= Solution
Choose one circular convention as $\mathbf e_R=(\hat{\mathbf z}+i\hat{\mathbf y})/\sqrt2$ with time dependence $e^{-i\omega t}$. Reversing the handedness convention changes the sign of $i$ but not the result here. The scattered complex amplitude is
$$
\mathbf E_{s,R}\propto\frac{u_z+iu_y}{\sqrt2}\mathbf v.
$$
The scalar factor contributes a common phase to both output components. Consequently the scattered wave is \b[linearly polarized along $\mathbf v$], rather than circular or elliptical, with
$$
\boxed{I_R=\frac{KI_0}{2}(u_y^2+u_z^2)(1-u_y^2)=\frac{I_a+I_b}{2}.}
$$
The dominant longitudinal needle response has rank one, so both incident circular components emerge in the same real output direction.
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