Question

A plastic light pipe has an index of refraction of 1.45. For total internal reflection, what is the maximum angle of incidence to the wall of the pipe if the pipe is in (a) air? (b) water? Be careful: The problem asks for the angle with the ???????????????? of the pipe. This is not the angle in Snell's law. Use ????=1.333 for the index of refraction of water.

Answer 1

(a) 43.6°

(b) 66.5°

Step-by-step explanation:

Refractive index for plastic np = 1.45

refractive index of air na = 1

(a)

Let the maximum value of angle of incidence in air is i and the angle of refraction is 90° for total internal reflection.

By using Snell's law

`^(p)n_(a)=(n_(air))/(n_(plastic))=(Sin i)/(Sin 90)`

`(1)/(1.45)=(Sin i)/(1)`

Sin i = 0.6897

i = 43.6°

(b)

Let the maximum value of angle of incidence in water is i and the angle of refraction is 90° for total internal reflection.

By using Snell's law

`^(p)n_(w)=(n_(water))/(n_(plastic))=(Sin i)/(Sin 90)`

`(1.33)/(1.45)=(Sin i)/(1)`

Sin i = 0.917

i = 66.5°