## Abstract

The cumulative retardance
${\Delta}_{t}$ introduced between the *p* and the *s* orthogonal linear polarizations after two successive total internal reflections (TIRs) inside a right-angle prism at complementary angles ϕ and
$90\xb0-\varphi $ is calculated as a function of ϕ and prism refractive index *n.*
Quarter-wave retardation (QWR) is obtained on retroreflection with minimum angular sensitivity when
$n={\left(\sqrt{2}+1\right)}^{1/2}=1.55377$ and
$\varphi =45\xb0$. A QWR prism made of N-BAK4 Schott glass
($n=1.55377$ at
$\lambda =1303.5\text{\hspace{0.17em} nm}$) has good spectral response ($<5\xb0$ retardance error) over the 0.5–$2\text{\hspace{0.17em}}\mu \text{m}$ visible and near-IR spectral range. A ZnS-coated right-angle Si prism achieves QWR with an error of
$<\pm 2.5\xb0$ in the 9–$11\text{\hspace{0.17em}}\mu \text{m}$ (${\text{CO}}_{\text{2}}$ laser) IR spectral range. This device functions as a linear-to-circular polarization transformer and can be tuned to exact QWR at any desired wavelength (within a given range) by tilting the prism by a small angle around
$\varphi =45\xb0$.
A PbTe right-angle prism introduces near-half-wave retardation (near-HWR) with a $\le 2\%$ error over a broad
$\left(4\le \lambda \le 12.5\text{\hspace{0.17em}}\mu \text{m}\right)$ IR spectral range. This device also has a wide field of view and its interesting polarization properties are discussed. A compact (aspect ratio of
2), in-line, HWR is described that uses a chevron dual Fresnel rhomb with four TIRs at the same angle
$\varphi =45\xb0$.
Finally, a useful algorithm is presented that transforms a three-term Sellmeier dispersion relation of a transparent optical material to an equivalent cubic equation that can be solved for the wavelengths at which the refractive index assumes any desired value.

© 2008 Optical Society of America

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### Equations (23)

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(1)
$${\Delta}_{t}\left(\varphi \right)=\Delta \left(\varphi \right)+\Delta \left(90\xb0-\varphi \right),$$
(2)
$$\Delta \left(\varphi \right)=2\text{\hspace{0.17em}}{\text{tan}}^{-\text{1}}\left[{\left({n}^{2}\text{\hspace{0.17em}}{\text{sin}}^{\text{2}}\text{\hspace{0.17em}}\varphi -1\right)}^{1/2}/\left(n\text{\hspace{0.17em} sin \hspace{0.17em}}\varphi \text{\hspace{0.17em} tan \hspace{0.17em}}\varphi \right)\right],$$
(3)
$${\text{sin}}^{-\text{1}}\left(1/n\right)={\varphi}_{c}\le \varphi \le 90\xb0-{\varphi}_{c},$$
(4)
$${\Delta}_{t}\left(\text{45}\xb0\right)=4\text{\hspace{0.17em}}{\text{tan}}^{-\text{1}}\left[{\left({n}^{2}-2\right)}^{1/2}/n\right].$$
(5)
$${\Delta}_{t}\left(\text{45}\xb0\right)=\pi /2$$
(6)
$$\left[{\left({n}^{2}-2\right)}^{1/2}/n\right]=\text{tan}\left(\pi /\text{8}\right)=\sqrt{2}-1,$$
(7)
$$n={\left(\sqrt{2}+1\right)}^{1/2}=\mathrm{1.55377.}$$
(8)
$${\Delta}_{t}=90\xb0\text{,}\partial {\Delta}_{t}/\partial \varphi =0.$$
(9)
$$g\left(x\right)=f\left(x\right)+f\left(a-x\right),$$
(10)
$$g\prime \left(x\right)=f\prime \left(x\right)-f\prime \left(a-x\right).$$
(11)
$$g\prime \left(a/2\right)=0.$$
(12)
$${n}^{2}-1=\frac{{B}_{1}{\lambda}^{2}}{{\lambda}^{2}-{C}_{1}}+\frac{{B}_{2}{\lambda}^{2}}{{\lambda}^{2}-{C}_{2}}+\frac{{B}_{3}{\lambda}^{2}}{{\lambda}^{2}-{C}_{3}}\text{.}$$
(13)
$$n=n\left(\text{Si}\right)/n\left(\text{ZnS}\right)\text{.}$$
(14)
$$RE\approx \pi -\left(2/{n}^{2}\right)\text{,}$$
(15)
$$e=\left(RE/2\right)\text{sin \hspace{0.17em} 2}{\theta}_{i}\text{.}$$
(16)
$${\theta}_{r}=-45\xb0\text{,}\phantom{\rule[-0.0ex]{2.0em}{0.0ex}}{\epsilon}_{r}=45\xb0-\left(RE/2\right)\xb0\text{.}$$
(17)
$$\eta ={n}^{2}-1\text{,}\phantom{\rule[-0.0ex]{2.0em}{0.0ex}}x={\lambda}^{2}\text{,}$$
(18)
$${a}_{3}{x}^{3}+{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}=0\text{,}$$
(19)
$${a}_{3}={S}_{B}-\eta \text{,}\phantom{\rule[-0.0ex]{1.0em}{0.0ex}}{a}_{2}=\left(\eta -{S}_{B}\right){S}_{C}+{S}_{BC}\text{,}$$
(20)
$${a}_{1}={P}_{C}\left({S}_{B/C}-\eta {S}_{1/C}\right)\text{,}\phantom{\rule[-0.0ex]{1.0em}{0.0ex}}{a}_{0}=\eta {P}_{C}\text{.}$$
(21)
$${S}_{B}={\displaystyle \sum _{i=1}^{3}{B}_{i}}\text{,}\phantom{\rule[-0.0ex]{1.0em}{0.0ex}}{S}_{C}={\displaystyle \sum _{i=1}^{3}{C}_{i}}\text{,}$$
(22)
$${S}_{1/C}={\displaystyle \sum _{i=1}^{3}\left({1/C}_{i}\right)}\text{,}\phantom{\rule[-0.0ex]{1.0em}{0.0ex}}{S}_{BC}={\displaystyle \sum _{i=1}^{3}\left({B}_{i}{C}_{i}\right)}\text{,}$$
(23)
$${S}_{B/C}={\displaystyle \sum _{i=1}^{3}\left({B}_{i}/{C}_{i}\right)}\text{,}\phantom{\rule[-0.0ex]{1.0em}{0.0ex}}{P}_{C}={C}_{1}{C}_{2}{C}_{3}\text{.}$$