47.6k views
3 votes
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

User Rcourtna
by
8.4k points

1 Answer

2 votes

Answer:

a) θ = 65º , b) the light is refracted

Step-by-step 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

User Augustine Kim
by
7.8k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.