6) challenge problem: in searching the bottom of a pool at night, a

watchman shines a narrow beam of light from the 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?

7.08 m

Explanation:

(If we ignore the refraction of light)

There will form two triangles one at the top of pool (torch, edge of pool and point of incidence at top of pool) and one at the bottom of pool (torch, edge of pool and point of incidence at bottom of pool) .

first triangle at the top of pool (torch, edge of pool and point of incidence at top of pool)

First we need to find the angle of incidence:

Height = perpendicular =  1.3 m

Point of incidence = base= 2.5 m

now

tan∅= perpendicular / base

∅= tan⁻¹ (perpendicular / base)

∅=tan⁻¹ (1.3/2.5)

∅= 27.5°

Now 2nd triangle at the bottom of pool (torch, edge of pool and point of incidence at bottom of pool)

Height = perpendicular =  1.3 m + 2.1 m= 3.4 m (torch height + depth of pool)

∅= 27.5°

Spot of light hit bottom of pool= base=?

tan∅ = perpendicular / base

base = perpendicular / tan∅

base = 3.4/tan(27.5°)

base= 7.08 m

Spot of light hit bottom of pool= base= 7.08 m