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

Question

An object with a height of 33 $\mathrm{cm}$ is placed 2.0 $\mathrm{m}$ in front of a concave 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 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.

Nick Auwerda · 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.

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

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.

Rodger Claar · Answer

we&#x27;re gonna use the mirror equation one of breast to some see objects. This will never has pride this insane image because one of our focal, like focal length was positive. So we rearrange the cadre. So for distance to the image that&#x27;s prime equals yes, times that focal length five by yes myself. So cool. No plug and values from the problem. So this prime equals 27 years. Times 67 years, divided by 27 years Arness 67 years So with all mad Motorbike divide Simple five as prime This is the image. It&#x27;s negative. 30 Sam years For the first part For the Heidi image, he&#x27;s magnification equation. Education equals Heidi image. Hi. Object equals negatives. Distance image is prime distance The apt to rearrange solve rate prime its prime equals bankers h times as prime over s So we plug in values here. So what? We got another And problem so h prime equals negative. 17 year tolatimes make 37 years over 27 years Gives us a height of on 0.5 Sims Positive son m equals 3/2 thanks casing. It was 1.5 or three halves telling us that the images upright

Rodger Claar · Answer

draw radar going to solve this problem. Here&#x27;s your Khan gave mayor and then we troll our principal access. It&#x27;s a reference line through the sound of the mirror. So for this problem, focal length focal point his 27 years away. The object is twice is for 47 years away, so that right era is the object. And then the object ISS, 47 years away. We want to locate image to turn works up First rule. You draw your ray from top the object parallel a principal axis, and then it reflects through the focal point. Then you draw your second ray through the so called point, hits a mirror and then reflects pail of the prince waxes where the two razor sect. It&#x27;s where the image for me. So if you draw this to exact scale, things will line up perfectly. But you could see the object and image is pretty close. Eso the distance to the image distance of the images 47 years away. Same is the object. It&#x27;s located upside down, so the images inverted

Problem

An object with a height of 33 $\mathrm{cm}$ is pl…

Problem 26 Easy Difficulty

An object with a height of 33 $\mathrm{cm}$ is placed 2.0 $\mathrm{m}$ in front of a concave 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

(a) $d_{i}=120 \mathrm{cm}$
(b) see graph

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James S. Walker

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Zachary M.

Aspen F.

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Wave Optics - Intro

Multiple Aperture Interference - Overview

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(a) $d_{i}=120 \mathrm{cm}$(b) see graph

Physics

Andy C.

Zachary M.

Aspen F.

Jared E.

Wave Optics - Intro

Multiple Aperture Interference - Overview

James S. WalkerPhysics

Andy C.

Zachary M.

Aspen F.

Jared E.

(a) $d_{i}=120 \mathrm{cm}$
(b) see graph

James S. Walker

Physics