Physics
posted by confused on .
What is the angle of refraction of a light ray that enters the surface of a calm lake from air at an angle of 45o to the normal? The refractive index of water is 1.333.
I don't even know how to start this or what the formula is. Thanks for your help in advance. It really is appreciated.

Snell's law:
n1 Sin(theta)=n2 Sin(thea)
Index of refraction of air (n1)=1.00
angle of incidence is 45 to the normal (sin theta)
Index of refraction of lake (n2)=1.33
solve for theta 
Okay so then 1(sin45) = 1.33(sinx)
I got 32.04degrees? 
n1 Sin(theta)=n2 Sin(theta)
Index of refraction of air (n1)=1.00
angle of incidence is 45 to the normal (sin theta)
Index of refraction of lake (n2)=1.33
solve for theta
(1.00/1.33)sin45=sin(theta)
0.544=sin(theta)
sin1 (0.544)=theta
33.0 degrees=theta 
I got 32.1 degrees. The problem worked out should read
0.532=sin(theta)
sin1 (0.532)=theta
32.1 degrees=theta 
Thank you so much