. A flat piece of glass covers the top of a vertical cylinder that is completely filled with water. If a ray of light traveling in the glass is incident on the interface with the water at an angle of ua = 36.2°, the ray refracted into the water makes an angle of 49.8o with the normal to the interface. What is the smallest value of the incident angle ua for which none of the ray refracts into the water?

        θ = 50.6

Explanation:

This is a refraction exercise, which is described by the equations

          n₁ sin θ₁ = n₂ sin θ₂

where in index 1 we will use it for the incident ray and subscript 2 for the refracted ray

The refractive index of water is tabulated and is n = 1.33, let's calculate the refractive index of the glass

          n₁ = n₂ sin θ₂ /sin θ₁

          n₁ = 1.33 sin 49.8 /sin 36.2

          n₁ = 1.72

With the values of the two refractive indices we can calculate the angle for the total internal reflection

           sin θ = n₂ / n₁

             θ = sin⁻¹ (n₂ / n₁)

            θ = sin⁻¹ (1.33 / 1.72)

            θ = 50.6