Find the critical angle for water-to-air refraction, with n_water = 1.33 and n_air = 1.

Solution & explanation

1. At critical incidence, the refracted angle is 90°.

2. Write Snell’s law for the boundary.

   $$1.33\times \sin \left(C\right)=1\times \sin \left(\frac{\pi }{2}\right)$$

3. Use sin90° = 1 and divide by 1.33.

   $$\sin \left(C\right)=\frac{1}{1.33}$$

4. Evaluate the ratio.

   $$\sin \left(C\right)=0.7518796992$$

5. Take inverse sine in degrees.

   $$C=\frac{180\times \arcsin \left(\frac{1}{1.33}\right)}{\pi }$$

6. TIR occurs for incidence greater than this value, toward air.

C ≈ 48.75°