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.

Question

I don't even know how to start this or what the formula is. Thanks for your help in advance. It really is appreciated.

Answer 1

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

Answer 2

Okay so then 1(sin45) = 1.33(sinx)

I got 32.04degrees?

Answer 3

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)

sin-1 (0.544)=theta

33.0 degrees=theta

Answer 4

I got 32.1 degrees. The problem worked out should read

0.532=sin(theta)

sin-1 (0.532)=theta

32.1 degrees=theta

Answer 5

Thank you so much