5 2 D O C U M E N T 2 9 O P T I C A L E X P E R I M E N T . . . . (3) For cm and , a sulpho-carbonic acid layer of 25-cm length and yields a deflection of the order of magnitude . [This page is missing.][15] Let there be an open (or closed) line on the plane with the coordinates run- ning from a reference point along its measured arc length. Such a line then has solu- tions to the wave equation of the type: . . . (6) The integral should extend over the curve. r is a function of and and therefore indirectly of s A and α are given (real) functions of s. (6) solves the wave equation at distances from the curve that are large against the wavelength. For the following we want to confine ourselves to the case where r is large against the length of the curve and the latter is large against the wavelength of the radiation Then we can set to sufficient approximation: A and α are gradually constantly variable along the curve in such a way that if one depicts the curve length as of the order of magnitude 1. ( and are finite.) Because under these conditions the curve appears to be a small angle from the starting point, the surfaces of equal phase at the starting point are nearly vertical to the connection of an arbitrary point on the curve with the starting point. as can easily be gathered. Any choice of the curve and of the functions A and α yields, in a part of space that excludes the curve, the solution of a planar problem of deflection (for light sources at rest). We, however, are not interested here in the phenomenon of deflec- tion but in the ray’s path, neglecting the deflection. How do we find the course that can this be derived from (6)? For a given starting point, is a gradually A dn ν ------ ----------- ------ q c - l Δ = Δ 1= q c] [ ------- 1 300 -------- -= 10 2– [p. 6] [p. 7] ξ η, s , ϕ A r ------e t r ν -–-- α + sd = ξ η ϕ e t r0 V ---- r0 -------------------- - A Ae –jω Δ V --- α + sd = s------ dA ds - s------ ds - A r ------
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