The configuration of the optical system is shown in Fig. 1. Three (or more) lasers are fused in a beam spliter and expand by a beam expander then modified by a LC-SLM. After that the modified light is imaged on the screen through the lens.
The light field in the focal plane of an SLM can be obtained with the Fourier transform of the emitted light. Thus, I=|FFT(U)|2, where I is the intensity of beam in the far field, and U is the light field of emitted beam. If the phase of U is randomly distributed, then beams are scattered to spots with a random intensity in the far field. To obtain random spots with a large bandwidth, the size of the SLM should be large. With a larger scattering angle, the spatial correlation of the random phase should be weak. If a particular device is selected, the optimal values of these parameters are fixed.
When an SLM is used, it is important to reduce the intensity of the zeroth-order of the far-field beam because it is often very strong, which seriously deteriorates the performance of structured light illumination and phase retrieval. Because of the wavelength-dependent response of the SLM, strong zeroth-orders will appear in each spectrum if arbitrary voltages are applied to the device. In this study, we are aiming to design a control voltage to eliminate these zeroth-order spots.
For the k-th beam spectrum, the zeroth-order intensity is related to the phase of the modulated beam; that is,
$$ I_{0k} = A_{k}\left|\sum_{m,n}^{N}\exp\left(i\phi_{k}\left(m,n\right)\right)\right|^{2}. $$
(1)
Where I0k denotes the zeroth-order intensity of the k-th spectrum; Ak is a fixed intensity factor; N is the total number of rows (and columns), i.e., the resolution of the SLM is N×N; and ϕk denotes the phase retardation of the corresponding spectrum. Accounting for the crosstalk effect in liquid crystal SLM, we have
$$ \phi_{k}\left(m,n\right) = f_{k}\left[x\left(m,n\right)\right]\otimes h\left(m,n\right). $$
(2)
Where fk[] denotes the voltage-phase response of the k-th spectrum, x(m,n) is the control voltage. h(m,n) is the point spread function (PSF) of the crosstalk [14], which has non-zero values only in a small region around (0,0). ⊗ represents the convolution operator.
To reduce every I0k, we need to solve for the control voltages. An apparent way is to optimize x(m,n) with the objective function \(J = \sum _{k}\ I_{0k}\). However, it is very hard.
In order to generate random illumination, x(m,n) should be independent random variables to broaden the scattering angle. Assuming that they are identically distributed, as N→∞, the expectation of the zeroth-order intensity is
$$ \overline{I_{0k}}=A_{k}N^{2}\left|\overline{\exp\left(i\phi_{k}\right)}\right|^{2}. $$
(3)
The objective function can then be converted to
$$ J^{\prime} = \sum_{k}\ \left|\overline{\exp\left(i\phi_{k}\right)}\right|^{2} $$
(4)
and
$$ \overline{\exp\left(i\phi_{k}\right)} = \overline{\exp\left(i\sum_{a,b}^{N}f_{k}\left[x\left(m-a,n-b\right)\right] h\left(a,b\right)\right)}. $$
(5)
To convert the two-dimensional convolution to one-dimensional vector multiplication, we reshape h(m,n) into a S×1 vector h′, and x(m,n) into a N2×S matrix x′. Here, S equals to the number of non-zero elements in h(m,n). We assign a new index t to represent the t-th row of x′. Further, we denote the PDF of x by p(x). Thus, we have
$$\begin{array}{*{20}l} \overline{\exp\left(i\phi_{k}\right)} &= \overline{\exp\left(i\sum_{t}^{N^{2}}f_{k}\left[x^{\prime}_{t}\right] h^{\prime}\right)} \end{array} $$
(6)
$$\begin{array}{*{20}l} &= \int_{-\pi}^{\pi}\cdots\int_{-\pi}^{\pi}\prod_{t}^{N^{2}} \left(\exp\left(if_{k}\left[x^{\prime}_{t}\right]h^{\prime}\right) p\left(x^{\prime}_{t}\right)\right) dx^{\prime}_{1} \cdots dx^{\prime}_{N^{2}} \end{array} $$
(7)
$$\begin{array}{*{20}l} &= \prod_{t}^{N^{2}} \int_{-\pi}^{\pi}\exp\left(if_{k}\left[x^{\prime}_{t}\right]h^{\prime}\right) p\left(x^{\prime}_{t}\right) dx^{\prime}_{t} \end{array} $$
(8)
$$\begin{array}{*{20}l} &= \prod_{i}^{S}\prod_{t}^{N^{2}} \int_{-\pi}^{\pi}\exp\left(if_{k}\left[x^{\prime}_{t}(i)\right]h^{\prime}(i)\right) p\left(x^{\prime}_{t}(i)\right) dx^{\prime}_{t}(i) \end{array} $$
(9)
$$\begin{array}{*{20}l} &\ \dot{=} \prod_{i}^{S}J^{\prime\prime}(i). \end{array} $$
(10)
Since p(x)>0 and the range of fk[] is only a few multiples of 2πs, reducing any of J′′(i) will reduce the J. To simplify the optimization, let hmax=max(h′), and choose the corresponding term \(J^{\prime \prime }_{max} = J^{\prime \prime }(i_{max})\) as the objective function. Where, imax is the index of the maximum of h′. We denote \(x^{\prime }_{t}(i_{max})\) as \(x_{t}^{\prime \prime }\), then the objective function becomes
$$ J^{\prime\prime}_{max} = \sum_{k}\ \prod_{t}^{N^{2}}\left|\int_{-\pi}^{\pi}\exp\left(if_{k}\left[x_{t}^{\prime\prime}\right]h_{max}\right)p\left(x^{\prime\prime}_{t}\right) dx^{\prime\prime}_{t}\right|^{2}. $$
(11)
Further, \(x^{\prime \prime }_{t}\) is independent and identically distributed, then
$$\begin{array}{@{}rcl@{}} J^{\prime\prime}_{max} &=& \sum_{k}\ \left|\int_{-\pi}^{\pi}\exp\left(if_{k}\left[x_{t}^{\prime\prime}\right]h_{max}\right)p\left(x^{\prime\prime}_{t}\right)dx^{\prime\prime}_{t}\right|^{2N^{2}}. \end{array} $$
(12)
In order to solve this problem with a computer, p(x) is discretized. Finally, by ignoring the N2-th power,the objective function becomes
$$\begin{array}{@{}rcl@{}} J^{\prime\prime}_{max} &=& \sum_{k}\ \left|\int_{-\pi}^{\pi}\exp\left(if_{k}[x^{\prime\prime}]h_{max}\right)p\left(x^{\prime\prime}\right)dx^{\prime\prime}\right|^{2} \end{array} $$
(13)
$$\begin{array}{@{}rcl@{}} &=& \sum_{k}\ \left|\sum_{l}^{M}\exp\left(if_{k}\left[x_{l}^{\prime\prime}\right]h_{max}\right)p_{l}\right|^{2}. \end{array} $$
(14)
Here, p(x) is a discrete PDF that has a probability of pl at \(x^{\prime \prime }_{l}\). Then, only 2M parameters need to be optimized to minimize \(J^{\prime \prime }_{max}\), thereby suppressing the zeroth-order of every spectrum.