Posted by **confused** on Thursday, January 24, 2013 at 3:22pm.

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.

- Physics -
**Anonymous**, Thursday, January 24, 2013 at 3:37pm
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

- Physics -
**confused**, Thursday, January 24, 2013 at 3:45pm
Okay so then 1(sin45) = 1.33(sinx)

I got 32.04degrees?

- Physics -
**Anonymous**, Thursday, January 24, 2013 at 3:49pm
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

- Physics -
**Devron**, Thursday, January 24, 2013 at 3:55pm
I got 32.1 degrees. The problem worked out should read

0.532=sin(theta)

sin-1 (0.532)=theta

32.1 degrees=theta

- Physics -
**confused**, Thursday, January 24, 2013 at 4:22pm
Thank you so much

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