in searching the bottom of a pool at night a watchman shines a narrow beam of light from his flashlight 125 cm above the water level to a point 185 cm from his foot at the edge of the pool. where does the spot of light hit the bottom of the pool, relative to the egde, if the pool is

Complete question is;

In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight 125 cm above the water level to a point 185 cm from his foot at the edge of the pool. Where does the spot of light hit the bottom of the pool, relative to the edge, if the pool is 215 cm deep

356.2 cm

Explanation:

First of all, let's find the angle of incidence from the distances ;

tan θ1 = l1/h1

tan θ1 = 185/125

tan θ1 = 1.48

θ1 = tan^(-1) 1.48

θ1 = 55.95°

For the refraction from air into water, we have

n_air × sin θ1 = n_water × sin θ2,

sin θ2 = (n_air × sin θ1)/n_water

Where,

n_water is refractive index of water = 1.33

n_air is refractive index of air = 1

Thus;

sin θ2 = (1 × sin 55.95)/1.33

sin θ2 = 0.62297

θ2 = sin^(-1) 0.62297

θ2 = 38.53°

So, the horizontal distance from the edge of the pool from l is;

l = l1 + l2 = l1 + h2 tan θ2

where;

l1 = 185cm

h2 = 215 cm

So,

l = 185 + 215(tan 38.53)

l = 185 + 171.2045

l = 356.2045 cm ≈ 356.2 cm

true tan speed = ang vel x radius