43.

The most adequate method of acquiring more detailed information as to the origin and the paths of the different rays of which the radiations $I_1$, $I_2$, $I_3$, $\dots I_n$ consist, is to pursue the opposite course and to inquire into the future fate of that pencil, which travels exactly in the opposite direction to the pencil $I$ and which therefore comes from the first medium in the cone $d\Omega$ and falls on the surface element $d\sigma$ of the second medium. For since every optical path may also be traversed in the opposite direction, we may obtain by this consideration all paths along which rays can pass into the pencil $I$, however complicated they may otherwise be. Let $J$ represent the intensity of this inverse pencil, which is directed toward the bounding surface and is in the same state of polarization. Then, according to [sect:40.] Sec. 40, $J=I.\mathrm{\backslash Label}[eqn](45)\text{upshape (45)}$

At the bounding surface $d\sigma$ the rays of the pencil $J$ are partly reflected and partly transmitted regularly or diffusely, and thereafter, travelling in both media, are partly absorbed, partly scattered, partly again reflected or transmitted to different media, etc., according to the configuration of the system. But finally the whole pencil $J$ after splitting into many separate rays will be completely absorbed in the $n$ media. Let us denote that part of $J$ which is finally absorbed in the first medium by $J_1$, that which is finally absorbed in the second medium by $J_2$, etc., then we shall have $J=J_1+J_2+J_3+\dots +J_n.$

Now the volume-elements of the $n$ media, in which the absorption of the rays of the pencil $J$ takes place, are precisely the same as those in which takes place the emission of the rays constituting the pencil $I$, the first one considered above. For, according to Helmholtz's law of reciprocity, no appreciable radiation of the pencil $J$ can enter a volume-element which contributes no appreciable radiation to the pencil $I$ and vice versa.

Let us further keep in mind that the absorption of each volume-element is, according to [eqn:(42)] (42), proportional to its emission and that, according to Helmholtz's law of reciprocity, the decrease which the energy of a ray suffers on any path is always equal to the decrease suffered by the energy of a ray pursuing the opposite path. It will then be clear that the volume-elements considered absorb the rays of the pencil $J$ in just the same ratio as they contribute by their emission to the energy of the opposite pencil $I$. Since, moreover, the sum $I$ of the energies given off by emission by all volume-elements is equal to the sum $J$ of the energies absorbed by all elements, the quantity of energy absorbed by each separate volume-element from the pencil $J$ must be equal to the quantity of energy emitted by the same element into the pencil $I$. In

other words: the part of a pencil $I$ which has been emitted from a certain volume of any medium is equal to the part of the pencil $J$ ($=I$) oppositely directed, which is absorbed in the same volume.

Hence not only are the sums $I$ and $J$ equal, but their constituents are also separately equal or $J_1=I_1,J_2=I_2,\dots J_n=I_n.\mathrm{\backslash Label}[eqn](46)\text{upshape (46)}$