## Abstract

Formulas of partially spatial coherent light are derived and its corresponding design algorithm is proposed for generating computer-generated holograms based on partially spatial coherent light. The partially coherent light is divided into two terms: spatially absolute coherent part and incoherent part. The former is propagated by angular spectrum method, while the latter is based on the optical transfer function. The related formulas are derived where the coherent function (degree of coherence) is related. A modified iterative algorithm is further developed for optimizing the phase distributions. Numerical simulations and optical experiments are both performed to verify the proposed algorithm. The results obtained by the proposed method and the traditional method are compared, and it is clear that the speckle contrasts in optical experiments are improved more than 46%, and the image quality is obviously improved. This method could also provide new applications for three-dimensional holographic display, beam shaping, and other wave-front modulation techniques.

© 2018 Optical Society of America under the terms of the OSA Open Access Publishing Agreement

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### Equations (19)

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(1)
$$I={I}^{\left(1\right)}\left(Q\right)+{I}^{\left(2\right)}\left(Q\right)+2\sqrt{{I}^{\left(1\right)}\left(Q\right)}\sqrt{{I}^{\left(2\right)}\left(Q\right)}\left|{\gamma}_{12}\right|\mathrm{cos}\left({\phi}_{1}-{\phi}_{2}\right),$$
(2)
$$E=\frac{1}{2}\sqrt{1-\left|{\gamma}_{12}\left(\tau \right)\right|}\sqrt{{I}^{\left(1\right)}\left(Q\right)+{I}^{\left(2\right)}\left(Q\right)}{e}^{i{\phi}_{in}}+\sqrt{\left|{\gamma}_{12}\left(\tau \right)\right|}\sqrt{{I}^{\left(1\right)}\left(Q\right){I}^{(2)}(Q)}{e}^{i{\phi}_{co}},$$
(3)
$${E}_{i}=\frac{\text{1}}{\text{2}}\sqrt{1-\left|{\gamma}_{12}\left(\tau \right)\right|}\sqrt{{I}^{\left(1\right)}\left(Q\right)+{I}^{\left(2\right)}\left(Q\right)}{e}^{i{\phi}_{in}},$$
(4)
$${E}_{c}=\sqrt{\left|{\gamma}_{12}\left(\tau \right)\right|}\sqrt{{I}^{\left(1\right)}\left(Q\right){I}^{(2)}(Q)}{e}^{i{\phi}_{co}},$$
(5)
$$S\left(\alpha ,\beta \right)={D}^{2}tri\left(\frac{\lambda d\alpha}{D}\right)\left(\frac{\lambda d\beta}{D}\right),\left|\frac{\lambda d\alpha}{D}\right|\le 1,\left|\frac{\lambda d\beta}{D}\right|\le 1,$$
(6)
$${I}_{i}\left(\xi ,\eta \right)={\displaystyle \underset{\infty}{\iint}{\tilde{I}}_{i}\left(\alpha ,\beta \right)\cdot S\left(\alpha ,\beta \right)\mathrm{exp}\left[i2\pi \left(\xi \alpha +\eta \beta \right)\right]d\alpha d\beta},$$
(7)
$${a}_{0}\left(\alpha ,\beta \right)={\displaystyle \underset{\infty}{\iint}{A}_{\text{0}}\left(\xi ,\eta \right)\mathrm{exp}\left[-i2\pi \left(\xi \alpha +\eta \beta \right)\right]d\alpha d\beta}.$$
(8)
$${A}_{c}\left(\xi ,\eta \right)={\displaystyle \iint {a}_{0}\left(\alpha ,\beta \right)\mathrm{exp}\left[ikd\sqrt{1-{\lambda}^{2}\left({\alpha}^{2}+{\beta}^{2}\right)}\right]\mathrm{exp}\left[i2\pi \left(\xi \alpha +\eta \beta \right)\right]d\alpha d\beta}.$$
(9)
$${I}_{0}\left(\xi ,\eta \right)\stackrel{OTF}{\to}{I}_{i}\left(\xi ,\eta \right).$$
(9)
$${A}_{0}\left(\xi ,\eta \right)=\sqrt{{I}_{0}\left(\xi ,\eta \right)}\cdot \mathrm{exp}\left(i{\phi}_{r}\right),$$
(11)
$${A}_{0}\left(\xi ,\eta \right)\stackrel{ASM}{\to}{A}_{c}\left(\xi ,\eta \right).$$
(10)
$${I}_{c}\left(\xi ,\eta \right)={\left|{A}_{c}\left(\xi ,\eta \right)\right|}^{2},$$
(11)
$${\phi}_{c}\left(\xi ,\eta \right)=angle\left[{A}_{c}\left(\xi ,\eta \right)\right].$$
(12)
$$A\left(\xi ,\eta \right)=\mathrm{exp}\left\{i\left[\left(1-\gamma \right){I}_{i}\left(\xi ,\eta \right)+\gamma {I}_{c}\left(\xi ,\eta \right)\right]\right\}\cdot \mathrm{exp}\left[i{\phi}_{c}\left(\xi ,\eta \right)\right],$$
(15)
$${\left|A\left(\xi ,\eta \right)\right|}^{2}\stackrel{OT{F}^{-1}}{\to}{I}_{i}^{\text{'}}\left(\xi ,\eta \right),$$
(16)
$$A\left(\xi ,\eta \right)\stackrel{ASM}{\to}{E}_{c}\left(\xi ,\eta \right).$$
(13)
$$I={I}_{i}^{\text{'}}+{\left|{E}_{c}\right|}^{2},$$
(14)
$$SC=\frac{\sqrt{\frac{1}{N}{\displaystyle \sum _{i=1}^{N}{\left({p}_{i}-\overline{I}\right)}^{2}}}}{\overline{I}},$$
(15)
$$PSNR=10\mathrm{lg}\left(\frac{{\left({2}^{n}-1\right)}^{2}}{MSE}\right),$$