Answer:
θ₂ = 40.5º
Explanation:
For this exercise we must use the law of refraction
n₁ sin θ₁ = n₂ sin θ₂
where index 1 is for the incident ray and index 2 is for the refracted ray
in this case the incident ray has an angle of θ₁ = 60º and the refractive index of the water is
n₂ = 1,333
sin θ₂ = [tex]\frac{n_1}{n_2} \ sin \theta_1[/tex]
let's calculate
sin θ₂ = 1 / 1.3333 sin 60
sin θ₂ = 0.64968
θ₂ = sin⁻¹ (0.64968)
θ₂ = 40.5º
