Answer:
The answer to the question is
The distance d, which locates the point where the light strikes the bottom is 29.345 m from the spotlight.
Explanation:
To solve the question we note that Snell's law states that
The product of the incident index and the sine of the angle of incident is equal to the product of the refractive index and the sine of the angle of refraction
n₁sinθ₁ = n₂sinθ₂
y = 2.2 m and strikes at x = 8.5 m, therefore tanθ₁ = 2.2/8.5 = 0.259 and
θ₁ = 14.511 °
n₁ = 1.0003 = refractive index of air
n₂ = 1.33 = refractive index of water
Therefore sinθ₂ = [tex]\frac{n_1sin\theta_1}{n_2}[/tex] = [tex]\frac{1.003*0.251}{1.33}[/tex] = 0.1885 and θ₂ = 10.86 °
Since the water depth is 4.0 m we have tanθ₂ = [tex]\frac{4}{x_2}[/tex] or x₂ = [tex]\frac{4}{tan\theta_2 }[/tex] =[tex]\frac{4}{tan(10.86)}[/tex] = 20.845 m
d = x₂ + 8.5 = 20.845 m + 8.5 m = 29.345 m.
