# Video: Refraction Angle of a Light Ray

The path of a light beam in air goes from an angle of incidence of 35.0° to an angle of refraction of 22.0° when it enters a rectangular block of plastic. What is the index of refraction of the plastic?

02:20

### Video Transcript

The path of a light beam in air goes from an angle of incidence of 35.0 degrees to an angle of refraction of 22.0 degrees when it enters a rectangular block of plastic. What is the index of refraction of the plastic?

We can highlight the angle of incidence, 35.0 degrees, and refer to that as 𝜃 sub 𝑖. Likewise the angle of refraction, 22.0 degrees, we can refer to as 𝜃 sub 𝑟. We want to know the index of refraction of the plastic, which we’ll call 𝑛 sub 𝑝. Let’s begin our solution by drawing a diagram of the situation.

When a ray of light crosses the air and plastic interface, the ray is refracted based on the index of refraction of the plastic. We’ll assume that the index of refraction of air, 𝑛 sub 𝑎, is equal to 1.00. To solve for the index of refraction of the plastic block, we can recall Snell’s law which says that when light crosses an interface moving from a material with index of refraction 𝑛 sub one to one with index of refraction 𝑛 sub two, then 𝑛 one times the sine of the incident angle equals 𝑛 two times the sine of the refracted angle.

Applying Snell’s law to our particular situation, we have 𝑛 sub 𝑎, the index of refraction of air, times the sine of the incident angle 35.0 degrees equals 𝑛 sub 𝑝, the index of refraction of the plastic, times the sine of the refracted angle 𝜃 sub 𝑟.

Dividing both sides by the sine of 𝜃 sub 𝑟, 𝑛 sub 𝑝 equals 𝑛 sub 𝑎 times the sine of 𝜃 sub 𝑖 divided by the sine of 𝜃 sub 𝑟. We can now plug in for 𝑛 sub 𝑎, 𝜃 sub 𝑖, and 𝜃 sub 𝑟. When we enter these values on our calculator, we find that, to three significant figures, 𝑛 sub 𝑝 is 1.53. That’s the index of refraction of the block of plastic.