### Video Transcript

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.

Weβre told in the statement that the angle of incidence of the light beam on the air-water boundary is 30 degrees; weβll call that π sub π. Weβre also told that the index of refraction of water is 1.33, which weβll call π sub π€.

We want to solve for the angle of refraction in air, which weβll call π sub π. To get started on our solution, letβs 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 π sub π? In this example, weβll assume that the index of refraction of air, π sub π, is exactly 1.00. To solve for π sub π, letβs recall Snellβs law which says that if a ray of light travels from one material with index of refraction π one into another material with index of refraction π two, then π one times the sine of the incident angle equals π 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 π sub π equals the index of refraction of air times the sine of the refracted angle π sub π. If we divide both sides of the equation by π sub π and take the inverse sine of both sides, we find that π sub π equals the arc sine of π sub π€ over π sub π times the sine of the incident angle π sub π.

Plugging in for π sub π€, π sub π, and π sub π, when we enter this value on our calculator, we find that π sub π to two significant figures is 42 degrees. Thatβs the refracted angle of light crossing this air-water interface.