Problem 1. Show that the Jones matrix of the E4 field is jjknL+p) cOs OeknL-) 2 sin OeknL+) e -sinek,L) +e cOs OeL+) 2 The Jones matrix representation is: E E. 0 0 Where -K m m 1-KI 7 K m sin cos PC] F -sin0 cos0 (647ux) 0 eikn,L 0 [F]= jkn,L 0 (a47*ux) Where IJ is the identity matrix, Cm is the coupling ratio, K and K2 is 0.5, e is the rotation angle of the propagation light through PC1, L is the length, n, and ny are the indices of the two axes, k is wave vector and is the phase difference between the light on the fast axis and that on the slow axis

Problem 1. Show that the Jones matrix of the E4 field is j j(kn_L+9) cos Dej(kn_L-o) sin de j(kn, L+0) [E]= + 2 - sin de j(kn, L-0) cos ejlkn L+4) + jj(kn L+) 2 The Jones matrix representation is: [[F][PC][PMF] 0 0 Where LI-K4 WK, [Cm]= K. VI-K. [1] [PC] = [PMF] - [ Teiknu Teilkn_L+() [E]= ej 0 cos e sin - sin cos 0 ejo 0 jk, e .(4) 0 j(kn L+) 0 0 Where [I] is the identity matrix, Cm is the coupling ratio, K; and Ky is 0.5, e is the rotation angle of the propagation light through PC1, L is the length, Ny and Ny are the indices of the two axes, k is wave vector and p is the phase difference between the light on the fast axis and that on the slow axis. Problem 1. Show that the Jones matrix of the E4 field is j j(kn_L+9) cos Dej(kn_L-o) sin de j(kn, L+0) [E]= + 2 - sin de j(kn, L-0) cos ejlkn L+4) + jj(kn L+) 2 The Jones matrix representation is: [[F][PC][PMF] 0 0 Where LI-K4 WK, [Cm]= K. VI-K. [1] [PC] = [PMF] - [ Teiknu Teilkn_L+() [E]= ej 0 cos e sin - sin cos 0 ejo 0 jk, e .(4) 0 j(kn L+) 0 0 Where [I] is the identity matrix, Cm is the coupling ratio, K; and Ky is 0.5, e is the rotation angle of the propagation light through PC1, L is the length, Ny and Ny are the indices of the two axes, k is wave vector and p is the phase difference between the light on the fast axis and that on the slow axis