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

В начало   Другие форматы (PDF, DjVu)   <<<     Страница 376   >>>  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-Tj in place of ß. If the mirror is inclined at an angle of 450 to the direction of the rays of light, h = 1 and 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 450 to its path, at A. A portion is reflected to a mirror at B, parallel to 5^4, from which it is again reflected to pass through the plate at Af and finally into a telescope at T. Another portion is transmitted through the glass plate at A to a mirror at C, perpendicular to 5^4, from which it is returned to the glass plate at Ar and from there a further portion is reflected into the telescope at T. When the mirrors are set as described, with absolute accuracy, we call the experiment the “ideal Michelson-Morley experiment.” We wish to compute the angle T'A'T. ' 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 BrMf (a distance r in 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' *