From Steve Scott To Howard Yuh Bob Granetz Re

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From Steve Scott To Howard Yuh Bob Granetz Re Powered By Docstoc
					From: Steve Scott
To: Howard Yuh, Bob Granetz
Re: MSE Memo #3: How to Compute MSE Polarization Fraction
Date: June 17, 2003


This memo describes how to compute the polarization fraction of light collected by the MSE
diagnostic.


    The following expression gives the measured intensity Itot for a polarized light source of
intensity Ip at angle λ and an unpolarized light source of intensity Iup :

                               Ip
            2Itot = Iup + Ip + √ [cos(2λ) cos A + sin(2λ)(cos B − sin A sin B)]           (1)
                                2
where

                                       A = A1 cos(ω1 t)
                                       B = A2 cos(ω2 t)                                   (2)

where ω1 and ω2 are the two PEM frequencies. Note that the last term in Eq. 1 varies
rapidly in time. Much of the variation is at harmonics of the PEM frequencies, i.e. in the
range of 10’s of kHz, but there is a component due to the sin A sin B term that varies at the
difference of the PEM frequencies, which can be just a few kHz.
   One can show that the magnitude of cos(2λ) cos A + sin(2λ)(cos B − sin A sin B) in Eq. 1
              √
never exceeds 2, which is reassuring since otherwise the intensity would go negative at
some time points. A little work in idl indicates that Eq. 1 can be rewritten as

                                2Itot = Iup + Ip (1 + f (λ)g(t))                          (3)

where g(t) is a time-varying function with magnitude unity and f (λ) can be approximated
by
                                          1
                      f (λ) = 1. − (1. − √ ) exp(−0.013λ2 ) (for λ < 45o )
                                           2
                                   1
               f (λ) = 1. − (1. − √ ) exp(−0.013(90 − λ)2 ) (for λ > 45o )                (4)
                                    2

    We can extract the maxima and minima of the intensity and the time-average intensity
directly from Eq. 3:
                                         1          1
                              I max =      Iup +      Ip (1 + f (λ))
                                         2          2
                                         1          1
                               I min   =   Iup +      Ip (1 − f (λ))                      (5)
                                         2          2

                                                1
                           Figure 1: Numerically-computed f (λ).

Thus by examining the minimum and maximum ofthe measured intensity, we can determine
the polarization fraction Pf = Ip /(Ip + Iup ):

                                            1 I max − I min
                                   Pf =                                                   (6)
                                          f (λ) I max + I min

    Interestingly, even if the polarization fraction is 100%, i.e. Iup = 0, the minimum signal
intensity will not go all the way to zero unless we’re at a polarization angle λ for which
f (λ) ≈ 1. If our pems were oriented in the usual configuration, i.e. at 0o and 45o then the
measured λ would be just a few degrees, and so f (λ) ≈ 0.7 − 0.8. But in our configuration,
the pems are oriented somewhat past vertical, and so our measured angles are typically in
the range 20o − 30o . In this case, we can approximate f (λ) = 1 in Eq. 5. In some calibration
shots we might possibly get measured polarization angles less than 20o for which it may be
necessary to include the correction due to f (λ).




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