Question Video: Calculating the Critical Angle Given 𝑛 Values Physics

Video Transcript

What is the critical angle for a light ray traveling in water with a refractive index of 1.33 that is incident on the surface of water above which there is ice with a refractive index of 1.31? Answer to the nearest degree.

In this situation, we have some water, on top of which is a layer of ice. A ray of light propagates through the water until it reaches the interface between these materials. We want to know what is the critical angle for this light ray. The critical angle for this ray, we’ll call it 𝜃 sub c, is the angle of incidence for which the angle of refraction of this ray is 90 degrees. Refracted like this, we see that the ray of light travels along the boundary between the ice and the water.

In order to solve for 𝜃 sub c, we’ll want to recall a law of optics called Snell’s law. This law says that if we have a ray of light that’s incident on a boundary between materials of index of refraction 𝑛 sub i and 𝑛 sub r, then 𝑛 sub i times the sine of the angle of incidence of the ray equals 𝑛 sub r times the sine of the angle of refraction. Knowing this, if we consider again our given scenario, we can see the angle of incidence in this case is 𝜃 sub c, the critical angle.

Furthermore, the angle of refraction is 90 degrees. If we label the index of refraction of the water 𝑛 sub i and that of the ice 𝑛 sub r, then, by Snell’s law, we can write that 𝑛 sub i times the sin of 𝜃 sub c equals 𝑛 sub r times the sin of 90 degrees. But then, the sin of 90 degrees is exactly one. Therefore, 𝑛 sub i times the sin of 𝜃 sub c equals 𝑛 sub r. And since it’s 𝜃 sub c, the critical angle, that we want to solve for, let’s divide both sides of this equation by 𝑛 sub i, canceling that factor on the left. And then, with the remaining equation, let’s take the inverse or arcsine of both sides.

We do this because taking the inverse sine of the sine of an angle gives us just that angle itself. Therefore, the critical angle we want to solve for is the inverse sin of 𝑛 sub r divided by 𝑛 sub i. 𝑛 sub i, we remember, is the index of refraction of water. That’s given as 1.33. 𝑛 sub r, the index of refraction of ice, is 1.31. We now have an expression for the critical angle that we can evaluate. When we do this and round our result to the nearest degree, we get 80 degrees. This is the critical angle for a light ray traveling in water with a refractive index 1.33 incident on a boundary with ice with a refractive index 1.31.