## Abstract

A conical mirror offers a means of magnifying the image of an object on an opaque screen, so that it may be viewed from a position directly in front of the screen, and the projecting distance necessary to produce the magnification is much less than that required by present methods. An experiment shows that the conical mirror gives high resolution when used in a specific projecting system. It may provide a convenient means with which to obtain high magnification for many optical systems. The magnification is shown by theory and experiment to be equal to the secant of the conical mirror angle.

© 1965 Optical Society of America

Full Article |

PDF Article
### Equations (20)

Equations on this page are rendered with MathJax. Learn more.

(1)
$$\frac{O{P}^{\prime}}{OP}=\frac{V{P}^{\prime}}{VP}.$$
(2)
$$\frac{O{P}^{\prime}}{OP}=\text{sec}\hspace{0.17em}(\text{conical}\hspace{0.17em}\text{mirror}\hspace{0.17em}\text{angle}).$$
(3)
$$\frac{-{r}^{\prime}}{r}=-M.$$
(4)
$$s={\int}_{{\theta}_{1}}^{{\theta}_{2}}\sqrt{{r}^{2}+{\left(\frac{dr}{d\theta}\right)}^{2}}d\theta .$$
(5)
$${s}^{\prime}={\int}_{{\theta}_{1}}^{{\theta}_{2}}\sqrt{{(-Mr)}^{2}+{\left[\frac{d(-Mr)}{d\theta}\right]}^{2}}d\theta .$$
(6)
$${s}^{\prime}=(-M)\hspace{0.17em}{\int}_{{\theta}_{1}}^{{\theta}_{2}}\sqrt{{r}^{2}+{\left(\frac{dr}{d\theta}\right)}^{2}}d\theta .$$
(7)
$$AE={(AM)}^{2}(QA)/[{(QM)}^{2}-{(QA)}^{2}].$$
(8)
$${(QM)}^{2}={(QV)}^{2}/(1-{\text{sin}}^{2}\varphi ).$$
(9)
$$AE=\frac{{(AM)}^{2}(QA)}{[{(QV)}^{2}/(1-{\text{sin}}^{2}\varphi )]-{(QA)}^{2}}.$$
(10)
$$\frac{V{P}^{\prime}}{QV}=\frac{A{P}^{\prime}}{QA}=\frac{V{P}^{\prime}-VF}{VF}=M,$$
(11)
$$\frac{V{P}^{\prime}}{QV}=\frac{V{P}^{\prime}-VF}{VF}$$
(12)
$$\frac{1}{QV}+\frac{1}{V{P}^{\prime}}=\frac{1}{VF}.$$
(13)
$$2x-(\text{tan}\alpha /2)y=0.$$
(14)
$$\frac{\text{sin}\beta}{QM}=\frac{\text{sin}2\omega}{QA+AE},$$
(15)
$$\frac{\text{sin}\beta}{2\hspace{0.17em}\text{cos}\omega (QM)}=\frac{\text{sin}\omega}{QA+AE}.$$
(16)
$$\frac{\text{sin}\beta}{AM}=\frac{\text{sin}\omega}{AE}.$$
(17)
$$\frac{2\hspace{0.17em}\text{cos}\omega (QM)}{AM}=\frac{QA+AE}{AE}.$$
(18)
$$2\hspace{0.17em}\text{cos}\omega =[{(QM)}^{2}+{(AM)}^{2}-{(QA)}^{2}]/(QM)(AM).$$
(19)
$$\frac{{(QM)}^{2}+{(AM)}^{2}-{(QA)}^{2}}{{(AM)}^{2}}=\frac{QA+AE}{AE}$$
(20)
$$\frac{{(QM)}^{2}-{(QA)}^{2}}{{(AM)}^{2}}+1=\frac{QA}{AE}+1.$$