a-converging-lens-of-focal-length-9000-mathrmcm-is-180-mathrmcm-to-the-left-of-a-diverging-lens-of-f

Question

A converging lens of focal length 9.000 $\mathrm{cm}$ is 18.0 $\mathrm{cm}$ to the
left of a diverging lens of focal length $-6.00 \mathrm{cm} .$ A coin is placed
12.0 $\mathrm{cm}$ to the left of the converging lens. Find (a) the location
and (b) the magnification of the coin's final image.

Nathan Silvano · Accepted Answer

Okay, so we know that we have in this problem coin. That was a place to the left off. Ah, cold verging lens. And this land&#x27;s is left off a diverging lance. We know that the separation between the converging lens and the diverging lens iss Let&#x27;s see, 18 centimeters. So this distance here is 18. Same team enters. Okay, what else do we know? We know that the coin displaced 12 centimeters from the converging lens. So this distance here is 12. Same team enters. We also know that the focal lens off the converging lens he&#x27;s nine team enters, and we know that the focal and off the diverging lands is minor. Six send team enters. Okay? We have any other information? I don&#x27;t think so. Okay. So what we want to calculate, we want to know what is the location and the magnification off this coin in this Toolan systems. We know that every time we have a tool and system which the two lands are separated by a distance, we have to work with one lens at a time and discovered the distance off the image off each one of those lenses. So let&#x27;s begin the calculation. What do we have here? We know that the distance off the object to the first lens, the converging lens is just to al ve think team enters. We know that the focal lens is not. I don&#x27;t need to rewrite this. We know that the focal lens off the converging lens is 1910 meters. So we can use the Finland&#x27;s equation to discover the position off their image that the first lens reduce. So this is going to be one divided by F one miners, one divided by these. You know, one This is going to be to the power off minus one. So this is just one divided by nine miners, one divided by 12 Oh, to the power of minus one. So this is just 36 centimeters. Okay, Now the distance off the or project off the second lens is just the difference between this separation and the position off the image off the first lens. So this is just miners 18 70 metres. Okay, so we have the distance off the object of the succulence. We have the focal lens off the second lands. So using again the Finland&#x27;s equation, we can discover the distance off the image off. The second lens is produced by the second lands. This is just one divided by I have two miners, one divided by the zero to oh R minus one. This is just one minor. Six miners, one divided by miners. 18 Oh, the power minus one. Finally, this gives us miners nine same team members. So the answer to the first itin what is the location of the image? The image is located at nine same team enters to the left off the diverging lens off die verging lens. Okay. And what is the magnification? The magnification off this system that&#x27;s try to write the M in here we have that. The magnification is just the multiplication off the magnification of the first lens with the 2nd 1 And we know this is defined us miners. The one I divided by d 01 multiply by minors did, too. I divided by the zero to So we have all this and this is going to be that is six times miners ninth divided by 12 times minus 18. So the magnification he&#x27;s positive 1.5. And that&#x27;s the final answer to this problem. Thanks for watching

Vishal Gupta · Answer

hi. In the given problem, there is a combination off two lenses oneness, diverging to the left and another is converging lens the right doctors. I want you lives the focal in Paul Diverging lens. It was given as F one is equal to minus 6.0 centimeter and an object has been kept at a distance off B you see Quinto minus 12.0 centimeter toe the left off this converging lens. So using can opinions from LA using pin lance formula only creation against it. This is one like you. Minus one by B. We accused the initial distance from the lens and peace. The object missions is equal to one by it. Putting all these values which are known que is missing to us minus one by Pete. This is minus will. Yes, it will one by minus six. So it&#x27;s one bike. You here comes so pretty one by minus six, minus one by well, or we can say when bike you Dizzy Porto or else injustice. Well, this is why the stool minus one minus three by great So we can say the distance off in age from this diverging lens is minus old centimeter means it is somewhere It was the left side off the lens and I stencil four centimeters in married you? No. The distance between the converging and diverging lens has been given. Asked D. It is missing. We have to find this distance. So we want to walked in the final image Art impunity So food The final image by Doc and Virgin lands to be at infinity. The object, the virtual object for it or converging lines should be at its spoke us and they spoke. A land f toe has been given us so well 70 m. So this image formed by the conch a violence will behave like a virtual object. So this should be at a distance equal to the focal in so business. Dance off the image wrong. This converging lens should be equal toe vocal and just 12 centimeter in magnitude. So clearly B plus four should be quality 12 centimeter in magnitude. Or we can say they should be well when zero minus 4.0 centimeter or we can say be should be 810 Santini. Thank you

Alexandra Nastasa · Answer

let&#x27;s start by writing down the information available to us. So we have that The focal length of our first converging ones is three centimeters, and the distance from our object to that first lens is four centimeters. We also know that the focal length of our second lens is negative five centimeters because it&#x27;s a diverging ones. And we know that the separation between the two lenses the distance between them is 17 centimeters. Now we want to know how far behind that first lens are we gonna get our image formed. So we have our object here we have our focal points, focal lengths, and we want to know how far away are we gonna form this object here, Q one this length. How long is that We&#x27;re here? We have P one. We can use the Finland&#x27;s formula, which is one over F equals one over P plus one over Q to solve. For that, we re arrange it. We get that Q one equals one over F one minus one over P one to the negative one. So everything is gonna be one over the stuff in the brackets. And if you plug in the values you have. You get that Q one is 12 centimeters, so this is 12 centimeters away. Now, how much further until we hit that second diverging lens Well s is 17 centimeters, so s minus. Q one minus 12 will give us five centimeters, which is also the distance from our new object, or are real image produced by our first length to our second lens. So that&#x27;s gonna meet P two. This happens to be the focal length of our second lens, but regardless, we now have a really image that&#x27;s going to be going through a diverging months. So all of the libraries, they&#x27;re gonna diverge from here, and we&#x27;re gonna have only a virtual image, so we can&#x27;t put a screen down and view that image. So the Onley riel image in this problem right now is that 1st 1 So at 12 centimeters, the right of the converging ones. Now what if our two lenses are only separated by 10 centimeters? So once again, F one is three centimeters. F two is negative. Five centimeters P one is four centimeters, but s is only 10 centimeters. Q one will remain the same because F one and p one stayed the same. Same formula. We still get 12 centimeters, but now what&#x27;s R P two p two, which is s minus Q one his negative two centimeters. That means that after our first lines here, we&#x27;re actually going to for our image behind that second ones. So we are actually going to yet a virtual image back here, not a real image. So we can&#x27;t lay down the screen and view that one, however, because it&#x27;s a virtual image that&#x27;s diverging. Here we have a really image conv urging due to that diverging lens. So we do have one rial image here. Where is it? Well, we can figure out the exact location. Using somewhere over here can figure out using the Finland&#x27;s formula to solve for Q two. So once again, rearranging we have Q two is equal to one over F to minus one over P two to the negative one. And if you plug in all those numbers, make sure to keep the sign. The same P is negative. F is negative here. We&#x27;re going to get that Q two is 3.33 centimeters, and that&#x27;s measured this way. So 3.33 centimeters to the left of the diverging ones

Salamat Ali · Answer

But if the given problem the position off the final image and the identification is given by, uh, writing the lenses formula that is one of our P plus one work you. Is it cool to win a war If, where magnification is minus key were p so that our total magnification will be totem, Unification will be a more magnification one times magnification to. So we&#x27;ll use these formulas and get our totem unification in the position off the image. So we&#x27;ll fire first point. Q. So, cube, we confined by some student value for P. That&#x27;s one over 40 plus one of acumen. Use a cold one or 30 from here or what you will be 1 20 centimeters. And our medication one will be minus 1 20 Divided by Folke. This gives us more in history. Um, and then our need position off the object will be minus mine. 1 20 minus 11 Sure Hope excuse us minus 10 cent. Reacher. If he&#x27;s minus 20 then our nuke you will be substituting those values in them lenses equation or nature will be 20 centimeter and then the edification and two will be minus 20. Derided by minus 10. This uses medication off, too. Then I don&#x27;t know. Magnification will be in one times in two ways we said Elia. So when I find those two numbers minus three times to excuse us, minus six. So the final image is positioned at the 20 century JER to the right, right off the second dive orgy urging lens and the magnification is off course minus six. But to be the final image we can, we can get the state of the final image by knowing the courage in that. In the interrogation is the height of the object imagery. You have a heart of the object rhythm. Unification regard here is a minus ese, which is Ah, less than zero. So that means our inches is greater than zero. And that means it&#x27;s dish will be interesting to know. Therefore, on final image is in working. Been working boxy. We have to do this step. Hey, tend to be on so doing in step A. We already know where one will be minus three men efficient one where we can find him in two are using a new P that is a modernist 10 and the new F that it&#x27;s f. It&#x27;s a cold, too, UH, plus 20. Using a sign convention for conversion lenses. There&#x27;s a student as well. Use one divide minus Stan Ross with Q. This is 1 to 1 20 Excuse us. Have a cute to be 6.67 centimeter. So knew him to be one in 6.67 Divided by more honest. In this gives a 0.667 Then our duty medication will be them one times and to sew, and one we have is only three were in the street. Story times 0.667 This gives his medication off minus two. From here. Uh, you see that? The doorman identification is less than zero, and our pitch is granted, the girls or his dish will be less than zero. This means the final image is in worded. You want it

Daniel Alva · Answer

in this programme, we have another object that is in front, off commercial ins at 16 era mirrors. So we&#x27;re gonna put the ocean right here and the provinces the the focal length off these first lens east of serum ears. So one of with 12 here until here. Also, the Berlin says the exists at second Linds, uh, 27 mirrors from the first lands. So this is 27 mirrors in the focal length. Off this seconds is minus tensor mirrors. And then at if because it&#x27;s a diversion with thought this thing in the rolling tusk and compute where the final image is produced because of the system. Okay, uh, in orderto a to compute worthy finally much east with firstly compute where the image produced by the first Land is for that we&#x27;re gonna use the A lance equation here for the first name. So for the first Lance way, we can compute worthy MITI&#x27;s well over you for its equal toe, um, one over f one. So did 1/12 minus one over B one B one in 16 is ah, object distance. So is what number 16. So here we go compute the value off you one. And this is for you. A for a a seven mirrors, right. So they wouldn&#x27;t produce. Yes, by the first lands is, um, 48 mirrors from the center of the first land. So it would weigh maybe at this point over here, look, so yeah, because he would have a the distance between dalliances is training. So more 20 a center mirrors X, it may be here, right? We can a seek. The image produced by the first land is over there using race. So let&#x27;s see a how to compute that. Well, first, we have array that comes from the image in bus through the focal point of the first length. Something like these. Okay, something they kiss and you merge Barlow to the principle exists. So you don&#x27;t like, okay? Yeah, this it should be. Barrow in a sickle re is a rain that comes barreling from the excess. Oh, sorry. Something like this ending bus it through the focal point for reasons that is this point in red here. Okay, help in peace. Maybe. OK, maybe like this. And this is the location off the image produced by the first. Let these long, right? Okay, Once we have computers, you much we must remember that this image is ah, it be becomes a non object for the seconds for the second there&#x27;s this image is on object and we can compute. Where is the object distance using second equation here. And this is just a separation between the lenses that is 20 minus the, uh, the image distance for the first lands, that is for yea. And this is just mind 28 30 meters in its negative. Because in this object start for the second list, lens is on the right off the lands were gonna be a that from the from the end here. That is on the on the right. Right. Okay, Yes. Objects us for the second lands is your full compute word. The final images. But if it were, the final image is where it&#x27;s gonna use the lance equation for the seconds before the sickness. Is this Okay? Well, over que two. That&#x27;s well over P to 80. War is equal toe over. So from here we can replace the values that we have well, over Q two is equal one over miles in his mind off. Well, over that is well over minus 28. And from here, we can easily compute the value off Q two. Yeah, this is miles 15.8 13 years. So finally, image that is, the image produced by the seconds is negative. 15.6. If its name defecate means that is on the left side. So it will be. Maybe you&#x27;re all here. Maybe here. So it is, uh, 15. I don&#x27;t think many here. Right. We could also see that using the race. And so this is the okay Now for the for, for for the second thing. And in order to see where the images we use, the first raven, Um, Spar Lou and then it us through the We&#x27;ll go going off the second. That is the focal point. Something like this from Sorry. Okay. In a single ray, indeed, to see poor cunts. A. When we have pull Ince&#x27;s. The second ray for the seconds is, um, raid the bus on deflecting for the thinker off the second. So is this Okay? Lets see. So these passed through the center of the Collins in us on the flags. So is this one made me. Okay, So he&#x27;s usually so The final image is at the intersection. So maybe here. Right. Okay. And so this it should be the point. So we we can see where the final image is. Also using the race right looking.

Elizabeth Clark · Answer

All right, We&#x27;ve got a situation here where we&#x27;ve got one lenses converging here. Emerging has a focal length of 12 centimeters and then 16 centimeters. Over here we have our object, vo one at 16 centimeters away. And then over here, 20 centimeters away. We have a diverging Lynn were Clinton&#x27;s, right? 20 centimeters over here. And so we&#x27;re going to try to figure out where does the object end up and what size is it and said It&#x27;s all right. So the deal is that we&#x27;re going to start off by taking this object right here, and we&#x27;re gonna find out where it&#x27;s images using the focal. Okay, equation. So 1/12 because of this converging lens equals 1/16 plus one over d I. So that means our image we&#x27;re gonna convert this into 40 eights for over 48 equals three or 48. Let&#x27;s one for 48. So that means our d I equals 48 centimeters. And then also, that&#x27;s we can look at this week by making you ray tracing. Make your center Ray come through here on the way over here. We&#x27;re going to get a pretty good sized inverted image up here and D I or d o minus d I over Dio it&#x27;s gonna give us our magnification is gonna be a minus four. That&#x27;s why it&#x27;s inverted. It&#x27;s negative and it&#x27;s large. All right, so now we have I&#x27;m going to scroll down and we&#x27;re going to start looking at when we have this or our our object for our second for second lens. All right, so here is our diverge. England&#x27;s right here, and we&#x27;re told that it&#x27;s focal length equals minus 10. And we&#x27;re now we have an object over here, which is aren&#x27;t inverted at 28 centimeters, because the entire before, we had 48. But there was a distance of 20 between the two lenses, 28 centimeters. And so whenever we can use, we can do a little ray tracing here. Also, here&#x27;s our focus. At minus 10 is two of them one on either side and this guy here, if we have a parallel ray, come in and hit this lens will go back through the focus and this rate. We will also have a center ray coming through here, and this center Ray will just keep going through but we can see right here is where our image is going to be, So we can use that to make sure that we&#x27;re getting the right answer. So let&#x27;s sit together, put together the equation now, uh, one of her f equals one over D o me plus one over G I. So we can turn all these into 280. It&#x27;s my S O minus 20 age. Over 280 equals ton over 280 plus one over D I. I&#x27;m going to subtract the 10 over 280 get minus 30 eight or 280 equals one over d i. Onda. We can use a calculator to get an answer for that. And that&#x27;s going to give us, um 7.3. So d I is gonna equal 7.3 and it&#x27;s going to be to the right. It&#x27;s the neck of 7.3. So it&#x27;s gonna be to the right 7.3 centimeters to the right, off the diverge England&#x27;s and that sensor

Problem

Repeat Problem $28,$ this time with the coin plac…

Problem 28 Medium Difficulty

A converging lens of focal length 9.000 $\mathrm{cm}$ is 18.0 $\mathrm{cm}$ to the
left of a diverging lens of focal length $-6.00 \mathrm{cm} .$ A coin is placed
12.0 $\mathrm{cm}$ to the left of the converging lens. Find (a) the location
and (b) the magnification of the coin's final image.

Answer

a) 9.0 $\mathrm{cm}$
b) $-1.5$

Related Courses

Physics

Related Topics

Discussion

Top Physics 103 Educators

Elyse G.

Christina K.

Andy C.

Marshall S.

Physics 103 Courses

Wave Optics - Intro

Multiple Aperture Interference - Overview

Recommended Videos

Watch More Solved Questions in Chapter 27

Video Transcript

James S. Walker

Physics

Elyse G.

Christina K.

Andy C.

Marshall S.

Wave Optics - Intro

Multiple Aperture Interference - Overview

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

a) 9.0 $\mathrm{cm}$b) $-1.5$

Physics

Elyse G.

Christina K.

Andy C.

Marshall S.

Wave Optics - Intro

Multiple Aperture Interference - Overview

James S. WalkerPhysics

Elyse G.

Christina K.

Andy C.

Marshall S.

a) 9.0 $\mathrm{cm}$
b) $-1.5$

James S. Walker

Physics