2&#x03C0; ambiguity-free digital holography method for stepped phase imaging: erratum

Duo Zhang; Tuo Li; Wenxiu Lei; Qian Zhang; Shuai Wang; Jun Dong; Jun Dong

doi:10.1364/JOSAA.509956

Our method is only suitable for stepped samples, which was also indicated in the title of the original paper [1]. Only if the phase value corresponding to the stepped sample is discrete does Eq. (3) in the original paper have a unique solution. Here, based on stepped samples as prior knowledge, a mathematical proof is also given in Appendix A to further prove that it is possible to reconstruct the sample theoretically if the sample is step-shaped.

Through the proof, the feasibility of our method can be verified theoretically; that is, the stepped objects with phase interval greater than ${2}\pi$ can be measured. Due to negligence, there was an unintended error in which the phase step of the sample did not exceed ${2}\pi$ in Fig. 3(a2) and Fig. 4(a2) of the original paper. However, this method is applicable when the sample phase step exceeds ${2}\pi$. Here, a sample with phase interval which is more than ${2}\pi$ for measurement is shown in Fig. 1. In the simulation, the step size is set as 1 nm. Since the thickness of the stepped sample is unknown, usually the initial thickness $a$ is set as zero. Fig. 1(a) represents the original phase distribution. Fig. 1(b) represents the result retrieved by the proposed method. To analyze the measurement results intuitively, we draw the phase profile curve as shown in Fig. 1(c). In Fig. 1(c), the phase result recovered by the proposed method (indicated by blue dotted line) fits the original phase distribution (indicated by red line) well. It demonstrates that the proposed method can successfully retrieve a phase interval of more than ${2}\pi$.

Fig. 1. (a) The original phase distribution. (b) The result is recovered by the proposed method. (c) Comparison of phase profile information along the horizontal dotted lines in (a) and (b).

Download Full Size | PDF

The analysis of anti-noise performance is also shown here. We use the awgn function in MATLAB to add different degrees of Gaussian white noise in the simulations. The thickness of the sample affects the anti-noise performance. The simulation results indicate that a small step size requires a higher signal-to-noise ratio and a large step size requires a lower signal-to-noise ratio. For example, the phase of the sample to be tested is distributed in [4,18]. If the step size is set as 0.28 nm, the sample information can be reconstructed when the SNR value is more than 90. If the step size is set as 1 nm, the sample information can be reconstructed when the SNR value is 70. In the experiments, a He-Ne laser (DH-HN250, ${{\rm TEM}_{00}}$ mode), a beam expander (DH-GCO-2501, mean square error of wave front ${\lt}{0.2}\lambda$), a Fourier transform lens (GCO-0202M), and sophisticated three-dimensional translation tables (DH-GCX-F301MR, moving ${\rm accuracy} = {0.003}\;{\rm mm}$) are applied to ensure high SNR. At the same time, it is necessary to further remove the noise by using the high pass filtering method and median filtering method.

To sum up, we found and proved that the sample with stepped distribution can guarantee the equation to have a unique solution, and thus the phase wrapping problem can be avoided. Therefore, it may have potential application in measuring samples with stepped distribution.

The corresponding source code is provided in Appendix B.

APPENDIX A

Mathematical proof of unique solution of Eq. (3). Eq. (3) in the original paper has a cosine function. If the thickness of the sample is continuous, one cosine value will correspond to multiple thickness solutions, so phase wrapping will occur. We find that Eq. (3) will have a unique solution only when the thickness of the sample is discrete, and here we give the process of the mathematical proof.

Equation (3) in the original paper can be finally simplified to Eq. (1):

(1)$$\cos \left(\frac{{4\pi nd}}{\lambda}\right) = \frac{{{A^2} - {{\left({\frac{{n \;-\; 1}}{{n \;+\; 1}}} \right)}^2} \cdot {A_0}^2 - {{\left({\frac{{n \;-\; 1}}{{n \;+\; 1}}} \right)}^2} \cdot {{\left({\frac{2}{{n \;+\; 1}}} \right)}^4} \cdot {A_0}^2}}{{2{{\left({\frac{{n \;-\; 1}}{{n \;+\; 1}}} \right)}^2}{{\left({\frac{2}{{n \;+\; 1}}} \right)}^2} \cdot {A_0}^2}}.$$

Since the thickness of the sample is discrete, Eq. (1) is a constraint equation. And the mathematical thickness of $d$ is ${s_n}\Delta$ (${s_n} = {1},{2},{3 \ldots }$). The step size is $\Delta$, and ${s_n}\Delta$ represents the thickness of the ${{s}_n}$ search.

Next, we use proof by contradiction to prove that Eq. (1) has a unique solution when $d$ is discrete.

Assuming that Eq. (1) has two solutions ${s_1}\Delta$ and ${s_2}\Delta$, there are two possible scenarios, which we will discuss one by one.

Case 1: When the interval between the two solutions ($\frac{{4\pi n{s_1}\Delta}}{\lambda}$ and $\frac{{4\pi n{s_2}\Delta}}{\lambda}$) is not an integral multiple of ${2}\pi$, we can get Eq. (2) according to Eq. (1):

(2)$$\begin{split}\frac{{4\pi n{s_1}\Delta}}{\lambda} &= \varphi + 2{m_1}\pi \\\frac{{4\pi n{s_2}\Delta}}{\lambda} &= 2\pi - \varphi + 2{m_2}\pi \end{split},$$

where $\varphi$ is the wrapped phase, $n$ is the refractive index, and ${m_1}$ and ${m_2}$ are two integers (${m_1}$ and ${m_2}$ have values ranging from 0,1,2,3…). By adding the two equations of Eq. (2), we can get

(3)$$\frac{{2n({s_2} + {s_1})\Delta}}{\lambda} = {m_2}+ {m_1}+ 1.$$

Case 2: When the interval between $\frac{{4\pi n{s_1}\Delta}}{\lambda}$ and $\frac{{4\pi n{s_2}\Delta}}{\lambda}$ is an integral multiple of ${2}\pi$, we can get Eq. (4) according to Eq. (1):

(4)$$\begin{split}\frac{{4\pi n{s_1}\Delta}}{\lambda} &= \varphi + 2{m_1^\prime} \pi \\\frac{{4\pi n{s_2}\Delta}}{\lambda} &= \varphi + 2{m_2^\prime} \pi \end{split},$$

where $\varphi$ is the wrapped phase, $n$ is the refractive index, and $m_1^\prime$ and $m_2^\prime$ are two integers ($m_1^\prime$ and $m_2^\prime$ have values ranging from 0,1,2,3…). By subtracting the two equations of Eq. (4), we can get

(5)$$\frac{{2n({s_2} - {s_1})\Delta}}{\lambda} = {m_2^\prime} - {m_1^\prime} .$$

In Eq. (3) and Eq. (5), since ${s_1}$ and ${s_2}$ are integers, ${s_2} + {s_1}$ is an integer, and ${s_2} - {s_1}$ is an integer. ${2}n/\lambda$ is not an integer, and $\Delta$ is set to a fixed value. Therefore, the left sides of Eq. (3) and Eq. (5) are non-integers. However, the right sides of Eq. (3) and Eq. (5) are integers. Thus, Eq. (3) and Eq. (5) are not valid and the premise that the thickness of the step sample is discrete ensures that Eq. (1) has a unique solution. It is proved that a sample with step thickness can guarantee the existence of a unique solution to Eq. (1).

APPENDIX B

%%%%%%%% the phase exceed ${2}\pi$ %%%%%%%%%%%

clear

close all

clc

lamda1=650e-6; %The unit of the length in the program is mm.

N=201;%Pixel size of the sample

n1=1.52; % The refractive index of the specimen

x=0:1:N-1;

y=0:1:N-1;

delta_1=0.28e-6; %Step

%%%%%%%%%%%%%Specimen generation %%%%

ori_d(1:50.25,1:50.25)=500$^{\ast}$delta_1;

ori_d(1:50.25,50.25:100.5)=2200$^{\ast}$delta_1;

ori_d(1:50.25,100.5:150.75)=500$^{\ast}$delta_1;

ori_d(1:50.25,150.75:201)=2200$^{\ast}$delta_1;

ori_d(50.25:100.5,1:50.25)=2200$^{\ast}$delta_1;

ori_d(50.25:100.5,50.25:100.5)=500$^{\ast}$delta_1;

ori_d(50.25:100.5,100.5:150.75)=2200$^{\ast}$delta_1;

ori_d(50.25:100.5,150.75:201)=500$^{\ast}$delta_1;

ori_d(100.5:150.75,1:50.25)=500$^{\ast}$delta_1;

ori_d(100.5:150.75,50.25:100.5)=2200$^{\ast}$delta_1;

ori_d(100.5:150.75,100.5:150.75)=500$^{\ast}$delta_1;

ori_d(100.5:150.75,150.75:201)=2200$^{\ast}$delta_1;

ori_d(150.75:201,1:50.25)=2200$^{\ast}$delta_1;

ori_d(150.75:201,50.25:100.5)=500$^{\ast}$delta_1;

ori_d(150.75:201,100.5:150.75)=2200$^{\ast}$delta_1;

ori_d(150.75:201,150.75:201)=500$^{\ast}$delta_1;

%%%%%%%%%%%%%%%

phase1=4$^{\ast}$pi.$^{\ast}$n1.$^{\ast}$ori_d/lamda1; % Phase distribution of the laser when pass though the specimen.

figure,imshow(phase1,[])

% Colormap parula

%%

r=(n1-1)./(n1+1);%Reflection coefficient

t=2./(n1+1);% Transmission coefficient

A_s=r$^{\ast}$1;

A_st=t.$^{\ast}$r.$^{\ast}$t.$^{\ast}$1;

U1=A_s.$^{\ast}$exp(sqrt(-1)$^{\ast}$pi)+A_st.$^{\ast}$exp(sqrt(-1)$^{\ast}$4$^{\ast}$pi.$^{\ast}$n1. $^{\ast}$ori_d/lamda1);

A1=abs(U1);

%%%%%%%%%%%%%%%%%%%%%%%%%%%

k1=0;

count=0;

step_1=1e-6;% step size

d=zeros(N,N);

height=0.84e3/(step_1$^{\ast}$1e6);

d0=linspace(0,0,height);

fai=linspace(0,0,height);

m1=linspace(0,0,height);

for p=1:N

for q=1:N

for j=1:height %Change the size of j allows you to

flexibly set the maximum thickness of the object

x=(n1-1)./(n1+1);

d0(j)=j$^{\ast}$step_1;

fai(j)=4$^{\ast}$pi.$^{\ast}$n1.$^{\ast}$d0(j)/lamda1;

m1(j)=abs((x)^6-4$^{\ast}$(x)^5+6$^{\ast}$(x)^4-4$^{\ast}$(x)^3+2$^{\ast}$(x)^2-(A1(p,q)).^2-(2$^{\ast}$(x)^4-4$^{\ast}$(x)^3+2$^{\ast}$(x)^2).$^{\ast}$cos(fai(j)));

end

[k,M_1]=min(m1);

d(p,q)=(M_1)$^{\ast}$step_1;

end

count=count+1

end

phase=4$^{\ast}$pi.$^{\ast}$n1.$^{\ast}$d/lamda1;

figure,imshow(phase,[]) %Reconstructed phase

% Colormap parula

%%%%%%%% Noise analysis %%%%%%%%%%%

clear

close all

clc

lamda1=650e-6; %The unit of the length in the program is mm.

N=201;%Pixel size of the sample

n1=1.52; % The refractive index of the specimen

x=0:1:N-1;

y=0:1:N-1;

delta_1=0.28e-6; %

%%%%%%%%%%%%%Specimen generation %%%%

ori_d(1:50.25,1:50.25)=500$^{\ast}$delta_1;

ori_d(1:50.25,50.25:100.5)=2200$^{\ast}$delta_1;

ori_d(1:50.25,100.5:150.75)=500$^{\ast}$delta_1;

ori_d(1:50.25,150.75:201)=2200$^{\ast}$delta_1;

ori_d(50.25:100.5,1:50.25)=2200$^{\ast}$delta_1;

ori_d(50.25:100.5,50.25:100.5)=500$^{\ast}$delta_1;

ori_d(50.25:100.5,100.5:150.75)=2200$^{\ast}$delta_1;

ori_d(50.25:100.5,150.75:201)=500$^{\ast}$delta_1;

ori_d(100.5:150.75,1:50.25)=500$^{\ast}$delta_1;

ori_d(100.5:150.75,50.25:100.5)=2200$^{\ast}$delta_1;

ori_d(100.5:150.75,100.5:150.75)=500$^{\ast}$delta_1;

ori_d(100.5:150.75,150.75:201)=2200$^{\ast}$delta_1;

ori_d(150.75:201,1:50.25)=2200$^{\ast}$delta_1;

ori_d(150.75:201,50.25:100.5)=500$^{\ast}$delta_1;

ori_d(150.75:201,100.5:150.75)=2200$^{\ast}$delta_1;

ori_d(150.75:201,150.75:201)=500$^{\ast}$delta_1;

%%%%%%%%%%%%%%%

phase1=4$^{\ast}$pi.$^{\ast}$n1.$^{\ast}$ori_d/lamda1; % Phase distribution of the laser when pass though the specimen.

figure,imshow(phase1,[])

% Colormap parula

%%

r=(n1-1)./(n1+1);%Reflection coefficient

t=2./(n1+1);% Transmission coefficient

A_s=r$^{\ast}$1;

A_st=t.$^{\ast}$r.$^{\ast}$t.$^{\ast}$1;

U1=A_s.$^{\ast}$exp(sqrt(-1)$^{\ast}$pi)+A_st.$^{\ast}$exp(sqrt(-1)$^{\ast}$4$^{\ast}$pi.$^{\ast}$n1. $^{\ast}$ori_d/lamda1);

A1=abs(U1);

A1=awgn(A1,70);%Add Gaussian white noise

%%%%%%%%%%%%%%%%%%%%%%%%%%%

k1=0;

count=0;

step_1=1e-6;% step size

d=zeros(N,N);

height=0.84e3/(step_1$^{\ast}$1e6);

d0=linspace(0,0,height);

fai=linspace(0,0,height);

m1=linspace(0,0,height);

for p=1:N

for q=1:N

for j=1:height %Change the size of j allows you to

flexibly set the maximum thickness of the object

x=(n1-1)./(n1+1);

d0(j)=j$^{\ast}$step_1;

fai(j)=4$^{\ast}$pi.$^{\ast}$n1.$^{\ast}$d0(j)/lamda1;

m1(j)=abs((x)^6-4$^{\ast}$(x)^5+6$^{\ast}$(x)^4-4$^{\ast}$(x)^3+2$^{\ast}$(x)^2-

(A1(p,q)).^2-(2$^{\ast}$(x)^4-4$^{\ast}$(x)^3+2$^{\ast}$(x)^2).$^{\ast}$cos(fai(j)));

end

[k,M_1]=min(m1);

d(p,q)=(M_1)$^{\ast}$step_1;

end

count=count+1

end

phase=4$^{\ast}$pi.$^{\ast}$n1.$^{\ast}$d/lamda1;

figure,imshow(phase,[]) %Reconstructed phase

Acknowledgment

We thank all the experts for their valuable suggestions on this work.

Disclosures

The authors declare no conflicts of interest.

REFERENCE

1. D. Zhang, T. Li, W. Lei, et al., “2π ambiguity-free digital holography method for stepped phase imaging,” J. Opt. Soc. Am. A , 39, 2376–2382 (2022). [CrossRef]

2π ambiguity-free digital holography method for stepped phase imaging: erratum

Abstract

APPENDIX A

APPENDIX B

Acknowledgment

Disclosures

REFERENCE

Cited By

Figures (1)

Equations (5)

Journal of the Optical Society of America A