Abstract

We intend to correct the typographical errors that occurred in our recent Letter [Opt. Lett. 39, 6422 (2014) [CrossRef]  ].

© 2015 Optical Society of America

In what follows, we correct the errors that appeared in [1].

The integrals in Eqs. (6) and (20) are each missing a normalization factor of 1/2. Equation (6) should read

120πa1(Θ)sinΘdΘ=1,
and Eq. (20) should read
αj(2k)=120π(1cosΘ)kaj(Θ)sinΘdΘ.

The second and third diagonal entries of the 4×4 matrix L0 given in Eq. (16) are incorrect. The correct result for L0 is

L0=[000001a¯2(0)+a¯3(0)200001a¯2(0)+a¯3(0)200001a¯4(0)].

The second term in the definition of L2 given in Eq. (18) contains several errors: there is a missing scalar factor of 2, and the (2, 2), (2, 3), and (3, 2) entries are incorrect. The correct definition of L2 is

L2=12ΔΩ^[1g0000c¯(2)0000c¯(2)0000a¯4(2)]+2c¯(2)[000001cot2θcosθsin2θφ0cosθsin2θφ1cot2θ00000].

As a consequence of the errata in Eq. (18), which are addressed above, Eqs. (23) and (24) have errors. The corrected equations are

Ω^·Q+μaQ+μs(1γ(0))Q12μsγ(2)ΔΩ^Q2γ(2)(1cot2θ)Q2γ(2)cosθsin2θφU=0,
and
Ω^·U+μaU+μs(1α3(0))U12μsc¯(2)ΔΩ^U+2γ(2)(1cot2θ)U2γ(2)cosθsin2θθφQ=0.

In these corrected equations, we have introduced γ(k)=(α2(k)+α3(k))/2 for k=0, 2.

Reference

1. J. Clark and A. D. Kim, Opt. Lett. 39, 6422 (2014). [CrossRef]  

References

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  1. J. Clark and A. D. Kim, Opt. Lett. 39, 6422 (2014).
    [Crossref]

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Equations (6)

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1 2 0 π a 1 ( Θ ) sin Θ d Θ = 1 ,
α j ( 2 k ) = 1 2 0 π ( 1 cos Θ ) k a j ( Θ ) sin Θ d Θ .
L 0 = [ 0 0 0 0 0 1 a ¯ 2 ( 0 ) + a ¯ 3 ( 0 ) 2 0 0 0 0 1 a ¯ 2 ( 0 ) + a ¯ 3 ( 0 ) 2 0 0 0 0 1 a ¯ 4 ( 0 ) ] .
L 2 = 1 2 Δ Ω ^ [ 1 g 0 0 0 0 c ¯ ( 2 ) 0 0 0 0 c ¯ ( 2 ) 0 0 0 0 a ¯ 4 ( 2 ) ] + 2 c ¯ ( 2 ) [ 0 0 0 0 0 1 cot 2 θ cos θ sin 2 θ φ 0 cos θ sin 2 θ φ 1 cot 2 θ 0 0 0 0 0 ] .
Ω ^ · Q + μ a Q + μ s ( 1 γ ( 0 ) ) Q 1 2 μ s γ ( 2 ) Δ Ω ^ Q 2 γ ( 2 ) ( 1 cot 2 θ ) Q 2 γ ( 2 ) cos θ sin 2 θ φ U = 0 ,
Ω ^ · U + μ a U + μ s ( 1 α 3 ( 0 ) ) U 1 2 μ s c ¯ ( 2 ) Δ Ω ^ U + 2 γ ( 2 ) ( 1 cot 2 θ ) U 2 γ ( 2 ) cos θ sin 2 θ θ φ Q = 0 .

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