You are sitting at the bottom of a pool, a depth 103.9 cm below the surface, looking directly up. If the water has a refractive index of 1.33, what is the diameter of the circle above you for which you can view objects above the water?

To solve this problem, we need to use the concept of total internal reflection. When light travels from a medium with a higher refractive index to a medium with a lower refractive index, there is an angle of incidence for which the refracted ray will be at 90 degrees to the normal. This is called the critical angle. For angles of incidence greater than the critical angle, total internal reflection occurs and the light is not transmitted into the second medium.

The critical angle can be calculated using Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the velocities of light in the two media, or equivalently, to the inverse ratio of the refractive indices.

In this case, we are looking for the critical angle in water with respect to air. The refractive index of water is given as 1.33 and the refractive index of air is approximately 1.00. So, we can set up the equation as follows:

sin(critical angle) = 1.00 / 1.33

Solving for the critical angle gives us a value of approximately 48.8 degrees.

Now, imagine a cone with its tip at your eye, the sides going out at the critical angle, and the base at the water's surface. The diameter of the circle you can see is the diameter of the base of this cone.

The depth of the water is given as 103.9 cm. We can use simple trigonometry to find the radius of the base of the cone:

tan(48.8 degrees) = radius / 103.9 cm

Solving for the radius gives us a value of approximately 103.9 cm. The diameter is twice the radius, so the diameter of the circle you can see is approximately 207.8 cm.