= Solution
The <method of images> uses the real <line charge> at the origin and the opposite <image charge> at $(2d,0)$. Let
$$
r_1=\sqrt{x^2+y^2},\qquad r_2=\sqrt{(x-2d)^2+y^2}.
$$
Superposing their logarithmic <electrostatic potentials> gives, in the physical region $x<d$,
$$
\boxed{\phi(x,y)=\frac{\mu}{2\pi\varepsilon_0}\log\frac{r_2}{r_1}.}
$$
At $x=d$, $r_1=r_2$, so $\phi=0$. Its tangential derivatives also vanish on that plane, giving $E_y=E_z=0$, exactly the <electrostatic boundary conditions at a conductor>. The image lies outside the physical domain, so it creates no extra physical source there. Together with the appropriate behaviour at infinity, <uniqueness of Poisson equation> identifies this field with the grounded-conductor solution.
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