Is diffraction empirically derived or based off something else?

Roselyn Daniel
2022-07-16
Answered

Is diffraction empirically derived or based off something else?

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Tolamaes04

Answered 2022-07-17
Author has **12** answers

Diffraction can be derived from the wave equation and basic principles, but the empirical observations came first, historically. So, as stated, the answer is: "both"

Historically, diffraction was observed before it was explained with the wave equation. Young's double slit experiment started the idea of light as a wave more than 50 years before Maxwell's equations (the "base principles" behind diffraction).

The wave equation also predicts diffraction. The math can be somewhat involved, but it can be summarized as follows: In 3 dimensions, only a plane wave can travel in just one direction - if the wave has any limits (i.e. if it has any cross-sectional shape), those limits (its shape), it must change as it travels. If you fix the wave shape at one point in the direction of travel (say, by using the shape of Young's double slit), then you can very accurately calculate a diffraction pattern somewhere further along the direction of travel.

Historically, diffraction was observed before it was explained with the wave equation. Young's double slit experiment started the idea of light as a wave more than 50 years before Maxwell's equations (the "base principles" behind diffraction).

The wave equation also predicts diffraction. The math can be somewhat involved, but it can be summarized as follows: In 3 dimensions, only a plane wave can travel in just one direction - if the wave has any limits (i.e. if it has any cross-sectional shape), those limits (its shape), it must change as it travels. If you fix the wave shape at one point in the direction of travel (say, by using the shape of Young's double slit), then you can very accurately calculate a diffraction pattern somewhere further along the direction of travel.

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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?