# Conference on the Michelson-Morley experiment held at the Mount Wilson observatory Pasadena, California February 4 and 5, 1927

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 CONFERENCE ON MICHELSON-MORLEY EXPERIMENT 377 Fig. 14 the direction 9) while the portion of the wave front at E advances to M\ then BMr is the position of the equivalent fixed mirror. Denote the angle MBM! by p; then GM' where GM' is perpendicular to BM. GM' — MM' sin 2a = r sin 0; BG—BM-\-MM' cos 2a; EM r cos 2a ; P EM+MM' * 378 E. R. HEDRICK Therefore _ r sin 0 _ rß sin 0 P~BM+MM' cos 2a~ EMß , ' 1 MM’ß cos 2a cos 2 a rß sin 0 cos 2a r—rß sin 0 sin 2a * But we have tan a = i — £; hence, to terms of the second order, . £2 sin a=i--, cos 2a = H—. 2 2 Substituting these values in the expression for tan p, we have tan p = ß2 sin 0 (cos 0 —sin 0) , q.p. Now if we denote the angle CAfT by <£ and the angle CArTr by \f/, we have (remembering that 0 and yf/ are negative angles) <£+p=2a —p or \$=2(p— a) ^ = 27 — 180° . Thus the positive angle T'A'T = (j) — \f/=2p— 2a— 27+180° . To determine the tangent of this angle, we find tan ( — 2a) = ^ tan (27 — 180°) i-(i+£)2’ = 2(i+0 i — (i+£)2 ’ and therefore *£2 tan ( — 2a — 27+180°) =-— q.p. 4 — f4 From this we obtain * /, /\ £2+2ß2 sin 0 (cos 0 —sin 0) i — 2ß2^2 sin 0 (cos 0 — sin 0) ? since tan 2p = 2ß2 sin 0(cos 0 —sin 0) to terms of the second order. Substituting for £ and reducing, we have finally tan (cj)—\p)=ß2 cos 20 .