4O . Geometrical Propositions applied 



fraction represented in the figure ; let a perpendicnlar at to the 

 face of the crystal cut these tangents in the points P /? Jf,, p 4 , m 4 ; 

 and let the lines OP", OM", Op", Om', respectively parallel to 

 PP,, MM pp mm ^, cut the line SJRs in the points P", M", p" , m". 

 The length of the path which a ray P describes within the 

 crystal is equal to the thickness 9 of the crystal divided by 

 the cosine of the angle P / OP / which the path of the ray makes 

 with a perpendicular to the faces of the crystal ; and the velocity 



O Sf 

 of P is equal to V x - -, (37) : dividing therefore the length 



of the path by the velocity, we find that the time in which a ray 

 P crosses the crystal is equal to 



ex op / 



Fx OS* cosP'OP/ 

 But as OP' is perpendicular to the tangent plane at P, we have 



^- _ = OP = PP" 

 cosP'OP, ^ 



Therefore the time is equal to ^= ^-^ . Similarly, the times in 



y x c/o 



which rays M, p, m, pass from one surface of the crystal to the 

 other are equal to 



6 x MM" x pp" x mm" 



42. Now suppose the path of a ray P to be projected per- 

 pendicularly on a right line having any proposed direction in 

 space. Through conceive a right line OL parallel to the pro- 

 posed direction, and meeting in L the tangent plane at P. The 

 length of the projection is equal to the length of the path multi- 

 plied by the cosine of the angle POL which the ray P makes 



with OL ; that is, the projection is equal to - . But 



cos Jr (JJr 



because OP is perpendicular to the tangent plane at P, we have 



OP* 

 cos POL = ~, and cos FOP, 



