Refraction and Lenses Legacy Problem #17 Guided Solution

In this three-step problem, we have a ray of light that is approaching a triangular prism that is made of a material with an index of refraction value of 2.41. In the diagram, it is shown that the light ray is approaching the boundary at an angle of incidence of 30 degrees. That's the angle between the normal line and the light ray. In Part A, we're asked to calculate the angle of refraction as it enters the material. This is a Snell's Law problem. We substitute for the index of refraction of the incident material, 1.00 for air, and the sine of 30 degrees is the angle of incidence. This is equal to 2.41 times the sine of the angle of refraction. We can divide each side of the equation by 2.41 and evaluate the left side. It comes out to be 0.2075. Then we can take the second sine, or inverse sine, of 0.2075, and this gives us the angle of refraction. It comes out to be 11.9741. We can round it to three significant digits, and it rounds to 12.00, or 12.0. Now in Part B of this problem, we're asked to calculate the angle of incidence at the opposite side of the triangle. This will involve a good deal of geometry. You'll notice in the graphic that's provided here that a triangle is shown in pink. This is the triangle that is formed by the refracted ray, the red ray on the diagram, in the top of this equal angular triangle, and the location where the refracted ray reaches the opposite face of the prism. This triangle has as an angle at the top, 60 degrees, and has the angle lower left, 78 degrees. The 78 degrees was determined by subtracting the angle of incidence from 90 degrees. After all, this angle at the lower left is simply the complementary angle of 12 degrees, since 12 degrees and the 78 degrees must add up to 90. What we wish to calculate is the angle theta B that you're shown in the triangle. That's the angle between the normal line at this opposite face and the refracted ray, or the incident ray. In order to calculate theta B, we can find theta A, which is shown in the pink triangle to the right side of the large prism. This theta A, along with 60 degrees and 78 degrees, must add up to 180 degrees. And so I go 60 degrees plus 78 degrees plus theta A equal 180 degrees, and I solve for theta A. It comes out to be 42 degrees, roughly, 41.9741 degrees exactly. Now that's theta A. What I really want to find in part B of this problem is theta B. Theta B and theta A are complements of one another. They should add up to 90 degrees. This 90 minus this 41.9741 should give me the value for theta A, for theta B. It comes out to be 48.0259, and again, I can round that to three significant digits, it comes out to be 48 degrees. Now in part C of this problem, we're asked to answer the question, will the light ray at this opposite face undergo TIR, or will it refract out of the prism? The answer to the question can be found if we can do a Snell's Law calculation in order to determine the angle, the critical angle at this boundary. So in order to find the critical angle, I apply the critical angle definition, which is it's the angle of incidence which causes the angle of refraction to be 90 degrees. So I say 2.41 times the sine of the critical angle equal 1.00 for air at this boundary times the sine of 90 degrees, and I solve for the critical angle. It comes out to be 24.5 degrees. Total internal reflection will occur whenever the angle of incidence is greater than the critical angle. Here we've already calculated the angle of incidence to be about 48 degrees. That's much greater than the critical angle. So light will not refract out of the prism, but will instead undergo total internal reflection.