30.

Let a certain volume-element of the pencil be bounded by two cross-sections at distances equal to $r_0$ (of arbitrary length) and $r_0+d{r}_0$ respectively from the vertex $O$. The volume will be represented by $d{r}_0·r_0^{2}d\Omega$. It emits in unit time toward the focal plane $d\sigma$ at $O$ a certain quantity $E$ of energy of monochromatic plane polarized radiation. $E$ may be obtained from [eqn:(1)] (1) by putting $dt=1,d\tau =d{r}_0{r}_0^{2}d\Omega ,d\Omega =\frac{d\sigma }{r_0^2}$ and omitting the numerical factor $2$. We thus get $E=d{r}_0·d\Omega d\sigma {ϵ}_{\nu }d\nu .\mathrm{\backslash Label}[eqn](31)\text{upshape (31)}$

Of the energy $E$, however, only a fraction $E_0$ reaches $O$, since in every infinitesimal element of distance $s$ which it traverses before reaching $O$ the fraction $({\alpha }_{\nu }+{\beta }_{\nu })s$ is lost by absorption and scattering. Let $E_r$ represent that part of $E$ which reaches a cross-section at a distance $r$ ($<r_0$) from $O$. Then for a small distance $s=dr$ we have $E_{r+dr}-E_r=E_r({\alpha }_{\nu }+{\beta }_{\nu })dr,$ or, $\frac{d{E}_r}{dr}=E_r({\alpha }_{\nu }+{\beta }_{\nu }),$ and, by integration, $E_r=E{e}^{({\alpha }_{\nu }+{\beta }_{\nu })(r-r_0)}$ since, for $r=r_0$, $E_r=E$ is given by equation [eqn:(31)] (31). From this, by putting $r=0$, the energy emitted by the volume-element at $r_0$ which reaches $O$ is found to be $E_0=E{e}^{-({\alpha }_{\nu }+{\beta }_{\nu })r_0}=d{r}_{0}d\Omega d\sigma {ϵ}_{\nu }{e}^{-({\alpha }_{\nu }+{\beta }_{\nu })r_0}d\nu .\mathrm{\backslash Label}[eqn](32)\text{upshape (32)}$ All volume-elements of the pencils combined produce by their emission an amount of energy reaching $d\sigma$ equal to $d\Omega d\sigma d\nu {ϵ}_{\nu }{\int }_0^{\infty }d{r}_0{e}^{-({\alpha }_{\nu }+{\beta }_{\nu })r_0}=d\Omega d\sigma \frac{{ϵ}_{\nu }}{{\alpha }_{\nu }+{\beta }_{\nu }}d\nu .\mathrm{\backslash Label}[eqn](33)\text{upshape (33)}$