Question

You are standing at the edge fo a swimming pool and shining a laser pointer onto the water surface. The laser beam hits the water at an angle of 50.8 deg above the horizontal. Some of the lights reflects off the water surface, and some enter the water. Find the angle between the reflected and refracted rays.

From Snell's law I got the refracted angle of the laser beam= 35.64 deg. Then I thought it was just the difference between that value and the part of the beam that's relfected off the water (which would be 50.8 deg right? Because the incident angle=reflected angle).

Answer 1

here,

angle of incidence , i = 50.8 degree

let the angle of refraction be r

using the snell's law

sin(r) /sin(i) = 1/refractive index of water

sin(r) /sin(50.8) = 1/1.33

r = 35.64 degree

angle of reflection = angle of incident

r' = i = 50.8 degree

the angle between the reflected and refracted ray, theta = 180 - r - r'

theta = 180 - 35.64 - 50.8

theta = 93.56 degree

the angle between the reflected and refracted rays is 93.56 degree