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Ch 33: The Nature and Propagation of Light
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All textbooksYoung & Freedman Calc 14th EditionCh 33: The Nature and Propagation of LightProblem 33

Chapter 33, Problem 33

Light enters a solid pipe made of plastic having an index of refraction of 1.60. The light travels parallel to the upper part of the pipe (Fig. E33.15). You want to cut the face AB so that all the light will reflect back into the pipe after it first strikes that face. (a) What is the largest that u can be if the pipe is in air?

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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, find the greatest value of angle theta at which the edge PQ of the solid glass can be shaped. So that a light wave moving parallel to the vertical part of the glass will bounce back into the glass when it hits the edge PQ, take the glass in the air and the refractive index of the glass is 1.70. OK. So our ang goal is to find the greatest value of angle data. So we're given a little picture here to help us visualize this problem and here's the edge PPQ. OK. So we're given some multiple choice answers here. Let's read them off to see what our final answer might be. A is 36 degrees. B is 54 degrees C is 90 degrees and D is 60 degrees. OK. So first off, let us recall and use the critical angle for the total internal reflection, which we're gonna denote it as tir the total internal reflection, which is represented in by theta subscript I, which is the incident angle, which gives us the refractive angle, which is theta subscript capital R which is equal to degrees as stated in Snell's law. So the angle of incidents which is theta I is related to angle, theta R due to the falling relationship that states that theta I the incident angle plus theta is equal to 90 degrees. Now we need to determine the value of the incident angle that gives us the refractive angle which is equal to 90 degrees. So note that the refractive index of glass, which I'm gonna call it NG is equal to 1.70. And that the refractive index of air, which I'm gonna do is N subscript A and A is equal to 1.00. OK. So now at this point, we can recall and New Snails Law Equation to help us find the angle. And that states that the, that NG, the refractive index of glass multiplied by sign of the incident angle is equal to N A, the refractive index of air multiplied by sign of the angle of refraction. A refractive angle I should say. So now we can plug in our known variables. So NG is 1. sign beta I be subscript I is equal to N A which is equal to 1. multiplied by sign of 90 degrees because the refractive angle is 90 degrees. So we can then write that sine theta I the incident. So the sign of the incident angle is equal to 1.00 divided by 1.70. And this is when we simplify a little bit. OK, then we could take it even further. When we simplify even more, we can say that the incident angle is equal to inverse sign of 1.00 divided by 1.70 which is equal to 36.03 degrees. Now we can solve for a final answer to find the greatest value of angle theta. So that to help us find out, we need to say data is equal to 90 degrees minus the angle of incidents or the incident angle. So let's plug in our known values to solve the beta. The 90 degrees minus 36.03 degrees is equal to 54. degrees, which is our final answer. So this is the awesome. So let's go look at our multiple choice answers. So that means that the correct answer has to be the letter B 54.0 degrees, 54 degrees. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.
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Light enters a solid pipe made of plastic having an index of refraction of 1.60. The light travels parallel to the upper part of the pipe (Fig. E33.15). You want to cut the face AB so that all the light will reflect back into the pipe after it first strikes that face. (b) If the pipe is immersed in water of refractive index 1.33, what is the largest that u can be?
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