Question

15. You make VLA images of a point source at all 4 stokes and find 1=50.0,Q=3, U-5.2, and V--0.1 mly. (a) Calculate the linearly polarized intensity, percentage of linear polarization, and give the polarization angle in degrees. (b) Calculate the percentage of circular polarization. (c) Note that Q, U, and V can be positive or negative. What about stokes I, can that ever be negative?

Answer 1

Since Ex and Ey are instantaneous time functions, the polarization ellipse is valid only for a given instant of time. For an un-polarized wave, it is not possible to obtain the instantaneous tilt angle T or the axial ratio. To overcome these limitations Stokes parameters are used, which is a set of four time averaged components (I, Q, U, V) obtained from the wave and can be related to the polarization ellipse. From its properties, (i) if two waves have identical Stokes parameters, the waves are identical, and (ii) if several independent waves propagating in the same direction are superimposed, the Stokes parameters of the resultant wave is the sum of Stokes parameters of individual waves.

pola mization is the ratio of the fractiona intensity وار the Total The degree of polarization or of polarized. emission to Vareuzer ; oepsi p= I intens sity. Ipol I tot Here 1 - 5019-3, US -5.2, r=-orim Jy 32 +/-5-2)(-0:02 50 6 50 P = 0.12 CS Scanned with CamScanner

Plin v Qlpo? 1 pira h / -01 -0.002 so 6 ? YN 2 fractional linear polarization jo splint, Plin 0.104 Pin in % is 10.4% fractional Circulas polarizados -1 epoor el of polarization angle position angle. Алкан роf vi24. 0°C X Liso tan Ca X I tant 1653 Stokes I is total intensity. It is of other intensities of any orthogonal polarization components and does not store any polarization informaton. I= I tot I tot = Ioot Igo I + 450f I uso á IRCP f Iccp Stoke is defined as & Io. so that Ioni usually nalize norm un polanzed light half the half the light intensidy transmitted through each CS Scanngo with then وحل the sum parametr I intensity for is Camb cabinet

I=/ so and B-U-V 20 a for horizon tal polarizador 1, Q=1, 0:0, 1:0 6 for verticale polar 1, Z I I-, a I 2 1 e arization. -1,020, V = 0 6 polar ап 2 са. at 450 0,0=1 V=0 @ polarized at - 45° Q-0, 0- -1, V-o © Right corcalar polarization I=1,Q=0,O=, V=/ Lese circular polanzaton "Q=0,0=o, vel is power Since of auto- from som Correlation of a and cwill always have a positive sign. Eu (2,4)= Fon Pcos (kz-wt) Eğ (2,4)= Eogh cos (kz-wte <Ex>+<E> CS Scanned with I a I y components 20 it If tr) then I: CamScanner