166.

We shall now use the laws of radiation we have obtained to calculate the temperature of a monochromatic unpolarized radiation of given intensity in the following case. Let the light pass normally through a small area (slit) and let it fall on an arbitrary system of diathermanous media separated by spherical surfaces, the centers of which lie on the same line, the axis of the system. Such radiation consists of homocentric pencils and hence forms behind every refracting surface a real or virtual image of the emitting surface, the image being likewise normal to the axis. To begin with, we assume the last as well as the first medium to be a pure vacuum. Then, for the determination of the temperature of the radiation according to equation [eqn:(274)] (274), we need calculate only the specific intensity of radiation ${𝖪}_{\nu }$ in the last medium, and this is given by the total intensity of the monochromatic radiation $I_{\nu }$, the size of the area of the image $F$, and the solid angle $\Omega$ of the cone of rays passing through a point of the image. For the specific intensity of radiation ${𝖪}_{\nu }$ is, according to [eqn:(13)] (13), determined by the fact that an amount $2{𝖪}_{\nu }d\sigma d\Omega d\nu dt$

of energy of unpolarized light corresponding to the interval of frequencies from $\nu$ to $\nu +d\nu$ is, in the time $dt$, radiated in a normal direction through an element of area $d\sigma$ within the conical element $d\Omega$. If now $d\sigma$ denotes an element of the area of the surface image in the last medium, then the total monochromatic radiation falling on the image has the intensity $I_{\nu }=2{𝖪}_{\nu }\int d\sigma \int d\Omega .$ $I_{\nu }$ is of the dimensions of energy, since the product $d\nu dt$ is a mere number. The first integral is the whole area, $F$, of the image, the second is the solid angle, $\Omega$, of the cone of rays passing through a point of the surface of the image. Hence we get $I_{\nu }=2{𝖪}_{\nu }F\Omega ,\mathrm{\backslash Label}[eqn](302)\text{upshape (302)}$ and, by making use of [eqn:(274)] (274), for the temperature of the radiation $T=\frac{h\nu }{k}·\frac{1}{\log ({\displaystyle \frac{2h{\nu }^{3}F\Omega }{c^2{I}_{\nu }}}+1)}.\mathrm{\backslash Label}[eqn](303)\text{upshape (303)}$ If the diathermanous medium considered is not a vacuum but has an index of refraction $n$, [eqn:(274)] (274) is replaced by the more general relation [eqn:(300)] (300), and, instead of the last equation, we obtain $T=\frac{h\nu }{k}\frac{1}{\log ({\displaystyle \frac{2h{\nu }^{3}F\Omega n^2}{c^2{I}_{\nu }}}+1)}\mathrm{\backslash Label}[eqn](304)\text{upshape (304)}$ or, on substituting the numerical values of $c$, $h$, and $k$, $T=\frac{0.479·{10}^{-10}\nu }{\log ({\displaystyle \frac{1.43·{10}^{-47}{\nu }^{3}F\Omega n^2}{I_{\nu }}}+1)}\text{ degree Centigrade}.$ In this formula the natural logarithm is to be taken, and $I_{\nu }$ is to be expressed in ergs, $\nu$ in "reciprocal seconds," i.e., $(\text{seconds}{)}^{-1}$, $F$ in square centimeters. In the case of visible rays the second term, $1$, in the denominator may usually be omitted.

The temperature thus calculated is retained by the radiation considered, so long as it is propagated without any disturbing

influence in the diathermanous medium, however great the distance to which it is propagated or the space in which it spreads. For, while at larger distances an ever decreasing amount of energy is radiated through an element of area of given size, this is contained in a cone of rays starting from the element, the angle of the cone continually decreasing in such a way that the value of $𝖪$ remains entirely unchanged. Hence the free expansion of radiation is a perfectly reversible process. (Compare above, [sect:144.] Sec. 144.) It may actually be reversed by the aid of a suitable concave mirror or a converging lens.