A Spherical Fish Bowl. A small tropical fish is at the center of a water-filled, spherical fish bowl 28.0 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.

Video transcript

Welcome back, everyone. We are making observations about a rubber duck that is submerged in a bot round bottom flask containing water. Now, we are told a couple of different things about this situation here, we are told that the duck is located 15 centimeters from the wall of the flask. Now, we are also told that the spherical section of the flask has a 44 centimeter diameter from this. If we divide this by two, we can see that the radius is 22 centimeters or another way to interpret this is that it is 22 centimeters from the outside to the ins in order to reach the inside. So we'll say that our radius is native 22 centimeters going from outside to inside. Now, we are tasked with finding what is going to be the distance of our image or the position of our image of the duck as well as the magnification, lateral magnification if a person is standing right next to the flask. Now, before I get started, I do wish to acknowledge our multiple choice answers here on the left hand side of the screen, those are going to be the values that we strive for. So without further ado let us go ahead and begin. What are we gonna use here? Well, we have this relation for spherical surfaces, we have N A divided by the position of our object plus NB divided by the position of our image is equal to NB minus N A divided by R. Now, what is N A? And what is NB? Well, we are told that for water which I will assign to N A, we are told that the end for our water is 1.33. Now NB, we are going to be referring to the end value for air, which is simply just one. Now with all of this in mind, let me go ahead and change colors here so that we can start part one and start filling in what we know to this equation. So what we have here is we have 1.33 divided by 15 plus one over si is equal to one minus 1.33 divided by negative 22 minus 1.33 divided by 15. Now, where I'm getting this value from is I'm subtracting 1.33 divided by 15 from both sides of our equation so that it cancels out on the left hand side. So then solving the right hand side of our equation, what we get is that one over si is equal to negative 0.07367 centimeters to the negative first power which one we solve for si gives us negative 13.6 centimeters. Great. Now changing color here one more time. For part two, we need to find our lateral magnification for this. For a spherical surface, it's just gonna be the negative of N A times SI divided by NB times. So we have all of these values. So let's go ahead and plug them in. We have negative 1.33 times 13. divided by one times 15, giving us a lateral magnification of 1.2. So as you can see, we have found both the position of our image as well as the lateral magnification, both of which correspond to our final answer. Choice of C Thank you all so much for watching. I hope this video helped. We will see you all in the next one.