## Abstract

A method for testing the profiles of spherical surfaces is presented. It consists of measuring the transversal deflection of a reflected He–Ne laser beam when the surface is rotated around an axis located near its center of curvature. A set of formulas that enables us to calculate the shape of the profile as well as the decentering of the rotation axis is obtained. By using a simple experimental setup, we found the differences between the experimental profile with respect to the ideal one; the accuracy that was obtained is ~3 μm. The method may be improved and is useful for convex as well as for concave surfaces. With minor modifications it is possible to test large surfaces and weak aspherics.

© 1993 Optical Society of America

Full Article |

PDF Article
### Equations (20)

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

(1)
$$y\left(x\right)={\displaystyle {\int}_{x0}^{x}\text{tan}\left[\frac{\varphi \left(x\right)}{2}\right]\mathrm{d}x},$$
(3)
$$\begin{array}{cc}\text{tan}\phantom{\rule{0.2em}{0ex}}i& =\frac{{\mathrm{P}}^{\prime}\mathrm{B}}{\text{PB}}\\ & =\frac{1}{r}\frac{\mathrm{d}r}{\mathrm{d}\theta}.\end{array}$$
(4)
$$r\left({\theta}_{2}\right)=r\left({\theta}_{1}\right)\text{exp}\left\{{\displaystyle {\int}_{{\theta}_{1}}^{{\theta}_{2}}\text{tan}\left[\varphi (\theta )/2\right]\mathrm{d}\theta}\right\}.$$
(5)
$$\text{tan}\phantom{\rule{0.2em}{0ex}}\varphi =\frac{yf}{l\left({l}^{\prime}-f\right)-{l}^{\prime}f}.$$
(6)
$$\text{tan}\phantom{\rule{0.2em}{0ex}}\varphi =\frac{y}{f}.$$
(7)
$${\left(x-a\right)}^{2}+{\left(y-b\right)}^{2}={{r}_{0}}^{2},$$
(8)
$$\begin{array}{ll}r\left(\theta ,a,b,{r}_{0}\right)=& a\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta +b\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \\ & \pm \phantom{\rule{0.2em}{0ex}}{\left[{{r}_{0}}^{2}-{\left(a\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta -b\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta \right)}^{2}\right]}^{1/2}.\end{array}$$
(9)
$$\begin{array}{l}a=-0.200\pm 0.060\phantom{\rule{0.2em}{0ex}}\text{mm},\\ b=0.017\pm 0.009\phantom{\rule{0.2em}{0ex}}\text{mm},\end{array}$$
(10)
$$\begin{array}{cc}r(\theta )& =r\left({\theta}_{1}\right)\text{exp}\left\{{\displaystyle \sum _{i=1}^{N}\left[\text{tan}\left({\varphi}_{i}/2\right)+\text{tan}\left({\varphi}_{i+1}/2\right)\right]\left(\alpha /2\right)}\right\}\\ & ={r}_{1}e\left({\varphi}_{i},\alpha \right),\end{array}$$
(11)
$$\delta r={r}_{1}\delta e+e\delta {r}_{1}.$$
(12)
$$\begin{array}{ll}\delta e& ={\displaystyle \sum _{i=1}^{N}\left|\frac{\partial e}{\partial {\varphi}_{i}}\right|\delta {\varphi}_{i}+\left|\frac{\partial e}{\partial \alpha}\right|\delta \alpha}\\ & =e(\theta )[{\displaystyle \sum _{i=1}^{N}\left|{\text{sec}}^{2}\left({\varphi}_{i}/2\right)+{\text{sec}}^{2}\left({\varphi}_{i+1}/2\right)\right|\times \frac{\alpha \delta {\varphi}_{i}}{4}}\\ & \phantom{\rule{0.2em}{0ex}}+{\displaystyle \sum _{i=1}^{N}\left|\text{tan}\left({\varphi}_{i}/2\right)+\text{tan}\left({\varphi}_{i+1}/2\right)\right|\times \frac{\delta \alpha}{2}}].\end{array}$$
(13)
$$\delta \varphi =\left|\frac{\partial \varphi}{\partial y}\right|\delta y+\left|\frac{\partial \varphi}{\partial f}\right|\delta f,$$
(14)
$$\frac{\partial \varphi}{\partial y}=\frac{f}{{f}^{2}+{y}^{2}},$$
(15)
$$\frac{\partial \varphi}{\partial f}=\frac{y}{{f}^{2}+{y}^{2}}.$$
(16)
$$\Delta r={r}_{T}-{r}_{E}={r}_{1}\left({e}_{T}-{e}_{E}\right)={r}_{1}\Delta e,$$
(17)
$$\begin{array}{ll}\delta \left(\Delta r\right)& ={r}_{1}\delta \left({e}_{T}-{e}_{E}\right)+\left|\Delta e\right|\delta {r}_{1}\\ & ={r}_{1}\delta {e}_{E}+\left|\Delta e\right|\delta {r}_{1},\end{array}$$
(18)
$$\delta \varphi \cong \frac{\delta y}{f}.$$
(19)
$$\delta e=\frac{N\alpha}{2f}\delta {y}_{\text{max}}+\frac{N{y}_{\text{max}}}{2f}\delta \alpha .$$
(20)
$$\delta e=\frac{1}{2f}\left[\Delta \theta \delta {y}_{\text{max}}+N{y}_{\text{max}}\delta \alpha \right].$$