A pencil flashlight submerged in water sends a light beam toward the surface at an angle of incidence of 30 degrees from the air-water boundary. Find the angle of refraction in air. Use a value of 1.33 for the refractive index of water.
We’re told in the statement that the angle of incidence of the light beam on the air-water boundary is 30 degrees; we’ll call that 𝜃 sub 𝑖. We’re also told that the index of refraction of water is 1.33, which we’ll call 𝑛 sub 𝑤.
We want to solve for the angle of refraction in air, which we’ll call 𝜃 sub 𝑟. To get started on our solution, let’s draw a diagram of this situation. In this scenario, an underwater flashlight shines light up to an air-water interface.
The ray of light is refracted or bent at the interface, and we want to find out how much; that is, what 𝜃 sub 𝑟? In this example, we’ll assume that the index of refraction of air, 𝑛 sub 𝑎, is exactly 1.00. To solve for 𝜃 sub 𝑟, let’s recall Snell’s law which says that if a ray of light travels from one material with index of refraction 𝑛 one into another material with index of refraction 𝑛 two, then 𝑛 one times the sine of the incident angle equals 𝑛 two times the sine of the refracted angle, where those angles are measured with respect to a line perpendicular to the interface.
Applying this relationship to our scenario, we see that the index of refraction of water times the sine of the incident angle 𝜃 sub 𝑖 equals the index of refraction of air times the sine of the refracted angle 𝜃 sub 𝑟. If we divide both sides of the equation by 𝑛 sub 𝑎 and take the inverse sine of both sides, we find that 𝜃 sub 𝑟 equals the arc sine of 𝑛 sub 𝑤 over 𝑛 sub 𝑎 times the sine of the incident angle 𝜃 sub 𝑖.
Plugging in for 𝑛 sub 𝑤, 𝑛 sub 𝑎, and 𝜃 sub 𝑖, when we enter this value on our calculator, we find that 𝜃 sub 𝑟 to two significant figures is 42 degrees. That’s the refracted angle of light crossing this air-water interface.