A beam of light is emitted in a pool of water from a depth of 64.0 cm. How far away, relative to the spot directly above it, must it strike the air-water interface in order that the light does not exit the water?

Answer 1

theta 2 must be at least 90 degrees so 1*sin90=1.333sin(theta2) theta2 = 48.61 degrees. if distance from spot above the light to this point = x then (x/64)=tan48.61 x= 72.619 cm