Here's why that is the case.
The idea here is that an aqueous solution, i.e. a solution that has water as the solvent, can only conduct electricity if it contains ions.
Keep in mind that pure water does not conduct electricity very well because it contains very, very low concentrations of hydronium cations,
#2"H"_ 2"O" _ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) ->#the auto-ionization of water
Now, sulfuric acid is a strong acid, which means that it dissociates completely in aqueous solution to produce cations and anions.
You can assume that both dissociations are strong, meaning that a solution of sulfuric acid will contain
#"H"_ 2"SO"_ (4(aq)) + 2"H"_ 2"O" _ ((l)) -> 2"H"_ 3"O"_ ((aq))^(+) + "SO"_ (4(aq))^(2-)#
Since you now have a significant amount of hydronium cations and of sulfate anions,
#"Ba"("OH") _ (2(s)) ->"Ba"_ ((aq))^(2+) + 2"OH"_ ((aq))^(-)#
When you add barium hydroxide to the sulfuric acid solution, the hydronium cations and the hydroxide anions will neutralize each other to produce water
#"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l))#
This will reduce the number of ions present in solution, which implies that the solution's capacity to conduct electricity will also decrease.
Moreover, barium sulfate,
#"Ba"_ ((aq))^(2+) + "SO"_ (4(aq))^(2-) -> "BaSO"_ (4(s)) darr#
This will reduce the number of ions floating around in solution even more, further decreasing its ability to conduct electricity.
So there you have it. The neutralization reaction that occurs between barium hydroxide and sulfuric acid and the possible precipitation reaction that produces barium sulfate will reduce the number of ions present in solution.
This will in turn decrease the solution's ability to conduct electricity, which is why the light dims.