an-object-with-a-height-of-33-mathrmcm-is-placed-20-mathrmm-in-front-of-a-convex-mirror-with-a-focal

Question

An object with a height of 33 $\mathrm{cm}$ is placed 2.0 $\mathrm{m}$ in front of a convex mirror with a focal length of $-0.75 \mathrm{m}$ . (a) Determine the
approximate location and size of the image using a ray diagram.
(b) Is the image upright or inverted?

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.

Rodger Claar · Answer

who don&#x27;t draw a ray diagram. To solve this problem, you can enjoy Reference Line Principal Axis three p. A. It&#x27;s how we measure everything from and reference to You&#x27;re convex Mirror is here, said Reflective side silver here. Non reflective size over here. Focal point. Technically, it&#x27;s right here. It&#x27;s negative. 20. Samir&#x27;s focal point Focal looks right here. Your object is approximately 37 years in front of me here, three severs, So draw raise to figure out where the image will be. So first roll is enjoy your first ray parallel principal axis And then it would reflect as if we will go through the focal point. That&#x27;s your first rate. Then you draw a second Ray ASAP overthrew the focal point, and then it reflects, apparently our principal access. So this is still your object here. First rate reflects up here. Secondary reflects over here. Since the rays do not converge on the side. Amir, we extend the rays back and see where they would intersect. So right here when it cross, is where you will see the image. So the images virtual it is within the focal point before the focal point, and it is smaller. That&#x27;s why Convex Mirror gives you smaller images because the way it&#x27;s curved Oh, so the image will be inside, and if a drawing to scale, you can physically measure everything up.

Matthew Kegley · Answer

So we have an object of two centimeters high in front of a convex mirror with a focal length of 10 centimeters, and we want to find the location of the image, the magnification of the image and the orientation of the image. The first thing you want to do for these optics problems is to draw out your ray diagram, identify the components, and i it yourself as to where these components are. So for our first step, we&#x27;ve drawn out this diagram. We have her object here, which is located to our left. We have our mirror, which is okay right here. And for this kind of problem, we see that we have to raise once you have our object in our mirror going to draw one ray of light that is parallel to our principal access. What you see, which is down here? This rail line is parallel tour principal access, and then it hit serve mirror and that it deflects diverges. And for this mayor, since the convex mirror, you wanna have a virtual it ray of light that connects to the rial deflection. The real deflection off this light and that&#x27;s that&#x27;s are going to be our first race, so I&#x27;ll just label. This is Ray number one. Yeah, just a little. This number one for our second Ray. We have to have at least two ways to determine our image. Um, the second raise one that goes toward the focal point of our mirror. And since this is a convex mirror, a focal point is going to be behind them. Here. So are our second round flight goes toward a mirror, and then it reflects from this mirror and turns out parallel to the principal access. So it&#x27;s the opposite of our first ray of light, which was incident parallel to our ex principal axes and then turned out, are deflected. And this deflection for everyone actually intersex with our focal point. Um, when it comes to our virtual part of this rain, so I&#x27;ll label are second rate number two. So we have our object here. I&#x27;m sure I&#x27;ve really well that you never convex mirror. We have our two rays of light. And from this reason, white, you can see that there&#x27;s, um, virtual raise that intersect at this point. Um, in order to connect the rial, raise that form when they deflect off of the mirror. So at this point of intersection, we have our image. That&#x27;s where our image forms, and my next to it is our focal point. So that&#x27;s the point that is intersected by our first ray. And that&#x27;s also but are second rate, but it reflex to the mirror first and turns out parallel. So now that we have everything labeled, we need the Orient ourselves. So with all these problems, we always know that our mirror will be placed at zero for our number line, and then anything that&#x27;s labelled to the left of it would be a positively anything to the right will be labeled as a negative life. So any object length or image length or focal length to the left or in front of the mirror, it&#x27;s going to be positive. And then anything behind it, this one to be a negative length. So since we have all this in mind, we can start using what we have given to us to solve this problem. So we&#x27;re given our focal length, and since our focal length is behind the convex mirror, that&#x27;s usually the case when we have our comics mirror drawn like this. Our focal length is going to be to the right of the convex mirror. Anything to the right of the mirrors on the being negative. So our focal length is going to be minus 10 centimeters, and then we also have the height of the object. We&#x27;ll just give that, uh, as h in its two centimeters. Okay, so we have this and the question is going to ask us what is our location of the image and whether the images upright or inverted and also the height of damage. Oh, and also and I forgot about this. The image is located 30 centimeters from the convex spherical near. So we have hr height of their object. And we also have our object location Just 30 centimeters. 30 centimeters is positive, since object is already located in front of the year. So now that we have are given, we can use the mirror equation to solve a problem. So our Mary equation and I&#x27;ll use a different color here, Mary equation is one overpay. My, uh, plus one over que is equal to one over f where P is our object length or location. Sees our Q is our image, location and F is our focal. So we want to find where the location of our images, which is que So let&#x27;s rearrange this. We have won over Q equal to one over F minus one over P. That&#x27;s are given information that&#x27;s minus 1/10. Linus, one over Ernie. And that gives us minus 4/30. So cue, um, for our image location is going to be seven or minus 7.5 senators. So that is our image location. It&#x27;s minus 77.5 centimeters. So now we wanted to find out whether the images upright or inverted, and you can see in our diagram that we made it so that it&#x27;s upright, but we want to double check with our magnification. We can double check that with the magnification equation. So our magnification equation chocolate in the I&#x27;ll put in red. His n equals two minus que overpay are image is negative since it&#x27;s actually located to the right to or behind the mirror. Um, so we gonna have 7.5 centimeters on top. I&#x27;m in. They were gonna have 30 centimeters on the bottom in this. This leaves us minus I&#x27;m sorry because that&#x27;s a plus 0.25 So you can see from our magnification here that we have a plus sign plus line means that our image is upright. So our image is upright. Okay, Because we know our plus sign is for our magnification, where you know that our image is going to be up right here. So now we want to do is we want to find the height of the image. So to find the height of the image, we can use another form of the of the magnification be formula. The magnification is also equal to the height of the object. I&#x27;m sorry. The height of the image which would live as h prime over the height of the object. We know her height of the objects, which was two centimeters. And we know our magnification from our previous problem. So all we have to do is plug those in and then we can solve for our image height. So let&#x27;s solve our image. Higher image height is age prime able to end Time H which is equal to positive 0.5 or 0.25 times two centimeters. So you can see, that gives us half a centimeter for our image height. So from what we were able to do, we had to draw our great diagram. Um, label. All of our components are raised of like and determined orientation that we were working in. Once we determine that we were able to use are given information the focal length, the height of the object in the location of the object, and then use our Mary equation to get the are the location of our image. And since this is a convex mirror, we knew that our focal length would be negative. And also as a result, um, our cue, our image location is also negative, based on a verdant very quick. So from there we could determine the magnification which was upright, which is positive, and that told us that our image was up right. And then we also use the magnification to find the height of the image, which we found to be half a centimeter. So thanks. And, uh, this was the answer to this problem.

Shoukat Ali · Answer

the first part of this problem, we&#x27;re going to matter. Um, education, that is a do you. So before going to draw the diagram, let&#x27;s get the radius of curvature as they you know that articles to to have the castle fat can kill you. So when setting where you off into the square weaken white articles to two in tow here, very for F as a 10 centimeters. So this would be where the value for the honor of articles, too. 36 nt. So from the figure, we can make the medication, which is at this point, So we can say that D I prime musicals too, Uh, minus four point 69 20 as it images behind that mirror. So the actual education will be the I entities equals toe. That&#x27;s cabbages. A close to six in tow, minus 4.69 So this will give us the California City I d icicles to 28 centimeters now, in part via this problem, we&#x27;re going to kill her down you match height. That is a it tight. So let&#x27;s my hair height and the figure we can see that the Met height is approximately 1.2 little sentiment that is, the height off image in the figure is at a prime. Entities equals toe 1.20 sentiment. So the actual height will be HIV, and it is a close to six into one point 20 centimeter. So this will give us a value for energize H I equals two. So one point six that sentiment zero centimeters. So the image height in the figure is 1.2 centimeters, so the actual ICT will be 7.6 centimeter. Thank you.

Rodger Claar · Answer

use a mirror equation. One arrest plus one over. That&#x27;s prime. He goes one over after SoCal like solved for s prime in the equation you re arrange You want a rest? Prime equals one of breath wise. One of rest take us to next step. You get one over as prime equals s mice s over sf final step you get That&#x27;s prime equals s times f over s myself as prime Miss Distance image assistance since the object f iss focal in its now we plug things in to get the distance to the image as prime, so as prime it would get device from the problem. 27 years times 67 years fired by 27 years minus and negative. 60 seven years and simplify of item on quest through And you can you of negative team Sam. Years equals distance. Image s problem of the magnification application equals hi image. Were I the object which is Samos negative distance image. This is the object. So they&#x27;re solved for h prime here. High image so high the image equals they have hate That&#x27;s prime over us. Plug in values now at age prime equals they have one point of centimeters times, It&#x27;s just tamed. Samir&#x27;s divided by 27 years at equals. 0.75 Sanders. So one some five centimeters that means equals ***. 15. Samir&#x27;s divided by 27 years. Sequels. Three. Force that tells us that the images 0.757 years Toll. Three force And this upright is 15 7 years behind, which is not right. 0.757 years tall. 15. Samir&#x27;s Behind Mirror.

Problem

Find the location and magnification of the image …

Problem 27 Medium Difficulty

An object with a height of 33 $\mathrm{cm}$ is placed 2.0 $\mathrm{m}$ in front of a convex mirror with a focal length of $-0.75 \mathrm{m}$ . (a) Determine the
approximate location and size of the image using a ray diagram.
(b) Is the image upright or inverted?

Answer

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