Page:Elementary Text-book of Physics (Anthony, 1897).djvu/450

436 direction given by the formula, or, in other words, light of only one wave length is found in any one direction. If white light, or any light consisting of waves of various lengths, fall on the grating, the light corresponding to different wave lengths will make different angles with $$AC,$$ that is, the light is separated into its several constituents, and produces a pure spectrum. Since different values of $$n$$ will give different values of $$\theta$$ for each value of $$\lambda,$$ it is plain that there will be several spectra corresponding to the several values of $$n.$$ When $$n$$ equals 1 the spectrum is of the first order; when $$n$$ equals 3 the spectrum is of the second order, etc. The grating furnishes the most accurate and at the same time the most simple method of determining the wave lengths of light. Knowing the width of an element of the grating, it is only necessary to measure $$\theta$$ for any given kind of light.

Hitherto the spaces from which the elementary waves proceed have been considered infinitely narrow, so that only one system of waves from each space need be considered. In practice, these spaces must have some width, and it niay happen that the waves from two parts of the same space may cancel each other. Let the openings, Fig. 130, be equal in width to the opaque spaces, and let the direction $$am$$ be taken such that $$ae$$ equals $$2\lambda.$$ Then $$ae'$$ equals $$\tfrac{1}{2}\lambda,$$ or the waves from one half of each opening are opposite in phase to those from the other half, and there can be no light in the direction $$am.$$ In general, if $$d$$ equal the width of the opening, there will be interference, and light will be destroyed in that direction for which $$\sin \theta = \frac{n\lambda}{d},$$ if the incident light be normal to the grating. Let $$f$$ represent the width of the opaque space. Then $$d + f = s,$$ and light occurs in the direction given by $$\sin \theta = \frac{n\lambda}{d + f},$$ provided that the value of given by this equation does not satisfy the first equation also.

If $$d$$ equal $$f,$$ we have $$\sin \theta = \frac{n\lambda}{d + f} = \frac{n\lambda}{2d}\cdot$$ When $$n$$ is even,