It was 114 minus water Plenty Waas It was five. Secondly, now we come back to them stew. So we have you Was its location 19.8 centimeters minus by segregated It was 44 A Certainly we don&#x27;t And so this is the distance. So this minus 16.5 centimeters. We did acted Object distance 16.5 centimeters. So Queudrue comes out to be 40.8. My 64 5. It was negative. 1.7. Sending me now you&#x27;re trying to find the fulfillment of the second friends. So, yes, it was one of us. We do. That&#x27;s one like you do in most of that equals one of our four minor plus one of one negative 1.7 in the north. It was negative. Three centimeter on the negative signifies that these are diver. Being that&#x27;s on top of that is three centimeters. Now, if we want to draw the red, I don&#x27;t get through the first census. Converged events draw the optical access. This is the focus, and you start from here. Then just to raise will do it. But let&#x27;s start a tawdry as well. This way, we&#x27;ll be going straight on a series where the first provides with Fox to be as far or first interview get images fund over here. Now, this was the object for the second Vance which came out to be diverting lands on this. This is the focus would one day through the focus would want today straight on the other way through the optical axis, the way that go through the optical axes go shred on. Here is the focus. So this way we keep going in this direction. Since this is a diverging dance you with you should not. With the rate through the first four push you should put a way through. The second focuses is over here. Do they need At the second focus on this, Ray will become parable Now we have gone Virginians after this. So that was just decided. Different colors. We have fun both events over here. That&#x27;s terrible. We have a fun virgin That&#x27;s over here. What happened at the convoluted Etsy&#x27;s from here? Director wanted A that is going through this football born. So here you have your second object. So from the second object you want today this way one today through the center on the other way process through its focus. So the day that passes sweets focus we&#x27;ll shred on these goes in this direction on this one passes still its focus and it makes a Lapeyre. So this will be find other image. This is the second intermediate you manage on. This is the first intermediate image

Averell Hause · Answer

so we can say for part A. We know that the first the lens forms a real image of the object. This is located at I sub one equaling won over F sub one minus one over piece of one, all to the negative First power. This would equal one over after minus one. Over. Rather ex of one want minus one over to F sub one alternated first power. This is equaling two times f sub one uh, this would be two times refs of f someone to the right of the lens ah, week and then say piece of two is equaling two times after one plus f sub too minus two f sub one. This is equaling two at sub, too. This would be in front of Mir and we can say that the subsequent image formed by the mirror is located, as I said to this would be located at one over F sub two minus one over piece of to alter negative first power. This is equaling won over F sub too minus one over to F sub, too. Uh, this is equaling. Also negative First power. This is equaling two times. EFS up to and then finally this would be to the left of the mirror. And then finally, we know that p sub one crime would be equal to two times efs of one plus f sub two minus two f sub two. This is equal in two efs of one and weakened. Then say this is to the right of the lens. We can then say that here, the final image formed by the lenses at a distance I sub one crime to the left of the lens. So this would be equal to one over efs of one divided rather minus one over a piece of one prime to the negative. First, this is equaling one over EPS of one minus one over to F sub one to the negative first, equaling two times efforts of one again to the left. And this would be our answer for ice of one prime note that this is going to be, uh, the location of the original object for part B. We know that then, uh, the lateral magnification, um, would be equal to negative eyes. Someone over a piece of one most glide by negative ice up to over a piece of two multiplied by negative I sub one prime over piece of one crime. This is equaling. Essentially just negative 1.0. So I&#x27;d be the ladder ladder on the magnification for part C. We know that the image Israel this is because the final image distance is positive. Ah, we know that here the magnification sorry for party. We know that the distance I said one prime two is to the left of the lens part e. We know that M is going to be, ah, less than zero. Which means that, um, image is inverted. That is the end of the solution. Thank you for watching.

Mayukh Banik · Answer

here we would have Q one equals one over f by this wall over B one in verse. It was one of a 30 minus one were 40 in verse. It was 1 20 centimeter. Now I m one is negative G one over the one which is negative off 1 20/40 It was nearly three. Now, in the second case, you would have we do It was 1 10 minus 1 20 It was negative. 10 centimeter. Using that you can find cute do which would be one of our F minus one of the left prime minus one of what we do in yours. It was negative. 1/20 we do is they&#x27;ve 10. So that because whatever. 10 with a positive sign and that comes out to be 20 centimeters. Well, magnification in this case is 20 over negative. Off negative. 10. So that&#x27;s to suppress. And Brian, So don&#x27;t mind. Irrigation is initially it&#x27;s native three. Now it is supposed to do so fascinating. Six. So that means the final images invited. Now if the dentist where converging then we would have queued week was if the seconds has done the margin generally we would have queued Week was 11 20 minus one of minus Dan In the worse it was 6.67 centimeter and two will come out to be negative 6.67 divided by a negative 10 in Cwa&#x27;s zero poor in 67 So in this particular guests and daughter will come out to be negative. Three times 0.67 It was negative two. So now final image will still be in the

Problem

A converging lens with a focal length of 4.0 $\ma…

Problem 46 Hard Difficulty

Answer

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

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James S. Walker

Physics

Elyse G.

Christina K.

Farnaz M.

Zachary M.

Wave Optics - Intro

Multiple Aperture Interference - Overview

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

Physics

Elyse G.

Christina K.

Farnaz M.

Zachary M.

Wave Optics - Intro

Multiple Aperture Interference - Overview

James S. WalkerPhysics

Elyse G.

Christina K.

Farnaz M.

Zachary M.

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

James S. Walker

Physics