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 is air with a refractive index of 1.00? Answer to the nearest degree.
In this situation, we have an interface between water and air. And we’re told that we have a ray of light coming through the water and reaching this boundary. We want to solve for what’s called the critical angle for this ray of light. Here is how the critical angle is defined. If our ray of light approaches this interface at the angle that we will call 𝜃 sub c to represent the critical angle, then at that interface, the ray of light has an angle of refraction of 90 degrees.
Snell’s law helps us understand how this works mathematically. It tells us that if we have a ray of light passing through some material with index of refraction 𝑛 sub i, and incident on an interface with another material at angle of incidence 𝜃 sub i, then the product of 𝑛 sub i and the sin of 𝜃 sub i is equal to the product of the index of refraction of the material the light is refracted into and the sine of the angle of refraction 𝜃 sub r.
Snell’s law is generally true. And when the angle of incidence 𝜃 sub i is equal to the critical angle, we can think of that as a special case of this law. It’s a special case because the angle of refraction 𝜃 sub r is 90 degrees. And since the sin of 90 degrees equals one, Snell’s law simplifies to an expression like this when the angle of incidence is the critical angle. In our case, since the ray of light is originally traveling through water, the index of refraction of water is 𝑛 sub i and the index of refraction of air is 𝑛 sub r. We’re told that water’s refractive index is 1.33 and that of air is 1.00. Substituting in these values, our equation becomes 1.33 times the sin of 𝜃 sub c equals 1.00.
Our goal here is to solve for that critical angle 𝜃 sub c. To do that, let’s start by dividing both sides by 1.33, canceling that factor on the left. That gives us this expression. And since we wanna solve for 𝜃 sub c itself, rather than the sine of this angle, let’s take the inverse sine or the arc sine of both sides. The inverse sine of the sine of an angle equals the angle itself. 𝜃 sub c then equals the inverse sin of 1.00 divided by 1.33. Entering this expression on our calculator and rounding the result to the nearest degree, we get 49 degrees. This is the critical angle for light that travels through water and is incident on a water–air interface.
