At the very end of Wagner's series of operas Ring of the Nibelung, Brünnhilde takes the golden ring from the finger of the dead Siegfried and throws it into the Rhine, where it sinks to the bottom of the river. Assuming that the ring is small enough compared to the depth of the river to be treated as a point and that the Rhine is 10.0 m deep where the ring goes in, what is the area of the largest circle at the surface of the water over which light from the ring could escape from the water?

Video transcript

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A few young boys are practicing diving in a swimming pool while doing so, one of the boys loses a silver coin in his pocket which falls into the pool. The coin reaches the bottom of the pool at a depth of 5. m. Treating the coin as a point, determine the circumference of the largest circle at which the water surface through which the light ray reflected from the coin will pass out of the water. So that is our end goals determine the circumference of the largest circle. OK. So we're given some multiple choice answers here. Let's read them off to see what our final answer might be. And let's note that all of the answers are in the same units of meters. So A is 5.70 B is 35.81 C is 48.75 D is 98.44 and E is 4.50. OK. So to help us visualize this problem, let us look at the following figure. OK. So we know that the coin reaches a depth of 5.0 m. So that's DD represents the depth and R is the radius of the circle. And then we have our critical angle and then we have the light ray which from it hits the surface of the water to the coin at the bottom of the pool. OK. So the figure here shows the incident ray of light at a critical angle. Thus it passes back at the edge of the circle circle of light. So let us say that R is the radius of the circle and that D is the depth of the pool. Like I mentioned, it's 5.0 m. So using the figure, we can write that that tangent of the critical angle, let's call it theta subscript capital C is equal to the radius divided by the depth. So we can rearrange this equation to isolate and solve for R the radius. And when we do that, we get that the radius is equal to the depth multiplied by tangent of the critical angle. Let's call this equation one. OK. So to determine the critical angle, we can use the Smells Law Equation. So Snell's Law Equation states that for this particular problem using what we know that the reflective index of water and subscript lowercase W multiplied by sine beta of subscript W is equal to the reflective refractive index of air. It's called it N subscript A multiplied by sign of theta A. OK. Where NW is the refractive index of water. And N A is the refractive index of air and N or sorry. And then theta subscript W so theta W is the is equal to the critical angle. So let's write that down. So theta subscript W is equal to the critical angle and beta A is equal to 90 degrees. OK. So now now that we know all of that, let's plug in our known variables to solve for the critical angle. So sign they see kind of the of the angle, the critical angle I should say is equal to 1. multiplied by sine of 90 degrees divided by 1.33. So 1.00 is the refractive index of air and 1.33 is the refractive index for water. OK. And we get this from rearranging Snell's Law equation using a little algebra. So like we did so like we should know to do, we need to isolate the angle, the critical angle by itself. So to do that and to solve specifically for the critical angle, all we have to do is take the inverse of sign. So the inverse of sign and noting that the sign of 90 degrees is one, all we have to worry about is the is the refractive index of air divided by the refractive index of water. So when we plug that into a calculator, we should get that. Our critical angle is equal to 48. degrees. So now we can use the critical angle value in equation one to solve for the radius. So let's do that. So the radius R is equal to 5. m, which is the depth multiplied by tangent of the critical angle, which we determined to be 48.75 degrees. So when we plug it into a calculator, we should get 5.70 meters. So now we need to solve for the circumference. And let's remember and recall that the circumference equation states that the circumference is equal to two pi multiplied by the radius. So let's plug in our known values. So the circumference equals two pi multiplied by the radius which was 5.70 m. So when we plug that into a calculator, we should get the circumference is 35. m and that is our final answer. OK. So let's look at our multiple choice answers. So that means the correct answer has to be the letter B 35.81 m. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.