## Abstract

Parhelic circles due to plate-oriented crystals (hence, with main
axes vertical) and 120° parhelia change in position when viewed
through a rotating polarizer. The parhelic circle moves vertically;
its largest shift is found at an azimuthal distance between 90° and
120° from the Sun. The 120° parhelia move both vertically and
horizontally. The magnitudes of the shifts are between 0.1° and
0.3°, depending on solar elevation. The mechanism is
polarization-sensitive internal reflection by prism faces of the ice
crystals. We outline the theory and present three visual and one
instrumental observation of the displacements of these halos in
polarized light.

© 1998 Optical Society of America

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### Equations (15)

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(1)
$$\frac{1}{{{n}_{\mathrm{eff}}}^{2}}=\frac{{sin}^{2}\mathrm{\gamma}}{n_{o}{}^{2}}+\frac{{cos}^{2}\mathrm{\gamma}}{n_{e}{}^{2}},$$
(2)
$${n}_{\mathrm{eff}}sini={n}_{o}sini\prime \left(e\to o\right),$$
(3)
$${n}_{o}sini={n}_{\mathrm{eff}}sini\prime \left(o\to e\right).$$
(4)
$$\left(1\right)o\left(3\right)o\left(2\right)\equiv \mathit{oo},\left(1\right)e\left(3\right)o\left(2\right)\equiv \mathit{eo},\left(1\right)o\left(3\right)e\left(2\right)\equiv \mathit{oe},\left(1\right)e\left(3\right)e\left(2\right)\equiv \mathit{ee}.$$
(5)
$${h}_{p}={h}_{\mathrm{sun}}\hspace{1em}\left(\mathit{oo},\mathit{ee}\mathrm{paths}\right),{n}_{o}cos{h}_{p}={n}_{e}cos{h}_{\mathrm{sun}}\hspace{1em}\left(\mathit{oe}\mathrm{path}\right),{n}_{e}cos{h}_{p}={n}_{o}cos{h}_{\mathrm{sun}}\hspace{1em}\left(\mathit{eo}\mathrm{path}\right).$$
(6)
$${n}_{e}sin{i}_{p}={n}_{o}sin{i}_{p}\prime \left(e\to o\right),{n}_{o}sin{i}_{p}={n}_{e}sin{i}_{p}\prime \left(o\to e\right),$$
(7)
$$tan{i}_{p}=tan\mathrm{\eta}cos\mathrm{\gamma}.$$
(8)
$${R}_{o\to e}={R}_{e\to o}=\frac{{sin}^{2}\left(2{i}_{p}\right){cos}^{2}\mathrm{\gamma}}{1-\left(1+1/{n}^{2}\right){cos}^{2}{i}_{p}{sin}^{2}\mathrm{\gamma}},{R}_{o\to o}={R}_{e\to e}=1-{R}_{o\to e},$$
(9)
$$cos{h}_{\mathrm{sun}}=nsin\mathrm{\gamma}.$$
(10)
$$\mathrm{Az}=180\xb0-2{i}_{p}.$$
(11)
$${T}_{e}\left(\mathrm{basal}\mathrm{face}\right)=1-\frac{{sin}^{2}\left(90\xb0-{h}_{\mathrm{sun}}-\mathrm{\gamma}\right)}{{sin}^{2}\left(90\xb0-{h}_{\mathrm{sun}}+\mathrm{\gamma}\right)},{T}_{o}\left(\mathrm{basal}\mathrm{face}\right)=1-\frac{{tan}^{2}\left(90\xb0-{h}_{\mathrm{sun}}-\mathrm{\gamma}\right)}{{tan}^{2}\left(90\xb0-{h}_{\mathrm{sun}}+\mathrm{\gamma}\right)}.$$
(12)
$${T}_{e}{R}_{e\to o}{T}_{o}+{T}_{e}{R}_{e\to e}{T}_{e}+{T}_{o}{R}_{o\to o}{T}_{o}+{T}_{o}{R}_{o\to e}{T}_{e}=T_{e}{}^{2}+T_{o}{}^{2}-{\left({T}_{e}-{T}_{o}\right)}^{2}{R}_{o\to e}\cong T_{e}{}^{2}+T_{o}{}^{2}.$$
(13)
$$\begin{array}{l}\mathrm{upper}\mathrm{and}\mathrm{lower}\mathrm{components}\left(\mathit{eo},\mathit{oe}\mathrm{paths}\right)=\frac{{T}_{e}{T}_{o}}{T_{e}{}^{2}+T_{o}{}^{2}}{R}_{o\to e}\cong 1/2{R}_{o\to e},\\ \mathrm{middle}\mathrm{component}\left(\mathit{ee}+\mathit{oo}\mathrm{path}\right)\cong 1-{R}_{o\to e}.\end{array}$$
(14)
$$\mathrm{\delta}=2{R}_{o\to e}\left[{h}_{p}\left(\mathit{oe}\right)-{h}_{\mathrm{sun}}\right],$$
(15)
$$F\left({i}_{p},\mathrm{\gamma}\right)=cos\left(|{i}_{p}-60\xb0|+60\xb0\right)\times cos\left(|{i}_{p}-60\xb0|-60\xb0\right)\mathrm{sin}\mathrm{\gamma}\mathrm{tan}\mathrm{\gamma}.$$