# Phase Difference for Light Passing Through Two Slabs

1. Apr 3, 2012

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

A beam of light of wavelength 600 nm passes through two slabs of material of identical thickness d= 1.40 micrometers, as shown in the figure. The slabs have different indices of refraction: n1= 1.41 and n2= 1.58. What is the phase difference between the two parts of the beam after it passes through the slabs?

2. Relevant equations

$\varphi$ = (2$\pi$/$\lambda$)*d*sin$\theta$

3. The attempt at a solution

I've calculated what I believe to be the difference in the number of wavelengths that fit in to each slab by applying the equation 2nt/ $\lambda$, where t is the given thickness of the slabs. The difference in question, I have calculated to be 0.793 m.

My only question is how might I convert this difference into a phase difference in radians? My text gives the equation I have given under 'relevant equations', but I cannot figure out how to incorperate any angle or separation distance d, because the problem does not involve slits. The beam of light simply hits two blocks of differing indices of refraction at the same time?

Can anyone offer any help? It would be much appreciated!

2. Apr 3, 2012

### collinsmark

I'm not quite sure what to make of that equation, but it looks reminiscent of something to do with diffraction. But diffraction doesn't apply to this problem.
Ummm. :uhh: I'm not following you there. The "thinkness" of the slabs is given in the problem statement as 1.40 micrometers.
Let's start over. The problem statement, as it's worded, doesn't have anything to do with interference or diffraction (unless there is a part 2 that comes later).

When the light enters a given slab, its wavelength becomes smaller. The following relationships might help. Here, $\lambda_0$ is the wavelength in a vacuum (approximately the same for air).
$$\frac{\lambda_0}{\lambda} = \frac{c/f}{v/f} = \frac{fc}{fv} = \frac{c}{v} = n$$
where $n$ here is the index of refraction. Determine $\lambda_1$ and $\lambda_2$. In each slab (of thickness 1.40 micrometers), determine how many wavelengths pass through each individual slab. Then calculate the difference in the number of wavelengths between the two cases.

Then you can calculate the difference in radians. A difference of 1.0 wavelength equals a phase difference of 2π radians.