## Abstract

For many years channel spectra, caused by multiple reflections of light between the faces of flat samples of optical material, have been used to determine refractive indices. Interferometers are excellent for this measurement, particularly in the far ir spectral region where their superior sensitivity and spectral resolution are required. The theory of the method is developed and the limitations are discussed. Experimentally determined refractive indices of silicon, germanium, and fused quartz are presented. These indices have been determined by these methods from data obtained with the Aerospace lamellar grating interferometer.

© 1967 Optical Society of America

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### Equations (12)

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(1)
$$A(\nu )={\widehat{t}}^{2}\{1+\sum _{l=1}^{\infty}{\widehat{r}}^{2l}\hspace{0.17em}\text{exp}[i(2\pi l\times 2\widehat{n}\nu h\hspace{0.17em}\text{cos}\beta )]\}\hspace{0.17em}\text{exp}(2i\pi \nu \widehat{n}h\hspace{0.17em}\text{cos}\beta )$$
(2)
$$T(\nu )=A\times {A}^{*},$$
(3)
$$T(\nu )=\tau (\nu )(1+2\sum _{l=1}^{\infty}{\rho}^{l}\hspace{0.17em}\text{cos}l\theta ),$$
(4)
$$\begin{array}{l}\tau (\nu )=\mid 4\widehat{n}/{(\widehat{n}+1)}^{2}{\mid}^{2}\text{exp}(-\alpha h\hspace{0.17em}\text{cos}\beta )/(1-{\rho}^{2}),\\ =16({n}^{2}+{k}^{2})\hspace{0.17em}\text{exp}(-\alpha h\hspace{0.17em}\text{cos}\beta )/\{{[{(n+1)}^{2}+{k}^{2}]}^{2}(1-{\rho}^{2})\},\end{array}$$
(5)
$$\rho =[{(n-1)}^{2}+{k}^{2}]\hspace{0.17em}\text{exp}(-\alpha h\hspace{0.17em}\text{cos}\beta )/[{(n+1)}^{2}+{k}^{2}],$$
(6)
$$\theta =4\pi nh\nu \hspace{0.17em}\text{cos}\beta +\delta =4\pi nh\nu \hspace{0.17em}\text{cos}\beta +{\text{tan}}^{-1}[2k/({n}^{2}+{k}^{2}-1)],$$
(7)
$$T(\nu )=\sum _{l=0}^{\infty}{A}_{l}\hspace{0.17em}\text{cos}l\theta ,$$
(8)
$$F(x)={\int}_{0}^{\infty}{\tau}^{\prime}(\nu )(1+2\sum _{l=1}^{\infty}{\rho}^{l}\hspace{0.17em}\text{cos}l\theta )\hspace{0.17em}\text{cos}2\pi \nu xd\nu ,$$
(9)
$$2n{\nu}_{\text{max}}h\hspace{0.17em}\text{cos}\beta +\delta /2\pi =m.$$
(10)
$${e}^{-2\alpha h}>\frac{\u220a{(1+n)}^{6}}{64{n}^{2}{(n-1)}^{2}}>\frac{\u220a{(1+n)}^{6}}{64{n}^{2}{(n-1)}^{2}\hspace{0.17em}[1+{(1-{n}^{2})}^{2}\u220a/64{n}^{2}]}.$$
(11)
$$T(\nu )=\tau (\nu )[1+2\sum _{l=1}^{\infty}\frac{{\rho}^{l}}{(1-\text{cos}{\beta}_{m})}{\int}_{\beta}\text{sin}\beta \hspace{0.17em}\text{cos}l\theta (\beta )d\beta ].$$
(12)
$${\int}_{\beta}\text{sin}\beta \hspace{0.17em}\text{cos}l\theta (\beta )d\beta =\left[[\text{cos}l4\pi nh\left(\frac{\text{cos}{\beta}_{m}+1}{2}\right)\right]\frac{\text{sin}}{y,}y,$$