error of 6 per cent., it can only be said that these ratios are only slightly different, if at all. The curves are all of the same form, and the method of excitation appears to have no marked effect on the intensity distribution. In particular, the wave-length of maximum energy is found to bear a constant ratio to the minimum wave-length, as shown below: The mean value of the ratio is 1.55. In conclusion, I should like to express my great indebtedness to Professor J. A. Crowther for his valuable advice and for the energetic interest he has shown in the progress of this work. Department of Physics, CI. The Amount of Uniformly-diffused Light that will go in Series through Two Apertures Forming Opposite Faces of a Cube. By LEWIS F. RICHARDSON, D.Sc., F.R.S. * FOR example, the radiation might come out of a blackened enclosure at a uniform temperature through one aperture and thence across the cube and through the other aperture into a calorimeter; the problem being to find Stefan's constant. We have to evaluate the integral † in which dA1, dA, are elements of area of the apertures, r is the distance between the centres of dA1, dAg, and S1, S2 *Communicated by the Author. Planck Vorlesungen ueber die Theorie der Wärmestralung' (Barth, 1913), p. 19. are the angles which the normals to dA1, dA, make with the line joining the centre of dA, to that of dA. Let the edge of the cube be of length l. Take two systems of rectangular coordinates 1, 1 and 2, y2, one in the plane of each aperture, the origins being at the centres of the apertures and the two x-axes parallel to one another and to edges of the cube. Then each integration ranging from -1/2 to +1/2. In order to make the calculation, consider instead of (3) a problem in finite differences. Regard each aperture as a window cut into no panes, each pane being square. Take each pane in the first window in turn with each pane in the second, forming n pairs of panes. Then dæ1 dy, dx, dy is analogous to (area of first pane) (area of second pane), and is the same for all pairs being /n. So instead of (3) we calculate 76 n pairs n1 1 (separation of mid-points of panes) = Qul, say. (4) Q2 varies as 12 because the distance of mid-points is proportional to 7. Let us therefore make the calculation for = 1. First approximation. One pane in each window : 1 Q1 = 1. Second approximation, n = 2. Four panes in each window: Third approximation, n = 3. Nine panes in each window: 1 (9 Q3 + + 8×2 811 (1 + 1/9)2 + (1 + 4/9) (1+1/9+1/9) 2 + 4 16 + = 0.649, 821, 9. Fourth approximation, n = 4. Sixteen panes to each window. To keep account of 256 pairs would be perplexing were it not that we can arrange them in a simple pattern. The blank squares can be filled in because the distribution of numbers is symmetrical both about the diagonals of the large square and also about the lines joining the mid-points of its sides. In fig. 1 let the central square represent any pane in one window. All possible relative positions of the panes in the other window are represented by the squares in fig. 1. In each square the upper number is the number of pairs of panes so situated, the lower number is the square of their distance apart multiplied by n3. To form Q we have to divide each upper number by the square of the lower number, and to sum the quotients. Presumably there is a corresponding transformation from a fourfold to a double. integral. However that may be, the result is Q4=0.639, 910, 5. To continue in this way until the accuracy of experiment were surpassed would be insufferably tedious. Instead, to make the "deferred approach to the limit *", it is assumed that, as there are neither discontinuities nor frills †, in (3) Qu=q+Bn2+ Cn-'+ Dn ̄ε,. ... (5) (n = 1, 2, 3, 4). -4 On eliminating B, C, D, from this set of four equations it is found that = q=0.627, 75. 2520 q 8192 Q4-6561 Q+896 Q2-7 Q1, (6) the coefficients in which are useful in other problems. Whence This may be called an ho-extrapolation. As a rough check Q2, Q3, Q4 are plotted against n-2 in fig. 2‡, and a circle drawn through the points is found to cut the axis, n2=0, at Phil. Trans. Roy. Soc. A. vol. ccxxvi. pp. 300, 311. + Loc. cit. p. 315. Fig. 2 was shown at the British Association, Glasgow, 1928. Q=0.628. To judge how many digits are reliable, we can neglect various parts of the four equations (5), retaining just enough of B, C, D to give a unique solution, B being retained first, then C, then D. Thus we form two sequences of extrapolations both leading to that already obtained : from Q4, Q3, Q2... 0.627, 95 from Q1, Q2, Qз... 0·631, 04 from Q1, Q2, Q3, Q4 ... 0.627, 75. From these numbers it seems safe to conclude that 0.6278 is right to within ±0001. That accuracy is, however, barely enough. An estimate of the error could be obtained. by the use of general formulæ expressing the difference between the integral and the sum as a series involving derivatives of the integrand. But it seems more profitable to compute Qe by the aid of a diagram like fig. 1. The law of these diagrams is found to be expressed by Qn H ΣΣ s 8=1-n_t=1-n (n2+s2+t2)2 It is thus found that Qe 0·633, 086, 28. = Extrapolating, as above, from Q and Qe gives 0·627, 63. Extrapolating from Q2, Q4, Qe gives 0·627, 81. Conclusion. The amount of light, uniformly diffused with brightness I, that will go in series through two apertures forming opposite faces of a cube of edge lis (0·6278±0·0001)2I. The physical meaning of the above result may be understood by comparison with the statement of Stefan's constant. Instead of two squares we have one square of side 1, drawn on a flat radiating solid of uniform brightness I. The second square is replaced by a hemisphere-at-infinity, drawn with its centre in the first square and its basal plane containing the radiating surface. The amount of light that will go through the square and hemisphere is known to be 12I, a result deducible from (1). The method will obviously extend to parallel rectangular apertures. The calculations were performed on a machine lent by the Government Grant Committee of the Royal Society. |