Abstract

There are many applications in biology and metrology where it is important to be able to measure both the amplitude and phase of an optical wave field. There are several different techniques for making this type of measurement, including digital holography and phase retrieval methods. In this paper we propose an analytical generalization of this two-step phase-shifting algorithm. We investigate how to reconstruct the object signal if both reference waves are different in phase and amplitude. The resulting equations produce two different solutions and hence an ambiguity remains as to the correct solution. Because of the complexity of the generalized analytical expressions we propose a graphical-vectorial method for solution of this ambiguity problem. Combining our graphical method with a constraint on the amplitude of the object field we can unambiguously determine the correct result. The results of the simulation are presented and discussed.

© 2012 Optical Society of America

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Figures (5)

Fig. 1.

Experimental setup illustrated schematically. $D1$ and $D1$, matrix detectors; $AO$, unknown complex amplitude of object wave; $AR1$ and $AR2$, complex amplitude distributions of two arbitrary reference waves; BS, beam splitter.

Fig. 2.

Graphical vectorial illustration of Eq. (4) at the complex plane for some possible values of the object value. (a) Illustration of some possible values which are an answer to Eqs. (4a) and (4b). (b) Illustration of some possible values which are an answer to Eqs. (4c) and (4d). $A¯R1$ is a known reference complex value, $A¯O$object complex value, $c1$ is a circle with radius $|A¯1|=[I1]1/2$ and center at the coordinate origin. $c2$ is a circle with radius $|A¯1|=[I1]1/2$ and center at the coordinate origin.

Fig. 3.

Graphical solution of Eq. (4). Example.

Fig. 4.

Mach–Zehnder interferometric setup for two-step PSI. BS1, BS2, beam splitters; $λ/4$, quarter-wave plate; $P$, polarizer; M1, M2, mirrors; GG, ground glass; Ob, object; D1, D2, detectors; MO, micro-objective; L1–L4 , lenses; WP, Wollaston prism; IP1, IP2, imaging planes.

Fig. 5.

Simulation results. (a) Amplitude object; (b), (c) interferograms $I1$, $I2$ accordingly; (d) reconstructed amplitude of object wave; (e) error regions by recovering object amplitude; (f) error regions by recovering object phase.

Equations (28)

${I1=IR1+IO+2IR1IO cos(φR1−φO)I2=IR2+IO+2IR2IO cos(φR2−φO),$
$[IR1+IR2−2 cos(d)IR1IR2]IO2+[2 cos(d)(J2+J1)IR1IR2−2J1IR2−2J2IR1−4 sin2(d)IR1IR2]IO+J12IR2−2 cos(d)J1J2IR1IR2+J22IR1=0,$
${φO′=φR1±arccos(I1−IR1−IO′2IR1IO′)φO′=φR2±arccos(I2−IR2−IO′2IR2IO′).$
${U¯1=U¯R1+U¯OI1=|U¯1|U¯2=U¯R2+U¯OI2=|U¯2|.$
$h=AO,max=2|SΔO1RO2||U¯R1−U¯R2|=|U¯R1×U¯R2||U¯R1−U¯R2|=|U¯R1||U¯R2||sin(δR)||U¯R1|2+|U¯R2|2−2|U¯R1||U¯R2|cos(δR).$
${I1=IR1+IO+2IR1IO cos(φR1−φO)I2=IR2+IO+2IR2IO cos(φR2−φO).$
$φR1−φO=x;φR2−φR1=d;I1−IR1=J1;I2−IR2=J2.$
${2IR1IO cos(x)=J1−IO2IR2IO cos(x+d)=J2−IO,$
${cos(x)=J1−IO2IR1IOcos(x+d)=J2−IO2IR2IO.$
${cos((x+d2)−d2)=J1−IO2IR1IOcos((x+d2)+d2)=J2−IO2IR2IO,$
${cos(d/2)cos(x+d/2)+sin(d/2)sin(x+d/2)=J1−IO2IR1IOcos(d/2)cos(x+d/2)−sin(d/2)sin(x+d/2)=J2−IO2IR2IO.$
${2 cos(d/2)cos(x+d/2)=J1−IO2IR1IO+J2−IO2IR2IO2 sin(d/2)sin(x+d/2)=J1−IO2IR1IO−J2−IO2IR2IO,$
${cos(x+d/2)=J1−IO4 cos(d/2)IR1IO+J2−IO4 cos(d/2)IR2IOsin(x+d/2)=J1−IO4 sin(d/2)IR1IO−J2−IO4 sin(d/2)IR2IO,$
$d≠0;d≠π.$
$1=[J1−IO4 cos(d/2)IR1IO+J2−IO4 cos(d/2)IR2IO]2+[J1−IO4 sin(d/2)IR1IO−J2−A24 sin(d/2)IR2IO]2,$
$16IO=[J1−IOcos(d/2)IR1+J2−IOcos(d/2)IR2]2+[J1−IOsin(d/2)IR1−J2−IOsin(d/2)IR2]2,$
$16IO=(J1−IO)2cos2(d/2)IR1+2(J1−IO)(J2−IO)cos2(d/2)IR1IR2+(J2−IO)2cos2(d/2)IR2+(J1−IO)2sin2(d/2)IR1−2(J1−IO)(J2−IO)sin2(d/2)IR1IR2+(J2−IO)2sin2(d/2)IR2,$
$16IO=(J1−IO)2sin2(d/2)+cos2(d/2)(J1−IO)2cos2(d/2)sin2(h)IR1+2(J1−IO)(J2−IO)sin2(d/2)−2(J1−IO)(J2−IO)cos2(d/2)cos2(d/2)sin2(d/2)IR1IR2+(J2−IO)2sin2(d/2)+(J2−IO)2cos2(d/2)cos2(d/2)sin2(d/2)IR2,$
$16IO=4(J1−IO)2sin2(d)IR1−8 cos(d)(J1−IO)(J2−IO)sin2(d)IR1IR2+4(J2−IO)2sin2(d)IR2,$
$4IO=(J1−IO)2sin2(d)IR1−2 cos(d)(J1−IO)(J2−IO)sin2(d)IR1IR2+(J2−IO)2sin2(d)IR2.$
$4 sin2(d)IR1IR2IO=(J1−IO)2IR2−2 cos(d)(J1−IO)(J2−IO)⁢IR1IR2+(J2−IO)2IR1$
$4 sin2(d)IR1IR2IO=(J12−2J1IO+IO2)IR2−2 cos(d)(J1J2−(J2+J1)⁢IO+IO2)IR1IR2+(J22−2J2IO+IO2)IR1$
$4 sin2(d)IR1IR2IO=J12IR2−2J1IR2IO+IR2IO2−2 cos(d)J1J2IR1IR2+2 cos(d)(J2+J1)IR1IR2IO−2 cos(d)IR1IR2IO2+J22IR1−2J2IR1IO+IR1IO2$
$[IR1+IR2−2 cos(d)IR1IR2]IO2+[2 cos(d)(J2+J1)IR1IR2−2J1IR2−2J2IR1−4 sin2(d)IR1IR2]IO+J12IR2−2 cos(d)J1J2IR1IR2+J22IR1=0$
${I1−IR1+IO=2IR1IO cos(φR1−φO)I2−IR2+IO=2IR2IO cos(φR2−φO),$
${cos(φR1−φO)=I1−IR1+IO2IR1IOcos(φR2−φO)=I2−IR2+IO2IR2IO.$
${φR1−φO=±arccos(I1−IR1+IO2IR1IO)φR2−φO=±arccos(I2−IR2+IO2IR2IO),$
${φO=φR1±arccos[I1−IR1+IO2IR1IO]φO=φR2±arccos[I2−IR2+IO2IR2IO].$