4O Geometrical Propositions applied 



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

 face of the crystal cut these tangents in the points P /? M^ p,, m t ; 

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

 PP,, MM ft pp ft mm 4 i cut the line SRs 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 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 



f) 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 



e x op- 



V x 08 x cos 

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



x pp" 



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 



8 x MM" 6 x pp" 6 x mm" 

 Fx OS ' TITOS' TITOS' r 



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 P'OL which the ray P makes 



with OL : that is, the projection is equal to 9 - ,.._. But 



cos P'OP 



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



OP OF 

 cos POL = r-, and cos P / OP / = -- 



