Answer:
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
