Question

A ray of light is incident on an air/water interface.

The ray makes an angle of θ1 = 32 degrees with respect to the normal of the surface. The index of the air is n₁ = 1 while water is n₂ = 1.33.
Choose an expression for the angle (relative to the normal to the surface) for the ray in the water, θ₂.

a) θ₂ = sin (θ₁).n₁/n₂
b) θ₂ = a sin (n₁/n₂)
c) θ₂ = a sin (sin(θ₁).n₂/n₁)
d) θ₂ = a sin (sin(θ₁).n₁/n₂)

Answer 1

The correct expression for the angle θ2 of a light ray in water after passing from air, based on Snell's Law, is θ2 = sin⁻¹(sin(θ1) * n₁ / n₂).

Step-by-step explanation:

The question involves Snell's Law, which describes the relationship between the angles of incidence and refraction when a light ray passes from one medium to another with different refractive indices.

Given the angle of incidence θ1 = 32 degrees, the index of refraction for air n₁ = 1, and water n₂ = 1.33, Snell's Law is written as n₁ * sin(θ1) = n₂ * sin(θ2).

To find the angle θ2 in the water relative to the normal to the surface, we rearrange the equation to solve for θ2, resulting in θ2 = sin⁻¹(sin(θ1) * n₁ / n₂). Therefore, the correct expression is d) θ2 = sin⁻¹(sin(θ1) * n₁ / n₂).