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°

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