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



$$Ff$$ in the opposite direction; and take

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

and join $$A$$ with $$f, g$$ and $$h$$. Then $$fA, Ag, Ah$$ will be the directions of the incident, reflected and refracted rays. Draw $$F D, H E$$ perpendicular to $$D E$$, and join $$fD, h E$$. Then $$f D F, h E H$$ will be the inclinations of the planes $$f A D, h A E$$ 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 $$A h$$ lies in the plane of incidence $$fA D$$. It is easy to see that the same is true of the reflected ray $$Ag$$. Also $$\angle gAD=fAD$$; and the angles $$f A D, h A E$$ are sensibly equal to $$F A D, H A E$$ 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 $$A H, A K, A L$$ 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 $$N A$$ 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 $$A H$$ is $$q\sin(\psi+\alpha)$$. Hence the wave $$A H$$ travels with the velocity $$V+q\sin(\psi+\alpha)$$; and consequently the line of its intersection