## Abstract

The present problem, in geometrical optics, deals with a more general case of inclined mirrors than the simpler and familiar ones of the kaleidoscope, the “hinged mirrors,” and the sextant. Here each of the two plane mirrors has its own independent axis in its own plane, the two axes being, however, parallel to each other, which requires that the intersection line of the two reflecting planes be always parallel to the two axes, but it may otherwise be located anywhere in space. The problem is stated so as to involve three parameters, in addition to the two independent variables. The main problem is that of the locus of the secondary image, *O*″, under these conditions. Derivations are given also for the locus of the primary image, *O*′, and for the intersection, *C*, of the mirror planes. From the geometrical standpoint the problem is that of two related isosceles triangles, *O*′*AO, O*′*A*_{0}*O*″, where *O* is the object point and *A, A*_{0}, are the points where the mirror axes cut the principal plane, all five points being coplanar. The problem is made more general physically by supposing that the mirrors are half-silvered. The geometrical situation is expressed also in terms of intersecting circles, useful for graphical construction.

© 1928 Optical Society of America

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### Equations (20)

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(1)
$$\begin{array}{c}\begin{array}{cc}cos\mathrm{\lambda}=({d}^{2}+{D}^{2}-{r}^{2})/(2Dd);& cos\mu =({r}^{2}+{d}^{2}-{D}^{2})/(2rd);\end{array}\\ cos\nu =({r}^{2}+{D}^{2}-{d}^{2})/(2rD);\end{array}$$
(2)
$$\begin{array}{ll}cos\mathrm{\lambda}\hfill & \begin{array}{cc}={x}_{1}/{({{x}_{1}}^{2}+{{y}_{1}}^{2})}^{1/2};& cos\mu =(d-{x}_{1})/{[{(d-{x}_{1})}^{2}+{{y}_{1}}^{2}]}^{1/2};\end{array}\hfill \\ cos\nu \hfill & =[({{x}_{1}}^{2}+{{y}_{1}}^{2})-{x}_{1}d]/{\{({{x}_{1}}^{2}+{{y}_{1}}^{2})\xb7[{(d-{x}_{1})}^{2}+{{y}_{1}}^{2}]\}}^{1/2},\hfill \end{array}$$
(3)
$${\rho}^{2}={D}^{2}+4rsin\alpha [rsin\alpha +Dsin(\nu -\alpha )],$$
(4)
$$\eta -\beta =\eta -\u220a-\alpha +\psi =\xi +\psi -\alpha $$
(5)
$$\varphi =\eta +\beta =\eta +\u220a+\alpha -\psi =2\eta -\xi +\alpha -\psi .$$
(6)
$$\begin{array}{ll}{x}_{2}=d+rcos(\xi -\alpha ),\hfill & {y}_{2}=rsin(\xi -\alpha ).\hfill \end{array}$$
(7)
$$\begin{array}{ll}cos\alpha \hfill & =cos[\pi -(\mu +\xi )]=-cos(\mu +\xi )=(sin\mu sin\xi -cos\mu cos\xi )\hfill \\ \hfill & =[{y}_{1}sin\xi -(d-{x}_{1})cos\xi ]/{[{(d-{x}_{1})}^{2}+{{y}_{1}}^{2}]}^{1/2},\hfill \end{array}$$
(8)
$$sin\alpha ={[1-{cos}^{2}\alpha ]}^{1/2}=[{y}_{1}cos\xi +(d-{x}_{1})sin\xi ]/{[{(d-{x}_{1})}^{2}+{{y}_{1}}^{2}]}^{1/2}$$
(9)
$$\alpha =\pi -\mu -\xi =\pi -\xi -{cos}^{-1}\{(d-{x}_{1})/{[{(d-{x}_{1})}^{2}+{{y}_{1}}^{2}]}^{1/2}\},$$
(10)
$${x}_{2}=d+{y}_{1}sin2\xi -(d-{x}_{1})cos2\xi ,$$
(11)
$${y}_{2}=-[{y}_{1}cos2\xi +(d-{x}_{1})sin2\xi ].$$
(12)
$${x}_{0}={\rho}_{1}cos(\pi +\eta )=-{\rho}_{1}cos\eta ,$$
(13)
$${y}_{0}={\rho}_{1}sin(\pi +\eta )=-{\rho}_{1}sin\eta ,$$
(14)
$${\rho}^{2}={{x}_{2}}^{2}+{{y}_{2}}^{2}=({{x}_{1}}^{2}+{{y}_{1}}^{2})+2d[{y}_{1}sin2\xi +2(d-{x}_{1}){sin}^{2}\xi ],$$
(15)
$$\varphi =\eta +\beta ,$$
(16)
$$\begin{array}{ll}sin\psi \hfill & =[dsin(\alpha -\xi )]/\rho \hfill \\ \hfill & =d[{y}_{1}cos2\xi +(\text{d}-{x}_{1})sin2\xi ]/[\rho {[{(d-{x}_{1})}^{2}+{{y}_{1}}^{2}]}^{1/2}]\hfill \end{array}$$
(17)
$${R}^{2}={\rho}^{2}+{{\rho}_{1}}^{2}-2\rho {\rho}_{1}cos[\varphi -(\pi +\eta )],$$
(18)
$${R}^{2}={\rho}^{2}+{(dsin\xi )}^{2}/{sin}^{2}\u220a+2\rho dcos(\varphi -\eta )sin\xi /sin\u220a.$$
(19)
$${R}^{2}=\overline{C{O}^{2}}={{\rho}_{1}}^{2}+{D}^{2}+2{\rho}_{1}Dcos(\mathrm{\lambda}-\eta ).$$
(20)
$${R}^{2}=\left\{{D}^{2}+{(dsin\xi /sin\u220a)}^{2}+[(2Ddsin\xi )/sin\u220a]\xb7\left[\left(\frac{{D}^{2}+{d}^{2}-{r}^{2}}{2Dd}\right)cos\eta +sin{cos}^{-1}+\left(\frac{{D}^{2}+{d}^{2}-{r}^{2}}{2Dd}\right)sin\eta \right]\right\}.$$