105.

Let us for future use solve also the more general problem of calculating the entropy radiation of a ray consisting of an arbitrary number of plane polarized non-coherent components ${𝖪}_1$, ${𝖪}_2$, ${𝖪}_3,\dots$, the planes of vibration (planes of the electric vector) of which are given by the azimuths ${\psi }_1$, ${\psi }_2$, ${\psi }_3,\dots$. This problem amounts to finding the principal intensities ${𝖪}_0$ and ${𝖪}_0^{\prime }$ of the whole ray; for the ray behaves in every physical respect as if it consisted of the non-coherent components ${𝖪}_0$ and ${𝖪}_0^{\prime }$. For this purpose we begin by establishing the value ${𝖪}_{\psi }$ of the component of the ray for an azimuth $\psi$ taken arbitrarily. Denoting by $f$ the electric vector of the ray in the direction $\psi$, we obtain this value ${𝖪}_{\psi }$ from the equation $f=f_1\cos ({\psi }_1-\psi )+f_2\cos ({\psi }_2-\psi )+f_3\cos ({\psi }_3-\psi )+\dots ,$ where the terms on the right side denote the projections of the vectors of the separate components in the direction $\psi$, by squaring and averaging and taking into account the fact that $f_1$, $f_2$, $f_3,\dots$ are non-coherent $\mathsf{K}{\psi} &= \mathsf{K}} \cos^{2}(\psi_{1} - \psi) + \mathsf{K{2} \cos^{2}(\psi} - \psi) + \dots \ \LeftText[\qquad]{or} \mathsf{K{\psi} &= A\cos^{2}\psi + B\sin^{2}\psi + C\sin\psi \cos\psi \ \LeftText[\qquad]{where} A &= \mathsf{K}} \cos^{2}\psi_{1} + \mathsf{K{2} \cos^{2}\psi} + \dots \ B &= \mathsf{K{1} \sin^{2}\psi} + \mathsf{K{2} \sin^{2}\psi} + \dots \ C &= 2(\mathsf{K{1} \sin\psi} \cos\psi_{1} + \mathsf{K{2} \sin\psi + \dots).} \cos\psi_{2

\Label[eqn]{(145)}\tag*{\upshape (145)}$

The principal intensities ${𝖪}_0$ and ${𝖪}_0^{\prime }$ of the ray follow from this expression as the maximum and the minimum value of ${𝖪}_{\psi }$ according to the equation $\frac{d{𝖪}_{\psi }}{d\psi }=0\text{or,}\tan 2\psi =\frac{C}{A-B}.$ Hence it follows that the principal intensities are $\left.

$$\begin{gathered} \&{𝖪}_0 \\ \&{𝖪}_0^{\prime } \end{gathered}$$

\right} = \tfrac{1}{2}(A + B ± \sqrt{(A - B)^{2} + C^{2}}), \Label[eqn]{(146)}\tag*{\upshape (146)}$ or, by taking [eqn:(145)] (145) into account,

$\left.$

$$\begin{gathered} \&{𝖪}_0 \\ \&{𝖪}_0^{\prime } \end{gathered}$$

$\right\} = \frac{1}{2}\Biggl(\mathsf{K}_{1} + \mathsf{K}_{2} + \dots \\ ± \sqrt{$

$$\begin{gathered} ({𝖪}_1\cos 2{\psi }_1\&+{𝖪}_2\cos 2{\psi }_2+\dots {)}^2 \\ \&+({𝖪}_1\sin 2{\psi }_1+{𝖪}_2\sin 2{\psi }_2+\dots {)}^2 \end{gathered}$$

}\,\Biggr). \Label[eqn]{(147)}\tag*{\upshape (147)}

Then the entropy radiation required becomes: $𝖫({𝖪}_0)+𝖫({𝖪}_0^{\prime }).\mathrm{\backslash Label}[eqn](148)\text{upshape (148)}$