Apoint source of light is 1.24 m below the surface of a pool. what is the diameter of the circle of light that a person above the water will see? assume the water has an index of refraction of =1.33.

D = 2.828 m

Explanation:

given,

distance of source of light = 1.24 m below surface of pool

refractive index of the water  = n₁ = 1.33

refractive index of air = n₂ = 1

refraction angle be = 90°

let C be the critical angle

Radius = d tan C

d is the depth of the source

Using Snell's law

n₁ sin C = n₂ sin R

1.33 x sin C = 1 x  sin 90°

C = 48.75°

hence,

R = 1.24 x tan 48.75°

R = 1.414 m

Diameter = 2 x R

D = 2 x 1.414

D = 2.828 m

double the distance = divide strength by 4

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer