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MEASUREMENT OF THE WAVES OF LIGHT

The diffraction fringes described in Lecture II., instead of being formed on the retina, may be formed on a screen, or upon ground glass, when they can be looked at through a magnifying lens from behind, or they can be observed in the air when the ground glass is removed. Instead of permitting them to form on the retina, we will suppose them formed on a screen. This places us in a condition to understand, even without trigonometry, the solution of the important problem of measuring the length of a wave of light.

Fig. 57.

We will suppose the screen so distant that the rays falling upon it from the two margins of the slit are sensibly parallel. We have learned in Lecture II. that the first of the dark bands corresponds to a difference of marginal path of one undulation; the second dark band to a difference of path of two undulations; the third dark band to a difference of three undulations, and so on. Now the angular distance of the bands from the centre is capable of exact measurement; this distance depending, as already stated, on the width of the slit. With a slit 1.35 millimeter wide,²⁹ Schwerd found the angular distance of the first dark band from the centre of the field to be 1'38"; the angular distances of the second, third, fourth dark bands being twice, three times, four times this quantity.

Let A B, fig. 57, be the plate in which the slit is cut, and C D the grossly exaggerated width of the slit, with the beam of red light proceeding from it at the obliquity corresponding to the first dark band. Let fall a perpendicular from one edge, D, of the slit on the marginal ray of the other edge at d. The distance, C d, between the foot of this perpendicular and the other edge is the length of a wave of the light. The angle C D d, moreover, being equal to R C R', is, in the case now under consideration, 1'38". From the centre D, with the width D C as radius, describe a semicircle; its radius D C being 1.35 millimeter, the length of this semicircle is found by an easy calculation to be 4.248 millimeters. The length C d is so small that it sensibly coincides with the arc of the circle. Hence the length of the semicircle is to the length C d of the wave as 180° to 1'38", or, reducing all to seconds, as 648,000" to 98". Thus, we have the proportion—

648,000 : 98 :: 4.248 to the wave-length C d.

Making the calculation, we find the wave-length for this particular kind of light to be 0.000643 of a millimeter, or 0.000026 of an inch.