Question

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

Answer 1

y= y' 0.67

Step-by-step 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