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?

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, determine the maximum value of angle theta at which the upper face PQ of the glass rod can be cut. So that a light wave traveling parallel to the vertical part of the rod will be reflected back into the rod when it strikes face PQ consider the rod immersed in water with a refractive index of 1.33 and a refractive index of rod material is 1.70. OK. So our end goal is to determine the maximum value of angle theta. OK. So we're given some multiple choice answers and they're all in the same units of degrees. Let's read them off to see what our final answer might be. A is 54.0. B is 51.5 degrees and C is 38.5 and D is 36.0. OK? And they're all degrees. OK. So we're given a figure here by from. So the problem gives us a figure here to help us visualize the problem. And here is the face PQ and then the angle theta. OK. So first off, let us recall that the critical angle for the total internal reflection is angle subscript, capitalize the incident angle which gives the refractive. So the angle of the refractive angle is 90 degrees. So let's write that down. So the total internal reflection is equal to theta subscript capital I which is the incident angle. So the total internal reflection is equal to the incident angle, which also gives us that the refractive angle beta subscript capital R is equal to 90 degrees. OK. Due to smells lock. So the angle of incidence beta I is we can write that theta I the angle of incidence plus theta is equal to 90 degrees and it is related to the refractive angle due to this relationship. So now we can determine the incident angle that gives us the value of the refractive angle being 90 degrees. So we need to make following note here really quick. So we don't forget that the refractive index of the rod and R is equal to 1.70 and that the refractive index of water is equal to 1.33. OK. So we can use Snow's Law equation to solve for the angle of incidents, the incident angle. So let's write down Snell's law to help us figure that out. So the in the refractive index for the rod multiplied by the sign of the angle of incidence. The incident angle is equal to the refractive index of water multiplied by sign of the refractive angle. So now we can plug in our known variables. So we know that the refractive index for the rod is 1.70 multiplied by sine of the incident angle which we're trying to solve for is equal to the refractive index of water which is 1. multiplied by sine of angle degrees. So sine is so the refractive angle is 90 degrees. So we can rearrange using a little bit of algebra to find that the angle or a little bit of algebra. And some geometry, we get that the incident angle is equal to the inverse of sine of 1.33 divided by 1.70. And note that sine of 90 degrees is one. So when we plug that into a calculator, we should get 51.47 degrees. So now we can solve for the maximum value of angle theta. So to do that theta must be equal to degrees minus the incident angle. So let's plug in our known values to solve for theta. So 90 then let's add a 0.0 degrees minus the value we just found for the incident angle which is 51.47 degrees is equal to 38. degrees. And that is our final angle So theta is equal to 38.5 degrees. So let's look at our multiple choice answers. So that means that our correct answer has to be C 38. degrees. Thank you so much for watching. Hopefully that helps and I can't wait to see you in the next video. Bye.