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A spherical fishbowl. A small tropical fish is at the center of a water-filled spherical fishbowl $28.0 \mathrm{~cm}$ in diameter.(a) Find the apparent position and magnification of the fish to an observer outside the bowl. The effect of the thin walls of the bowl may be ignored. (b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl?

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(a) Image is formed at centre of the bowl and magnification is 1.333(b) $s^{\prime}$ is larger than the diameter, hence focal point is outside the bowl

Physics 102 Electricity and Magnetism

Physics 103

Chapter 24

Geometric Optics

Electromagnetic Waves

Reflection and Refraction of Light

Simon Fraser University

University of Sheffield

University of Winnipeg

McMaster University

Lectures

02:30

In optics, ray optics is a geometric optics method that uses ray tracing to model the propagation of light through an optical system. As in all geometric optics methods, the ray optics model assumes that light travels in straight lines and that the index of refraction of the optical material remains constant throughout the system.

10:00

In optics, reflection is the change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated. Common examples include the reflection of light, sound and water waves. The law of reflection says that for specular reflection the angle at which the wave is incident on the surface equals the angle at which it is reflected. Reflection may also be referred to as "mirror image" or "specular reflection". Refraction is the change in direction of a wave due to a change in its speed. The refractive index of a material is a measure of its ability to change the direction of a wave. A material with a higher refractive index will change the direction of a wave to a greater degree than a material with a lower refractive index. When a wave crosses the boundary between two materials with different refractive indices, part of the wave is refracted; that is, it changes direction. The ratio of the speeds of propagation of the two waves determines the angle of refraction, which is the angle between the direction of the incident and the refractive rays.

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okay. And this problem, we're given that we have a fish bowl which we can approximate as a sphere with a diameter of 28 centimeters. And so it has a radius of 14 centimeters, and the fish is located here. So, um so the ass is out. It is equal toe 14 um, centimeters. And we can say, Let's call this region a and then this region B so and a is equal to its end of the, um Oh, it's the index of refraction of water. So I think that's 1.33 These I will check 1.33 So let's use that number. And then the next over fraction of air is, which is Region B is one. So we want to use the equation, um, to get so we want to get that where the image is formed. So we know and a over passes plus on dhe be over as prime sequel to and B minus and a divided by our so we can solve for s prime. And huh that's you can bring this, Entei, My over s to the other side. And then you can do cross multiplication. After that to get the, um at to solve for s prime. So that's gonna be, um, on and be times andi b minus on a divided by our minus and a over ass is gonna do that math in my head, this and then to the negative one here. That's why I'm equals. Mm. Pause the video. Okay. Yeah, I agree with my formula. So now I'm gonna plug in all these numbers, and I guess I'll pause the video again unless I do that. Okay. And then when I put that into a calculator, I ended up getting the wrong answer. Uh, and I knew it was wrong because it's just thinking about these rays. They should come out and then kind of get focused back here, like you could do a ray diagram if you wanted to. Um, but you would get that the images is located on the side, and I actually got a positive number. I realized it's because this is from the perspective of a fish here. The object, This is a convex lens. And so that is actually the race of courage is negative. 14 So that you get that s prime is equal toe negative, 14 centimeters. And so that means that the fish actually does appear to be at the center of the bullet is kind of interesting. And then he gets the magnification we want to evaluate. Negative. Um, what is the formula and a times s prime problems Interesting. Divided by and be times us. So if I plug in all those numbers, basically the fourteen's cancel and I get 1.33 So the fish, that's why the fish looks larger and, yeah, So, um, for B, we want to know, uh, if basically the set, like sun the sunlight from the other side it let me remind we we want to know if the sunlight gets focused at the center of the bull. And so I'm just gonna answer that. Um, without I like to answer things with math sometimes, if I can, even though I could probably answer it if I think about it. So I'm just gonna kind of use the same formula that we had used in one. So this s prime formula. Um, except I wantto have an A is equal to ah is one and then and B is equal to 1.33 And because we're imagining okay, this the sun, the image of the sun getting focused Where does the image of the sun get focused? Sounds basically infinity away. So he said ass equal to infinity and let's see what we get according to this formula. So I wonder if I could do it in my head. So if asked us to infinity that would basically kill this Entei over s turn. So we're left with the formula as prime is equal to and, um and B um times and be minus and a divided by our to the negative one power groups. And let's see. So then n b minus and a that's 0.33 f b I can't really do this in my head, so I'm gonna go ahead and type it into Does most so 1.33 if one So that's 0.33 abetted by are on. Then would we say our our waas 14 negative 14. Ah, but if the object is on Yeah, I guess I would still be, um I would still be negative because, um yeah, because it still would be con Oh, no, it would be con cave. If the object was on the left side, would be it would be, Ah, the left side of the mirror is Kong cave. And so that would be a positive are this is kind of interesting. Um, so for a positive are 14. What do we got? Um, I got that it was located at 56 centimeters, so that would be on the other side of the bowl. Um, and so, Yeah, I got that. That's 56 centimeters home, which is, like I said on the other side of the bull, I suppose kind of backing up in trying to do this without math. Um, you know that the rays are focusing. The focal point is that, um are over too. So, huh, Then I would think it would actually be inside the bull. But maybe it's something to do with the index of refraction. That's changing my intuition. Um, yeah. So maybe it is hard to do, actually, intuitively if you have to, because, I mean, you can't really into it. The index of refraction. Um, you are, like the strength of the index of refraction. So Yeah, I got it was outside the bowl. Therefore, we don't have to worry about sunlight focusing into the bowl. It's an interesting question, though.

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