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 40.0° with the normal, what is the incident angle at the oil-to-water surface?

Applying Snell's law of refraction;

n1*Sin Ф1 = n2*Sin Ф2

Where, n1 = Air index of refraction = 1; n2 = Oil index of refraction = 1.466; Ф1 = Air-oil incident angle; Ф2 = Air-oil refraction angle

Substituting;

1*Sin 40 = 1.466*Sin Ф2

Sin Ф2 = (1*Sin 40) / 1.466

Ф2 = Sin ^-1[ (Sin 40) / 1.466] = 26°

Angle of refraction for air-oil is equal to the angle of incidence for oil-water boundary.

Therefore, angle of incidence for oil-to-water = 26°