Light inside glass of refractive index 1.6 meets a glass–air boundary. Find the critical angle.

Solution & explanation

1. Given: n_glass = 1.6, n_air ≈ 1.

2. At the critical angle the refracted angle is 90°.

   $$1.6\times \sin \left(C\right)=1\times \sin \left(\frac{\pi }{2}\right)$$

3. Since sin90° = 1, divide by 1.6.

   $$\sin \left(C\right)=\frac{1}{1.6}=0.625$$

4. Take inverse sine in degrees.

   $$C=\frac{180\times \arcsin \left(0.625\right)}{\pi }$$

5. The value is about 38.682°.

6. State the result with units: C ≈ 38.68°; TIR requires incidence greater than C.

C ≈ 38.68°