A British company makes optical fibers that are 13.6 km in length. If a beam of light travels along one of those fibers, always reflecting at the critical angle, it will travel 20.3 km. What is the index of refraction of the fiber material?
Physics - drwls, Sunday, February 20, 2011 at 2:03pm
The ratio of wave travel distance (along a ray)to the distance the wave tralels long the fiber is 13.6/20.3 = 0.770
The angle that a ray makes with the side of the fiber is therefore
cos^-1 0.770 = 47.9 degrees. The angle of incidence is the complement of that, or 42.1 degrees.
Knowing that 42.1 deg is the critical angle for total internal reflection, you can conclude that (if there is no cladding)
1/N = sin 42.1 = 0.670
N = 1.49
Usually optical fiber is "clad" with a material of intermediate index between N and 1, to protect the fiber and reduce losses. This would change the answer.