Question

Example 5 Total Internal Reflection A beam of light is propagating through diamond (n, = 2.42) and strikes a diamond-air interface at an angle of incidence of 28°. (a) Will part of the beam enter the air (n2 = 1.00) or will the beam be totally reflected at the interface? (b) Repeat part (a), assuming that the diamond is surrounded by water (n2 = 1.33) instead of air.

Answer 1

a) The condition for total internal reflection is:

$$\theta \ge \arcsin\left( \dfrac{n_2}{n_1} \right) = \arcsin\left( \dfrac{1}{2.42} \right)\approx24\degree$$

Since $\theta = 28\degree>24\degree$, the beam will be totally reflected at the interface.

b) Now the critical angle is $\arcsin\left( \dfrac{1.33}{2.42}\right) \approx 33\degree$. Thus, $\theta = 28\degree<33\degree$ and part of the beam will enter the water.