36. The Bounding Surface of the Media

We shall now suppose, to begin with, that the bounding surface of the media is smooth ([sect:9.] Sec. 9).

Then every ray coming from the first medium and falling on the bounding surface is divided into two rays, the reflected and the transmitted ray.

The directions of these two rays vary with the angle of incidence and the color of the incident ray; the intensity also varies with its polarization.

Let us denote by $\rho$ (coefficient of reflection) the fraction of the energy reflected, then the fraction transmitted is $(1-\rho )$, $\rho$ depending on the angle of incidence, the frequency, and the polarization of the incident ray.

Similar remarks apply to

${\rho }^{\prime }$ the coefficient of reflection of a ray coming from the second medium and falling on the bounding surface.

Now according to [eqn:(11)] (11) we have for the monochromatic plane polarized radiation of frequency $\nu$, emitted in time $dt$ toward the first medium (in the direction of the feathered arrow upper left 3 hand in [fig:3]Fig. 3), from an element $d\sigma$ of the bounding surface and contained in the conical element $d\Omega$, $dtd\sigma \cos \theta d\Omega {𝖪}_{\nu }d\nu ,\mathrm{\backslash Label}[eqn](34)\text{upshape (34)}$ where $d\Omega =\sin \theta d\theta d\varphi .\mathrm{\backslash Label}[eqn](35)\text{upshape (35)}$ This energy is supplied by the two rays which come from the first and the second medium and are respectively reflected from or transmitted by the element $d\sigma$ in the corresponding direction (the unfeathered arrows). (Of the element $d\sigma$ only the one point $O$ is indicated.) The first ray, according to the law of reflection, continues in the symmetrically situated conical element $d\Omega$, the second in the conical element $d{\Omega }^{\prime }=\sin {\theta }^{\prime }d{\theta }^{\prime }d{\varphi }^{\prime }\mathrm{\backslash Label}[eqn](36)\text{upshape (36)}$ where, according to the law of refraction, ${\varphi }^{\prime }=\varphi \text{and}\frac{\sin \theta }{\sin {\theta }^{\prime }}=\frac{q}{q^{\prime }}.\mathrm{\backslash Label}[eqn](37)\text{upshape (37)}$

If we now assume the radiation [eqn:(34)] (34) to be polarized either in the plane of incidence or at right angles thereto, the same will be true for the two radiations of which it consists, and the radiation coming from the first medium and reflected from $d\sigma$ contributes the part $\rho dtd\sigma \cos \theta d\Omega {𝖪}_{\nu }d\nu \mathrm{\backslash Label}[eqn](38)\text{upshape (38)}$ while the radiation coming from the second medium and transmitted through $d\sigma$ contributes the part $(1-{\rho }^{\prime })dtd\sigma \cos {\theta }^{\prime }d{\Omega }^{\prime }{𝖪}_{\nu }^{\prime }d\nu .\mathrm{\backslash Label}[eqn](39)\text{upshape (39)}$ The quantities $dt$, $d\sigma$, $\nu$ and $d\nu$ are written without the accent, because they have the same values in both media.

By adding [eqn:(38)] (38) and [eqn:(39)] (39) and equating their sum to the expression [eqn:(34)] (34) we find $\rho \cos \theta d\Omega {𝖪}_{\nu }+(1-{\rho }^{\prime })\cos {\theta }^{\prime }d{\Omega }^{\prime }{𝖪}_{\nu }^{\prime }=\cos \theta d\Omega {𝖪}_{\nu }.$

Now from [eqn:(37)] (37) we have $\frac{\cos \theta d\theta }{q}=\frac{\cos {\theta }^{\prime }d{\theta }^{\prime }}{q^{\prime }}$ and further by [eqn:(35)] (35) and [eqn:(36)] (36) $d{\Omega }^{\prime }\cos {\theta }^{\prime }=\frac{d\Omega \cos \theta q^{\prime 2}}{q^2}.$

Therefore we find $\rho {𝖪}_{\nu }+(1-{\rho }^{\prime })\frac{q^{\prime 2}}{q^2}{𝖪}_{\nu }^{\prime }=𝖪$ or $\frac{{𝖪}_{\nu }}{{𝖪}_{\nu }^{\prime }}·\frac{q^2}{q^{\prime 2}}=\frac{1-{\rho }^{\prime }}{1-\rho }.$