How wvlnsh.f affect trapping efficiency of WLSF

Dear @horvathd ,
It is known that the trapping efficiency of the wavelength shifting fiber is 5.4%. In order to obtain the neutron count(N1) with photon number greater than 100 at the surface from “regwsfxx” to “regblkxx”, I calculated that neutron count(N2) with photon number greater than 1852(100/5.4%=1852), from “regznsxx” to “regwsfxx” , so I want to adjusted “PRO” at wvlnsh.f that let N1 approximately equal to N2, but it seems “PRO” does not affect neutron count , is there anything wrong with my wvlnsh.f ?
L+Z.flair (4.5 KB)
r-wvl.f (3.0 KB)
Look forward to your help!

Dear @xiongbp,

Could you clarify you question? The values count(N1) and count(N2) are for number of neutrons? The wavelength shifting routine will not have any effect on the propagation of neutrons, only optical photons. How do you define these counts of neutrons?

Dear @blefebvre ,
Here I use a user routine , just see the attachment.
r-sur.f (2.7 KB)
My structure have 4 parts, reglifxx, regznsxx, regwsfxx, regblkxx, so it have 3 interfaces. I use 1000000 neutrons incident and every neutron will generated photon at regznsxx, the photon count may from 0 to a large number, at the interface from “regznsxx” to “regwsfxx”, if the photon count >=1852, I get the neutron number +1, finally I can get how many neutron generated photons that >=1852, this is N1 .N2 is almost the same explanation .

Dear @blefebvre ,
From “regznsxx” to “regwsfxx”, I call it interface A, from “regwsfxx” to “regblkxx”, I call it interface B.
If interface A have 4000 neutrons that their generated photon number above 1852, the wavelength shifting fiber’s trapping efficiency is 5.4%, then at interface B, the neutron count that photon number is greater than 100(1852*5.4%=100) should be approximately 4000 too,right? How should I adjust the user routine to achieve it ?

Dear @blefebvre ,
The topic seems not express my question clearly, I will open another topic about it, please delete this topic, thank you.