80.

The rays which at the time $t$ belong to the interval $(\nu ,d\nu )$ and leave this interval in the time $\delta t$ on account of reflection from the moving reflector, are simply those rays which strike the moving reflector in the time $\delta t$. For the change in color which such a ray undergoes is, from [eqn:(83)] (83) and [eqn:(91)] (91), large compared with $d\nu$, the width of the whole interval. Hence we need only calculate the energy, which in the time $\delta t$ is transmitted to the reflector by the rays in the interval $(\nu ,d\nu )$.

For an elementary pencil, which falls on the element $d\sigma$ of the reflecting surface at the angle of incidence $\theta$, this energy is, according to [eqn:(88)] (88) and [eqn:(5)] (5), $I\delta t=2{𝖪}_{\nu }d\sigma \cos \theta d\Omega d\nu \delta t=2{𝖪}_{\nu }d\sigma \sin \theta \cos \theta d\theta d\varphi d\nu \delta t.$ Hence we obtain for the total monochromatic radiation, which falls on the whole surface $F$ of the reflector, by integration with respect to $\varphi$ from $0$ to $2\pi$, with respect to $\theta$ from $0$ to ${\displaystyle \frac{\pi }{2}}$, and with respect to $d\sigma$ from $0$ to $F$, $2\pi F{𝖪}_{\nu }d\nu \delta t.\mathrm{\backslash Label}[eqn](92)\text{upshape (92)}$ Thus this radiant energy leaves, in the time $\delta t$, the interval of frequencies $(\nu ,d\nu )$ considered.