90 H. A. Rowland— Concave Gratings for Optical Purposes. 
focus without telescopes would be solved. But an ordinary 
dividing engine rules equal spaces and so we shall further in- 
vestigate the question whether there is any part of the circle 
where the spaces are equal. We can then write 
Wa 
And the differential of this with regard to an arc of the circle 
must be zero. Differentiating and reducing by the equations 
daz: y—y ab 
dys xa” dy —C (x—2’) 
we have , 
p 2ab (y—y’) —2yb (x—a')—F [60°—(a* + y*+a")] ' 
+0} y-Y)LO—2*)(y-y) Py] —(e—2)[ (0a?) 20-2) 
2b : 
+f fey) —y (e—e)] | =0. 
It is more simple to express this result in terms of R, 7, p and 
the angles between them. 
Let y be the angle between p andr, and » that between p 
and R. Let us also put 
: : ie 
2 
Let f, 7 and 6 also represent the angles made by 7, R and 
is oe with the line joining the source of light and focus, 
and let : 
a= and ¢é 
_utY 
att. 
eee a 
Then we have 
__Reosyt+reossf Rsiny+rsinf _rcosi—Reosy 
: eae 2 Secs YS ae aan aii 2 
(0'—a’) (y—y')° + 0(e—a)'=p'(6—a sin* 8) 
b’—a’=Rr cos’ a 
sin Wy icone a: cos? BE Sa ot 
Ga : maT . 
a ... 
R=b——x; r=b+—% 
cos sin y sin Rr . 
2=6 rent —o coat eo = —- sin 7 cos @ 
cos a sin a@ cos a@ b 
bRrp 
By (y—y) +2 (B—a") (ea!) = 
2b°—(a*+y'?+a’)=Rr 
(cos 4 +cos Vv) 
