376 E. R. HEDRICK b) If the direction of motion of the mirror makes an angle with the direction of the rays, then from Figure 12 it is clear that the mirror really advances with a velocity _ v sin 0 V cos 0--=— , h so that the formulae for this case may be obtained from those of the previous case by putting / _ sin 0\ 0(cos e- in place of ß. If the mirror is inclined at an angle of 45 tan a=i — ß(cos 0 —sin 0) , while tan 7=i+jS(cos 0 —sin 0) . 3. APPLICATION TO THE MICHELSON-MORLEY EXPERIMENT In the Michelson-Morley experiment a ray of light from a source S (Fig. 13) meets a half-silvered glass plate, inclined at 45 We assume that the earth and the apparatus are moving through the ether in a direction making an angle 0 with the path of the rays SA. It will be necessary to determine the position of the equivalent fixed mirror at B. For convenience denote ß(cos 0 —sin 0) by £. Then the angle CAB = 2a where tan a = 1 — £. In Figure 14, if BE is the wave front of the ray reflected from A and if the mirror at B advances from BM to B | CONFERENCE ON MICHELSON-MORLEY EXPERIMENT 377 Fig. 14 the direction 9) while the portion of the wave front at E advances to M\ then BM GM' where GM' is perpendicular to BM. GM' — MM' sin 2a = r sin 0; BG—BM-\-MM' cos 2a; EM r cos 2a |