40 



PROFESSOR KELLAND ON THE POLARIZATION OF LIGHT 

 X 



in the plane AT, and of the other perpendicular to the plane AT'. Denote the vi- 

 bration in the plane AT by T, and that perpendicular to AT' by T'. Also let R, R' 

 be the resolved parts of the reflected vibration, in and perpendicular to the plane 

 XOY respectively. Denote further by R, T, the reflected and transmitted vibra- 

 tory motion put in play at the surface, whose type does not contain x. 



Let XI =<, XT=0 XT'=<', and call a, (3, y the resolved parts of the vibra- 

 tions without the crystal, and a t , /? 7,, those within. 



Conceive that at the same instant and for the same point, I tends downwards, 

 R upwards, I' outwards, and R' inwards, and the following will be the resolved 

 parts of the vibrations without the crystal. 



ct = I-R)sin-f R, 



Again, if we denote ATY by 6, and AT'Y by 6', we may resolve the vibra- 

 tions T and T' parallel to the directions of x, y, and z, thus : 



Draw OM perpendicular to OT in the plane AT, and ON perpendicular to OT 

 and to the plane OT': then turning the figure round OZ through 180 to bring it 

 into its proper position, or (which is the same thing) changing the direction of y, 

 we get 



But 



a,=T sin (j>, Cos 6-T sin <' sin & + T, 



/?,= T cos 0, cos 6 - T cos <' sin & (2.) 



7,=T sin0 + T'cos0'. 



It may be remarked that the second vibration may, in each direction, be ob- 

 tained from the first by writing <' for < /5 and 90 + & for 6. 



0,s= -T cos MY-T' cos NY 

 7,= T cos MZ + T' cos NZ. 



cos MX = sin (f>, cos 6, cos MY = cos (f>, cos 6, cos MZ sin 6, 

 cos NX= sin <p' sin &, cos NY = cos </>' sin &, cos NZ=cos 8'. 

 Therefore, by substitution, 



