= Solution
For $0<v<c$, the <Lorentz transformation> with coincident origins is
$$
\boxed{t'=\gamma\left(t-\frac{vx}{c^2}\right),\qquad x'=\gamma(x-vt),\quad\gamma=(1-v^2/c^2)^{-1/2}.}
$$
For two simultaneous events in $S$, $\Delta t=0$, so $\Delta t'=-\gamma v\Delta x/c^2$. Distinct spatial positions therefore generally give different times in the boosted frame: this is <relativity of simultaneity>.
Choose the emission events at $(t,x)=(0,0)$ and $(0,d)$. After emission, the <photon> world lines are $x=ct$ and $x=d+ct$. Applying the <Lorentz transformation> to either gives
$$
x'-ct'=\gamma(1+v/c)(x-ct).
$$
Thus the two <photons> move at <speed> $c$ in $S'$ on parallel lines whose intercepts differ by $\gamma(1+v/c)d$. Measuring their positions at the same $t'$ after both emissions gives the <photon separation under a collinear Lorentz boost>:
$$
\boxed{\Delta x'=\gamma(1+v/c)d=d\sqrt{\frac{c+v}{c-v}}.}
$$
This separation is constant. It is not the contracted distance between the stationary sources: the two emission events are not simultaneous in $S'$, and one <photon> has already moved when the other is emitted.
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