To be opaque to pair production

[spossatamente]

What does "to be opaque to pair production" mean?
(I'm talking about fireball).
I know what a pair production is but I can't get the crucial point: does it mean the photons don't collide?
If it is so that means an opaque-to-p.p.- source have a greater luminosity (emits more photons) than a non opaque source ...
but it can't be because the optical depth is usually an increasing function of the photon energy, and therefore, a large optical depth would prevent the escape of high-energy photons from the source...

 

[Bob S]

From : http://press.princeton.edu/chapters/s9092.pdf

"Ultravioletand X-ray fluxes from compact sources are often attenuated by
photoelectric absorption from intervening neutral or ionized matter, including absorption by Milky Way gas. At TeV energies, the Ultraviolet and X-ray fluxes fromcompact sources are often attenuated by photoelectric absorption from intervening neutral or ionized matter, including absorption by Milky Way gas. At TeV energies, the universe
becomes opaque to pair production attenuation of very high-energy photons
(γ ) on cosmic infrared photons(γ'). This process, represented by the reaction
γγ' → e⁺ e^-
prevents us from seeing extragalactic sources of multi-
TeVradiation at distances 1 Gpc (redshifts z 0.2). The highest-energy
cosmic photon yet detected was at ≈90 TeVfrom a Milagro source [9]. At
100 TeV energies and higher, telescopes yet lack sufficient sensitivity to
detect cosmic sources or the plane of our Galaxy."

Bob S