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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 theapproximate location and size of the image using a ray diagram.(b) Is the image upright or inverted?

(a) Thus, from the above graph the image is located at a distance $\left[d_{i}=55 \mathrm{cm}\right]$ behind the mirrorand it is virtual.From the ray diagram, the size of the image from the above graph is approximately $[9.0 \mathrm{cm}]$(b) From the ray diagram, the image is above the principal axis and facing upwards.Hence, the image is upright

Physics 103

Chapter 26

Geometrical Optics

Wave Optics

University of Washington

Simon Fraser University

Hope College

McMaster University

Lectures

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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's back here, so it's fairly common to mess that up. Don'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't know. It'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'll do the math. Ah, and then is it upright or inverted? Well, it's upright. Convex mere always produces a virtual upright reduced image. So let'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'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.

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