Calculate the power of lens, which has positive focal length of 20 cm

detineerlf
2022-07-17
Answered

Calculate the power of lens, which has positive focal length of 20 cm

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tiltat9h

Answered 2022-07-18
Author has **14** answers

focal length,

$f=+20\text{}cm=+0.20\text{}m$

Power of lens is given by

$p=\frac{1}{f}$

where f is in meters

$\Rightarrow p=\frac{1D}{+0.20}=+5D$

So power g given lens is +5D

$f=+20\text{}cm=+0.20\text{}m$

Power of lens is given by

$p=\frac{1}{f}$

where f is in meters

$\Rightarrow p=\frac{1D}{+0.20}=+5D$

So power g given lens is +5D

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

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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