2. The glass core of an optical fiber has an index of refraction 1.60. The index of refraction of the cladding is 1.48. What is the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber?

Explanation:

Using the Snell's law formula to solve this question which states that the ratio of the sine of angle of incidence to the sine of angle of refraction is a constant for a given pair of media. This constant is known as the refractive index for the given pair of media. Mathematically,

n = sin(i)/sin(r) where;

i is the angle of incidence

r is the angle of refraction.

n is the refractive index.

Given the refractive index of the optical fibre n₁ = 1.60 and that of cladding n₂ = 1.48

n₂/n₁ = sin(i)/sin(r)

The light ray can make with the wall of the core when its angle of refraction is 90⁰. The angle of incidence at this maximum point is known as the critical angle.

On substitution:

1.48/1.60 = sin(i)/sin90

1.48/1.60  = sin(i)/1

sin(i) = 1.48/1.60

sin(i) = 0.925

i = sin⁻¹0.925

i = 67.66⁰

Hence the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber is 67.66⁰.

im guessing the west of station b

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west