# Homework Help: Angles and refraction

1. Sep 11, 2007

### aliaze1

1. The problem statement, all variables and given/known data

A 1.0-cm-thick layer of water stands on a horizontal slab of glass. A light ray in the air is incident on the water 60 degrees from the normal.

What is the ray's direction of travel in the glass?

2. Relevant equations

n΄sinθ΄ = n¹sinθ¹

or, if that is confusing,

n1sinθ1 = n2sinθ2

3. The attempt at a solution

The index of refraction for water is 1.33, and the index of refraction for air is 1

therefore i set up the problem as:

(1)sin(60) = (1.33)sinθ

θ should = about 40.628 degrees, but the computer says it is incorrect... where am I going wrong?

thanks

Last edited: Sep 11, 2007
2. Sep 11, 2007

### learningphysics

What exactly does the question ask for?

3. Sep 11, 2007

### aliaze1

oh sorry about that, it asks this

'What is the ray's direction of travel in the glass?'

i just edited it also

4. Sep 11, 2007

### learningphysics

So you did the first part to get the angle across the air/water interface... but now you need to handle the water/glass interface... actually I'm thinking since:

n1sin(theta1) = n2sin(theta2)... n2sin(theta2) = n3sin(theta3)...

You can simply ignore the water part...

n1sin(theta1) = n3sin(theta3)

unless I'm missing something...

5. Sep 11, 2007

### aliaze1

Lol!!!!!!!!!!!

LOL oh i see!!!!!!!! hahaha i was thinking that the water itself was the reflective surface and ignored the word 'glass'!! haha this should work

6. Sep 11, 2007

### aliaze1

yay!

yes you were correct, it works!!

using the other index as 1.5 (ignoring the water entirely), i get 1sin(60)=1.5sin(x), and then i just had to solver for x, which gave me the correct value of ~35.3 deg

thanks!

7. Sep 11, 2007

### learningphysics

no prob! you're welcome!