A small light fixture is on the bottom of a swimming pool 1.00 m below the surface The light emerging from the water forms a circle on the still water surface What is the diameter of this circle?

Answer 1

So we know that Θc=arcsin(n1/n2) (or now you do.lol). So n1=index of refraction of air=1, n2=index of refraction of water=1.333 (these values obtained from a table in a physics book, readily available online as well). Thus, arcsin(1/1.333)=48.6°

So now you know your angle of incidence between a line perpendicular to the bottom of the pool at the center of the light and your ray of light. So from your vertical line, move your ray 48.6° to the right (or left). You should see an upside down right triangle formed between your vertical and your ray. The ray is the hypotenuse, the vertical is a side adjacent, and the top of the water is the opposite to your angle Θ. In order to find the diameter, follow some basic trig steps and simply double the side opposite. Thus you have; 1meter*tan(48.6°)=1.134meter*2=2.26meterswhich is your diameter.