We have now to consider the amplitude due to a single element, which we may conveniently regard as composed of a transparent part a bounded by two opaque parts of width ½d. The phase of the resultant effect is by symmetry that of the component which comes from the middle of a. The fact that the other components have phases differing from this by amounts ranging between ± amπ/(a + d) causes the resultant amplitude to be less than for the central image (where there is complete phase agreement). If Bm denote the brightness of the mth lateral image, and B0 that of the central image, we have
| Bm : B 0 = [∫ | + amπ/(a+d) | cosx dx ÷ | 2amπ | ] | ² | = ( | a + d | ) | ² | sin² | amπ | (1). |
| − amπ/(a+d) | a + d | amπ | a + d |
If B denotes the brightness of the central image when the whole of the space occupied by the grating is transparent, we have
B0 : B = a² : (a + d)²,
and thus
| Bm : B = | 1 | sin² | amπ | (2). |
| m²π² | a + d |
The sine of an angle can never be greater than unity; and consequently under the most favourable circumstances only 1/m²π² of the original light can be obtained in the mth spectrum. We conclude that, with a grating composed of transparent and opaque parts, the utmost light obtainable in any one spectrum is in the first, and there amounts to 1/π², or about 1⁄10, and that for this purpose a and d must be equal. When d = a the general formula becomes
| Bm : B = | sin² ½mπ | (3), |
| m²π² |
showing that, when m is even, Bm vanishes, and that, when m is odd,
Bm : B = 1/m²π².