Respuesta :
Answer:
The distance of the goggle from the edge of the pool is 4.726 m
Explanation:
The given information are;
The depth of the swimming pool = 3.0 m deep
The height of the laser pointer above the swimming pool edge = 1.0 m
The distance from the pool edge the laser pointer enters the water = 2.0 m
The angle between the pool and the laser = ∅ = tan⁻¹(1/2) = 26.57°
Therefore, the angle of incidence to the vertical [tex]\theta_i[/tex], = θ - ∅ = 90 - 26.57° = 63.43°
By Snell's law we have;
The ratio of the sin of the angle of incidence to the sin of the angle of refraction is a a constant equal to the ratio of the refractive indices as follows;
[tex]\dfrac{sin (\theta_i)}{sin (\theta_r} =\dfrac{n_r}{n_i}[/tex]
Where:
[tex]n_r[/tex] = Refractive index of the refractive medium which is water = 1.33
[tex]n_i[/tex] = Refractive index of the incidence medium which is air = 1.00
Therefore;
[tex]\theta_r = sin^{-1} \left (\dfrac{n_i \times sin (\theta_i)}{n_r} \right) = sin^{-1} \left (\dfrac{1\times sin (63.43)}{1.33} \right) = 42.26 ^{\circ}[/tex]
We have that tan([tex]\theta_r[/tex]) = (Distance of the goggles from the point directly above the point of incidence of the beam)/(The water depth)
tan(42.26) = (The horizontal distance of the goggles from the point of incidence of the laser on the water surface)/(3.0)
∴ The horizontal distance of the goggles from the point of incidence of the laser on the water surface = 3.0 × tan(42.26) = 2.726 m
The distance of the goggle from the edge of the pool = The horizontal distance of the goggles from the point of incidence of the laser on the water surface + The distance from the edge of the water surface the laser enters the water
The distance of the goggle from the edge of the pool = 2.726 + 2 = 4.726 m
The distance of the goggle from the edge of the pool = 4.726 m.
