## Abstract

The paper describes an experiment on the absorption of the 3*s*_{2}–3*p*_{4} helium–neon laser emission at 2947.903 cm^{−1} (3.39 *μ*) by methane. The emission frequency coincides closely to one of the components of the *P*(*F*^{+}) branch of the *ν*_{3} band of methane. Methane and nitrogen in different mixing ratios were introduced into an absorption cell and the transmittance as a function of pressure was determined. By relating the measured absorption coefficient with the known interaction of collision and Doppler effects on the broadening of the absorption line, the separation of the emission line and the nearest absorption line was deduced to be 0.003±0.002 cm^{−1}.

The collision broadened full-width at half-maximum of the absorption line was determined to be 0.13±0.04 cm^{−1} at atmospheric pressure. At 1 atm in the earth’s atmosphere, the transmittance can be calculated to be *T*=exp(−1.1 *L*) by using the published value of the concentration of methane where *L* is the path length in kilometers. The effects of the laser emission in several possible cavity modes and of the several absorption lines in the methane group which overlap each other at high pressures are discussed.

© 1965 Optical Society of America

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### Equations (11)

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(1)
$$T(\nu )={e}^{-k(\nu )\hspace{0.17em}u}.$$
(2)
$$\omega =[2(\nu -{\nu}_{0})/\mathrm{\Delta}{\nu}_{\text{D}}]{(\text{ln}2)}^{{\scriptstyle \frac{1}{2}}},$$
(3)
$$a=[(\mathrm{\Delta}{\nu}_{\text{C}}+\mathrm{\Delta}{\nu}_{\text{N}})/\mathrm{\Delta}{\nu}_{\text{D}}]{(\text{ln}2)}^{{\scriptstyle \frac{1}{2}}}$$
(4)
$$k(\nu )=CP/[{(\nu -{\nu}_{0})}^{2}+{C}^{\prime}{P}^{2}]\to (C/{C}^{\prime}P).$$
(5)
$$T={\sum}_{n}{A}_{n}\hspace{0.17em}\text{exp}[-{k}_{n}(\nu )u].$$
(6)
$$T={e}^{-ku}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\text{or}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}k=-\text{ln}T/u,$$
(7)
$$\begin{array}{l}\mathrm{\Delta}\nu =\nu -{\nu}_{0}=\omega \mathrm{\Delta}{\nu}_{\text{D}}/2{(\text{ln}2)}^{{\scriptstyle \frac{1}{2}}}\\ \cong 0.003\pm 0.002\hspace{0.17em}{\text{cm}}^{-1}.\end{array}$$
(8)
$$\begin{array}{ll}\hfill 2947.903\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}{\text{cm}}^{-1}& (\text{laser}\hspace{0.17em}{\text{emission}}^{1,2})\hfill \\ \hfill \underset{\_}{2947.888\pm 0.015\hspace{0.17em}{\text{cm}}^{-1}}& (\text{methane}\hspace{0.17em}{\text{line}}^{3-5})\hfill \\ \hfill 0.015\pm 0.015\hspace{0.17em}{\text{cm}}^{-1}& (\text{difference}).\hfill \end{array}$$
(9)
$$\mathrm{\Delta}{\nu}_{\text{C}}=a\mathrm{\Delta}{\nu}_{\text{D}}/{(\text{ln}2)}^{{\scriptstyle \frac{1}{2}}}=P\mathrm{\Delta}{\nu}_{\text{D}}/63\hspace{0.17em}{(\text{ln}2)}^{{\scriptstyle \frac{1}{2}}}({\text{cm}}^{-1}\hspace{0.17em}{\text{Torr}}^{-1}),$$
(10)
$$\mathrm{\Delta}{\nu}_{\text{C}}=0.13\pm 0.04\hspace{0.17em}{\text{cm}}^{-1}.$$
(11)
$$k(\nu )=\text{const}/\{1+{[2(\nu -{\nu}_{0})/\mathrm{\Delta}{\nu}_{\text{C}}]}^{2}\}.$$