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##### Andy C.

University of Michigan - Ann Arbor

##### Marshall S.

University of Washington

##### Aspen F.

University of Sheffield

### Video Transcript

our question wants us to consider. Think you're P 22.16 which shows this light ray traveling around inside a slab of crown glass. So it says that the ray is incident on the surface at an angle of 55 degrees. So data a one is 55 degrees with the normal, and it reflects from point A to B to C for party. It says, at which of these points is the part of the rate enter the air. Okay, so the, um, first thing we can do is figure out all the different angles. So since data a one is equal to 55 degrees and we can use the law of refraction, which states that beta 8th 1 is equal to beta a two Okay, so the triangle A e b. And the figure would have to add up to equal 180 degrees, which would mean that they ate too, plus they to be one. And then there's a right angle there, so that would be 90 degrees. So plus 90 degrees has to equal 180. So if we saw for there to be one, we find that this is equal to 90 degrees, 180 minus. And I need just 90. This is 90 degrees minus data A two which we found any 55 degrees. So if you carried out it out this expression we find that data et tu is equal. Excuse me, They'd be one is equal to a 35 degrees. But again, using the law of refraction, they'd be one has to be equal today to be too. So they did. B two is also equal to 35 degrees. All right, we'll get a new page going here. So now that we know they to be too, we can use the fact that the triangle B C. D. Adds up to 180 degrees to find they to see one. So they should be too. Plus data. See one. Well, since there's also a right triangle, this has to be plus 90 degrees is equal to 1 80 or they had a C one is equal to 90 degrees minus stated. Be one excuse me minus stay to be too. Since we played in 35 degrees for theta. Be too. We find that this is equal to 55 degrees and again using the law every fraction. This is equal to fade A C two. So the expression which gives the relationship between the critical angle and the refractive index, is given by sign they. Then this is the critical angle. Angle will call this see um CR for critical. So this sign of the critical angle is equal to the index of refraction of air to the index of refraction of glass. So, therefore, the critical angle. Fate. A CR is equal to the inverse sine of ratio of the index refraction of air two, the index of refraction of glass. Carrying out this expression, we find that this angle is equal to 41.14 degrees. So if the angle of incident is less than this angle, it can travel through. But if it's greater than this angle, it can't travel through. And since data see one and data see, two are equal to 55 degrees and they do a one and data a tour equal to 55 degrees. Those were both greater than the critical angle. So let's start a new page here, so the critical angle is less than data. A one and a, too. So we'll just call that data, eh? And it's also less sand. They had a C one of beat they to see to the only thing the critical angle is greater than is. They'd be one and a two b two, which will call for you to be, Uh, therefore, the ray will enter the air at point D. So we'll say this is for par. A ray enters air at point be. We can go ahead and box all this and is their solution for part a part B is it? The glass slab is surrounded by carbon. Don't fuselage. It's an heir. At what point does Ray enter the flute? Few silly. Well, since the, uh, index of refraction of the carbon excuse me, not diffuse late carbon die sulfide. Since the index of refraction of carbon die sulfide is 1.6 to 8, that is greater than glass. So our reasoning here is since the index every fraction of the carbon die soulful eight, which will disc all see, is greater than the index every fraction of the glass. Therefore, it's going to enter at all of those because the critical angle is gonna be large enough for all of us. So we say thus enters carbon So fight. Yeah, a be and see which weakened boxing is their solution for part B.

University of Kansas
##### Andy C.

University of Michigan - Ann Arbor

##### Marshall S.

University of Washington

##### Aspen F.

University of Sheffield