Question

An LED mounted in the wall of a pool sits 1.6 m below the surface and emits light rays in all directions. Some rays move forward and upward towards the water/air interface. Approximate the LED as a small source. For the following, explain your reasoning with a diagram. a. What is the critical angle in degrees for total internal reflection of the rays at the water/air interface? Should be ~48. b. For light rays that escape the pool, what is the smallest angle between the light rays and the normal to the LED surface? Should be ~42 c. What is the greatest distance from the pool wall that light rays escape the surface of the water, in m? The answer is not 2.39 or 2.15

Answer 1

a)

$\theta _{c}$ = critical angle = ?

$\theta __{r}$ = angle of refraction = 90

using snell's law at the air-water interface

n_water Sin$\theta _{c}$ = n_air Sin$\theta __{r}$

n_water Sin$\theta _{c}$ = n_air Sin90

(1.33) Sin$\theta _{c}$ = 1

Sin$\theta _{c}$ = 1/1.33 = 0.752

$\theta _{c}$ = Sin^-1(0.752)

$\theta _{c}$ = 48 deg

b)

In triangle ABC

$\theta$ = 90 - $\theta _{c}$

$\theta$ = 90 - 48

$\theta$ = 42 deg

c)

d = distance

h = height of LED below the surface = 1.6 m

In triangle ABC

tan$\theta$ = BC/AB

tan42 = h/d

d = 1.6/tan42

d = 1.8 m