Page:StokesAberration1846.djvu/5

Theory of the Aberration of Light. with the refracting surface travels along $$A B$$ with the



velocity $$\mathrm{cosec}\ \psi\left\{ V+q\sin(\psi+\alpha)\right\}$$. Observing that $$\tfrac{q}{\mu^{2}}$$ is the velocity of the æther within the refracting medium, and $$\tfrac{V}{\mu}$$ the velocity of propagation of light, we shall find in a similar manner that the lines of intersection of the refracting surface with the reflected and refracted waves travel along $$A B$$ with velocities

$\mathrm{cosec}\ \psi_{'}\left\{ V+q\sin(\psi_{'}+\alpha)\right\} ,\ \mathrm{cosec}\ \psi'\left\{ \frac{V}{\mu}+\frac{q}{\mu^{2}}\sin(\psi'+\alpha)\right\}$|undefined

But since the incident, reflected and refracted waves intersect the refracting surface in the same line, we must have

{{MathForm2|(A)|$$\left.\begin{align} \sin\psi_{'}\left\{ V+q\sin(\psi+\alpha)\right\} = & \sin\psi\left\{ V+q\sin(\psi_{'}+\alpha)\right\} ,\\ \mu\sin\psi'\left\{ V+q\sin(\psi+\alpha)\right\} = & \sin\psi\left\{ V+\frac{q}{\mu}\sin(\psi'+\alpha)\right\}. \end{align}\right\} $$}}

Draw $$H S$$ perpendicular to $$A H$$, $$ST$$ parallel to $$N A$$, take $$S T : H S : : q : V$$, and join $$H T$$. Then $$H T$$ is the direction of the incident ray; and denoting the angles of incidence, reflexion and refraction by $$\phi,\phi_{'},\phi^{'}$$, we have

$$\phi-\psi=SHT=\frac{ST\sin S}{SH}=\frac{1}{V}\times$$ resolved part of $$q$$ along $$AH$$

$$=\frac{q}{V}\cos(\psi+\alpha)$$. Similarly,

$\phi_{'}-\psi_{'}=\frac{q}{V}\cos\left(\psi_{'}-\alpha\right),\ \phi'-\psi'=\frac{q}{\mu V}\cos\left(\psi'+\alpha\right):$

whence

$\begin{align} \sin\psi= & \sin\phi-\frac{q}{V}\cos\phi\cos(\phi+\alpha),\\ \sin\psi_{'}= & \sin\phi_{'}-\frac{q}{V}\cos\phi_{'}\cos(\phi_{'}-\alpha),\\ \sin\psi'= & \sin\phi'-\frac{q}{V}\cos\phi'\cos(\phi'+\alpha). \end{align}$