Analysis of a new phase and height algorithm in phase measurement profilometry

Abstract

Traditional phase measurement profilometry adopts divergent illumination to obtain the height distribution of a measured object accurately. However, the mapping relation between reference plane coordinates and phase distribution must be calculated before measurement. Data are then stored in a computer in the form of a data sheet for standby applications. This study improved the distribution of projected fringes and deducted the phase-height mapping algorithm when the two pupils of the projection and imaging systems are of unequal heights and when the projection and imaging axes are on different planes. With the algorithm, calculating the mapping relation between reference plane coordinates and phase distribution prior to measurement is unnecessary. Thus, the measurement process is simplified, and the construction of an experimental system is made easy. Computer simulation and experimental results confirm the effectiveness of the method.

Appendix

Fringe function for cycle correction of grating fringe:

In $\Delta OAM$, the following relationships are obtained using the sine law,

$$\frac{{\overline {{OA}} }}{{\sin \angle AMO}}=\frac{{\overline {{OM}} }}{{\sin \angle OAM}}.$$

(12)

Additionally, the following relationships are shown in Fig. 1.

$$\sin \angle AMO=\cos \angle O{I_2}M=\cos (\beta +\theta ),$$

(13)

$$\sin \angle OAM=\sin (90^\circ +\beta ).$$

(14)

In $\Delta POA$and $\Delta PO^{\prime}M^{\prime}$,

$$\overline {{OA}} =\frac{{\overline {{PO}} }}{{\overline {{P{O^\prime }}} }}\overline {{{O^\prime }{M^\prime }}} =\frac{s}{f}\overline {{{O^\prime }{M^\prime }}} ,$$

(15)

where f is the focal distance of the projector.

$$\tan \beta =\frac{{\overline {{OA}} }}{s}.$$

(16)

Generally, $\overline {{OM}} =x$ and $\overline {{{O^\prime }{M^\prime }}} ={x^\prime }$. Then, x can be calculated from Eqs. (12)–(16) as follows:

$$x=\frac{{s{x^\prime }}}{{f\cos \theta - {x^\prime }\sin \theta }},$$

(17)

where ${\text{sin}}\theta =\frac{d}{s}$ and $\cos \theta =\frac{{\sqrt {{s^2} - {d^2}} }}{s}$.

So,

$$x=\frac{{{s^2}{x^\prime }}}{{f\sqrt {{s^2} - {d^2}} - x\prime d}}.$$

(18)