Final answer:
The critical angle for the light ray traveling in water with a refractive index of 1.33 incident on the surface of water above air with a refractive index of 1.00 is approximately 48.6°.
Step-by-step explanation:
The critical angle can be calculated using Snell's law, which states that the critical angle (θc) is the angle of incidence for which the angle of refraction is 90 degrees. In this case, the light ray is traveling from water to air. The refractive index of water (n1) is 1.33 and the refractive index of air (n2) is 1.00. Using Snell's law, we have:
n1sin(θc) = n2sin(90°)
1.33sin(θc) = 1.00sin(90°)
sin(θc) = 1.00/1.33
θc ≈ 48.6°
Therefore, the critical angle for the light ray traveling in water with a refractive index of 1.33 incident on the surface of water above air with a refractive index of 1.00 is approximately 48.6°.