Page:StokesAberration1846.djvu/4

Theory of the Aberration of Light.

Draw $Gg,Hh$ perpendicular to the plane $P$, and in the direction of the resolved part $p$ of the velocity of the æther, and

Fig. 1.

$Ff$ in the opposite direction; and take

$Ff:Hh:FA::P:{\frac {p}{\mu ^{2}}}:V$, and $Gg=Ff$,

and join $A$ with $f,g$ and $h$. Then $fA,Ag,Ah$ will be the directions of the incident, reflected and refracted rays. Draw $FD,HE$ perpendicular to $DE$, and join $fD,hE$. Then $fDF,hEH$ will be the inclinations of the planes $fAD,hAE$ to the plane $P$. Now

$\tan FDf={\frac {p}{V\sin FAD}},\ \tan HEh={\frac {\mu ^{-2}p}{\mu ^{-1}V\sin HAE}}$

and $FAD=\mu \sin HAE$; therefore $FDf=\tan HEh$, and therefore the refracted ray $Ah$ lies in the plane of incidence $fAD$. It is easy to see that the same is true of the reflected ray $Ag$. Also $\angle gAD=fAD$; and the angles $fAD,hAE$ are sensibly equal to $FAD,HAE$ respectively, and we therefore have without sensible error, $\sin fAD=\mu \sin hAE$. Hence the laws of reflexion and refraction are not sensibly affected by the velocity $p$.

Let us now consider the effect of the velocity $q$. As far as depends on this velocity, the incident, reflected and refracted rays will all be in the plane $P$. Let $AH,AK,AL$ be the intersections of the plane $P$ with the incident, reflected and refracted waves. Let $\psi ,\psi _{'},\psi ^{'}$ be the inclinations of these waves to the refracting surface; let $NA$ be the direction of the resolved part $q$ of the velocity of the æther, and let the angle $NAC=\alpha$.

The resolved part of $q$ in a direction perpendicular to $AH$ is $q\sin(\psi +\alpha )$. Hence the wave $AH$ travels with the velocity $V+q\sin(\psi +\alpha )$; and consequently the line of its intersection