- #1
ask_LXXXVI
- 53
- 0
*I am using the conventions of circular polarisation according to electrical engineering , not the one used in optics*
Let us take a uniform plane TEM wave traveling in +z direction which is composed of two linearly polarised TEM waves , one whose electric field lies in X direction , the other whose electric field lies in Y direction . Let us take the case of circular polarisation
so we take ,
Ex = E0 cos([tex]\omega[/tex] t - [tex]\beta[/tex] z) ax
Ey = E0 cos([tex]\omega[/tex] t - [tex]\beta[/tex] z + [tex]\pi[/tex]\2) ay
Now the resultant TEM wave has the Electric field vector left handed circularly polarised .
On the Poincare Sphere this will be given by the north pole point.
Suppose we had same wave ,but traveling in -ve z direction.
Ex = E0 cos([tex]\omega[/tex] t + [tex]\beta[/tex] z) ax
Ey = E0 cos([tex]\omega[/tex] t + [tex]\beta[/tex] z + [tex]\pi[/tex]\2) ay
My doubt is :- is the wave left handed circularly polarised or right handed circularly polarised ? And where on Poincare sphere is it located?
Let us take a uniform plane TEM wave traveling in +z direction which is composed of two linearly polarised TEM waves , one whose electric field lies in X direction , the other whose electric field lies in Y direction . Let us take the case of circular polarisation
so we take ,
Ex = E0 cos([tex]\omega[/tex] t - [tex]\beta[/tex] z) ax
Ey = E0 cos([tex]\omega[/tex] t - [tex]\beta[/tex] z + [tex]\pi[/tex]\2) ay
Now the resultant TEM wave has the Electric field vector left handed circularly polarised .
On the Poincare Sphere this will be given by the north pole point.
Suppose we had same wave ,but traveling in -ve z direction.
Ex = E0 cos([tex]\omega[/tex] t + [tex]\beta[/tex] z) ax
Ey = E0 cos([tex]\omega[/tex] t + [tex]\beta[/tex] z + [tex]\pi[/tex]\2) ay
My doubt is :- is the wave left handed circularly polarised or right handed circularly polarised ? And where on Poincare sphere is it located?
Last edited: