1917]
AND ON THE THEORY OF FOUCAULT's TEST
463
If in (22) | be supposed to increase without limit, we find
/-**- + {log 0£}», ........................... (23)
becoming logarithmically infinite.
Since in practice f, or rather K%, is large, the edges of the field may be expected to appear very bright.
As may be anticipated, this conclusion does not depend upon our supposition that £ = 0. Reverting to (7) and supposing 0 = 6, we have
2) - Si 2
sin T [Si (20£2) - Si (20£)] + cos T[Ci (20fc) - Ci (20&) + log (&/&)], (24)
and / = oo , when £2 = oo . If £ vanishes in (24), we have only to replace Ci (20£) by y + log (20£) in order to recover (22).
We may perhaps better understand the abnormal increase of illumination at the edges of the field by a comparison with the familiar action of a grating in forming diffraction spectra. Referring to (5) we see that if positive values of £ be alone regarded, the vibration in the plane of the second aperture, represented by g~l sin (0£), is the same in respect of phase as would be clue to a theoretically simple grating receiving a parallel beam perpendicularly, and the directions $ = ± 6 are those of the resulting lateral spectra of the first order. On account, however, of the factor £-*, the case differs somewhat from that of the simple grating, but not enough to prevent the illumination becoming logarithmically infinite with infinite aperture. But the approximate resemblance to a simple grating fails when we include negative as well as positive values of f, since there is then a reversal of phase in passing zero. Compare fig. 2, where positive values are represented by full lines and
Fig. 2.
negative by dotted lines. If the aperture is symmetrically bounded, the parts at a distance from the centre tend to compensate one another, and the intensity at 0 = ± 0 does not become infinite with the aperture.
We now proceed to consider the actual calculation of I = (19)2 + (21)2 for various values of 0/0, which we may suppose to be always positive, since I is independent of the sign of 0. When £0 is very great and 0/0 is not nearly equal to unity, Si {(0 + 0) £} in (19) may be replaced by \TT and Si ((0 - 0) £} by ±£TT, according as 0/0 is less or greater than unity. Under the same conditions the Ci's in (21) may be omitted, so that
J=7T2(1, or 0) + I log
0-0ssume one of these values, we have simply