106 Mr. MacCullagh on the Laws of Reflexion 
IP, parallel to the axis of the crystal, meet the surface of the 
sphere in P. Let the great circle ZO E be 
the plane of incidence, containing both the 
direction IO of the ordinary refracted ray re) 
produced backwards, and the direction I E ~ P 
of a normal to the extraordinary wave; and R 
draw the great circles PZ, PO, PE. The 
angle Z will be the azimuth of the plane of incidence. Let 
ZO=4,ZE=9,PO0O=y),PE=¥Y, the angle ZOP 
= @, and the angle ZEP= 4. Call the angle of inci- 
dence i, and suppose 6 to be the reciprocal of the ordinary 
refractive index, and a the reciprocal of the extraordinary. 
Each of the refracted rays, in turn, may be made to dis- 
appear, by polarizing the incident ray in a certain direction 
assigned by theory. When the extraordinary ray disappears, 
the reflected ray is polarized in a plane inclined to the plane 
of incidence at an angle 6 determined by the formula 
a : igfit HV Bea sin? z : 
| tan 6 = cos (i+ ¢) tan§+2 (a®—2*) sin 4 sin cos) ay ers G3)" 
When the ordinary ray disappears, the plane of polariza- 
tion of the reflected ray is inclined to the plane of incidence 
at an angle 6’ determined by the formula 
! 
—tan f! = cos (7+ 4") cotan 6’ + (a*—6*) =a 
ait es Se 
sin Y cos V Sin (ia) (3.) 
And when the angles B, 8’, become equal, the plane of po- 
larization of the reflected ray becomes independent of the 
plane of polarization of the incident ray; and the angle of in- 
cidence 7, at which this equality takes place, is the polarizing 
angle of the crystal. Hence we have the equation of condi- 
tion 
; Cae ey __sin®z 
cos (i+ ¢) tan§+2 (a°—O*) sin§ sin cos ah 
2 if & = 0.(4.) 
aa a 1 ¢p27ey08 24. 1, , sin’ 2 
+cos (¢+ 9 )cotan# + (a*—D*) ee a sin J’ cos singel) 
to be fulfilled at the polarizing angle. 
Since 7+, in this equation, is nearly equal to a right an- 
gle, putz+$ = > + 8, and will be a small quantity. Draw 
P R an arc of a great circle perpendicular to ZO E, and let 
