## Abstract

The maximum is found for the deviation of light passing through a transparent wedge of refractive index *n* and wedge angle *α*. The methods are conceptual and geometric, and they require very little calculation. There turn out to be two qualitatively different ray path candidates for maximum deviation, and the geometric approach leads naturally to a criterion involving *n* and *α* that decides between the two candidates. Finding the maximum deviation is equivalent to finding the outer radius of a circular halo.

© 2008 Optical Society of America

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### Equations (13)

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(1)
$$\frac{d}{2}=\mathrm{sin}\frac{\mathrm{\Delta}}{2}\mathrm{.}$$
(2)
$$rD(t)=s(t)-s(t-\alpha ),$$
(3)
$$r{D}^{\prime}(t)={s}^{\prime}(t)-{s}^{\prime}(t-\alpha )\phantom{\rule{0ex}{0ex}}\{\begin{array}{l}>0\phantom{\rule[-0.0ex]{1em}{0.0ex}}\text{if \hspace{0.17em} \hspace{0.17em}}\alpha \le t<{t}_{0}\\ ={s}^{\prime}(t)-{s}^{\prime}(-t+\alpha )>0\phantom{\rule[-0.0ex]{1em}{0.0ex}}\text{if \hspace{0.17em} \hspace{0.17em}}\alpha /2<t\le \alpha \end{array}\mathrm{.}$$
(4)
$$\mathrm{tan}\frac{{\alpha}_{2}}{2}=\frac{1}{m}\mathrm{.}$$
(5)
$$m\mathrm{sin}\frac{\alpha}{2}=\mathrm{cos}\frac{\alpha}{2}$$
(6)
$$\mathrm{cos}(\alpha +{\mathrm{\Delta}}_{\mathrm{C}})=\mathrm{cos}\alpha -m\mathrm{sin}\alpha ,$$
(7)
$$0\le \alpha +{\mathrm{\Delta}}_{\mathrm{C}}\le \pi \mathrm{.}$$
(8)
$$\mathrm{sin}\frac{{\mathrm{\Delta}}_{\mathrm{T}}}{2}=m\mathrm{sin}\frac{\alpha}{2}\mathrm{.}$$
(9)
$${d}_{\mathrm{T}}\le {d}_{\mathrm{C}}\phantom{\rule[-0.0ex]{1em}{0.0ex}}\text{if \hspace{0.17em} \hspace{0.17em}}0\le \alpha \le {\alpha}_{1},$$
(10)
$${d}_{\mathrm{C}}\le {d}_{\mathrm{T}}\phantom{\rule[-0.0ex]{1em}{0.0ex}}\text{if \hspace{0.17em} \hspace{0.17em}}{\alpha}_{1}\le \alpha \le {\alpha}_{2},$$
(11)
$$m\mathrm{sin}\frac{\alpha}{2}=\mathrm{cos}\frac{3\alpha}{2}$$
(12)
$${\mathrm{\Delta}}_{\mathrm{max}}={\mathrm{\Delta}}_{\mathrm{C}}=-\alpha +{\mathrm{cos}}^{-1}(\mathrm{cos}\alpha -m\mathrm{sin}\alpha )\mathrm{.}$$
(13)
$${\mathrm{\Delta}}_{\mathrm{max}}={\mathrm{\Delta}}_{\mathrm{T}}=2{\mathrm{sin}}^{-1}(m\mathrm{sin}\frac{\alpha}{2})\mathrm{.}$$