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?

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.

To find the index of refraction of the fiber material, we can use the formula:

n = Lc / Lf

Where:
n is the index of refraction of the fiber material,
Lc is the length of the path the light travels in vacuum,
Lf is the length of the path the light travels in the fiber.

Given:
Lc = 20.3 km (kilometers)
Lf = 13.6 km (kilometers)

Substituting the given values into the formula:

n = 20.3 km / 13.6 km

n ≈ 1.4926

Therefore, the index of refraction of the fiber material is approximately 1.4926.

To find the index of refraction of the fiber material, we can use the formula:

n = distance traveled outside the fiber / distance traveled inside the fiber

In this case, the distance traveled outside the fiber is 20.3 km, and the distance traveled inside the fiber is 13.6 km. Therefore, the index of refraction (n) can be calculated as:

n = 20.3 km / 13.6 km

Now we can calculate the value of n.

n = 1.4926470588235294

So, the index of refraction of the fiber material is approximately 1.4926.