An oil layer that is 5.0 cm thick is spread smoothly and evenly over the surface of water on a windless day.What is the angle of refraction in the water for a ray of light that has an angle of incidence of 45° as it enters the oil from the air above? (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°
E) 52°

Question

Quizplus · Accepted Answer

To find the angle of refraction in the water, we 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), $\theta_1 = 45°$, and $n_2 = 1.15$ (for oil). Solving for $\theta_2$ gives the angle of refraction in the oil. Then, we use Snell's Law again for the oil-water interface, with $n_2 = 1.15$ (for oil), and $n_3 = 1.33$ (for water), and $\theta_2$ from the first calculation to find $\theta_3$, the angle of refraction in the water.First calculation for air to oil:$$1 \sin(45°) = 1.15 \sin(\theta_2)$$$$\sin(\theta_2) = \frac{\sin(45°)}{1.15}$$$$\theta_2 = \sin^{-1}\left(\frac{\sin(45°)}{1.15}\right)$$Then, for oil to water:$$1.15 \sin(\theta_2) = 1.33 \sin(\theta_3)$$Using $\theta_2$ from the first step, solve for $\theta_3$.Without doing the exact math here, the process involves using Snell's Law twice, first to find the angle of refraction in the oil, and then to find the angle of refraction in the water. Given the indices of refraction and the initial angle of incidence, the correct answer after performing the calculations is found to be 32°, which corresponds to choice B. This process illustrates the application of Snell's Law across two different interfaces (air to oil, then oil to water) to find the final angle of refraction.

An Oil Layer That Is 5