244 Prof. Lewis, On a crystal of Siephanite [May 15, 



the two faces, and &> and w l the angles which the reflected rays 

 make with A C. Let e and yjr be the angles (in circular measure) 



(fig. 3). 



*^A 



which iO^ subtends at the bright and faint signals respectively ; 

 and let the projections of the broken line NC^ perpendicular to 



A C and SG be a and t. Then e = — , and ty = j. It is clear these 



angles will have their greatest values if the parts of the faces used 

 be those immediately at the edge. This will also be the only 

 position in which the faint signal can be distinctly seen behind 

 the crystal. I shall therefore suppose the effective parts to be 

 at the edge at N and N v Now 



CN 2 sin 2 a l = r 2 + r 2 — 1rr x cos oc x ; 



where r and r l are the radii of the two circles enveloped by the 

 faces. Also, if 



„„_, EN »•— r cos or. r. , 



NCE = 7, tan 7 = y^ = — = - 1 cosec a, - cot a ; 



' ' CE r sin a 1 r 



where a can be introduced to give an approximate value of 7. 

 From the diagram we see that 



ECA=^-~^6, 



therefore NO A =7 - (^ -© -0) . 



Also NCN t -NCA+N x CA. 



