polarizationjonesvectormatrices(編輯修改稿)
2025-06-08 10:53 本頁面
【文章內容簡介】 0 and ?y=?, so that ?x ?y ? Then our Efields are ? The negative sign before ? indicates a lag in the yvibration, relative to xvibration ? ?? ?????????tiytixAeEAeE11 Circular polarization ? To see this lag (in action), take the real parts ? We can then write ? Remembering that ?=2?/T, the path travelled by the evector is easily derived ? Also, since E2=Ex2 + Ey2 = A2(cos2?t + sin2?t)= A2 ? The tip of the arrow traces out a circle of radius A. tAtAEtAEyx????s in2c o sc o s??????? ???12 Circular polarization ? The Jones vector for this case – where Ex leads Ey is ? The normalized form is, ? This vector represents circularly polarized light, where E rotates counterclockwise, viewed headon ? This mode is called leftcircularly polarized light ? What is the corresponding vector for rightcircularly polarized light? ???????????????????????iAAeAeEeEEiioyioxo yx 1~2?????????i121???????i121Replace ?/2 with ?/2 to get 13 Elliptically polarized light ? If Eox ? Eoy , . if Eox=A and Eoy = B ? The Jones vector can be written ?????????????iBAiBAType of rotation? Type of rotation? counterclockwise clockwise What determines the major or minor axes of the ellipse? Here AB 14 Jones vector and polarization ? In general, the Jones vector for the arbitrar
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