Question

A plastic light pipe has an index of refraction of 1.44. For total internal reflection, what is the minimum angle of incidence if the pipe is in the following media?

(a) air

Answer 1

1.

Answer 2

total internal reflection occurs when light travels from a denser medium to a less dense medium and the angle of refraction exceeds 90 deg

we use snell's law; I will use "1" to refer to the denser medium and "2" to refer to the less dense medium

n1 sinx1 = n2 sinx2

for total internal reflection, x2 (angle of refraction) = 90 so sinx2 =1



for air, n2=1 so we have

1.44 sin x1 = 1 (sin x2)
sin x1 = 1/1.44=0.6944
x1=arcsin(0.6944)= 43.98 deg and this is the critical angle for this plastic in air

for water, n2=1.33 and

1.44sinx1=1.33
sinx1=0.9236
x1= arcsin(0.9236)= 67.46 deg and this is the critical angle for plastic in water

Answer 3

Index of refraction of the pipe with respect to air = 1.44
Therefore minimum angle of incidence = sin-1(1/1.44) = 43.98 deg
Ans: 38.12 deg

b= Index of refraction of water = 1.33
Therefore index of refraction of the pipe with respect to water = 1.44/1.33 =1.08
Therefore minimum angle of incidence = sin-1(1/1.08)=67.80