## Abstract

We have made a precise comparison of the wavelength passed by a Hg^{198} Zeeman filter with that of the 2537-Å absorption line of an atomic beam. This was done by passing the light from the filter through the atomic beam and then comparing the amplitudes of the two peaks in the resulting reversed lines. A calibration was obtained by measuring the ratio of peak amplitudes with the beam tilted at various angles to the optic axis so as to introduce known Doppler shifts. The wavelengths of the Zeeman filter and atomic beam are shown to be equal to about one part in 10^{9}.

© 1963 Optical Society of America

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### Equations (15)

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(1)
$$\mathrm{\Delta}\nu /\nu =(v/c)\hspace{0.17em}\text{sin}\theta ,$$
(2)
$$\theta \approx (c/v)(\mathrm{\Delta}\nu /\nu )=1.5\times {10}^{6}\times {10}^{-8}=0.015\hspace{0.17em}\text{rad}.$$
(3)
$$\text{slope}=(1-R)/\text{shift},$$
(4)
$$\text{shift}=(1-R)/\text{slope}.$$
(5)
$$\text{shift}=(0.005\pm 0.006)/(0.019\pm 0.0012).$$
(6)
$$\begin{array}{l}\text{std}.\hspace{0.17em}\text{dev}.\hspace{0.17em}\text{of}\hspace{0.17em}\text{shift}=\text{shift}\times {\left[\frac{\text{var}R}{{(1-R)}^{2}}+\frac{\text{var}(\text{slope})}{{(\text{slope})}^{2}}\right]}^{{\scriptstyle \frac{1}{2}}}\\ =0.27\times {\left[\frac{{(0.006)}^{2}}{{(0.005)}^{2}}+\frac{{(0.0012)}^{2}}{{(0.019)}^{2}}\right]}^{{\scriptstyle \frac{1}{2}}}\\ =0.27\times \frac{6}{5}=\mathrm{0.324.}\end{array}$$
(7)
$$\text{shift}=(0.27\pm 1.0){10}^{-9}.$$
(8)
$$\mathrm{\Delta}\nu /\nu =(v/c)\hspace{0.17em}\text{sin}\theta ,$$
(9)
$$\mathrm{\Delta}\nu /\nu =(v/c)\hspace{0.17em}\text{sin}\theta \hspace{0.17em}\text{sin}\phi ,$$
(10)
$$R={T}_{1}/{T}_{2}=\text{exp}[-{\alpha}_{0}Nx({e}^{-{{\omega}_{1}}^{2}}-{e}^{-{{\omega}_{2}}^{2}})].$$
(11)
$${\omega}_{1}\equiv 2{({\nu}_{1}-{\nu}_{0})}^{2}/\mathrm{\Delta}{\nu}_{D}{(\text{ln}2)}^{{\scriptstyle \frac{1}{2}}}.$$
(12)
$$R\approx {e}^{-{\alpha}_{0}Nx/4}.$$
(13)
$${\alpha}_{0}Nx=\mathrm{0.02.}$$
(14)
$$N\approx 3\times {10}^{10}\text{atoms}/{\text{cm}}^{3},$$
(15)
$$N\approx 1.7\times {10}^{9}.$$