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 wa

32.1

Explanation:

NOTE: You did not state the angle of incidence, and thus, I will be using 45° as my angle of incidence, all you need to do is replace it with your own value if it's different.

To solve this question, we are going to be using Snell's Law.

Snell's law describes the relationship between the angles of incidence and refraction, when referring to light or other waves passing through a boundary between two different isotropic media, such as water, glass, or air.

Snell's law is mathematically given as

sin(A1)/sin(A2) = n2/n1, where

n1 = incidence index

n2 = refracted index

A1 = incidence angle

A2 = refracted angle

The refraction index of oil is 1.15, and that of water is 1.33, so

if we take oil first,

sin A2 = (n1.sinA1)/n2

sin A2 = (1 * sin 45)/1.15

sin A2 = 0.7071/1.15

sin A2 = 0.6149

A2 = sin^-1 0.6149

A2 = 37.9°

Then

sin A3 = (1.15 * sin 37.9) / 1.33

sin A3 = (0.6149 * 1.15) / 1.33

sin A3 = 0.7071 / 1.33

sin A3 = 0.5317

A3 = sin^-1 0.5317

A3 = 32.1

number of electrons for neutral cl