### Discussion

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##### Christina K.

Rutgers, The State University of New Jersey

LB
##### Aspen F.

University of Sheffield

### Video Transcript

grass to determine the maximum angle, which they called data for which the light rays incident on the end of the light pipe figure P 22.38 are subject to total international or internal reflection. So assume that the pipe has an index of refraction in some two of 1.36 coming out of air, which has in victory fraction of one. Okay, so we're going to apply Snell's law to point A and at 0.8 light travels from pipe Air. Thus, the index of reflection in one is equal to 1.36 and index refraction in to here is equal to one. So we're actually going to reverse these because of the direction of travel. So we're gonna call in two 1.0 and in 11.36 Therefore, the critical angle sign they had a C is equal to in one all right into divided by in one. So therefore, the critical angle they to see is equal to the inverse sine of the ratio into two in one. Using that, we've now found that critical angle, which is 47.3 degrees. So from the figure the angle of refraction can be calculated as follows, just using the fact that the total angle miss are the total inside angles of ah, triangle have to add up to 180 degrees We have stay to see. Plus, data are is equal to 90 degrees or they are just playing value for the critical angle that we just found is equal to 42.7 degrees. Okay, Now, um, we can go back to smells Law point. Beware. The light travels from pipe to air to say that, uh, Snell's law says that in one which we're gonna call, uh uh, pipe to air. So in one here is just going to be Let's go ahead and write out our values. Um, in one is going to be 1.0 times this sign of the angle, uh, fatal one here, which is the angle, uh, data that we're trying to find. This is going to be equal to into which is the one point, uh, 1.36 value that we used previously for the index refraction of the medium that is in the pipe, multiplied by the sine of the angle that we just found data are let's re write that they did look a little nicer here. Okay, so then solving for theta, we find that this is equal to 1.36 claims the sign data are divided by one. And that whole thing has, um, is being that whole ratio is being put inside the signed it a minus one here. So let's go ahead and indicate that's a theater is equal to signed the minus one of that ratio. We find that data is equal to 67.3 degrees box, and it is their solution.

University of Kansas
##### Christina K.

Rutgers, The State University of New Jersey

LB
##### Aspen F.

University of Sheffield
The passage contains a physics problem that involves calculating the maximum angle for which light rays incident on the end of a light pipe are subject to total internal reflection. The problem is solved step-by-step, including the application of Snell's law and geometric relationships.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
### Discussion

You must be signed in to discuss.
##### Christina K.

Rutgers, The State University of New Jersey

LB
##### Aspen F.

University of Sheffield

### Video Transcript

grass to determine the maximum angle, which they called data for which the light rays incident on the end of the light pipe figure P 22.38 are subject to total international or internal reflection. So assume that the pipe has an index of refraction in some two of 1.36 coming out of air, which has in victory fraction of one. Okay, so we're going to apply Snell's law to point A and at 0.8 light travels from pipe Air. Thus, the index of reflection in one is equal to 1.36 and index refraction in to here is equal to one. So we're actually going to reverse these because of the direction of travel. So we're gonna call in two 1.0 and in 11.36 Therefore, the critical angle sign they had a C is equal to in one all right into divided by in one. So therefore, the critical angle they to see is equal to the inverse sine of the ratio into two in one. Using that, we've now found that critical angle, which is 47.3 degrees. So from the figure the angle of refraction can be calculated as follows, just using the fact that the total angle miss are the total inside angles of ah, triangle have to add up to 180 degrees We have stay to see. Plus, data are is equal to 90 degrees or they are just playing value for the critical angle that we just found is equal to 42.7 degrees. Okay, Now, um, we can go back to smells Law point. Beware. The light travels from pipe to air to say that, uh, Snell's law says that in one which we're gonna call, uh uh, pipe to air. So in one here is just going to be Let's go ahead and write out our values. Um, in one is going to be 1.0 times this sign of the angle, uh, fatal one here, which is the angle, uh, data that we're trying to find. This is going to be equal to into which is the one point, uh, 1.36 value that we used previously for the index refraction of the medium that is in the pipe, multiplied by the sine of the angle that we just found data are let's re write that they did look a little nicer here. Okay, so then solving for theta, we find that this is equal to 1.36 claims the sign data are divided by one. And that whole thing has, um, is being that whole ratio is being put inside the signed it a minus one here. So let's go ahead and indicate that's a theater is equal to signed the minus one of that ratio. We find that data is equal to 67.3 degrees box, and it is their solution.

University of Kansas
##### Christina K.

Rutgers, The State University of New Jersey

LB
##### Aspen F.

University of Sheffield
user
The passage contains a physics problem that involves calculating the maximum angle for which light rays incident on the end of a light pipe are subject to total internal reflection. The problem is solved step-by-step, including the application of Snell's law and geometric relationships.

TYPE: SINGLE_QUESTION_WITH_ANSWER