An oil film floats on a water surface.The indices of refraction for water and oil,respectively,are 1.333 and 1.466.If a ray of light is incident on the air-to-oil surface at an angle of 44.3° with the normal,what is the angle of the refracted ray in the water?

Question

Quizplus · Accepted Answer

Using Snell's law, we have

n1 sin θ1 = n2 sin θ2,

where n1 and n2 are the indices of refraction of the media, and θ1 and θ2 are the angles of incidence and refraction, respectively, measured from the normal.

For the air-to-oil interface, we have

n1 = 1 (air)
n2 = 1.466 (oil)
θ1 = 44.3°

Solving for θ2, we get

sin θ2 = (n1/n2) sin θ1 = (1/1.466) sin 44.3° = 0.505

Since θ2 is the angle measured from the normal within the water, we have

sin θ2 = sin(90° - θ2water)

where θ2water is the angle of refraction within the water.

Therefore,

sin θ2water = sin(90° - θ2) = cos θ2 = √(1 - sin^2 θ2) = 0.862

Taking the inverse sine, we get

θ2water = 31.6°.

Therefore, the answer is C.

An Oil Film Floats on a Water Surface