How does an optical spatial filter work?

on2t1inf8b
2022-07-20
Answered

How does an optical spatial filter work?

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Bubbinis4

Answered 2022-07-21
Author has **8** answers

The principle behind the use of a pinhole in the way that you describe is simply to create a subresolvable source of light, thus destroying all wavefront information. You are correct that the exact diffraction pattern from a subresolvable point source is not Gaussian, but it is very nearly so and indeed Gaussian modes are the eigenfunctions of an approximation to the wave equation that holds for paraxial fields, i.e. fields of very small numerical aperture and with wave vectors all directed to within about 0.1 radians of the nominal beam propagation direction.

So in practice the beam will become more Gaussian after passage through a subresolvable pinhole, i.e. one of diameter that is smaller than approximately λ/NA, where NA is the beam's numerical aperture.

So in practice the beam will become more Gaussian after passage through a subresolvable pinhole, i.e. one of diameter that is smaller than approximately λ/NA, where NA is the beam's numerical aperture.

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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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