4.2 Effect of the SLM Certainly, to understand what phase is accumulated to the Electric field, it is necessary to correctly quantify the evolution of the Electric field while passing through the SLM. For this reason, we will construct the Jones matrix of the SLM [13]. And calculate the phase it imparts on two different circular polarization states of light. We know TNSLM is made up of many layers of birefringent layers. First, lets assume the twisting is linear, i.e. the twist φ can be interpreted as ψ(z) = αz (4.4) where z is the distance along the propagation direction and α is contant.For our TNSLM total twist angle, ψ(d ) = π2 ; d being the thickness of the SLM. And, we denote the total birefringence of the SLM to be δt = 2π (n e − n 0 )d λ (4.5) Now let there be N number of layers in the SLM. With these properties, each of the layer will be twisted at an angle φ N and the birefringence of that layer is given by δt N. So the full Jones matrix of the SLM is given as: 2πn0 d e i λN J = R(ψ) 0 N ψ R(− ) 2πn e d N e i λN 0 (4.6) Which can be further simplified as: h ψ iN J = R(ψ) ψ0 R(− ) N (4.7) where ψ0 = e i ⇒ J =e 2πn d δ i ( λ0 + 2t 2πn 0 d λN δt ψ cos N e −i 2N ) R(ψ) δt ψ si n N e i 2N ei δt 2N δt e −i 2N 0 δt ψ si n N e −i 2N δt ψ cos N e i 2N N 0 δt e i 2N =e (4.8) 2πn d δ i ( λ0 + 2t ) R(ψ) X −iY Z −Z X +iY (4.9) with X Y Z s = cos ψ2 + ( δt 2 ) 2 s δt /2 ψ2 + ( δt )2 ¶ = µq si n 2 ψ2 + ( δ2t )2 s ψ ψ2 + ( δt )2 ¶ = µq si n 2 ψ2 + ( δ2t )2 (4.10) Here, the phase factor in front of the rotation matrix gives the dynamical phase due to the SLM. So, dynamical phase = µ 2πn 0 d δt + λ 2 ¶ (4.11) Now the geometrical phase associated to a particular polarization state due to J’ is given as ar g ⟨ψ|J ′ |ψ⟩ for |ψ⟩ ∈ {|LC P ⟩, |RC P ⟩}. where the matrix J’ is given as: J ′ = R(φ) X −iY Z −Z X +iY = R(φ)U (δt , ψ) (4.12) Here, at this point we face an experimental difficulty that the parameter δt i.e. the total birefringence present in J’ that will be present in thus calculated geometrical phases, cannot be measured directly. For this reason, we record the Mueller matrices of the SLM extract the "effective" birefringence and rotation and calculate back the total birefringence[14]. 4.2.1 Rotator Retarder Model We employ the fact that any unitary matrix U has an optically equivalent system consisting of one retarder and one rotator [15]. So we write U (δt , ψ) = R(ψeq )Ret (δeq , θeq ) cosψeq si nψeq cosθeq = −si nψeq cosψeq si nθeq δeq −si nθeq e −i 2 cosθeq 0 0 ei δeq 2 cosθeq −si nθeq (4.13) si nθeq cosθeq From this relation and equation 4.12, we have : si n δeq 2 X = cos Y = si n Z = cos δeq 2 δeq 2 δeq 2 cosψeq cos(2θeq − ψeq ) si nψeq (4.14) si n(2θeq − ψeq ) = 0 4.13 we get δ in terms of δ and ψ as: comparing eqns. 4.10 and 4.15, t eq eq δt = 2 s · µ µ ¶ ¶¸2 δeq −1 cos cos − ψ2 cosψeq 2 (4.15)