= Solution
Differentiating the <image charge> potential gives
$$
E_x=\frac{\mu}{2\pi\varepsilon_0}\left(\frac{x}{x^2+y^2}-\frac{x-2d}{(x-2d)^2+y^2}\right).
$$
At the conducting plane, $E_x(d^-)=\mu d/[\pi\varepsilon_0(d^2+y^2)]$. With outward conductor normal $-\mathbf e_x$, the <surface charge density> is therefore
$$
\boxed{\sigma(y,z)=-\varepsilon_0 E_x(d^-,y,z)=-\frac{\mu d}{\pi(d^2+y^2)}.}
$$
It is independent of $z$ by translational symmetry. As a check, integrating across the plane gives $\int_{-\infty}^{\infty}\sigma(y,z)\,dy=-\mu$, the total induced charge per unit length, opposite to the real <line charge>.
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