(optics) phase shift definition

1. Jan 13, 2008

Folks,

I've got a question. Is this definition correct?

$$\delta _i = \frac{2\pi}{\lambda_i} * n_i * \cos \, \theta _i$$

where

$$\delta _i$$ is the optical thickness of medium $$i$$

$$\lambda _i$$ is the wavelength in medium $$i$$ (i.e. not $$\lambda _0$$, the vacuum wavelength)

$$n_i$$ is the refractive index of medium $$i$$

$$\theta _i$$ is the angle with respect to the normal inside medium $$i$$

2. Jan 14, 2008

Claude Bile

No.

First of all your units are out. The LHS has units m, while the RHS has units m^-1. You need to get rid of the factor of 2*pi/lamda altogether and multiply the RHS by some distance - be it the thickness of a medium, or a specified propagation distance.

Alternatively, if you want the phase difference rather than the optical path length (as it seems to indicate from the thread title), retain the factor of 2*pi/lambda - but it has to include the vacuum wavelength, not the wavelength in the medium, as you have already accounted for the medium via the n_i term. You still need to include a distance in metres on the RHS.

Claude.

3. Jan 14, 2008

You're right about the distance. I thought I had typed it in, but I missed it; that makes the optical path length.

Lambda has a big question mark. What if your incidence medium is not vacuum? Also, I did not include a phase offset term $$\xi$$, which would account for phase change upon reflection; it's an extra term: $$\pi$$ if you're reflecting off a higher index and 0 otherwise. So, how about this:

$$\delta _i = \frac{2\pi}{\lambda_{m}} * n_i * d* \cos \, \theta _i + \xi$$

I've been looking for the appropriate notation in many texts. They give lambda a subscript 0 and make you think that means vacuum wavelength; no! That's the wavelength in the medium of incidence "m". That happens to be vacuum in simple examples, but it is not a rule.

4. Jan 14, 2008

Claude Bile

You have already accounted for the fact that you are in a medium by including the factor n_i. The n_i term can be absorbed into lambda to give the wavelength in the medium.

Claude.

5. Jan 15, 2008

Okay

$$\delta _i = \frac{2\pi}{\lambda_{0}} * n_i * d* \cos \, \theta _i + \xi = \frac{2\pi}{\displaystyle \left( \frac{\lambda_{0}}{n_i} \right)} * d* \cos \, \theta _i + \xi = \frac{2\pi}{\lambda_{i}}* d* \cos \, \theta _i + \xi$$

That makes sense. However, there is a problem. If you take a look at Modern Optics by Guenther, page 121 (in the multilayer dielectric coatings appendix), you will find "where $$n_0$$ is the refractive index of the incident medium." That implies a medium that can be something else instead of free space. The only restriction seem to be transparency, otherwise the effective wavelength will take complex values. I'm not sure if that is possible. I know that for absorbing media, when you apply Snell's law, you do get complex angles and that's perfectly acceptable.

Continuing on the discussion about references, Laser Resonators and Beam Propagation by Hodgson and Weber, pages 205-206 (optical coatings section), has the same notation as Guenther and it suggests the same meaning. The list of symbols reads "wavelength" or "central wavelength" for $$\lambda_{0}$$. Finally, Thin Film Optical Filters by Macleod, page 41 (3rd ed.) or page 20 (American Elsevier Pub. Co., 1969), completely drops the subscript, and if you go to the list of symbols you will see "wavelength (normally in vacuo)". That doesn't mean always in vacuo.

So, yes, I'm still skeptical about either

$$\delta _i = \frac{2\pi}{\lambda_{0}} * n_i * d* \cos \, \theta _i + \xi$$

where $$\lambda_{0}$$ is the vacuum wavelength.

OR

$$\delta _i = \frac{2\pi}{\lambda_{0}} * n_i * d* \cos \, \theta _i + \xi$$

where $$\lambda_{0}$$ is the medium with index 0, which can be anything.

Authors make the confusion even greater for something that should be straight-forward. If you have a reference where this is clearly stated, please let me know - it would be help me clarify it. Thanks.

6. Jan 15, 2008

Claude Bile

You are trying to find the accumulation of phase over a given distance. Whether it is an EM wave traveling through glass, or a sine wave written on a piece of paper the same equation should apply. In your specific case, the wavelength depends on the refractive index via the following relation;

$$\lambda_i = \frac{\lambda_0}{n_i}$$

This is the ONLY reason why the refractive index pops up in this case.

I know authors can get sloppy with their notation, sometimes you have to back your own intuition!

Claude.