- #1

- 20

- 3

## Homework Statement

Show that

cos 2χ = ((Eb^2 - (Ea)^2) / ((Eb)^2 + (Ea)^2) = (-2(EL*ER)) / ((EL)^2 + (ER)^2)

and

sin 2χ = (2Ea*Eb) / ((Eb)^2 + (Ea)^2) = ((ER)^2 - (EL)^2) / ((EL)^2 + (ER)^2)

where EL and ER are the left and right circularly polarized field components of a wave.

## Homework Equations

The Stokes parameters are :

I = (ER)^2 + (EL)^2 I = (Ea)^2 + (Eb)^2

Q = 2(ER*EL)cos 2ψ Q = I*cos 2χ * cos 2ψ

U = 2(ER*EL)sin 2ψ U = I*cos 2χ * sin 2ψ

V = (ER)^2 - (EL)^2 V = I*sin 2χ

## The Attempt at a Solution

We can see that sin 2χ = V/I, from which it is straight forward to show the second part.

For the first part I get :

cos 2χ = Q/(I*cos 2ψ)

= (2(ER*EL)cos 2ψ)/(I*cos 2ψ)

= (2(ER*EL)) / I

= (2(EL*ER)) / ((EL)^2 + (ER)^2)

I am missing the negative part. I am not even sure if this is the correct method to use. Any help would be appreciated.