Question

Total Internal Reflection: A ray of light in glass strikes a water-glass interface at an angle of incidence equal to one-half the critical angle for that interface. The index of refraction for water is 1.33, and for the glass it is 1.43. What angle does the refracted ray in the water make with the normal

Answer 1

Answer:

θ₄ = 37.2º

Explanation:

For this exercise it must be solved in two parts, the first part we look for the critical angle, for this we use the law of refraction with the angle in the middle of transmission of tea = 90º

        n₁ sin θ₁= n₂ sin  90

        θ₁ = sin⁻¹ $\frac{n_2}{n_1}$

        θ₁ = sin⁻¹ (1.33 / 1.43)

        θ₁ = 68.4º

They indicate that the angle of incidence is half of the critical angle

        θ₃ = 68.4 / 2 = 34.2º

Let's use the law of refraction again

         n₁ sin θ₃ = n₂ sin θ₄

         sin θ₄ = $\frac{n_1}{n_2}$   sin θ₃

         sin θ₄ = $\frac{1.43}{1.33}$  sin 34.2

         θ₄ = sin⁻¹ 0.604345

        θ₄ = 37.2º