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

θ₄ = 37.2º

Step-by-step 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
`(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 =
`(n_1)/(n_2)` sin θ₃

sin θ₄ =
`(1.43)/(1.33)` sin 34.2

θ₄ = sin⁻¹ 0.604345

θ₄ = 37.2º