predictcalculate-a-convex-lens-f200-mathrmcm-is-placed-100-mathrmcm-in-front-of-a-plane-mirror-a-mat

Question

Predict/Calculate A convex lens $(f=20.0 \mathrm{cm})$ is placed
10.0 $\mathrm{cm}$ in front of a plane mirror. A matchstick is placed 25.0 $\mathrm{cm}$
in front of the lens, as shown in FlGURE $27-32$ . (a) If you lookthrough
the lens toward the mirror, where will you see the image of the
matchstick? Is the image real or virtual? Explain. (c) What is the
magnification of the image? (d) Is the image upright or inverted?

Nathan Silvano · Accepted Answer

So in this problem we have, um, object. There was a place Let&#x27;s put here at a distance off 25. Same team enters throne of Excellence with focal length off 20 centimeters. So we have a focal lens off. Apprenticing team enters. And this is between the lens a tensing Tim mater&#x27;s. We put a mirror. Okay, so problem ones to discover the location of the image if the images ritual or Rio, what is the magnification? So before we doing any calculation, let&#x27;s think about the physics of the light rays that we should know at this moment. We know that first of all, the light race that came parallel, it&#x27;s converge to the mirror. The mirror is going to reflect this light tree. You&#x27;re gonna pass this fruit, d&#x27;oh the lens and keep moving. Then we have a focus here, the second type of light race we had. He enters to the focus, then goes parallel. And we can say that this point in here is where the image is going to be produced. So let&#x27;s confirm all this doing the calculations. No. So, first of all, let&#x27;s just cover the image produced by the lens. The distance of the image lens is just one divided by F miners, one divided by zero, all to the power of miners. One. This is going to be equal one divided by 20 miners, one divided by 25 and this one This is equal to 100 ST Emitters. But now think about we know that the image is going to be produced 100 treatment 170 meters. That, too, behind the convex lens. But after tense and team enters, we have a mirror and the mirror that we can say that the image is going to be produced a 90 ST Emitters behind the mirror. But since the mirror reflects the image he produced, his ah parent object, he produces another object that has a position off miners. 80 Because miners miners 90. The image produced by the Mirror is going to be at the position minus 90 Sink Tim enters so we can say that the distance between the object and the cove excellence is just minus 90 80 minus. A decent team enters so moving on, we can calculate the rial position of the image. She&#x27;s going to be one divided by 20 miners, one divided by miners, 80 minus one. This is equals two 16. Same team enters two there, left off the lens. And we know that because the light presses the light passes for the image. It is a really image, so it&#x27;s a really image because the light buses fruit. There&#x27;s 16 thinkyou mentors to the left. What else? We have to calculate the magnification. Okay, so let&#x27;s see. The magnification is going to be magnification of on excellence and the magnification of the mirror riches. Just the I won d I too Divided by D 01 The zero to this is 100 16 give you divided by 25 minus 80. So multiplying all this we have minors, you know point eight off magnification. Since the magnification is negative, the image is inverted and that&#x27;s the final answer to this problem.

Mayukh Banik · Answer

the object distances given by one would be my last one over one over focus, then minus one over update distance at the first glance in once off that it was one of 15 minus one of what? Wendy in once off, that he was 16. Sending week means disease. We have the image on really making other ways in the running so involved, it is the first image. So, uh, second image may be invited with dress for the first image, which would make it No, they object. Distance that the second lands will be given by one F minus. Lotta must be too. In one of that, it was one of us, Ben minus 1 15. It was It was 30 centimeters to the left of the media. So the final bag division will be given by he want to do over the one big do. It was 60 days 30 or what? Ready dance? 15. It was

Adnan Gill · Answer

So we have a lens, which is a diverging lens, and there&#x27;s a mirror which is a concave mirror or convergent mirror, and then the focal length off the lens is negative. Point your center peters, whereas the focal in off the mirror is 12 points your centimeters and on the to our distance off 30 centimeters apart, and an object is placed 20 centimeters from the lens. And what we need to do here is we need to find the image location, the location off the final image in party. So since it&#x27;s a combination, will force, find where the image is formed by one, and then we will use that as the object for em. So we start with one or Diego plus one over D. I is equal to one over F. Dior for the lens is 20 centimeters were trying to figure out the I and the focal length for the lenses. Negative eight. So one over the I is equal to negative 1/8 minus 1/20 and that gives D. I is equal to negative 5.71 centimeters, so the image is formed 5.71 centimeters to the left off me. Let&#x27;s so that means that the oral distance from the mirror will be 30 plus 5.71 So now we can start talking about the mirror. So the distance object now is 30 plus 5.71 which is 35.71 and we again used the same Formula one over Dio plus one over the I is equal to one over if so d or in this case is 35.71 and you&#x27;re trying to figure out the I and the four collect for the mirror is 12. So one over the I is equal to 1/12 minus one over 35.71 and that gives D. I is equal to 18.1 centimeters, so this would be to the right off the mirror. Now part be asked whether the image Israel or virtual so since the I is greater than zero, Therefore, the images riel and part C as if the image is inverted or upright. So since the images real, all real images are, uh, work it so d image would be inverted

Daniel Alva · Answer

in this problem, we have conversion and with its focal length in a complex mirror with these radius. And also we have a non recently, its location. They&#x27;re in serum, uterus, impromptu of the land in the less in the mirror are gloves, centimeters a park. In the first part of the problem, we want who a compute where image produced by the linen ease. So, in order to companies, when when I use the let&#x27;s equation for the situation for the lance is this and we replaced the values for the folks earlier, we have that teaches three. The optical insistence is right here for the land sleaze. Well, over living last movie you want and from here we can compute the value off you. Well, who want is five point 25 centimeters, though is more or less at this point. This is the location off the ridge produced by that first length in in second for the program. We want to know this image that is producing this point here is ah, a really or ah view jewel image. But the US would have completely body off if you are in the s party. Did you mean that is a really image. And in earth, see is a ah in government or upright image where one of compute the magnification, the magnification because of delayed this mind you won over, Do you want? This is my five point when you find three mirrors holder, the one that is living mirrors. So it is just my oath 0.75 Because this is my note. It means that this a much getting Bert. We could also say that this image is Imbert because our conversion and produce an image that when these images in the open style GOP yes, it burns. But wait, I also see that from the tine off the magnification. In the next part of the problem, we want to know where the image produced by the Comebacks mirror is. So for the Olympics mirror, the all is located at this point so hard that we can compute the optic distance. Your business will be a 12 miners 5.25 in this is it sinks point being before mirrors. So this is the object, this us And in order to compute the image this us, we&#x27;re gonna use the mirror equation. So this is for the mural. So for the mirror, a focal land is it&#x27;s off. Oh, uh, radius. So this is just one of voting here, it miles. Wow. Very true. Are So this is just minus two. And this is mine. Not because is a convex mirror still replacing the value of B That is 1/6. Identify, plus own. While ordering you drew, we can completely value of cultural. So good to is this mind ALS? Well, im going 64 um, years. And because this is mine ALS in minutes that the image is located past the mirror. Still, at this point, maybe 11 before. So at this point, okay, one of body that did the point. Uh huh. And in the next part of the problem, you want to know this? It&#x27;s a really or image or a beautiful image, because this is his name. It is a teardrop. You know, the magnetic issue because of the mirror is just mind if you drool over being so and we already have all these values. So we just need to replace, uh, the results. And this is born to die. They&#x27;re finally the order magnification. We know that Arista brother off the magnification due to the Linz in the magnification produced by the mirror. And this is just mindless born 75 that is going through tonight. Indeed it is. Just mind off or wanted one. And because this is mine was invested, a final image is inverted. Okay, in respect to the original. Okay, so we can say that is imaginatively computing the part e off. The problem is embarrassed. We respect. So the original off.

Kristela Garcia · Answer

in this problem. You have two lenses, one convents when conch 8 50 centimeters apart. The complex has a focal in the 25 centimeters, and the con cave has a focal negative. 40 centimeters on were asked to calculate where the image is being produced. This object is two centimeters tall and the object is initially placed 30 centimeters in front the first lens. So to do this we can use the Finland&#x27;s equation, which is one over d o plus one over D. I equals one of her out and we can solve for one over D I now we want to get d I by itself. So we&#x27;re gonna go ahead, inverse, inverse the entire thing. And this gets us a distance of the image. One for the first limbs off. Ah, 150 senators. That means that it is somewhere over here. This is going to be labeled as 150 centimeters. And so if I take away the distance that is between these two, we&#x27;re going to sell for a new distance of the object so distance of the object to as, ah, 100 centimeters and this is going to be a negative value because it is now on the right side of the lens and that we&#x27;re going to repeat the same process. So d I equals one over one over s which is now negative 40 minus one over. And this gives us negative 100 and we get a new distance of the image as negative 66.667 centimeters. So that means that my image is being created to the left of the second blends and we can use that to determine our magnification Now are magnification is going to be de i one times d i to all over d 01 times d 02 So we get negative. 66.7 round up times 1 50 divided by 100 negative 100 times 30 gets the magnification about 3.33 So, like we said, this is going to the fact that this is a positive value. It means that my image will be up, right? So since the final image will be formed on the left of the convex lens, that is the same side is the original light bulb, which means that this final image will be virtual So this is a virtual image. Again, we got a distance of negative 66.7, which tells us, starting from our second lens are con que blintz that we are 60 about 67 centimetres in this direction. So my image will be located out here because that is beyond the 50 centimeters that separates the two lenses. So because the final image is to the left of my convex Maikon, Kate lends this is again a virtual image.

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

Problem

The diameter of a collimated laser beam can be ex…

Problem 95 Hard Difficulty

Answer

a) 16 $\mathrm{cm}$ to the left of the lens
b) real
c) $-0.80$
d) inverted

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James S. Walker

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Marshall S.

Farnaz M.

Aspen F.

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Wave Optics - Intro

Multiple Aperture Interference - Overview

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a) 16 $\mathrm{cm}$ to the left of the lensb) realc) $-0.80$d) inverted

Physics

Marshall S.

Farnaz M.

Aspen F.

Jared E.

Wave Optics - Intro

Multiple Aperture Interference - Overview

James S. WalkerPhysics

Marshall S.

Farnaz M.

Aspen F.

Jared E.

a) 16 $\mathrm{cm}$ to the left of the lens
b) real
c) $-0.80$
d) inverted

James S. Walker

Physics