Phase Difference for Light Passing Through Two Slabs

AI Thread Summary
A beam of light with a wavelength of 600 nm passes through two slabs of equal thickness (1.40 micrometers) but different indices of refraction (n1= 1.41 and n2= 1.58). The phase difference between the two parts of the beam can be calculated by determining the number of wavelengths that fit into each slab using the formula 2nt/λ. The calculated difference in wavelengths is 0.793 m, which needs to be converted into a phase difference in radians. The relationship between wavelength difference and phase difference is that one wavelength corresponds to a phase difference of 2π radians. Understanding these concepts is crucial for solving the problem accurately.
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Homework Statement



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?

Homework Equations



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

The Attempt at a Solution



I've calculated what I believe to be the difference in the number of wavelengths that fit into 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!
 
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Rome_Leader said:

Homework Statement



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?

Homework Equations



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

The Attempt at a Solution



I've calculated what I believe to be the difference in the number of wavelengths that fit into 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.
Ummm. :rolleyes: I'm not following you there. The "thinkness" of the slabs is given in the problem statement as 1.40 micrometers.
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?
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.
 
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