Question

In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight, 1.3 m above the water level, onto the surface of the water at a point d = 2.9 m from his foot at the edge of the pool. Part A Where does the spot of light hit the bottom of the h = 2.4-m-deep pool? Measure from the bottom of the wall beneath his foot.

Answer 1

2.am 1.3 m f e Wall 2.4m O, W ate

Refractive index of air = n₁ = 1.0003

Refractive index of water = n₂ = 1.333

Height of the flashlight above the water level = AB = 1.3 m

Distance where the light hits the water = BF = 2.9 m

Depth of the pool = BC = 2.4 m

The light hits the bottom of the pool at point E.

Tan$\theta$₁ = AB/BF

Tan$\theta$₁ = 1.3/2.9 = 0.448

$\theta$₁ = 24.145^o

$\theta$₂ = 90 - $\theta$₁

$\theta$₂ = 90 - 24.145

$\theta$₂ = 65.855^o

By Snell's Law,

n₁Sin$\theta$₂ = n₂Sin$\theta$₃

(1.0003)Sin(65.855) = (1.333)Sin$\theta$₃

Sin$\theta$₃ = 0.6847

$\theta$₃ = 43.21^o

BC = FD = 2.4 m

BF = CD = 2.9 m

Tan$\theta$₃ = ED/FD

Tan(43.21) =ED/2.4

ED = 2.254 m

EC = ED + DC

EC = 2.254 + 2.9

EC = 5.154 m

The spot of light hits the bottom of the pool at a distance of 5.154 m from the bottom of the wall beneath the foot of the watchman.