a scuba diver in a pool looks at his instructor. the angle between the ray in the water and the perpendicular to the water is 25. find the height of the instructors head

   y= y'  0.67

Explanation:

This is an exercise in refraction of light,

         n₁ sin θ₁ = n₂ sin θ₂

where subscript 1 is used for the incident medium and subscript 2 for the refracted medium

          sin θ₁ = n2 /n1   sin θ₂

the incident medium is air with refractive index n1 = 1 and the medium where the ray is refracted is water with n = 1.33

        let's calculate

        sin θ₁ = 1.33 / 1 sin  25

        θ₁ = sin⁻¹ (0.56208)

        θ₁ = 34.2º

when the ray is refracted we can assume that the adjacent leg (water surface) is the same for the two media

let's use the trigonometry relationship

          tan θ₁ = x / y

          tan θ₂ = x / y '

         tan θ₁ = y’ tan θ₂

         y = y ’ tanθ₂ / tan θ₁

to finish exercise you must know the depth of the object

         y =y'  tan 25/ tan 34.2

         y= y'  0.67