Question

Problem 3) Bob stands at the edge of the swimming pool holding a laser 1.5m above the ground. He shines the red laser beam onto the surface of the water that is 3.0m from the edge of the pool. Determine the distance, d, where the light hits the bottom of the pool if the pool is 2.5m deep

Answer 1

Answer:

d = 5.75m

Explanation:

Using snell's law, we have,

n₁ × sin(i) = n₂ × 2 × sin(r)

n1= refractive index of 1st medium= 1

n2=  refractive index of 2nd medium = 1.33

r= angle of reflection

therefore,

$r = \sin^{-1}\frac{n_1\sin i}{n_2}$

Here,

i = 90 - θ

$\theta = \tan^-^1(\frac{1.5}{3} )\\\\=26.56^\circ$

$r = \sin^{-1}\frac{n_1\sin i}{n_2}$

$r = \sin^{-1}\frac{(1)\sin (90-26.56)}{1.33}\\\\r = 42.26m$

$\tan r = \frac{2.5}{d_1}$

$d_1 = \frac{2.5}{\tan (42.26)} \\\\d_1 = 2.75m$

Therefore, the distance is

d = 3 + d₁

d = 3 + 2.75

d = 5.75m