The President's Address. By E. M. Nelson. 161 



Combining (i.) and (ii.), we obtain 



From (iv.) we have q — r = fi r ; but q — r = CY 



.-. CV = yur. 



Now, in the two triangles H C 0, H C V, we have the included 

 angle at C and the side H C = r common to both, and the sides 



C0 = - and C V = /* r both determined ; therefore the remaining 



Let H = a, then H V = /j. a, and r sin C = a sin 6, and r sin C = 

 fx a sin cf). 



.'. Sin $ = /x sin <£. 



When the lens is a hemisphere, then H U becomes perpendicular 

 to the axis, and tan 6 = /x, and cot <f> = /jl. 



Although it is outside the subject in hand, it is interesting to note 

 that these two triangles are similar,! therefore the angle C H V = 6, 

 and CHO = <j). Again, because CH is the normal, CHV is the 

 angle of incidence, and CHO that of retraction ; and as sin 6 has 

 been proved equal to /x sin </>, therefore sin C H V = /a sin CHO, 

 which proves the aplanatism of the lens, when the condition that 

 q = /jup is fulfilled, because it is true wherever the point H may be 

 placed on the curve A H. The limit is reached when V H is a tan- 

 gent to the curve, i.e. when 6 is a right angle. 



We will now take a practical example to illustrate the use of the 

 above formulas. Let it be required to construct an aplanatic front 

 for an oil-immersion objective of N.A. 1*3 ; let p = 1 *5. 



* The above is the common formula for the solution of the third side of a 

 rectilinear triangle, when two sides and the included angle are given, viz. — 



c 2 = a 2 + ¥ - 2 a b cos C. 

 t Euclid, Bk. vi. Prop. 6. 



