find-the-location-and-magnification-of-the-image-produced-by-the-mirror-in-problem-27-using-the-mirr

Name: find the location and magnification of the image produced by the mirror in problem 27 using the mirr
Uploaded: 2020-05-04
Duration: 3 min 17 s
Channel: Numerade
Description: Find the location and magnification of the image produced by the mirror in Problem 27 using the mirror and magnification equations.

Question

Find the location and magnification of the image produced by the mirror in Problem 27 using the mirror and magnification equations.

Nick Auwerda · Accepted Answer

question 27 says an object with the height of 33 centimeters, or 330.33 meters, is placed two meters in front of a convex mere, which, of course, means that has a focal length of negative points of five meters. So Con Cave near would be out here, Um, but a convex me remember that focal length is actually negative, so it&#x27;s back here, so it&#x27;s fairly common to mess that up. Don&#x27;t do that on then it says, determine the approximate location and size of the image using and ray diagram. So when we use a ray diagram on a con vex mere, we draw our light ray that straight off of the top of the object straight towards the mere pair a little the principal axis, and it reflects as though it had come from that focal point up like that. And then our 2nd 1 we draw straight towards the focal point, and then it reflects this way, as though it had come from directly behind the mirror. And so here is our image. And so, ah, the question asked us determine the approximate location and size of the image to the image distance is going to be negative and it looks like maybe 0.6 meters. I don&#x27;t know. It&#x27;s hard to say in this picture will do the math in a minute and then is the image location in size. And it also looks to be about, um, maybe about 0.15 meters high. Image height 0.15 meters again. We&#x27;ll do the math. Ah, and then is it upright or inverted? Well, it&#x27;s upright. Convex mere always produces a virtual upright reduced image. So let&#x27;s go ahead and do the math. So are we have are mere equation one over Dio plus one over D I equals one over half. Ah, And so I have 1/2 plus one over d. I equals one over negative 10.75 or D I equals ah one over negative 10.75 minus 1/2 and will invert all that. And so if I take negative 0.75 inverse minus two in verse and I invert that I get ah, an image distance d I is equal to negative 0.545 or negative 0.55 meters, which is fairly close to what I estimated in that picture and then we&#x27;re supposed to find the image. Height will remember the opposite of the object height to the image. Height is the same as, ah, the object Distance to the image distance. And so the image height h I is equal to negative object height h o times d I over Dio. And so what I get out of this is negative 0.33 times negative 0.5 5/2 and so negative 0.33 times negative 0.55 divided by two is 20.9 eso I did not do a very good job estimating their 0.9 meters high.

Nick Auwerda · Answer

question 26 says an object for the height of 33 centimeters is placed two meters in front of the con cave mere with the focal length of 20.75 meters. Ah, we&#x27;re supposed to diagram this to determine the location and size of the image. And so what&#x27;s do that quickly? I&#x27;m just going to sketch this. I&#x27;m not actually put it on graph paper, but would get reasonable with this. Ah, and then we can move on from there. So we have ah, focal length of 0.75 meters. So here&#x27;s one meter. Here&#x27;s two meters. Ah, this is our focal length. No, no. Our focal length of sorry. 0.75 So here&#x27;s our focal length 0.75 meters. Um, our object is placed two meters away. So here&#x27;s our object, and it&#x27;s got a high of 0.33 meters. Um, 0.33 meters. Now, if we were to sketch this, we&#x27;re gonna take a line straight off of the top of the object towards the mere, and that line will reflect through the focal point back here somewhere. And then we&#x27;ll go off of the object through the focal point towards the mirror, and that&#x27;s going to reflect back here. And so we see that our image is going to form right there. Um, this was one meter. This looks to be a little bit past it. And so our image distance is probably around 1.2 meters Will do the math to prove it in just a second. And then it says, What is the approximate size? Ah, well, if this is 0.33 meters and this is it&#x27;s also about 0.2 meters high, but it&#x27;s negative. It is an inverted thing. So let&#x27;s go ahead and actually do the math on this and then ah, see what we get. So remember, one over Dio plus one over D I is one over f And so in this case, I have, um, won over two plus one over d. I is equal to one over 10.75 or one over D. I equals one over 10.75 minus 1/2. And so I&#x27;m going to invert every side of this to get my answer. And on my calculator, I&#x27;m just gonna push 0.75 inverse minus two inverse push, enter and then invert that and get 1.2. And so my image distances in fact, 1.2 meters. Now. Remember the ratio of the image distance to the object distance is also the ratio of their heights. So 1.2 divided by two is 20.6 and 0.6 times 0.33 is 0.1 night eight and so our height, eyes 0.198 and it is inverted.

Sarah Mccrumb · Answer

everybody. We want to find a magnitude of an object that is three focal points away. Here one. And this is one. Okay, I saw the object here. Get halfway goes here, and it looks like the next one is gonna hear, like, here and this other one. So if I could get this trust, there we go. So they all need right here past the one focal point. And because of that, it looks like and we put two years or what? Okay, so the lens. So on the right hand side, the lens I distance of 1.5 1/2 hopeful way, okay. And its size, we used it. So it&#x27;s just half of itself. So the converging, but is 1/2 okay? Because this image is used to half of itself. Okay. And because of that magnitude of the commercial in Okay, Thank you.

Mayukh Banik · Answer

it is given to be Ah couldn&#x27;t give meter. So f Equus positive for content is given to be 27 centimeter and magnification is three and the images upright. So that&#x27;s positive. Three. So from here we have negative image. Distance over object distance. It was three. So s bribe equals negative three s. We will use this in the formula. One of our air equals one of our s. Thus one of s bribe iss and explain being object on the image distance one of s minus 1/3 s. It was to over three s, so we would have yes, Equus to over three times if we know the value off F which is 27 centimeter. So that makes us to be 18 centimeter. So the object he&#x27;s 18 centimeter Ari from the then

Zachary Warner · Answer

to figure out the magnification and focal length of a flat mirror. Let&#x27;s consider some properties, so the image formed by a flat mirror has the following properties. One. The images as far behind the mirror as the object is in front of the year. So if the object is two feet in front of the mirror, it appears as though the image of that object is two feet behind them. Here the images unmodified, virtual and upright. So by the definition of a flat mirror the images on magnified, meaning the magnification em is equal to one. It doesn&#x27;t increase or decrease. So that&#x27;s one of our answers. Let&#x27;s box it in. The other thing is, what is the focal length? Well, to find the focal length, let&#x27;s consider the radius of curvature. So for a flat, near the radius of curvature, by definition, isn&#x27;t pretty, is the nearest flat. This means that focal length also infinite. So infinite is the answer, its forehead, and let&#x27;s underline it so we can mark that as the answer

Mayukh Banik · Answer

if we have a con. Careful meter and we haven&#x27;t object over here. We haven&#x27;t image over there. Then this means that the light rail coming in this direction is getting reflected in that direction. Now we can see that this strangle and this strangle is similar because this is the perpendicular angle of incidence equals angular fifth section. So these two triangles are similar. This is the height of the object. This is the height of the image. This much is the object distance. And this much is the image distance big. We can use the similarity of the triangle and we can figure out that h not over the nart equals h I over de I. This is the image. So this is a church. I so from here magnification Israel that it do the rescue of the image height do object height which comes out to be immense distance over object distance. So if we know Emma&#x27;s distance, if we know object distance, then we should be able to find the magnification

Problem

During a daytime football game you notice that a …

Problem 29 Medium Difficulty

Find the location and magnification of the image produced by the mirror in Problem 27 using the mirror and magnification equations.

Answer

Thus, the image distance of the formed by the convex mirror is $[-0.55 \mathrm{m}]$
The magnification of the image produced by the convex mirror is $[0.27]$

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James S. Walker

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Christina K.

Andy C.

Liev B.

Zachary M.

Wave Optics - Intro

Multiple Aperture Interference - Overview

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Thus, the image distance of the formed by the convex mirror is $[-0.55 \mathrm{m}]$The magnification of the image produced by the convex mirror is $[0.27]$

Physics

Christina K.

Andy C.

Liev B.

Zachary M.

Wave Optics - Intro

Multiple Aperture Interference - Overview

James S. WalkerPhysics

Christina K.

Andy C.

Liev B.

Zachary M.

Thus, the image distance of the formed by the convex mirror is $[-0.55 \mathrm{m}]$
The magnification of the image produced by the convex mirror is $[0.27]$

James S. Walker

Physics