Abstract

This erratum gives corrections for the errors in a previously published paper [J. Opt. Soc. Am. A 36, 312 (2019) [CrossRef]  ].

© 2019 Optical Society of America

In [1], Eq. (5), which gives the output electric field after passing obliquely through the SEO, should read

$$\begin{split}{{\textbf{E}}_{{\rm{out}}}}(x,y)&={E_0}{\mathbb J} \cdot \left[{\begin{array}{*{20}{c}}1\\0\end{array}} \right]\\ &={E_0}\left[{\begin{array}{*{20}{c}}{\cos \left({\frac{{c{\rho_0}}}{4}} \right)\cos \left({\frac{{c{\rho_1}}}{4}} \right)-{e^{-{\rm{i}}{\phi_0}}}{e^{{\rm{i}}{\phi_1}}}\sin \left({\frac{{c{\rho_0}}}{4}} \right)\sin \left({\frac{{c{\rho_1}}}{4}} \right)}\\[6pt] {{\rm{i}}{e^{-{\rm{i}}{\phi_0}}}\cos \left({\frac{{c{\rho_1}}}{4}} \right)\sin \left({\frac{{c{\rho_0}}}{4}} \right)+{\rm{i}}{e^{-{\rm{i}}{\phi_1}}}\sin \left({\frac{{c{\rho_1}}}{4}} \right)\cos \left({\frac{{c{\rho_0}}}{4}} \right)}\end{array}} \right]\end{split}.$$
As a consequence of this, the Stokes parameters [originally reported between Eqs. (5) and (6)] are corrected to be
$$\begin{split}{s_1} &=\cos ({\phi_1})\sin \left({\frac{{c{\rho_0}}}{2}} \right)\sin ({\phi_0}-{\phi_1})+\sin ({\phi_1}) +\left[{\cos \left({\frac{{c{\rho_1}}}{2}} \right)\cos ({\phi_0}-{\phi_1})\sin \left({\frac{{c{\rho_0}}}{2}} \right)+\cos \left({\frac{{c{\rho_0}}}{2}} \right)\sin \left({\frac{{c{\rho_1}}}{2}} \right)} \right], \\ {s_2} &=-\sin ({\phi_1})\sin ({\phi_0}-{\phi_1})\sin \left({\frac{{c{\rho_0}}}{2}} \right)+\cos ({\phi_1}) +\left[{\cos \left({\frac{{c{\rho_1}}}{2}} \right)\cos ({\phi_0}-{\phi_1})\sin \left({\frac{{c{\rho_0}}}{2}} \right)+\cos \left({\frac{{c{\rho_0}}}{2}} \right)\sin \left({\frac{{c{\rho_1}}}{2}} \right)} \right], \\ {s_3} &=-\cos \left({\frac{{c{\rho_0}}}{2}} \right)\cos \left({\frac{{c{\rho_1}}}{2}} \right)+\cos ({\phi_0}-{\phi_1})\sin \left({\frac{{c{\rho_0}}}{2}} \right)\sin \left({\frac{{c{\rho_1}}}{2}} \right).\end{split}$$
There are sign errors in Eqs. (23) and (24). They are corrected to be
$${\alpha_3}=0,\quad {\unicode{x00B5}_3}=-\frac{{4{c^2}\cos [c{\cal L}\tan (\theta^\prime)/4]{\cal L}\tan \theta^\prime}}{{4{\pi^2}-{c^2}{{\cal L}^2}\mathop {\tan}\nolimits^2 \theta^\prime}},\quad {\gamma_3}=c,\quad {\eta_3}=0,$$
and
$${\alpha_4}=0,\quad{\unicode{x00B5}_4}=\frac{{4{c^2}\cos [c{\cal L}\tan (\theta^\prime)/4]{\cal L}\tan \theta^\prime}}{{4{\pi^2}-{c^2}{{\cal L}^2}\mathop {\tan}\nolimits^2 \theta^\prime}},\quad{\gamma_4}=c,\quad{\eta_4}=0.$$

REFERENCE

1. A. Ariyawansa, K. Liang, and T. G. Brown, “Polarization singularities in a stress-engineered optic,” J. Opt. Soc. Am. A 36, 312–319 (2019). [CrossRef]  

References

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  1. A. Ariyawansa, K. Liang, and T. G. Brown, “Polarization singularities in a stress-engineered optic,” J. Opt. Soc. Am. A 36, 312–319 (2019).
    [Crossref]

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Equations (4)

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E o u t ( x , y ) = E 0 J [ 1 0 ] = E 0 [ cos ( c ρ 0 4 ) cos ( c ρ 1 4 ) e i ϕ 0 e i ϕ 1 sin ( c ρ 0 4 ) sin ( c ρ 1 4 ) i e i ϕ 0 cos ( c ρ 1 4 ) sin ( c ρ 0 4 ) + i e i ϕ 1 sin ( c ρ 1 4 ) cos ( c ρ 0 4 ) ] .
s 1 = cos ( ϕ 1 ) sin ( c ρ 0 2 ) sin ( ϕ 0 ϕ 1 ) + sin ( ϕ 1 ) + [ cos ( c ρ 1 2 ) cos ( ϕ 0 ϕ 1 ) sin ( c ρ 0 2 ) + cos ( c ρ 0 2 ) sin ( c ρ 1 2 ) ] , s 2 = sin ( ϕ 1 ) sin ( ϕ 0 ϕ 1 ) sin ( c ρ 0 2 ) + cos ( ϕ 1 ) + [ cos ( c ρ 1 2 ) cos ( ϕ 0 ϕ 1 ) sin ( c ρ 0 2 ) + cos ( c ρ 0 2 ) sin ( c ρ 1 2 ) ] , s 3 = cos ( c ρ 0 2 ) cos ( c ρ 1 2 ) + cos ( ϕ 0 ϕ 1 ) sin ( c ρ 0 2 ) sin ( c ρ 1 2 ) .
α 3 = 0 , µ 3 = 4 c 2 cos [ c L tan ( θ ) / 4 ] L tan θ 4 π 2 c 2 L 2 tan 2 θ , γ 3 = c , η 3 = 0 ,
α 4 = 0 , µ 4 = 4 c 2 cos [ c L tan ( θ ) / 4 ] L tan θ 4 π 2 c 2 L 2 tan 2 θ , γ 4 = c , η 4 = 0.

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