Question

A plastic light pipe has an index of refraction of 1.48. For total internal reflection, what is the minimum angle of incidence if the pipe is in the following media? (a) air (b) water Need

Answer 1

Answer:

a) 42.52°

b) 63.98°

Explanation:

Refractive index of pipe = 1.48 = n₂

Refractive index of air = 1.0003 = n₁

$\theta_r=sin^{-1}\frac{n_1}{n_2}\\\Rightarrow \theta_r=sin^{-1}\frac{1.0003}{1.48}\\\Rightarrow \theta_r=sin^{-1}0.676\\\Rightarrow \theta_r=42.52^{\circ}$

∴ Minimum angle of incidence is 42.52°

Refractive index of water = 1.33 = n₁

$\theta_r=sin^{-1}\frac{n_1}{n_2}\\\Rightarrow \theta_r=sin^{-1}\frac{1.33}{1.48}\\\Rightarrow \theta_r=sin^{-1}0.899\\\Rightarrow \theta_r=63.98^{\circ}$

∴ Minimum angle of incidence is 63.98°