Light rays, initially in air, are normally incident on the vertical surface of a glass prism, as shown in the accompanying diagram. The prism has a refractive index of 1.50. The slanted face of the prism is aligned at an angle 𝜙 above the horizontal. What is the largest value of 𝜙 such that an incident ray is totally reflected at the slanted face?
The prism is immersed in water. Find the largest value for 𝜙 such that the ray is totally reflected at the slanted face. Use a value of 1.333 for the refractive index of water.
In this problem, we’ll assume that the refractive index of air, we’ll call it 𝑛 sub 𝑎, is equal to 1.00. We’re told in part one of the problem statement that the refractive index of the prism, which we’ll call 𝑛 sub 𝑝, equals 1.50. And finally, we read that the refractive index of water is 1.333, a value we’ll call 𝑛 sub 𝑤. We want to solve for two pieces of information in this problem. First, we want to solve for the maximum angle 𝜙 that allows total internal reflection of light within the prism, when the prism is surrounded by air. We’ll call that value 𝜙 sub 𝑎. In part two, we once more want to solve for 𝜙, the maximum angle allowing total internal reflection, this time with the prism surrounded by water. We’ll call this value 𝜙 sub 𝑤.
We can begin our solution by recalling a few facts about total internal reflection. We know that when light strikes an interface between two different materials, if the incident angle is an angle called the critical angle, 𝜃 sub 𝑐, that means the ray will be refracted at an angle of 90 degrees to the line normal to that interface. In other words, the ray won’t leave the first material 𝑛 sub one. When this happens, Snell’s law, which says that the index of refraction of the first material times the sine of the incident angle equals the index of refraction of the second material times the sine of the refracted angle, is simplified because 𝜃 sub 𝑟 is 90 degrees, which means that 𝑛 one times the sine of the critical angle is equal to 𝑛 two.
Before applying this relationship to our particular problem, let’s draw a side-on view of the prism. From the side, we have a ray of light reaching the edge of the prism at a 90-degree angle to the interface. The ray is totally internally reflected when it tries to leave the prism, moving along the prism air interface. If we draw a dotted line perpendicular to the slanted part of the prism, then the incident angle of the ray to that interface is equal to 90 degrees minus 𝜙 sub 𝑎.
So when we apply Snell’s law for the critical angle to our situation, we write that 𝑛 sub 𝑝, the index of refraction of the prism, times the sine of 90 degrees minus 𝜙 sub 𝑎 is equal to 𝑛 sub 𝑎, the index of refraction of air. If we divide both sides of this equation by 𝑛 sub 𝑝 and take the arc sine of both sides, we find that 90 degrees minus 𝜙 sub 𝑎 equals the arc sine of 𝑛 sub 𝑎 over 𝑛 sub 𝑝. Or equivalently, 𝜙 sub 𝑎 is 90 degrees minus the arc sine of 𝑛 sub 𝑎 over 𝑛 sub 𝑝.
We can now plug-in for the given values of 𝑛 sub 𝑎 and 𝑛 sub 𝑝. 𝑛 sub 𝑎 is 1.00 and 𝑛 sub 𝑝 is 1.50. When we calculate this angle, we find that, to three significant figures, 𝜙 sub 𝑎 is 48.2 degrees. That’s the largest angle of the prism wedge that allows total internal reflection.
Now that we’ve solved for 𝜙 sub 𝑎, we move on to solve for 𝜙 sub 𝑤. This is that same largest angle that still allows total internal reflection, but this time we imagine the prism is surrounded by water. This only changes one thing in our equation. Instead of 𝑛 sub 𝑎, we use 𝑛 sub 𝑤 since the prism is now surrounded by a different material. We move ahead as before plugging in for our indexes of refraction, 𝑛 sub 𝑤 and 𝑛 sub 𝑝.
When we calculate this value, we find it’s equal to 27.3 degrees. That’s the maximum angle of the prism wedge that still allows total internal reflection when the prism is immersed in water.