A light ray passing from air into water at an angle of 30º from the normal in air would

When light passes from one medium to another, such as from air to water, it bends or refracts. This phenomenon is described by Snell's Law, which states: n₁ * sin(θ₁) = n₂ * sin(θ₂), where:

• n₁ is the refractive index of the first medium (air),
• θ₁ is the angle of incidence (the angle made by the light ray with the normal in the first medium),
• n₂ is the refractive index of the second medium (water),
• θ₂ is the angle of refraction (the angle made by the light ray with the normal in the second medium).

The refractive index of air is approximately 1, and the refractive index of water is approximately 1.33. Given the angle of incidence in air is 30º:

Using Snell's Law:

1 * sin(30º) = 1.33 * sin(θ₂)

You will find:

sin(θ₂) = sin(30º) / 1.33

sin(θ₂) ≈ 0.5 / 1.33

sin(θ₂) ≈ 0.375

Now, solve for θ₂ by taking the inverse sine (arcsin):

θ₂ ≈ arcsin(0.375)

θ₂ ≈ 22.09º

Thus, when a light ray passes from air into water at an angle of 30º from the normal in air, it will make an angle less than 30º from the normal in water, approximately 22.09º. This is because the light ray bends toward the normal as it enters a denser medium (water).