# Question Video: Comparison Between Reflection and Refraction Angles

A light beam in air is incident on the surface of a pond with a refractive index of 1.333, making an angle of 20° with respect to the surface of the pond. What is the angle of reflection? What is the angle of refraction?

03:52

### Video Transcript

A light beam in air is incident on the surface of a pond with a refractive index of 1.333, making an angle of 20 degrees with respect to the surface of the pond. What is the angle of reflection? What is the angle of refraction?

In this problem statement, we’re told that the pond has a refractive index of 1.333, which we’ll call 𝑛 sub 𝑝. We’re also told that it makes an angle of 20 degrees with respect to the surface of the pond. We’ll call that angle 𝜃 sub 𝑝. When the light reaches the interface between the air and the pond, we want to know what is the angle of reflection, which we’ll call 𝜃 sub 𝑒. And we also want to know the angle of refraction, which we’ll call 𝜃 sub 𝑎.

Let’s start out by drawing a diagram of the situation. In this example, we have our incident ray coming into the air–pond interface. Part of the incident ray is reflected and bounces back into the air. And part is refracted into the pond. We want to solve for both the reflected angle 𝜃 sub 𝑒 and the refracted angle 𝜃 sub 𝑎.

To figure out 𝜃 sub 𝑒, let’s recall the law of reflection. This law says that when it comes to rays of light, the angle of incidence is equal to the angle of reflection. We can write that as an equation as it relates to our scenario, by writing that 90 degrees minus 𝜃 sub 𝑝 is equal to 𝜃 sub 𝑒.

Since 𝜃 sub 𝑝 is 20 degrees, that means that 𝜃 sub 𝑒 equals 90 minus 20 degrees or 70 degrees. That’s the angle at which light is reflected from the surface of the pond.

To solve for the angle of refraction 𝜃 sub 𝑎, we want to recall Snell’s law. This law says that, for incident and refracted rays, the index of refraction of the material the incident ray travels through times the sine of the incident angle equals the index of refraction that the refractive ray travels through times the sine of the refracted angle.

Applying Snell’s law to our situation, 𝑛 sub air, the index of refraction of the air, times the sin of 90 degrees minus 𝜃 sub 𝑝 is equal to the index of refraction of the pond times the sin of 𝜃 sub 𝑎, the angle of refraction. We’ll assume in our scenario that the index of refraction of air is exactly one.

We now rearrange this equation algebraically to solve for 𝜃 sub 𝑎. First, we divide both sides by 𝑛 sub 𝑝, canceling that out on the right-hand side. And then we take the inverse sine or arcsine of both sides of the equation, which leaves on the right-hand side 𝜃 sub 𝑎. So 𝜃 sub 𝑎 equals the inverse sine of the sin of 90 degrees minus 𝜃 sub 𝑝 all divided by 𝑛 sub 𝑝, the index of refraction of the pond.

Plugging in for 𝜃 sub 𝑝 and 𝑛 sub 𝑝, when we plug these values into our calculator, we find that 𝜃 sub 𝑎 is equal to 45 degrees. That’s the angle with respect to the normal made by the refracted ray.