CONFERENCE ON MICHELSON-MORLEY EXPERIMENT 383
V. PROFESSOR PAUL S. EPSTEIN (CALIFORNIA INSTITUTE OF technology)
The result of Professor Hedrick’s analysis is that the two beams of light acquire a difference of phase
d—8r = hß2 cos 2#
and a difference in direction
Aa = ß2 cos 2#
in which terms of the fourth order are neglected.
Let us now choose the plane in which we observe the fringes as x = o of a cartesian system (Fig. 16). We can then represent the two waves by the formulae (s = light-vector)
I 27r I
s = A cos — (x cos a+y sin a+c/)+5 ,
s' = A cos cos a'+^ sin a'+c/)+5'j .
The illumination of the screen is then (x = o, sin a = a)
(s-\-sr)2 = ^A2 c°s2 |^- y(a— a')+5 — ô'j cos2 ;y(a+a') + 2c/+5+<5/j == 2A.2 cos2 |^- y(a— a')+ô — 5'j .
We have maxima, where the argument of the cosine is a multiple of 7r.
The position of the central fringe is therefore given by
PAUL S. EPSTEIN
The distance between two maxima, or the width of the fringes, is given by the equation
Ay(a— a^+S — ô' = t ,
A T—(Ô — Ô') X f ^
A y =--—r-t---. (2)
Let us first consider the interferometer at rest. We cannot take the ideal adjustment, because then we should have no fringes. Formula (2) shows that we must have a finite difference a0 — o!0 in order to get a finite width of fringes. This width is of the order of i mm, so that (8 —S' = o) we have the order of magnitude
, X 5‘io_s „
&o 7 = -Z7=2. 5-IO 4.
2 A y 2*io 1
In the actual experiment, we have in addition to a0— af0 the rotation Aa:
a—cl' — cio — aé+Aa ,
= _ X_ Ô-Ô'
^ 2t cio — a0+Aa The order of magnitude is
Aa= - cos 2# = (22-- cos 2# = io“8 cos 2#
W \3 • io10/
Therefore an expansion is permissible:
\ / Ô-Ô' Ô-Ô' A
( / ( f\2
27T \cto a0 (a0— a0J2
X 5-5' A Aa
27T a0 — a0\ a0 — a,
The first term represents the shift due to the difference of phase; the second term is due to the rotation. We see that it is 0.4-io-4 of the first term, that is, quite outside the possibility of observation under the conditions of Michelson, Morley, and Miller’s experiment.