An oil layer that is 5.0 cm thick is spread smoothly and evenly over the surface of water on a windless day.A light ray from the air above enters the oil at 45° with the normal and then goes into the water.What is the angle of refraction of this ray in the water? The index of refraction for the oil is 1.15,and for water it is 1.33.
A)27°
B)32°
C)36°
D)39°

Question

Quizplus · Accepted Answer

First, use Snell's Law for the air-oil interface: $n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$, where $n_1 = 1$ (for air), $\theta_1 = 45^\circ$, and $n_2 = 1.15$ (for oil). Solving for $\theta_2$ gives the angle of refraction in the oil. Then, use Snell's Law again for the oil-water interface with $n_2 = 1.15$ (for oil), $n_3 = 1.33$ (for water), and the previously found $\theta_2$ to find $\theta_3$, the angle of refraction in the water. The correct calculation leads to an angle close to 32°.

An Oil Layer That Is 5