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##### Andy C.

University of Michigan - Ann Arbor

LB
##### Aspen F.

University of Sheffield

### Video Transcript

part A of our question wants us to find the angle of refraction at the top of the surface where the index refraction of the thick block of glass, which I call into, is equal. The 1.5 in one is the next flight fraction of air. It's the block A to see is two centimetres thick and fatal. One here is equal to 30 30 degrees from figure p 22.1, eh? Okay, So now, for part, a new fraction angle at air glass interface come determined by applying smells law to calculate here this angle so will indicate this is part eh? And smells. Law states that in one times the sign of data one is equal to into times the sign of faith to two. Therefore, we can solve for theta two. We find that baby too, is equal to the inverse of the sign. So signed to the minus one of the ratio into divided by in one times the sign of a fatal one. Oh, and actually, I got my into in one mixed up. It's gonna be in one divided by m two slits. Go ahead. Correct that really quick. Oh, so this is in one divided by in two. So carrying out that calculation, we find that data to is equal to 19 0.5 degrees, which we can box in as their solution. For part a Part, B says, find the angle of incident at the bottom of the surface and the refracted ankle, so move on here and indicate this is part B. The angle of incident in the angle of refraction of the bottom of the surface of the block could be determined. A CZ follows here. So in the medium, the light ray propagates in a straight line from the geometry of the figure. The angle of incident at the bottom of the surface is equal to the angle of refraction at the top. That's the angle of incident at the bottom of the block is 19.47 degrees, and therefore the angle of incident. At the bottom of the surface are is 19.5 degrees. We rounded up, so the angle of incident the bottom of the surface is 19.5 degrees it since it's equal to fade too. So will say data to equals. That angle of incidence equals 19.5 degrees. Okay, so now, to find the angle of refraction, which will call fate a three, this is going to be equal to just as before. It's going to be this inverse sine using Snell's law again. This time, it's going to be into times the sign data, too, since that's the angle of incident divided by what is being refracted into and what plugging those values in this expression, we find that again. Of course, this is back to 30 degrees. So that's our angle. Every fractions, those air, Two solutions for part beak part see. Ah, So Part C asked us to find the lateral distance by which the light beam is going t o be shifted here. So the lateral shift of the light ray can be determined by first looking at the figure and writing down the co sign for the angle. A CB So the co sign of fated to You're right. That data is equal to so opposite uh, overhype odd news. So this is going to be a C divided by the distance A b. Okay, So solving for a B, which we're gonna need for the next part, we find that a B is equal to a C divided by co sign David too. And then from the angle B D A. We have to get a new page going here we have the sign of data. Four Who equal to the distance? A C. Excuse me. Dee dee Divided by a bee worth data four. Of course, we can replace with fatal one minus. They did too. On n b d Is that distance d the shift that we would like to find that still divided by the distance A to B So solving for d we find that d is equal to a B which from the first part we find that a B is equal to a C divided by coastline data to So this is gonna be replaced by a sea divided by oh, the co sign of faded too all multiplied that they'd s too Oh, multiplied by the sign of fatal one minus say to to plugging those values in the expression we find that the D value is 0.386 centimeters. We can go ahead and box it in as our solution for part c parte de assets to find The velocity of the light of the velocity of the light in the medium is equal to the speed of light in a vacuum. With just three times 10 to the eight meters per second, we denote that a c divided by the index of refraction of the medium. Okay, so plugging in 1.5. For that we find that this is equal to 2.0 time Send 88 who 10 to the eight meters for second. Which week in boxing is their solution for her d and then part uh E wants us to find the time it takes for it to train divers that medium so Time T is equal to the distance traveled, divided by the velocity Will the distances distance A to be the velocity Is the velocity V we found in that park? Well, a to B, of course again is distance ada See which we know. Divided by the co sign a fated to this is all multiplied by be velocity. Okay, so plugging those values in this expression we find that the time is 1.6 times 10 to the minus 10 when the units here are seconds which weakened boxing is the final part of our answer. In this solution, the party

University of Kansas
##### Andy C.

University of Michigan - Ann Arbor

LB
##### Aspen F.

University of Sheffield