1903.] 



On Skew Refraction through a Lens, etc. 



521 



The form of these cusps, for a section of constant z, is exhibited 

 near the centre of Plate 10, on half the scale of the two cusps of the 

 secondary (which are for sections of constant £). 



33. The following examples illustrate the use of the harmonic 

 formula? : — 



(a) To find the points of the annulus at which the deviation is 

 exactly perpendicular to the plane of symmetry — 



We must put m = 0, giving 



900 cos 9 - 57 cos 20 = 75, whence = ±88° 51'. 



(b) To find the points at which the horizontal deviation is greatest, 

 we must put dl/dO - 0, giving 



9«*?- e= ±95- 46'. 

 cos 64 



(c) For the total deviation to be a maximum or minimum, its cosine 

 11 must be a minimum or maximum. We thus obtain — 



*J = -0-0013 sin + 0-0020 sin 20, 



dd 



= - 0-0001 sin (13 - 40 cos 6) = 0. 



The factor sin vanishes at 0° and 180°, at which there are maxima. 

 The factor 13 - 40 cos vanishes for = ±71*, and gives the points 

 of minimum deviation. 



The points on the " direction curve " which correspond to these three 

 determinations are indicated by the words " horizontal," " maximum," 

 "minimum," on the right hand of the curve. 



(d) To find at what distance z the radius of curvature of a cross 

 section becomes infinite at one of the two points 0° and 180° in the 

 section. 



From equations (4) we deduce 



dx a , d V 



dO d6 n 



dy ■ n d m' 



dO dO 11 



At 9 and 180°, dx/dO is finite, but dyjdO vanishes (since dm'/dO and 

 dn'/dO contain only sines). Hence the tangent is horizontal. The 

 condition of an infinite radius of curvature is d 2 y/d0 2 = ; or 

 d 2 m' 



- COS + Z -r^ — = 0, 



