POLARISATION OF LIGHT. 



27 



Having thus given a general account 

 of the phenomena exhibited by the 

 double system of rings, we shall now 

 proceed to explain tlu- law which regu- 

 lates the polarised tints m this class of 

 crystals. 



In all crystals without axes, the axis 

 coincides with some prominent line in 

 the crystals such as the axis of the 

 rhomboid, the axis of the octohedron 

 with a square base, &c. ; but as the re- 

 sultant axes, above described, do not 

 always, or even frequently, coincide 

 with/any fixed line in the crystals, Dr. 

 Brewster conceived that they were not 

 the real axes of the crystals, but only the 

 resultants of the real axes, or lines, in 

 which the opposite actions of the two real 

 axes compensated each other. Various 

 other considerations rendered this opinion 

 almost certain, and still more recent dis- 

 coveries have established it upon an im- 

 pregnable basis. Hence, he was led to 

 consider all the phenomena of the rings 

 and all those of double refraction as the 

 result of two rectangular axes, the prin- 

 cipal one of which was equally inclined 

 to the two resultant axes round which 

 the rings are formed. 



It is easy upon this principle to deter- 

 mine what will be the tint developed at 

 any given inclination, by each of the 

 axes acting separately, after we have as- 

 certained the relative'inclinations of each 

 axis. For this purpose, let ACBD, 

 fig. 33, represent any crystal, with two 

 axes or systems of rings ; and let us 

 suppose it turned into a sphere. Let P 



Fig. 33. 



be the pole or centre of one of the sys- 

 tems of rings, and P' the pole of the 

 other system. Join PP', and bisecting 

 PP in O, draw AOB at right angles to 

 PP', and continue PP' to C and D. We 

 shall call the axis or diameter passing 

 through the point of the sphere Oo, and 



the diameters drawn through P and P', 

 Pp and Py respectively. Now the 

 double system of rings round P, P' may 

 be produced by means of an axis O o, 

 and another axis A B, or CD perpendi- 

 cular to O o. If O o is a negative axis, 

 which we shall suppose it to be, then the 

 axis A B must also be a negative one; 

 but if we suppose the two axes to be Oo, 

 and CD, then CD must be positive. 

 We shall suppose, then, that the two axes 

 are Oo and AB, both negative. 



Since the action of the axis AB, or 

 the tint which it produces at P, 90 from 

 A, is destroyed or compensated by the 

 action of the axis O o or the tint pro- 

 duced by it ; and it is evident that the 

 ratio of the intensities of the axes A, B 



must be that of 1 to - ^-p.. For, as 

 sin. 2 O P 



the tint produced at P by A B is equal 

 to the tint produced at the same point by 

 C ; and as the tint produced at P by 

 AB is its maximum tint, AP being an 

 arch of 90, then the maximum tint pro- 

 duced by O o will be found by the ana- 

 logy, sin. s OP : rad.::i : sin ! Qp - 



Hence, the maximum tint of AB will be 

 to the maximum tint of O o as sin. a 90 



or 1 : - ^r^, the maximum tint 



sin. 



OP' 



which any axis produces being a proper 

 measure of its intensity. 



From a great number of observations 

 made at all points of the sphere, and 

 from measurements of the projected 

 rings, Dr. Brewster found that all the 

 phenomena of the rings, with all their 

 varieties of form, were represented by the 

 following law : 



The tint produced at any point of the 

 sphere, by the joint action of two axes, 

 is equal to the diagonal of a parallelo- 

 gram, whose sides represent the tints 

 produced by each axis separately, and 

 ivhose angle is double of the angle 

 formed by the two planes passing 

 through that point of the sphere, and 

 the respective axes. 



In order to explain the application of 

 this law, let it be required to determine 

 the tint produced at E, Jig. 33, by the 

 two axes Oo, A B, whose relative intensi- 

 ties are as 1 to . . Through the 



given point of the sphere E draw three 

 great circles A E F, C E, and O E: then let 

 'T=Tint required at the point E 



