Question

A point source of light is 1.39 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.

Answer 1

We can find the maximum diameter of the circle of light can have. Above the critical angle of incidence, the light rays are totally internally reflected. Assume that the maximum deviation that the light can have is the critical angle _c,
From geometry, we can find that
tan _c = (D/ 2) / d
Where D is the diameter of circle of light, d is the depth of the pool.
_c = tan^-1((D/ 2) / d )
Then substituting this angle as incident angle and = 90⁰ as refracted angle in the Snell's equation,
n_i sin i = n_r sin r
Where n_i is the refractive index of water, n_r is the refractive index of air.
1.33 x sin (tan^-1((D/ 2) / d ) ) = 1
sin (tan^-1((D/ 2) / d ) ) = 0.7518
tan^-1((D/ 2) / d ) = 48.75
(D/ 2) / d = 1.140
D = 3.17 m