Question

A small spotlight mounted in the bottom of a swimming pool that is 4.5 m deep emits light in all directions. On the surface of the pool, directly above the light, the area illuminated is a circle. Determine the maximum radius of this circle. The index of refraction of water is 1.33.

Answer 1

Maximum radius = 5.1m

Step-by-step explanation:

For us to get the radius of the circle of light, we have to first calculate the critical angle which is the angle of incidence above which total internal reflection occurs, i.e. the angle of incidence when θ2 = 90° .

At the point where the total internal reflection occurs, the light ray doesn't pass through the interface, thus, this point is on the edge of the circle of light.

From Snell's law, we have:

n_water * sin (θ1) = n_air * sin(θ2)

Thus;

sin(θ1) = n_air/(n_water * sin(θ2))

Since the critical angle is the value of θ1 when θ2 = 90° and that sin(90°) =1, thus;

sin(θ1) = (n_air/n_water) * sin(90°) = n_air/n_water

We are told the Refractive index of water is 1.33. meanwhile the Refractive index of air is not given but it has a constant value of 1.

Thus, we can determine θ1:

sin(θ1) = (nair/nwater) = 1/1.33

sin(θ1) = 1/1.33

(θ1) = sin^(-1)(1/1.33)

(θ1) = 48.75°

The question when looked at critically, depicts a right triangle with vertices including the light and the extremity of the circle, and we know one of its angles(θ1 = 48.75°) and one of its sides(4.5 m).

Thus, from trigonometric ratio, we can determine the radius as;

r/4.5 = tan(θ1)

r = 4.5tan(48.75°) = 5.1 m