uninfluenced by the motion of the earth. The method which I employ will, I hope, be found simpler than Fresnel’s; besides it applies easily to the most general case. Fresnel has not given the calculation for reflexion, but has merely stated the result; and with respect to refraction, he has only considered the case in which the course of the light within the refracting medium is in the direction of the earth’s motion. This might still leave some doubt on the mind, as to whether the result would be the same in the most general case. If the aether were at rest, the direction of light would be that of a normal to the surfaces of the waves. When the motion of the aether is considered, it is most convenient to define the direction of light to be that of the line along which the same portion of a wave moves relatively to the earth. For this is in all cases the direction which is ultimately observed with a telescope furnished with cross wires. Hence, if A is any point in a wave of light, and if we draw AB normal to the wave, and proportional to V oraccording as the light is passing through vacuum or through a refracting medium, and if we draw BC in the direction of the motion of the aether, and proportional to v or and join A C, this line will give the direction of the ray.Of course, we might equally have drawn A D equal and parallel to B C and in the opposite direction, when D B would have given the direction of the ray. Let a plane P be drawn perpendicular to the reflecting or refracting surface and to the waves of incident light, which in this investigation may be supposed plane. Let the velocity v of the aether in vacuum be resolved into p perpendicular to the plane P, and q in that plane; then the resolved parts of the velocityof the aether within a refracting medium will beLet us first consider the effect of the velocity p. It is easy to see that, as far as regards this resolved part of the velocity of the aether, the directions of the refracted and reflected waves will be the same as if the aether were at rest. Let BAC (fig. 1) be the intersection of the refracting surface and the plane P; DAE a normal to the refracting surface; AF, AG, A H normals to the incident, reflected and refracted waves. Hence A F, A G, A H will be in the plane P, and ∠GAD = FAD, μ sin HAE = sin FAD. Take AG = AF, | 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- |