water has an angle of incidence of 35.0 degrees. At what angle does the light enter the air?
Actually the plastic sheet does not enter into the problem.. Let n_w = n water = 1.33; n_p = n plastic = 1.500; n_a = n air = 1.00 Now using Snell's Law we have n_w*sin(thetaw) = n_p*sin(thetap) = n_a*sin(thetaa) See how the middle part of the eqn is not needed... So n_w*sin(thetaw) = n_a*sin(thetaa) Therefore thetaa = arcsin(n_w/n_a*sin(thetaw)) arcsin(1.33/1.0*sin(35)) = 49.7o
