You place a point source of light at the bottom of a pool of water 1.00 m deep. The source emits rays upward in all directions. You notice that a circle of light is formed by the rays that are refracted into the air and that the rays outside of this circle are reflected back into the water. The index of refraction of water is 1.33. The radius R of the circle at the surface of the water is approximately
A) 0.75 m.
B) 1.00 m.
C) 1.13 m.
D) 1.33 m.
E) infinite.

Question

You place a point source of light at the bottom of a pool of water 1.00 m deep. The source emits rays upward in all directions. You notice that a circle of light is formed by the rays that are refracted into the air and that the rays outside of this circle are reflected back into the water. The index of refraction of water is 1.33. The radius R of the circle at the surface of the water is approximately
A) 0.75 m. 
B) 1.00 m. 
C) 1.13 m. 
D) 1.33 m. 
E) infinite.

Quizplus · Accepted Answer

When light travels from water to air, it will refract away from the normal. For the light to appear to be coming from the circle at the surface, every ray that leaves the water at a certain angle has to be refracted at the surface to the same angle. Therefore, the angle of incidence with respect to the normal at the surface of the water must be the same for all rays that refract into the air. This angle can be determined using Snell's Law:

n1 sin θ1 = n2 sin θ2

where n1 is the refractive index of the initial medium (water), n2 is the refractive index of the final medium (air), θ1 is the angle of incidence with respect to the normal in water, and θ2 is the angle of refraction with respect to the normal in air.

We can first find the critical angle of total internal reflection, which is the angle of incidence at which light will refract at an angle of 90 degrees with respect to the normal at the surface. At this angle, the refracted ray will travel along the surface and not reach the air, causing total internal reflection. The critical angle can be found using Snell's Law with θ2 = 90 degrees:

n1 sin θc = n2 sin 90

sin θc = n2/n1 = 1/1.33

θc ≈ 49 degrees

For all rays that refract into the air, the angle of incidence with respect to the normal at the surface must be less than this critical angle, so that they do not undergo total internal reflection. Therefore, the maximum angle of refraction with respect to the normal in air is also 49 degrees, and the maximum angle of incidence with respect to the normal in water is:

sin θ1 = n2/n1 sin θ2 = (1/1.33) sin 49 ≈ 0.531

Now, we can consider a ray that leaves the point source at the bottom of the pool and passes through a point P on the surface of the water, forming an angle of incidence θ1 with respect to the normal. The ray will refract at angle θ2 with respect to the normal in air, and travel along a straight line tangent to the circle of light at point P. This tangent line creates a right angle with the radius of the circle that passes through point P, so we can use the Pythagorean Theorem to relate R, the depth of the point source d (equal to 1.00 m), and the distance x between the point P and the center of the circle:

R^2 = d^2 + x^2

We can solve for x using the trigonometric relationships:

sin θ2 = x/R

sin θ1 = (n2/n1) sin θ2 = (1/1.33) (x/R)

sin θ1 = sin arcsin(0.531) = 0.531

0.531 = (1/1.33) (x/R)

x/R ≈ 0.398

x ≈ (0.398) R

Substituting into the Pythagorean equation:

R^2 = d^2 + (0.398)^2 R^2

R ≈ 1.13 m

Therefore, the radius of the circle at the surface of the water is approximately 1.13 m, which is answer choice C.

You Place a Point Source of Light at the Bottom