152 Mr Swan*s Formulw for constructing 



point. For this purpose we have the equation to the circle 

 found by formula (2) 



(a;-2w)2+ (5^ + 4/7?) 2=r2, 



in which substituting for x its value, d cos % we obtain 



y = 2-86515. 

 But the correct value of y at that point is 

 rf sin I = 2-84634. 



Therefore the distance of the two arcs, reckoned along y is 

 •019 ; but since the ordinates are evidently inclined to the 

 normals to the circles at the point A at an angle 



45-i = 40°54', 

 4 



The true distance of the circles will be -019 sin 40° 54'= -012 

 — a difference perhaps within the limit of the errors of work- 

 manship in such apparatus. 



Again, to calculate the aberration of the rays reflected at 

 A, we have from the equation to the circle 



dy _ X — 2 we 



dx y -{' 4, m 



which should evidently be equal to tan d ; and substituting 

 the values of x and y, we find & = 41° 17' 7". Now, the 

 true value of 6 at the point A in the circle depends upon that 



of ^, and we have seen that 



^=45 —% also ^ = tan "~ --. 

 ' 4 2 X 



Therefore, substituting the values of x and y, we find 6 — 40° 

 52' 56", which shews an error in the inclination of the reflect- 

 ing surface at A amounting to 0° 24'. But since the error in 

 the direction of the reflected rays is double of this, and the 

 same error aff^ects the opposite reflecting side of the prism, 

 the final error in the direction of the rays will be 1° 36', and 

 this will give a lateral aberration of -56 inches when the rays 

 are finally returned to the flame. 



In this example we have taken a value of jti perhaps rather 

 higher than that of the ordinary flint-glasses for the least 

 refracted rays. We shall, therefore, in the next example 

 assume /* =: 1*55. 



Suppose the mirror intended for a light of the first 



