Total Internal Reflection — Question 1

Question 1

An angle measured from the surface can reverse a total-reflection diagnosis

Light in glass of index 1.50 reaches a glass-air boundary. One ray makes 30 degrees with the surface and a second makes 50 degrees with the surface. Air has index 1.00.

1. Find the critical angle from the normal and classify each ray.

2. For the ray that can refract, calculate its refraction angle from the normal.

Question 1 - Solution

An angle measured from the surface can reverse a total-reflection diagnosis

(a) Method and calculation

$$\theta_c=\sin^{-1}(1/1.50)=\boxed{41.8103^\circ}.$$$$\theta_{i,A}=90^\circ-30^\circ=60^\circ>\theta_c,\quad\theta_{i,B}=40^\circ<\theta_c.$$

(b) Interpretation and further calculation

$$\boxed{A:\ \text{total internal reflection};\quad B:\ \text{partial reflection and refraction}}.$$$$\sin\theta_{t,B}=1.50\sin40^\circ\Rightarrow\boxed{\theta_{t,B}=74.6186^\circ}.$$

Physical check

The critical-angle criterion always uses the angle from the normal; reflection at the boundary can still occur below that threshold.