### Discussion

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##### Andy C.

University of Michigan - Ann Arbor

##### Marshall S.

University of Washington

### Video Transcript

sore question says that we have an index of refraction in equal to 1.4 for these fiber optic cables, and the bundle has a diameter d of one times 10 to minus four. Um, so using this refractive index in this bundle size, it wants us to figure out the smallest radius permitted to bend all of the lights that none of the light escapes. So in order for that to be true, fado would have to be greater than or equal to our critical angle if they to see. And ah, that means that the radius minus today and we're divided by the radius is greater than or equal to one over the index of refraction. It We can rearrange this equation to try to get the radius by itself and for trying to get our men. We find that this is, uh, are times in minus in times D is equal to let's make that look a little better is equal to radius are rearranging again, just trying to get our by itself. We find that, um are times in minus one is greater than or equal to D minus index of refraction in. So then solving for our This would be our our men if we, uh, set things equal to each other, so our men would have to equal Ah, in times D divided by in minus one. Okay, so ah, well, and I wrote in minus D here, and this would just actually be in times d. Sorry about that. We'll fix that steak real quick. So de times in, Okay. So, solving for our men, we find that our men then is equal to 3.5 times 10 to the minus four meters, which is 3.5 times the diameter d. We can go ahead and box set in as our solution to the question.

University of Kansas
##### Andy C.

University of Michigan - Ann Arbor

##### Marshall S.

University of Washington