a-use-a-ray-diagram-to-determine-the-approximate-location-of-the-image-produced-by-a-concave-lens--2

Question

(a) Use a ray diagram to determine the approximate location of the image produced by a concave lens when the object is at a distance 2$|f|$ from the lens.(b) Is the image upright or inverted? (c) Is
the image real or virtual? Explain.

William Affel · Accepted Answer

All right. So for this problem, we have an object that&#x27;s sitting two times the focal length away from a con cave lens. And we have a couple questions that you want answered. The image is going to form do to this object being in front of this lens. So in order to do that, what we need to do is create a ray diagram for the image is going to be formed in order to create a ray diagram born to draw. To raise the first ray is going to be parallel to our central access here. And by drawing this ray, we know that the, uh, back tracing of the ray is going to follow through the focal point on the near side of legs. So if we back trace this ray, this is the way the rays going to diverge. It diverges that its rear end the back of it traces back towards the focal length. Then we&#x27;re all a second Ray, and this race is going to go through the middle of the lens. Now, the reason this regulars in the middle of the lens, because it goes through the middle of the lens that it will not be effected at all, will not be refracted or all. It is a a perfectly thin lens, ideally thin loans. It&#x27;s then where these to raise cross is where our image forms. That would be right here, so this would be our image. I So now we formed our image. We have a couple questions to answer about it. First of all, where is it located? It&#x27;s located About, I would say is about 1/3 2/3 away, 2/3 F away from the lens on the near side. Next we want to know, is the image upright for, uh, inverted? And it&#x27;s clearly upright since is going the same direction as our original object? So it is upright in the last thing we want to know is, Is it really or is it virtual? And from what we know from the A and B? Because it&#x27;s on the near side of the lens and it is also operate. It must be a virtual image looks, since all real images are inverted and they&#x27;re found on the opposite side of whatever lens or optical instrument there. Ah, they&#x27;re being used to create, and those are answers

Cory Olson · Answer

to solve this problem, we first need to create a ray diagram. And as you can see, we have some of those components here already. So the curved line is the con cave near the line. Passing through it is the principal access. The point labeled F is the focal point, or the point from which there is diverge and the point labelled C is the center of curvature. If you imagine that conclave mere going into a complete circle, that point labelled C would be the center of the circle. The object is represented by the Arrow and as stated in the problem, it&#x27;s half of the distance to the focal point from the mere now. To complete this ray diagram, we are going to need to plot out the path of each of the representative race. I like to start with the P ray because it has a very clear initial path, so the P ray or the parallel right will be represented on by Red, and it is going to go from the object in a line that is parallel to the axis, like so and then it&#x27;s going to reflect back through the focal point and We are going to extend that reflecting Ray on the other side of the Mir so that we can determine where it intersects with the other rays and will label that one P ray. Now the next one we will do is the F ray or use green for that one, the effort you could almost think of as the opposite of the Pierret. It&#x27;s going Teoh initially pass through the focal point and then reflect parallel to the axis, and I will show I would use a dotted line to show where it extends behind the object and then here to the mirror. And then it&#x27;s going to reflect back any wine parallel and again. We&#x27;re going to extend that parallel line on the other side here to find out where those to intercept and this is the effort. Now we could use the point at which of these two lines intersect to determine the location of the image. But we&#x27;re going to plot the sea ray as well to check our work, so we&#x27;ll use blue for that one. The sea ray, as its name suggests, is going to pass through the sea point with center of curvature and it&#x27;s actually going to reflect back in the same a path that it takes initially. So again I will use a dotted line to show where it&#x27;s extended behind the object, and then it&#x27;s gonna go to the mere like that and remember, it&#x27;s gonna extend back in the same, reflect back in the same path. But we need to extend it on the other side of the mirror to see where intersex. And so now we can see all three of these lines are intersecting at that same point will label that with Sea Ray, and now we can use that that point of intersection to determine the location of the image. So the image is going to extend up to that point, and we use an arrow for the images like we did the object. Only this time I&#x27;ll use a dotted line so we know it&#x27;s not the object itself, and we&#x27;ll label that one image, and we can now use this completed diagram to answer the rest of the question. So we see that the image is, in fact, upright omits has the same orientation of the object. It&#x27;s on the same side of the axis on. Then, when it comes to whether it is a real or virtual image, you can ask yourself this question. Is there any light passing through the image? In this case, the rays that are on the same side of the mirror with the image are just those extensions that we drew with the dotted lines, and therefore there is no light passing through the image, making this a virtual image.

William Affel · Answer

All right. So for this problem, we have an object that&#x27;s half a focal length away from a convex lens. And we have a couple questions that we need to answer what the image is going before. So in order to gains, my dears about what the image is going to look like Mitral ray diagram. Uh, this is how we&#x27;re gonna set our ray diagram up and the first Ray that we&#x27;re gonna draw Besides our object, Ray is going to be array. That&#x27;s parallel to the central access has raised that are parallel of the central axis that go through convex lenses, then get refracted to go through the focal point of the conduct. Sloan&#x27;s in the four side, The next ready wanna draw is one that goes to the center of the lens. Because when there goes in the center over, the ones won&#x27;t get refracted or all. And because we see that Thies to raise will never cross, we need to back trace them to see where the anedge got form. And while I haven&#x27;t drawn this extremely precisely, the image should be Ford. If you do so right about here above the focal length of the lens. Now we can check that using our lens equation, we can say that one over F is equal to one over the distance of the object, plus one over the distance of the image. Since this is a convex lens, so one over the focal length must be equal to distance of the object, which is just one over f over to plus one over D I one over f is going to be equal to two over f plus one over D I Someone over D I was gonna equal to negative one over half and we see that d I is gonna be negative f so that checks out. We see that our our image should show up negative f and using our ray diagram, we can answer the next two questions we have. 1st 1 being is an upright or inverted, and it&#x27;s clearly upright since the rays Ah, me on the same side of the central access as the object is. And the last part of the question is, this is a real or virtual image, and it has to be a virtual image is a virtual image because it shows up on the same side of the lens as the object. And it&#x27;s also a virtual image simply because we know for convex lenses. When we place objects within the focal length of thumb, we create virtual images. Finally, we know it&#x27;s a virtual image cause upright and real images are inverted, and those are answers.

William Affel · Answer

all right. So for this problem, we have an object that&#x27;s located two times the focal length away from a convex lens, and we have a couple questions we want answered. What the image is going to be formed in order to answer those questions were after quitting Ray diagram. So the first ray that run a draw is a Wraith parallel of the central access because we know that raised their parallel with central access and go through convex lenses end up crossing the focal point on the four side of that kind of excellence. Next we&#x27;re gonna draw is one that goes through the center of the lens because raising go through the center of the lens. Do not get refracted, redraw this other really. And then, by drawing these to raise, we should be able to see that they cross at about two F. So the distance of the image is about to F as far as we can tell from our ray diagrams that answers the first question. Next question. One answer is, Is the image upright or inverted? And it&#x27;s clearly inverted, since our object is, ah, upright in the ray diagram and it&#x27;s upside down on the other side of the lens. Uh, in the last question. The answer is, Is it a riel or a virtual image? And this is a real image. First of all, we know it&#x27;s a real image because it&#x27;s inverted. And secondly, we know it&#x27;s a real image because ending up on the fore end of a convex lens for end of ah, converging of a converging lens, so that tells us a real image, and those are answers.

Sarah Mccrumb · Answer

everybody is asking us. But in a real diverging Brecht etcetera on. You&#x27;re What about my line? I love you. Here we go. Here, we have a diversion. One. Okay. On the radar diagram. Looking image object close inside. The focal point of a divergent one. Yeah, that&#x27;s what asking us. And you can ask different questions that so here on. And we&#x27;re gonna have this here and this here in this course and the well and they have okay. And if we have this little guy, okay, so as image and this one here point. So I&#x27;ve been on the same side of the object, include the fertile. This is image I image on. So the image comes out and about. And that&#x27;s because if you remember classic de Burgin image, those like that, the light breeze coming in from right here make a rosy. Okay, so that&#x27;s the classic. And that&#x27;s what we&#x27;re seeing here. So this is the image from these bounced off right here. Okay, So the images Bertel, and if the raise their birds from the lands are extended in the location lens and they appear to were taped with tape and the patient gives a position, in a sense image. Bob Optic access it is is elected in this and say because about access taxes. Okay, Okay. And the imagine studios size because you see how it locally here and here it&#x27;s gonna be Burt sold about A what? But it&#x27;s also gonna be I see that. Right? Do you have a smaller original here? So it&#x27;s gonna get smaller. Okay. Are you even say with these inside? Hey, guys, that&#x27;s your answer. Thank you.

Kyle Godbey · Answer

for Question twenty. We have not too many really numbers to deal with, but just some facts. So for part ET, the fact that we&#x27;re given is that the object distance and image distance should be equal. So our standard equation of one over D II, plus one over the object it looks, ah, little more simple with two over the let&#x27;s say object is the one that we&#x27;re going to stick with being equal to one over F. Thus D object is equal to two times the focal length, which is are the radius of curvature. So the object for part A should be placed at the radius of curvature. Pretty interesting. Low for part B, it asked. Is that image really? Or is it virtual well if the images in front of the mirror D I is greater than zero, which means Dio is greater than zero. Thus it is riel. Let&#x27;s use all the colors. Why don&#x27;t we, for part, see it is asking about the magnification. So a serious party for Part C is asking if it&#x27;s inverted or upright. But we can, um, use the magnification to get there. So if it&#x27;s defined as minus D I over Dio which, since they are equivalent, it&#x27;s just minus Theo over the O. And that is obviously dis minus one. Since the magnification is negative, the image is going to be inverted. And finally, we kind of already spilled the beans here. With this calculation, we could drop that down here and say the magnification is minus one. So, uh, not any bigger or any smaller in a box just for good measure.

Problem

An object is a distance $f / 2$ from a convex len…

Problem 63 Medium Difficulty

(a) Use a ray diagram to determine the approximate location of the image produced by a concave lens when the object is at a distance 2$|f|$ from the lens.(b) Is the image upright or inverted? (c) Is
the image real or virtual? Explain.

Answer

(a) (2/3)f
(b) upright
(c) virtual

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James S. Walker

Physics

Andy C.

Aspen F.

Jared E.

Meghan M.

Wave Optics - Intro

Multiple Aperture Interference - Overview

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(a) (2/3)f (b) upright (c) virtual

Physics

Andy C.

Aspen F.

Jared E.

Meghan M.

Wave Optics - Intro

Multiple Aperture Interference - Overview

James S. WalkerPhysics

Andy C.

Aspen F.

Jared E.

Meghan M.

(a) (2/3)f
(b) upright
(c) virtual

James S. Walker

Physics