Question Video: Using the Critical Angle to Determine Refractive Index Difference Physics

Video Transcript

The critical angle required for total internal reflection at a boundary between two substances is 30 degrees. The refractive index of the medium that light is reflected from is 1.1. What is the difference between the refractive index of the medium from which the light is reflected and the refractive index of the medium in which the light propagates before it is reflected?

In this scenario, we have a boundary between two substances. We’ll say that the index of refraction of the material above the boundary is 𝑛 one and that of the substance below the boundary is 𝑛 two. A ray of light passes through the top substance, we’ll say, until it reaches the boundary at what’s called the critical angle required for total internal reflection. At this angle of incidence, what we’ve labeled 𝜃 sub 𝑐, the ray of light is reflected so that it travels like this, parallel to the boundary.

In our problem statement, we’re told that the refractive index of the medium that the light is reflected from is 1.1. In our diagram, we see the light is reflected from this material with index of refraction 𝑛 two. Therefore, 𝑛 two equals 1.1. We want to know the difference between this index of refraction and the index of refraction we’ve called 𝑛 one. To know what this difference is, we’ll need to solve for 𝑛 one.

We can start to do that by recalling a law known as Snell’s law. This law says that if we have a ray of light traveling through a material with index of refraction 𝑛 sub 𝑖, an incident on an interface, and an angle of incidence 𝜃 sub 𝑖, then 𝑛 sub 𝑖 times the sin of 𝜃 sub 𝑖 equals the index of refraction of the material the light is refracted into, 𝑛 sub 𝑟, times the sine of the angle of refraction, 𝜃 sub 𝑟.

Knowing this general law, we can then apply it to our particular scenario here. For us, the ray of light is originally traveling through a material with index of refraction we’ve called 𝑛 sub one. Then, the angle of incidence of this ray to the interface is the critical angle 𝜃 sub 𝑐. 𝑛 sub one times the sin of 𝜃 sub 𝑐 equals 𝑛 sub two times the sine of the angle of refraction, which we see is 90 degrees. This is always the case when our angle of incidence is the critical angle. We can recall that the sin of 90 degrees is exactly equal to one. The right-hand side of our expression is 𝑛 sub two times one or just 𝑛 sub two.

Let’s recall now that we’re told what the critical angle is. It’s 30 degrees. Substituting this in, our equation becomes 𝑛 one times the sin of 30 degrees equals 𝑛 two. The sin of 30 degrees is exactly one-half, so 𝑛 one times one-half equals 𝑛 two. Or if we multiply both sides of this equation by two, we find that 𝑛 one equals two times 𝑛 two. 𝑛 two, we know, is 1.1, and this means that 𝑛 one is twice that, 2.2.

We haven’t quite gotten to our final answer though, because remember we want to solve for the difference between the two indices of refraction, 𝑛 one and 𝑛 two. Knowing that 𝑛 one is 2.2 and 𝑛 two is 1.1, the difference between these, 𝑛 one minus 𝑛 two, equals 1.1. This is the difference between the refractive index of the medium from which the light is reflected and the refractive index of the medium in which the light propagates before it is reflected.