Question

9) A small light fixture is on the bottom of a swimming pool,0.75 m below the surface. The lightemerging from the water forms a circle on the still water surface.What is the diameter of this circle?
m

Answer 1

Depth h = 0.75 m

Critical angle of water –air surface C = sin ^–1 ( 1/ n )

Where n = index of refraction of water = 4/3

Plug the values we get C = 48.59 degrees

We know tan C = r / h

From this radius of the circle r = h tan C

                                                 = 0.8504 m

Diameter of the circle d = 2r = 1.7 m