Does diffraction of a coherent laser beam affect its polarization state?

Jayvion Caldwell
2022-07-22
Answered

Does diffraction of a coherent laser beam affect its polarization state?

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nezivande0u

Answered 2022-07-23
Author has **16** answers

Yes, diffraction can lead to polarization effects. However, this requires moving out of the regime of scalar optics. Scalar optics is the regime when you can neglect the polarization of light and only treat its spatial complex amplitude.

Scalar optics can break down when the paraxial limit breaks down. This happens around when you introduce spatial features into your field which are close to as small as or smaller than the wavelength of light you are considering.

A good example of this is an extremely tightly focused Gaussian beam. Suppose the beam is propagating along the z direction. To use some language from Fourier optics. If the focus is so small it means the beam is composed of a very large range of wavevectors kx and ky. The plane waves corresponding to these large wavevectors can have a large angle with the optical axis. That means it is possible for those plane waves with large transverse wavevectors to have large components in the direction of propagation. This is not what is typically the case for paraxial Gaussian beams which are purely transversely polarized. At the focus the superposition of these polarization vectors can lead to surprising polarization patterns. See for example:

Polarization of tightly focused laser beams

You ask specifically about diffraction. Well just reverse the effect I have described above. Shine a transversely linearly polarized plane wave onto an opaque screen which has a hole on it with dimensions on the order of λ. What you will find is that in the vicinity of the hole the diffracted light will have non-transverse polarization.

Scalar optics can break down when the paraxial limit breaks down. This happens around when you introduce spatial features into your field which are close to as small as or smaller than the wavelength of light you are considering.

A good example of this is an extremely tightly focused Gaussian beam. Suppose the beam is propagating along the z direction. To use some language from Fourier optics. If the focus is so small it means the beam is composed of a very large range of wavevectors kx and ky. The plane waves corresponding to these large wavevectors can have a large angle with the optical axis. That means it is possible for those plane waves with large transverse wavevectors to have large components in the direction of propagation. This is not what is typically the case for paraxial Gaussian beams which are purely transversely polarized. At the focus the superposition of these polarization vectors can lead to surprising polarization patterns. See for example:

Polarization of tightly focused laser beams

You ask specifically about diffraction. Well just reverse the effect I have described above. Shine a transversely linearly polarized plane wave onto an opaque screen which has a hole on it with dimensions on the order of λ. What you will find is that in the vicinity of the hole the diffracted light will have non-transverse polarization.

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Suppose you have source of coherent light, a laser beam, with +45 polarization. A polarizing beam splitter (PBS) splits the beam into its horizontal and vertical components. The second polarizing beam splitter recombines them into a single beam again. If the two paths between the two PBS are of equal length and there is nothing else in them that would cause a phase shift, I'd assume the output polarization is equal to the input polarization: +45.

What is the output polarization if you place an attenuator in the path of the horizontal component? How do you calculate it?

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Currently I have a gas with a density that follows and inverse square law in distance, $r$. Given that I know the mass attenuation coefficient of this gas, I wish to calculate an effective optical depth using a modified version of the Beer-Lambert Law that uses mass attenuation coefficients:

$\tau =\frac{\alpha {\rho}_{gas}(T)l}{\rho}=\frac{\alpha Mp(T)}{\rho RT}\int \frac{1}{{x}^{2}}dx$

Where $\alpha $ is the mass attenuation coefficient for the solid phase of the gas [cm${}^{-1}$], $\rho $ is the mass density of the solid phase of the gas, l is the path length, M is the molar mass of the gas, $p(T)$is the pressure of the gas as a function of temperature, R is the ideal gas constant and T is the temperature of the gas. ${\rho}_{gas}$ is the mass density of the gas itself and can be extracted from the ideal gas law:

${\rho}_{gas}=\frac{p(T)M}{RT}$

The integral emerges from my attempt at rewriting the first equation for a non uniform attenuation, that I have here due to the inverse square law effecting the density of the gas.

However, I am now concerned that units no longer balance here since τ should be unitless. Can anyone help guide me here?

$\tau =\frac{\alpha {\rho}_{gas}(T)l}{\rho}=\frac{\alpha Mp(T)}{\rho RT}\int \frac{1}{{x}^{2}}dx$

Where $\alpha $ is the mass attenuation coefficient for the solid phase of the gas [cm${}^{-1}$], $\rho $ is the mass density of the solid phase of the gas, l is the path length, M is the molar mass of the gas, $p(T)$is the pressure of the gas as a function of temperature, R is the ideal gas constant and T is the temperature of the gas. ${\rho}_{gas}$ is the mass density of the gas itself and can be extracted from the ideal gas law:

${\rho}_{gas}=\frac{p(T)M}{RT}$

The integral emerges from my attempt at rewriting the first equation for a non uniform attenuation, that I have here due to the inverse square law effecting the density of the gas.

However, I am now concerned that units no longer balance here since τ should be unitless. Can anyone help guide me here?

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