# Snell's Law and the Refraction of Light

1. Mar 30, 2006

### BoogieL80

I've gotten stuck on the following problem:

The drawing shows a rectangular block of glass (n = 1.52) surrounded by liquid carbon disulfide (n = 1.63). A ray of light is incident on the glass at point A with a = 34.0° angle of incidence. At what angle of refraction does the ray leave the glass at point B?

http://www.boomspeed.com/boogiel80/refraction.gif [Broken]

I was using the formula n1sin theta1 = n2 sin theta2 and manipulated the foruma for sin theta2 (I assumed the liquied was n1 and the glass was n2). I then found the inverse sin of my answer to get the angle. I got an answer of 36.8 degrees. Webassign told me that was incorrect. I thought that maybe since the angle was going back into the same liquid that maybe it would be the same as the angle of incidence, but that was also incorrect. Any feedbook would be appreciated.

Last edited by a moderator: May 2, 2017
2. Mar 30, 2006

### Physics Monkey

Snell's law is the appropriate way to attack the problem, but you need to apply it twice. Once for going into the glass and once for leaving it. Be careful to note that these surfaces are not parallel so you will have to do some geometry to relate angles.

3. Mar 31, 2006

### BoogieL80

My only thing with that method is that I get the beginning angle (33.9 in my calculations). That makes sense to me, because I learned from reading that if you have a transparent material emmersed in a material, n1 should equal n3 and thus theta1 should equal theta3, but webbassign is saying that answer is incorrect. Should the angle really be something else other than the original angle?

4. Mar 31, 2006

### BoogieL80

Actually I figurd out what you meant by "doing some geometry". Thanks again.