Question

What is the critical angle for the interface between water and diamond?

33 degrees
90 degrees
57 degrees
0 degrees
98 degrees

Answer 1

33 degrees

Step-by-step explanation:

The critical angle for a light ray moving from a denser medium with refractive index
`n_1` to a second medium with refractive index
`n_2` is given by

`\theta_c = sin^(-1) ((n_2)/(n_1))`

In this case, the critical angle occurs when light moves from diamond to water. The index of refraction of the two materials are:

`n_d = 2.42` for diamond

`n_w = 1.33` for water

So the critical angle is

`\theta_c = sin^(-1) ((1.33)/(2.42))=sin^(-1)(0.550)=33.3^(\circ)`