REFLECTED AT THE SURFACE OF A CRYSTAL. 45 



<h'r 3S 



and J=- . (7). 



r r b 



Let us then substitute in equation (6) these different quantities, and we 

 shall have 



2 (0 r + ZL x*) 2 sin 2 -^ = 2 | r + ^ (z 2 cos 2 (f> + 2ip sin (f) cos +)? sin 2 0) 1 2 sin 2 * 



= 2 (0 r + ^* 2 cos 2 0+jo 2 sin 2 0) 2 sin 2 y , 



</>'? /fe i 



because 2 psin*-2-=:0, from the cu-cumstance that every value of i has two 



equal values of p with opposite signs. 



Hence 20r + ^5a; 2 2sin 2 =2>-cos 2 ^) + sin 2 + 2-^z 2 cos 2 +;u 2 sm 2 ). 2 sin 2 



= 2 (0 r + ^jo 2 )2sin 2 ^ sin 2 + 2(0 r + ^-i 2 ) 2sin* y cos*0 

 * 2 ) 2 sin 2 cos 2 (p. 





But 2(r + - 2 2sin 2 = S2 - - 2 sin 2 



2 



- C 2 (8). 



Hence, finally, 2(0r + ^5r02sin 2 ^=^(sin 2 0- 



T 2i 2 



Similarly, 2 (0 r + 2-^-<5 a; 2 ) 2 sin 2 ^ = ^ (sin 2 0-2 cos 2 0) 



/* mm 



Again, ^((pr + d a?} sin A j = 2 (</> r + p 2 ) sin Ae sin 2 



rf>'r 

 + 2 (0 r + J ) sin ** cos 2 



and 2 ((j)r + y- r f) sin A = -22 (0 r + ^/> 2 ) sin A * 



as equation (8) shews. 



d)'r 

 Denote this quantity 2 (<pr + J-~ i 3 ) sin A e by M ; 



then 2(0r + 



similarly 2 (0 r + 5 a^ sin k r = M (sin 2 0-2 cos 2 0). 

 VOL. XV. PART I. N 



