Description of the method, and scans of the original paper:
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Initial state: nucleus A (spin=0) + electron in an S state (spin=1/2)
Final state: nucleus B* (spin=1) + neutrino (spin=1/2)
What happens at microscopic level, according to the modern view (unknown at the time): a spin=1 virtual W is exchanged, so we have A(s=0)+e(s=+1/2)->B*(s=+1)+ν(s=-1/2)
We want in fact to demonstrate that s(ν)=-1/2 and not +1/2.

Second step of the chain: B* (spin=1) decays to B (spin=0) + photon (spin=1)
Angular momentum conservation imposes the spin of the photon to be parallel to that of B*.
Spin of B* was opposed to neutrino, as said before.
Momentum of B* was opposed to neutrino, too.
So, helicity of B* was the same as neutrino helicity.
If the photon is selected in the same direction as B* recoil, its helicity is the same as B* and, then, as the neutrino.
Now the game is to determine photon helicity.

If B* emits the photon while at rest, this photon will be unable to give the inverse reaction γ+B -> B*, since part of the energy was carried away by the recoiling father nucleus.
But if the B* emitted the gamma while traveling at the right speed, the gamma can be able to do the inverse reaction.
So, A=Eu152 and B=Sm are very fortunate since the neutrino from Eu152 electron capture and the photon from Sm*->Sm have roughly the same energy!
Only photons emitted in the same direction of motion of the father B* will be able to give resonant scattering with B's in the distant target.

Experiment:

The resonantly scattered photons will go in all directions and can be detected by the NaI detector.

How do we know photon helicity?
This is why the source A is surrounded by magnetized iron: its transparency to the photons depend on the relative alignment of the photon and iron spins. By changing the latter, you see a difference in the number of counts in NaI.
(The Pb shield just prevents the photons to reach NaI if they have not been scattered by the B target.)