Form of Newton's Rings. 231 



i. e. when <9 = 54° 44' 8"'2 (about). 



For this value co = W° 28' 16"-4 (about). 



This is the greatest value of co, which varies from zero when 

 6 is zero to this greatest value, and then decreases to zero 

 again as 6 approaches a right angle. We notice that the 

 value of co is entirely independent of the shape, thickness, 

 and material of the lens and plate employed. 



The extreme points of any ring in this line are by (ii.) 

 x= +a. 



The actual diameter of the ring is therefore 



„ cos 6 sin 6 cos 6 



2a jq ^ = a = 



cos (u — co) sm co 



= a \/l + 3cos 2 (v.) 



The diameter, as seen projected on the horizontal, is the 

 distance from x=a, £ = 0, to x = — a, z = 0, i. e. 2a. 



Let us now consider the intersection of (i.) with y= +a. 



Since, so far as the rings are concerned, equation (ii.) is 

 also to be considered, so that a = 0, we see from (i.) that 



i. e. the intersection of this plane with the ring-system is also 

 a straight line, and, moreover, a line parallel to the surfaces 

 of the glass plate. 



We may write (i.) in the form 



_ 1 ^(^ 2 + ?/ 2 ) sin y 2 cos 2 , 



Z ~* x 2 +y i co^e ~ Zl x*+y* co&d ' ' ^ 



Thus, when x and y vanish, z lies between zero and — z lt 

 the exact value depending on the vanishing ratio of % to y. 



Thus this portion of the axis of z lies entirely in the 

 surface. 



Again, we may write the equation (i.) in the form 



{x 2 + y*) {x sin 6-2z) + 2y* sin 2 \z-z x cot 2 0] =0 ; 



and hence we see that the line 



z==z l cot 2 9 



_ 2z _ 2z x cos 2 6 

 — sin 0~ sin 3 6 



lies wholly in the surface. 



This is a line parallel to the axis of y. 



