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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

The critical angle for total internal reflection at a liquid– air interface is 42.5°. (a) If a ray of light traveling in the liquid has an angle of incidence at the interface of 35.0°, what angle does the refracted ray in the air make with the normal?

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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, determine the angle at which the light wave gets refracted when it passes from liquid to air with an angle of incidence of 32 degrees. The value of the critical angle for the total internal reflection at the liquid air interface is 41 degrees. OK. So that's our end goal is to determine the angle at which the light wave gets refracted. So we're given some multiple choice answers. They're all in the same units of degrees. Let's read them off to see what our final answer might be. A is 48.7 B is 50.7 C is 43.7 and D is 43.7. OK. So first off, we need to find the liquids refractive index. So to determine that value, we need to recall and use the critical angle note that the necessary condition for the total internal reflection is that light should not be incident on the material with a larger N value or N is the refractive index value. In this case, the liquid has a larger end value for the refractive index. So at this point, we need to be able to recall and use Snell's Law Equation which states that in subscript, A multiplied by sine beta subscript A is equal to N subscript B multiplied by sine data subscript B where N A is equal to the refractive index for the liquid. And N subscript B is equal to the air's refractive index. OK. So now we can make the following statements, we need to know that N A is equal to the N value and liquid. So N A is equal to N liquid and that NB is equal to N air and let's call it capital A N subscript capital A is equal to the refractive index of air which is the numerical value for that. So the numerical value for the refractive index of error is 1.00. We also need to know and state that beta A is equal to the critical angle theta subscript C when theta B is equal to 90 degrees. OK. So we can then write knowing all these statements that the N subscript liquid. So N liquid, so the refractive index of liquid multiplied by the sign of the critical angle is equal to the refractive index of air which the numerical value is 1. multiplied by sine of theta B which is 90 degrees. So we can rearrange this equation to isolate and solve for the refractive index of the liquid all by itself. So when we do that, we should get that, it's one divided by the sign of the critical angle which when we plug in our numerical values, it should be one divided by sine of 41 degrees which when we plug into a calculator, we should get that the refractive index of the liquid is 1.52. OK. So now we need to consider Snell's Law once again to help us find our fi like our final answer which is the angle at which the light wave gets refracted. So to do that, we need to rearrange Snell's law to solve for theta B. So sine theta B is equal to is equal to the refractive index of liquid. So NL multiplied by sine theta A all divided by N subscript A is it's the refracted index of air. So sine beta B is equal to the refractive index of the liquid multiplied by sine theta A all divided by the refractive index of air. So now we could plug in our known variables. So we know the refractive index for the liquid was 1.52 multiplied by sine of 32 degrees all divided by 1.00 which is the numerical value for the refractive index of air. So in order to solve first, theta B, all, we have to do is take the inverse of sine of 1.52 multiplied by sine of degrees all divided by 1.00. So when we plug that into a calculator, we should get that A theta B is equal to 53.7 degrees. So this is our final answer. So this is the angle at which the light wave gets refracted ray, we did it. So that means when we look at our multiple choice answers, the correct answer has to be the letter D 53. 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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