I don't think setting the photocathode “reflectivity” to be nonzero is a trivial fix. Basically, as far as I can tell Geant4 detects photons by doing the following:

1) Is the photon being possibly detected? Okay, apply the detection efficiency. 2) Was it detected? If it was, tag it with “detected” and store its information. 3) Otherwise, if it was not detected, tag it with “absorbed” (or some other thing; not sure what it is specifically but I would guess absorbed), kill the photon track, and report the final position of the photon.

Intervening in the middle of step 3 is nontrivial, and not a user-level thing we're expected to be able to easily access. I also don't think we can do something clever with the analysis afterwards (like setting the photocathode reflectivity to 1.00, counting the photons which reflect and then applying the detection efficiency ourselves afterwards) because you get way too many photons reflecting.

Honestly, I think the best answer to this question of “how many more photons might we expect if we had a nonzero photocathode 'reflectivity'” might be by just using existing ratios and doing it by hand.

Like, for example, let: N=total number of original-track photons generated by an incident muon [known] R=photocathode “reflectivity” [basically known; we choose this] I=“probability that an original photon hits the PMT” (original photons which end up at the photocathode/all original photons) [known] D=“probability that a photon which hit the PMT is detected” (quantum efficiency) [known] P=number of detected photons due to a nonzero photocathode reflectivity [not known] A=original number of detected photons (N*I*D) (just writing this here for clarity)

Finding P is simple: (I've also attached this equation so you can get a better look) P(R)=N*I*D*sum(n=0 to inf){(1-D)*R*I}^n=N*I*D/{1-(1-D)*R*I}=A/{1-(1-D)*R*I}

That is, to be a detected photon, it must: –be an original photon (N) –of those photons, become incident on the PMT the first time (I) –eventually be detected (D)

To be a photon which is reflected off the photocathode once before coming back and being detected on its second attempt, it must: –not be detected the first time (1-D) –be reflected successfully (R) –become incident on the PMT again (I)

This needs to be an infinite sum since a photon could possibly be reflected a second time, and go through this process over again, etc. We can estimate this:

D=~7% (going on the quantum efficiency we're inputting, that's a decent number to pick if we are assuming things are flat). I=~90% according to a table I showed in a meeting on 4/20/17. If we were to actually do this formally, of course, we'd want to get the exact fraction from new data (which is probably smaller than this, since this number is for PTFE reflectivity of 1.00), but this serves as an estimate. R=~5% (approximately true for normal glass at around perpendicular incidence)

Using this, the percentage increase in number of photons is A/(1-(1-D)*R*I)=1/(1-0.93*0.9*0.05)=1/0.958=1.044A, about a 4.4% increase in total light output. Also, calculating other possible values for R and P, we have:

P(R=0.05)=1.044A=4.4% increase P(R=0.1)=1.091A=9.1% increase P(R=0.6)=2.0A=100% increase P(R=1.0)=6.13A=513% increase

Basically, all the numbers say here is that this correction in overall light output is probably a higher-order effect (unless the photocathode “reflectivity” is something huge like R=0.5 or above). Also, we probably can't correct the simulation to do this explicitly (or at least, I would have no understanding of how to do it). I hope this gives some insight into the issue.