112 



(y+y). dp =s( — I- — j I rfx + pdz — (ci+yp){»dx-ir/3dy-i-ydz) > 



(y+y')-'??=(-+-^) ^d,/ +qdz — (^+yq)[udx+/idl/+vdz) ^ 



{' being the distance of the luminous point from the mirror ; and since these equations are to be 

 satisfied independently of the ratio between {dx, dy), they resolve themselves into the three fol- 

 lowing, 



(y+y). r = (- + A) . { 1 + /-(« + rp)' } 

 (y+yM= (i + 1). ^l+y^-(^-f-vy)^^ 



(v+vO- * = (-+ ^) . ^/'y - (*+'/?') (/s+yy) I (W) 



which contain the solution of the problem. 



To shew the geometrical meaning of these equations (W), let us take the vertex for origin, 

 the normal at that point for the axis of (z), and the tangents to the lines of curvature for the 

 axes of (x) and (y) ; we shall then have 



p = 0, 9 = 0, « = 0, »■ = ;^> ^— R'' '/ = y* = COS. /, 



/ being the angle of incidence, and R, R', being the two radii of curvature of the mirror ; and 

 the formulae (W) become 



h.cos.I=R{l—»')=R'(l-n «/3 = 0, (X') 



(h) being the harmonic mean between the conjugate focal distances, so that 



2 _ 2 }_ 



T-7 + 7- 



The equation niS = 0, shews that the plane of incidence must coincide either with the plane of 

 the greatest or the least osculating circle to the mirror ; and if we put ,3 = 0, that is, if we 

 choose the plane of incidence for the plane of (:i;, z) we shall have R' = iZ(I — a^), R > R', so 

 that it is with the plane of the greatest osculating circle that the plane of incidence coincides. 

 We have also 1 — «* = cos. */, 



h = R. cos. /= R'. sec. J } 



¥= RR', R' = R. COS. -iS ^ ' 



from which it follows, that the harmonic mean between the conjugate focal distances, is equal 

 to the geometric mean between the radii of curvature of the mirror ; and that the square of the 

 cosine of the angle of incidence is equal to the ratio of those two radii of curvature. It follows 



