Draw Gg, Hh perpendicular to the plane P, and in the direction of the resolved part p of the velocity of the aether, and Fig. 1. F f in the opposite direction; and take and join A with f, g and h. Then f A, Ag, Ah will be the directions of the incident, reflected and refracted rays. Draw FD, HE perpendicular to D E, and join f D, 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 sin FAD = μ sin HAE; therefore tan 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 ∠gAD = fAD; and the angles fAD, h A E are sensibly equal to FAD, HAE respectively, and we therefore have without sensible error, sin fAD = μ sin h A E. 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 ψ, ψ The resolved part of q in a direction perpendicular to A H is q sin (ψ + α). Hence the wave A H travels with the velocity V + q sin (ψ + α); and consequently the line of its inter- | section with the refracting surface travels along A B with the Fig. 2. Velocity cosecObserving thatis the velocity of the aether within the refracting medium, and 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 But since the incident, reflected and refracted waves intersect the refracting surface in the same line, we must have Draw H S perpendicular to A H, S T parallel to N A, take ST:HS::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 φ, φ resolved part of q along AH Similarly, whence |