If the width of an aperture is 0.Imm and the distance from the central fringe to the first fringe (z)is 5mm and the distance between the aperture and the screen (L)is 93cm determine the wavelength of light source used ?

Joel French
2022-07-15
Answered

If the width of an aperture is 0.Imm and the distance from the central fringe to the first fringe (z)is 5mm and the distance between the aperture and the screen (L)is 93cm determine the wavelength of light source used ?

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Dayana Zuniga

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

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This is basically a Single slit diffraction experiment.

d = width of the aperture or slit =0.1 mm=0.01 cm

L=distance between the aperture and the screen = 93cm

We know:

$\beta =\frac{\lambda L}{d}\equiv \lambda =\frac{0.01\times .5}{93}=0.000053763cm=5.3763\times {10}^{-5}cm\phantom{\rule{0ex}{0ex}}=5376.3\text{}Awhere$ \beta =5mm$(the\; distance\; between\; the\; central\; fringeand\; first\; fringe)$

The wavelength of light source used is 5376.3 Angstroms

d = width of the aperture or slit =0.1 mm=0.01 cm

L=distance between the aperture and the screen = 93cm

We know:

$\beta =\frac{\lambda L}{d}\equiv \lambda =\frac{0.01\times .5}{93}=0.000053763cm=5.3763\times {10}^{-5}cm\phantom{\rule{0ex}{0ex}}=5376.3\text{}Awhere$ \beta =5mm$(the\; distance\; between\; the\; central\; fringeand\; first\; fringe)$

The wavelength of light source used is 5376.3 Angstroms

asked 2022-07-23

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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