A New Method for Investigating the Refractive Index and the Thickness of Thin Interference Films on Glass

If plane polarized light falls on lead glass the surface of which had been chemically changed to be covered with an interference film, the reflected light is in general elliptically polarized. If, however, the path difference of reflected rays from the air-film boundary and from the film-glass boundary is $2n_{0}d cos r+\frac{1}{2}\lambda =k\lambda$, where $k$ gives the order of the maximum, the first maximum for $k=1\mathrm{}$ being $d{({n_0}^2-{sin}^2 \varphi )}^{\frac{1}{2}}=\frac{1}{4}\lambda$, and where $n_0$ is the refractive index of the film, $d$ its thickness, $\varphi$ the angle of incidence, $r$ the angle of refraction and $\lambda$ the wave-length of the monochromatic light, then the reflected light is plane polarized. For this angle of incidence, $\varphi$, the plane of polarization of the reflected plane polarized light will be the same as if the polarized light were reflected from an ideal surface of glass (not covered with a film). From the angle between the plane of polarization of the analyzer and the plane of incidence the refractive index of the glass, $n$, can be calculated according to the formula deduced from Fresnel's relations: $n^2={sin}^2 \varphi [1+t{g}^2\varphi t{g}^2 (\psi +{45}^{\circ }\left)\right]$, where $\varphi$ is the angle of incidence, $\psi$ the angle between the plane of polarization of the analyzer and the plane of incidence (the angle between the plane of polarization of the polarizer and the plane of incidence is 135°). We can calculate the refractive index, $n_0$, and thickness, $d$, of the surface film from the formula for the maxima if we determine two angles of incidence, ${\varphi }_1$ and ${\varphi }_2$ for two different wave-lengths. The refractive index of the film, $n_0$, is given by the formula: $n_0={\left(\frac{{{\lambda }_1}^2 {sin}^2 {\varphi }_2-{\lambda }_2 {sin}^2 {\varphi }_1}{{{\lambda }_1}^2-{{\lambda }_2}^2}\right)}^{\frac{1}{2}}.$