You have a beaker with a layer of olive oil floating on top of water. A ray of light travels through the oil and is incident on the water with an angle of 50.2°. Using the index of refraction of the oil as 1.470 and the index of refraction of water as 1.333, determine the critical angle in oil for the oil-water interface.

? c = °

Determine if the ray of light refracts into the water or reflects off the oil-water interface back into the oil.

a. refracts into the water

b. reflects back into the oil

a) θ = 65º, b) the light is refracted

Explanation:

When a ray of light passes from a material with a higher index to one with a lower index, the ray separates from the normal one, so there is an angle for which the ray is refracted at 90º, the refractive equation is

n₁ sin θ₁ = n₂

where θ₁ is the incident angle, n₂ and n₁ are the indexes of incident and refracted parts

let's calculate

sin θ = n₂ / n₁

sint θ = 1.3333 / 1.470

sin θ = 0.907

θ = sin⁻¹ 0.907

θ = 65º

For the incident angle of 50.2º it is less than the critical angle, so the light is refracted according to the refraction equation