108 Professor Airy on the 



VII. When Fresnel's rhomb is interposed in position 45°, we 

 must suppose AR, fig. 18, to be perpendicular to the plane of 

 internal reflection, and we shall have the vibrations 



- -j~ sin ~{vt-x) parallel to AR, and ~t= cos -^-(vt - x) 



perpendicular to AR. Resolving these in directions parallel and 

 perpendicular to AC, we find for the former 



sin — (vt- x) . cos a+<p+ 45° + — 7= . cos -^ (vt-x) .sin a + <p + 45°, 



—7=.sm---(vi;—x).cosa-ripT*i>-r fa ~\~ 



7i sin T 



or - -75 sin — (vt- x) - (a + <j>) - 45°: 



and for the latter, 



-7= sin — (vt-x).sin a+d> + 45°+ — 7=r cos—- (vt — x) .cosa+0+45°. 



or —7% cos — (v t — x) - (a + <p) — 45°. 



Now taking the same expressions as before, for the vibrations 

 in Oo„ kc. and comparing the sums in directions parallel and 



perpendicular to AC, (putting — (v t - x) - 45° = if), we have 



— y= sin ? — (a + d>) or — —7= cos a + d> . sin £ + —7= sin a + (p . cos £ 

 V2 V 2 v 2 



= (&&■ — ■% J sin £ + (f — *wj cos £ 



—7= cos £ - (a + <p) or —7= sin a + <£ . sin £ + —7= cos a+ # . cos £ 

 = (w + y) sin £ + (x + &) cos £ 



