A polystyrene pipe of index 1.49 is surrounded by air of index 1.00. What is the critical angle for light travelling inside the pipe?

1. The critical angle is theta_c = inverse sine of (n2 divided by n1), valid when n1 is greater than n2. 2. Here the light starts inside the plastic, so n1 = 1.49, and the surrounding air gives n2 = 1.00. 3. Substituting, theta_c = inverse sine of (1.00 divided by 1.49) = inverse sine of 0.671. 4. This gives theta_c = 42.2 degrees. 5. Any ray inside the plastic striking the wall at more than 42.2 degrees is totally reflected, making the inner surface a perfect mirror with no silvering. 6. The 48.6 degrees option is the water to air value and would need n1 = 1.333. _Source: OpenStax College Physics (CC BY 4.0), Ch 25 "Geometric Optics", section 25.4 Total Internal Reflection_