37°
ROY J. KENNEDY
The theory of the arrangement is as follows: The interference phenomena will be the same as if the mirror M2 were replaced by its image in Mz. Under the conditions of the experiment, where the paths are nearly equal, Mx is perpendicular to the beam incident on it, and the reflected beams are brought nearly to parallelism, the image of M2 will be nearly parallel and coincident with the face of Mx. Elementary theory shows that the resulting interference pattern then practically coincides with Mx. It would needlessly complicate this discussion to develop the general theory of interference for all inclinations of the mirrors; the experimentally realized case of near parallelism alone is necessary.
Let Figure 10 represent a greatly exaggerated cross-section of Mx and the image of M2, normal to their planes and to the dividing line in M2. Mx lies in the plane x = o, and the levels of M2 are at equal distances on opposite sides of a parallel plane at the distance x from Mx. Let a monochromatic wave, in which the displacement is given by
fall on Mx and M2 from the left. At the surface of Mx the displacement in the reflected wave is then given by
if we ignore the loss through imperfect reflection. The displacement in the plane of Mx in the wave reflected from the upper part of M2 is
£x — a cos co(/+e)
The square of the resultant displacement is then
= O'2 j cos GO(t“j-é) “f"cos co |~f-€---—-—-
This can be reduced to the form
CONFERENCE ON MICHELSON-MORLEY EXPERIMENT 371
Similarly, the square of the resultant displacement in the interfering beams below the dividing line is found to be
2#2 i-|-cos ^ (x+a)J cos2 co(/ — 8) .
The intensities, being proportional to the squares of the amplitudes, can be represented by
J^Æa^i+cos ^ (%— a)j
and
/2 = &a2£i+cos ^ (#+a)J .
Now co = 27tv where v = frequency of the light. Hence co/c=27r/X. Therefore
and
/i = &a2|^i+cos ^ (x— a) j J2 = &a2£i+cos —■ (#+a)j .
For values oi x = n\/4, where n is an integer,
7i = Æa2^i±cos >
the sign being positive for even values of n and negative for odd values. The same expression holds for Z2; hence, under these conditions,
Jx = /a.
To the observer, then, the field of view is equally intense on both sides of the dividing line when x = n\/4.
We have now to determine the least change in x from this value which will produce a perceptible difference in illumination in the two sides of the field. If x is given the variation dx while a is kept constant, the difference in intensity will be