A ray of light strikes a side of an equilateral glass prism at 45º as shown in the diagram. Find the angle that the light leaves the prism. Assume that n= 1.58.

Question

A ray of light strikes a side of an equilateral glass prism at 45º as shown in the diagram. Find the angle that the light leaves the prism. Assume that n= 1.58.

Answer 1

Although I cannot see a diagram in your question, I think the text gives enough information.

By Snell's law,sinθ₁/sinθ₂ = n₂/n₁

Solving for θ₂: θ_{2 =} sin^-1(sinθ₁ • n_1/n₂)

For the first refraction, θ_{2 =} sin^-1(sin45° • 1_/1.58) = 26.6°

We can find the angle that the light ray hits the other side of the prism from the geometry of the prism's equilateral triangle. If you draw this out, you will see that this angle is 63.4°.

Using Snell's law again, θ_{3 =} sin^-1(sin63.4° • 1.58_/1) = sin^-1(1.41)

This has no solution, meaning that the light will not emerge from the prism, but rather reflect off the second surface.

Now I wish I could see the diagram. Perhaps the correct interpretation is that the 45° angle is relative to the horizontal base of the prism, not relative to the side of the triangle. With that assumption, the light ray is incident at 75° to the glass.

θ_{2 =} sin^-1(sin75° • 1_/1.58) = 37.7°

This result means that the light ray will strike the opposite face at an angle of 22.3°

Now, θ_{3 =} sin^-1(sin22.3° • 1.58_/1) = sin^-1(.60) = 36.9°

This is the angle the light ray emerges from the glass relative to the prism face, which means it will make an angle of 6.9° relative to the horizontal base.