## Abstract

The conditions under which light interference in a transparent quarter-wave layer of refractive index ${n}_{1}$ on a transparent substrate of refractive index ${n}_{2}$ leads to 50% reflectance for incident unpolarized light at an angle *φ* are determined. Two distinct solution branches are obtained that correspond to light reflection above and below the polarizing angle, ${\phi}_{p}$, of zero reflection for *p* polarization. The real *p* and *s* amplitude reflection coefficients have the same (negative) sign for the solution branch $\phi >{\phi}_{p}$ and have opposite signs for the solution branch $\phi <{\phi}_{p}$. Operation at $\phi <{\phi}_{p}$ is the basis of a 50%–50% beam splitter that divides an incident totally polarized light beam (with *p* and *s* components of equal intensity) into reflected and refracted beams of orthogonal polarizations [Opt. Lett. **31**, 1525 (2006)
] and requires a film refractive index ${n}_{1}\u2a7e(\sqrt{2}+1)\sqrt{{n}_{2}}$. A monochromatic design that uses a high-index $\mathrm{Ti}{\mathrm{O}}_{2}$ thin film on a low-index $\mathrm{Mg}{\mathrm{F}}_{2}$ substrate at $488\text{\hspace{0.17em}}\mathrm{nm}$ wavelength is presented as an example.

© 2007 Optical Society of America

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### Equations (33)

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(1)
$${R}_{s}=\frac{{S}_{0}{S}_{2}-S_{1}{}^{2}}{{S}_{0}{S}_{2}+S_{1}{}^{2}},$$
(2)
$${R}_{p}=\frac{n_{1}{}^{4}{S}_{0}{S}_{2}-n_{2}{}^{2}S_{1}{}^{2}}{n_{1}{}^{4}{S}_{0}{S}_{2}+n_{2}{}^{2}S_{1}{}^{2}},$$
(3)
$${S}_{i}={(n_{i}{}^{2}-u)}^{1\u22152},\phantom{\rule{1em}{0ex}}i=0,1,2,$$
(4)
$$u={\mathrm{sin}}^{2}\phantom{\rule{0.2em}{0ex}}\phi .$$
(5)
$$P={S}_{0}{S}_{2}\u2215S_{1}{}^{2},$$
(6)
$${R}_{s}=\frac{P-1}{P+1},\phantom{\rule{1em}{0ex}}{R}_{p}=\frac{n_{1}{}^{4}P-n_{2}{}^{2}}{n_{1}{}^{4}P+n_{2}{}^{2}}.$$
(7)
$$R_{p}{}^{2}+R_{s}{}^{2}=1.$$
(8)
$${a}_{4}{P}^{4}+{a}_{3}{P}^{3}+{a}_{2}{P}^{2}+{a}_{1}P+{a}_{0}=0,$$
(9)
$${a}_{4}=n_{1}{}^{8},$$
(10)
$${a}_{3}=-2n_{1}{}^{4}(n_{1}{}^{4}+n_{2}{}^{2}),$$
(11)
$${a}_{2}=n_{1}{}^{4}(n_{1}{}^{4}-12n_{2}{}^{2})+n_{2}{}^{4},$$
(12)
$${a}_{1}=-2n_{2}{}^{2}(n_{1}{}^{4}+n_{2}{}^{2}),$$
(13)
$${a}_{0}=n_{2}{}^{4}.$$
(14)
$${P}^{2}=(1-u)(n_{2}{}^{2}-u)\u2215{(n_{1}{}^{2}-u)}^{2}.$$
(15)
$${b}_{2}{u}^{2}+{b}_{1}u+{b}_{0}=0,$$
(16)
$${b}_{2}={P}^{2}-1,$$
(17)
$${b}_{1}=(n_{2}{}^{2}+1)-2n_{1}{}^{2}{P}^{2},$$
(18)
$${b}_{0}=n_{1}{}^{4}{P}^{2}-n_{2}{}^{2}.$$
(19)
$$\phi =\mathrm{arcsin}\left({u}^{1\u22152}\right).$$
(20)
$${R}_{s}=(P-1)\u2215(P+1)=-1\u2215\sqrt{2},$$
(21)
$$P={(\sqrt{2}-1)}^{2}.$$
(22)
$${n}_{1}={({n}_{2}\u2215P)}^{1\u22152}.$$
(23)
$${n}_{1}=(\sqrt{2}+1)\sqrt{{n}_{2}}.$$
(24)
$$P=n_{2}{}^{2}\u2215n_{1}{}^{4}.$$
(25)
$${c}_{2}{u}^{2}+{c}_{1}u+{c}_{0}=0,$$
(26)
$${c}_{2}=n_{1}{}^{8}-n_{2}{}^{8},$$
(27)
$${c}_{1}=2n_{1}{}^{2}n_{2}{}^{4}-n_{1}{}^{8}(n_{2}{}^{2}+1),$$
(28)
$${c}_{0}=n_{1}{}^{4}n_{2}{}^{2}(n_{1}{}^{4}-n_{2}{}^{2}).$$
(29)
$${R}_{p}={R}_{s}=-1\u2215\sqrt{2}=-0.7071,\phantom{\rule{1em}{0ex}}{n}_{1}=\sqrt{{n}_{2}}=1.2247.$$
(30)
$${D}_{1}=(\lambda \u22152){(n_{1}{}^{2}-{\mathrm{sin}}^{2}\phantom{\rule{0.2em}{0ex}}\phi )}^{-1\u22152},$$
(31)
$${d}_{1}={D}_{1}\u22152,$$
(32)
$${\Delta}_{r}=\pi ,{\Delta}_{t}=0,$$
(33)
$${\psi}_{r}+{\psi}_{t}=\pi \u22152.$$