# Displacement of emerging light ray

1. May 22, 2012

### jssamp

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

A light ray is incident on a flat piece of glass with index of refraction n. Show that if the incident angle θ is small, the emerging ray is displaced a distance d = tθ(n - 1)/n form the incident ray, where t is the thickness of the glass and θ is in radians.

2. Relevant equations

n1sinθ1 = n2sinθ2

3. The attempt at a solution

Using Snell's law, right triangles, sin(a-b) identity, and small angle approximation, I have got it to:

d = tθ(n-(cosθ/cosθR))/n [θR is angle of refraction in glass]

how do I get cosθ/cosθR = 1?

Last edited: May 22, 2012
2. May 22, 2012

### Staff: Mentor

You should be able to get to the result in a more direct way. What is your understanding of "the small angle approximation"?

3. May 22, 2012

### jssamp

sin(theta) = theta for theta < .2 radians

4. May 22, 2012

### Staff: Mentor

Yes, anything else? (Hint: Another trig function has almost the same value for small angles)

5. May 22, 2012

### jssamp

oh, yeah, cos(theta) = 1 - theta/2

6. May 22, 2012

### jssamp

I might use that back when I used the sin(A-B)=sinAcosB-sinBcosA identity.

7. May 22, 2012

### jssamp

I'll try it from the start with this added info. thanks gneill.

8. May 22, 2012

### Staff: Mentor

There's yet ANOTHER trig function that's ~θ when θ is small. You may find it even more convenient...

9. May 22, 2012

### jssamp

OK, I got it solved! Thanks for the help, I'd never heard about small angle approximation for cos and tan until this problem. The tan approximation was the key piece I was missing.

Thanks for the help!

10. May 22, 2012