### Video Transcript

At what minimum angle will you get total internal reflection of light travelling in water and reflected in ice? Use a value of 1.309 for the refractive index of ice and use a value of 1.333 for the refractive index of water.

Weβll call the refractive index of ice, 1.309, π sub π and weβll call the refractive index of water, 1.333, π sub π€. We want to solve for the smallest angle possible where the light will be totally internally reflected when moving in water and reflected off ice. Weβll call that angle π sub π‘.

To begin working toward our solution, letβs draw a diagram of the situation. Say that we have a flat interface between water and ice and a ray of light comes in through the water and meets that interface. If we draw a line thatβs perpendicular to the interface, then the angle of the incoming ray of light, which we can call π sub π, is measured relative to that line. When the ray crosses the interface, it will be bent away from the normal line because the index of refraction of ice into which the ray is moving is less than the index of refraction of water.

Now if we were to increase the incident angle π sub π, then the reflected ray would also go out at a greater reflected angle. If we keep going increasing π sub π, eventually, the reflective ray will actually move parallel to the interface between the water and the ice. Meaning, none of the light actually left the water and went into the ice. It was all contained. This incident angle is known as the critical angle, often abbreviated π sub π. Itβs the smallest incident angle at which light is totally internally reflected.

In our case, weβve named this angle π sub π‘, but itβs the same thing. To solve for π sub π‘, weβll use a relationship called Snellβs law. Snellβs law says that if we consider the way light rays move across an interface between two different materials, then the index of refraction of the first material through which light moves, π sub π, times the sine of the angle at which the ray is incident on the interface, π sub π, is equal to the index of refraction of the second material times the sine of the reflected angle, π sub π.

So in our case, the index of refraction of water, π sub π€, times the sine of π sub π‘, the critical angle, is equal to the index refraction of ice, π sub π, times the sine of the reflected angle, π sub π. But what is that reflected angle? At the critical angle π sub π‘, π sub π is equal to 90 degrees. And the sine of 90 degrees is equal to one. So if we rearrange to solve for π sub π‘ β by first dividing both sides by π sub π€ which cancels that term on the left, and then if we take the arc sine or inverse sine of both sides which leaves us on the left side of our equation simply with π sub π‘ β we find that this critical angle π sub π‘ equals the inverse sine of π sub π divided by π sub π€. We know both π sub π and π sub π€. So we can plug in those values now.

When we enter this value on our calculator, we find that π sub π‘ is equal to 79.11 degrees. Thatβs the smallest possible incident angle for which light rays will be totally internally reflected back into water against an interface of ice.