518 



Prof. J. D. Everett. 



[Jan. 22, 



25. Applying these principles, we have 



Deviation at first refraction = 45° -- 28° 7 J' =16° 52 J', 



which is accordingly the downward slope of the rays in the glass. The 

 tangent of 16° 52 J' is 0*303, which agrees with the value above 

 deduced from the co-ordinates of the ends of the secondary line. 



Again, the distance OC of the centre of curvature from the origin 

 is ^/99 = 9*950; and the condition that C is collinear with the ends 

 of the secondary line is the vanishing of the determinant. 



12-50, 



10-15, 







13-43, 



9-04, 



-9-95 



1, 



1, 



i. 



Expanding, we find that the positive terms amount to 237 4, and the 

 negative to 237*3, a sufficiently close agreement. 



26. Let A in the annexed figures be the point where a line through C 

 parallel to the rays in the lens meets the plane of the annulus. If we 

 draw any straight line through A, cutting the annulus in two points 

 P, P', the rays at these two points are in the same plane of incidence 



OD = OD' = radius of circle = 1. AT is a tangent. 



OA = 5*32 in both figures. APP' any secant. 



DC = 10, giving CO = 9'95. 



CAPP', and will therefore meet ; the point in which they meet being 

 in the primary focal line. P and P' coincide at the point of contact T 

 of a tangent from A, and the ray from T passes through one end of the 

 primary focal line, its other end being symmetrically placed on the 

 other side of the plane of symmetry. 



We have OC = OA = OC tan 28° 7 J' = 5-318, and the 



angular position 6 of T is determined by cosec 6 = OA, giving 

 = 19" 10'. The rectangle AP. AF, or the' square of AT, is 27-2847. 

 For any given value of = AOP we can, from these data, calculate 

 6' - AOP'. The following pairs of points of emergence of intersecting 

 rays are thus found — 



