Answer:
5.2 m
Explanation:
from the question we are given the following
depth of pool (d) = 3.2 m
height of laser above the pool (h) = 1 m
point of entry of laser beam from edge of water (l) = 2.5 m
we first have to calculate the angle at which the laser beam enters the water (∝),
tan ∝ = \frac{1}{2.2}
∝ = 24.44 degrees
from the attached diagram, the angle with the normal (i) = 90 - 24.4 = 65.56 degrees
lets assume it is a red laser which has a refractive index of 1.331 in water, and with this we can find the angle of refraction (r) using the formula below
refractive index = \frac{sin i}{sin r}
1.331 = \frac{sin 65.56}{sin r}
r = 43.16 degrees
we can get the distance (x) from tan r = \frac{x}{3.2}
tan 43.16 = \frac{x}{3.2}
x = 3 m
To get the total distance we need to add the value of x to 2.2 m
total distance = 3 + 2.2 = 5.2 m