reciprocals of the wave velocities of plane waves travelling in the same direction. It was proved in Art. 90 that the axes of the elliptical 0 Fig. 114. section of the ellipsoid of elasticity are the bi sectors of OQ and OQ′ Fig. 114, Q and Q' being in the planes containing the wave normal ON and optic axes OH, OH'. The radii OP and OP' at right angles to OQ, OQ and therefore also at right angles to the optic axes, belong to the circular sections, and have therefore a length 1/. If v1 and v2 are the semiaxes of the ellipse, and the angle POP is denoted by 24, we have from the equation to the ellipse, reduced to polar coordinates, reciprocals of the The angle v2 cos2 + v2 sin2 = b2. between OQ and OQ', and the angle 26, are supplementary to each other, so that Another relation between the wave velocities is obtained by making use of the fact that in any ellipsoid the sum of the reciprocals of the squares of any three diameters at right angles to each other, is constant. The section we are considering has 1/v, and 1/v, for semiaxes, and 1, m, n for the direction cosines of its normal. Hence l'a2 + m2b2 + n2c2 is the reciprocal of the square of the radius vector, which is normal to the section, and 2 v12 + v22 + l2a2 + m2b2 + n2c2 = a2 + b2 + c2 or by making use of 2 12 + m2 + n2 = 1, (v12 + v22) = (a2 + c2) + l2 (b2 − a2) + n2 (b2 − c2) ; .. (v12 – v22) cos A = (a2 + c2 − 2b2) + l2 (b2 − a2) + n2 (b2 — c2) .(2), we The expressions are simplified still further if, instead of A, introduce the angles 0, and 0, between ON and OH, OH' respectively, and the angle σ included between the two optic axes. The spherical triangle formed by OH, OH', ON gives The optic axes lie in the plane az and if their positive directions are chosen so that the axis of X bisects the angle included between them, we find, writing l1, n1; l, n2 for their direction cosines : For the angle σ between the optic axes we have COS σ = ll1⁄2 + n ̧n2 = l12 − n22; .. sin 0, sin 02 cos A = cos σ cos e cos 02 and introducing the values of 4, and n1, Art. 85; Comparing this with (3), we finally obtain the simple equation Combining (4) and (5) we may express separately 2v1 = (a+c2)-(a2-c2) cos (01+02) 2v22 = (a2 + c2 ) − (a2 – c2) cos (01 – 02). If the difference between a and c is so small that its square may be neglected, we may write where v stands for the velocity to which both e, and 2 approach when a で c vanishes. For v we may therefore write either √ac or § (a + c), and for 2 we may write ac (a + c). or Introducing the values of v2 v2 from (4) we obtain The proposition 110. Relation between ray velocities. contained in the last article represents a theorem which may be applied to any ellipsoid of semiaxes 1/a, 1/b, 1/c; if v1, v2 are the reciprocals of the principal axes of a section which forms angles 6, and 02 with the circular sections. We may therefore write down at once the corresponding equations for the reciprocal ellipsoid, substituting the ray velocities s, and s2 for 1/v, and 1/2. We obtain in this S2 281 ̄2 = (a ̄2 + c ̄2) – (a−2 — c ̄2) cos (N1 + N2), 282 + S2 -2 -2 -2 (a ̄2 + c ̄2) — (a ̄2 — c2) cos n, cos 2, $12 — 822 = (α ̄2 — c−2) sin where 7 and 2 are the angles formed between the normal to the section and the axes of single ray velocities. 111. The surface of equal phase difference, or Isochromatic Surface. If we imagine a number of plane waves crossing at a point O (Fig. 145) in a crystalline medium, there being two wave velocities in each direction, we may construct a surface such that at any point P, belonging to the surface, the phase difference between the two wavefronts which have OP for wave normals is the same. If ρ be any radius vector OP, v1 and v2 the wave velocities, the two optical distances from 0 to P are p/v2 and p/v1, hence the required surface has for equation: It will be sufficient to confine the discussion to the case of a small difference between the two wave velocities. We shall consider therefore a c and a fortiori ab to be so small that their squares may be neglected. We may then apply equation (6) and by introducing the principal indices of 1 , Мз the equation to the surface of equal phase с Unless highly homogeneous light is used, & must not exceed a small multiple of a wave-length, if interference effects are to be observed. It follows that unless the observations are carried out close to one of the optic axes, in which case either sin 0, or sin 0, is small, μ-μ1 must be small. This justifies the simplification we have introduced in treating ac as a small quantity. In uniaxal crystals, there is only one axis, so that putting 0,0,0, the polar equation to the surface of equal phase difference or "isochromatic" surface then becomes = This surface is formed by the revolution about the optic axis of a family of curves for which the polar equation is represented by (8) and which is drawn to scale in Fig. 146. Only half of the curves is shown, there being symmetrical halves below the line PQ. The scale is such that if the substance is Iceland Spar, and the length marked AB represents one millimetre, the inner curve is the isochromatic surface of phase difference equal to 100 wave-lengths, the wave-length being that of sodium light; the phase difference belonging to the outer curve is five times as great. OC is the optic axis. The upper portions of the curves are sensibly parabolic, because when is small, the radius vector p is nearly equal to its projection a on the optic axis, so that the equation to the curve becomes 22/a = constant. In biaxal crystals the isochromatic surface has four sheets surrounding the optic axes. Their intersection with the plane containing these axes is represented in Fig. 147 for the case where the angle between the optic axes is 60°. When p is infinitely large, it follows from (7) that either 0, or 02 is zero. If 0, vanishes, 0, must be equal to σ, the angle included between the optic axes. For large values of p we may still take approximately 2 = σ and the equation to the isochromatic surface approaches therefore a surface the equation to which is by (7) This is the equation to a circular cylinder, having one of the optic axes as axis. The intersection of this cylinder with the plane of the paper gives two straight parallel lines, which are the asymptotes to the curve which forms the intersection of the isochromatic surface with the plane containing the optic axes. If p' be the distance of the asymptotes there are two similar asymptotes parallel to the second optic axis. These asymptotes are shown by dotted lines in the figure, and it will be noticed that each of them intersects one branch of the curve to which it is a tangent at infinity. The two distances Po and P1 of the vertices of the surface found by substituting 6,0,= 0 and 01 = We then find 112. M 2 2 = π+ σ may be respectively. Application of the Isochromatic Surface to the study of polarization. Let a doubly refracting plate, Fig. 148, receive light at different inclinations. An eye placed at E and looking towards a point S on the plate observes certain interference effects. Tracing the disturbance backwards from E, there will be two wave normals within the plate corresponding to SE. Let OS=p be that wave normal which forms the smaller angle with OM the normal to the plate. According to Art 104 the difference in path at S, and therefore at E, of the two waves which have traversed N T Fig. 148. 1 the crystal is P . A similar reasoning applies to the interference observed in the direction ET, OT being the direction of the wave normal inside the crystal. Draw OS parallel to O1T and EN at right angles to the plate. The interference seen at Tis the same as that due to the phase difference at S, for waves propagated through 0. If i be the inclination of the line of sight tan i = NS |