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 2.5 m from his foot at the edge of the pool. Where does the spot of light hit the bottom of the pool, measured from the bottom of the wall beneath his foot, if the pool is 2.1 m deep pool?

4.4 m

Explanation:

We are told the light from his flashlight, 1.3 m above the water level. Thus; h1 = 1.3m

Also,we are told that the light shone 2.5 m from his foot at the edge of the pool. Thus, L1 = 2.5 m

Angle of incidence θ1 is given by;

tan θ1 = L1/h1

tan θ1 = 2.5/1.3

tan θ1 = 1.9231

θ1 = tan^(-1) 1.9231

θ1 = 62.53°

Using Snell's law, we can find the angle of refraction from;

Sin θ2 = (η_air/η_water) Sin θ1

Where;

η_air is Refractive index of air = 1

η_water is Refractive index of water = 1.33

Thus;

Sin θ2 = (1/1.33) × sin 62.53°

Sin θ2 = 0.6671

θ2 = sin^(-1) 0.6671

θ2 = 41.84°

We want to find where the spot of light hit the bottom of the pool if the pool is 2.1 m deep. Thus, h2 = 2.1 m

Now, the spot can be found from;

L = L1 + L2

Where L2 = (h2) tan θ2

L = 2.5 + 2.1 tan 48.84

L = 2.5 + (2.1 × 0.8954)

L ≈ 4.4 m

answer; ///(74) degrees is your answer..