Problem

Predict/Calculate A glass paperweight with an ind…

12:03

Problem 56 Medium Difficulty

A ray of light enters the long side of a $45^{\circ}-90^{\circ}-45^{\circ}$ prism and undergoes two total internal reflections, as indicated in FiGURE $26-66$ The result is a reversal in the ray's direction of propation. Find
the minimum value of the prism's index of refraction, $n,$ for these
internal reflections to be total.

Answer

1.41

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Video Transcript

all right. So for this problem, we have a light ray that's entering 45 45 90 prison and the is undergoing to total internal reflections that result in a complete reversal of the direction of the late right. Taking all this into account, we must assume that because it's resulting in a total reversal of direction of the late Ray that is entering the 45 45 90 degree prism, like so at a 90 degree angle to the surface. So there's no refraction upon entry and it must exit at a 90 degree angle. There's no refraction exiting, so that gives us a lot of, ah, give us a lot of information because it tells us that as it bounces off here, I've only drawn the left side of our 1990 45 degree prison here. But that's all we need, because we can see that on the first total internal reflection. The light ray makes a 90 degree angle itself is gonna be useful because what we want to know is we want to know the index of a fraction of the glass that's going to cause or the minimum index of refraction of last. Excuse me, that's going to cause thes two total internal reflections. So let's find that out. So we know that because the light ray makes a 90 degree angle with itself. If we draw a normal line to the surface of the prism upon its first total internal reflection, we can deduce that it's angle of incidence. There. We'll see 45 degrees like improve this by extending this normal line down here and say that because the normal line must make a 90 degree angle with surface and this angles already 45 degrees, we've just created over 45 45 90 degree triangle and we're simply bisecting that. Ah, that 90 degree angle using the first incident Gray, I can tell you the incident angle is 45 degrees. It's nothing. Instead of snails along, we can say that the index of refraction of the class times a sign of the incident angle going to be equal to the index of refraction of air and because this is total internal reflection, sometimes a sign of 90 or just one now, because the into the index of refraction of Ares, also about one, we can say that the Indus reflection of class times Sign of the angle of incidence. They're just evil one, because we already calculated or geometrically proved that the angle of incidence is 45 degrees being. Say that the index of reflection of glasses just gonna be won over sign of 45 degrees, which, if you numerically plug in your calculator, should give you an index of refraction of one point for one. So all those problems seems pretty complicating the beginning. If you break it down simply 1/2 and use the information the problem gives you, it's pretty simple to solve. This is your final answer.

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