2k (1-k2) I=1 cos 2a + 2k (1 − k2) cos (8 — y − 2¥) (1+k2) {(1+k2)2 cos2 2¥+4k sin22)1··· (10). Let us now draw a line from the centre making an angle a with the principal section of the analyser. Then if we consider a series of points on this line, which are not very distant from one another, we may suppose that k is approximately constant for such points. From (10) we see that Ia is a maximum or minimum according as 8-y-2↓ = 2nπ oг (2n + 1) π ; and since 8 − y = Ar2 + B, the points of maximum intensity are determined by Ar2 + B = 2nπ +2¥............ (11). In the neighbourhood of the centre, k does not differ much. from unity, and we may therefore as a first approximation put a; whence writing 0 for a, the equation of the isochromatic curves becomes This equation represents a spiral curve which commences at the origin. The form of the curve when n=1 is shown in the figure; if we put n = 2, we obtain a second spiral which is derived from the former by turning it through two right angles. For values of n greater than two, the two spirals will be found to be reproduced. 162. The fourth case which we shall consider, arises when plane polarized light is incident upon two plates of quartz of equal thickness, one of which is right-handed, and the other left-handed; and the planes of polarization and analysation are parallel. The displacements in the two elliptically polarized waves on emergence from the first plate are given in § 156; and we must recollect that on emergence, we must write +x for in the values of (u, v). Since the second plate of quartz is left-handed, the sign of k must be reversed, and therefore on entering the second plate we must write The four quantities m, n, μ, v must be determined by equating the coefficients of sin, cos in the equation u+u' = U+ U', v + v′ = V + V'. Having obtained the values of m, n, p, v, we must write +x for in the expressions for U', V'. Since the planes of polarization and analysation are supposed to be perpendicular, the displacement on emergence from the analyser will be (U+ U') cos a + (V + V') sin a ; we must therefore form this expression, and then write down the sum of the squares of the coefficients of sin o, cos p, which will give the intensity. The actual calculations are somewhat tedious, but on performing the above operations, it will be found that 163. This expression can vanish in two ways. In the first case sin (8-7)=0, which requires that which represents a series of circular rings, which are black if homogeneous light be employed, but coloured if white light be used. In the second case the intensity will vanish when In the neighbourhood of the centre, k does not differ much from unity, and we may therefore take as a first approximation 8-y=4α + 2nπ; whence writing for a, the equation of the isochromatic curves are Ar2+ B = 40 + 2nπ, which is the equation of a spiral curve. Let us first suppose that n = 0; then it follows that when 0 = 1B, r = 0; so that the spiral commences at the origin, and the distances of successive points from the origin increase with 0. When 0 = π, r2 = (2π — B)/A. = Next let n = 1, then r0, when 01B-; and when 0 = 1B, r2 = (2π - B)/A. We therefore see that the spiral corresponding to n = 1 is equivalent to the spiral corresponding to n = 0, turned from left to right through a right angle. |