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#
Fig. 16
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
3&4
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
or
A T—(Ô — Ô') X f ^
A y =--—r-t---. (2)
a—a 27r
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
yo=-— -—-41
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.