464 Mr. R. A. Houstoun : 
the angular distance of an image in the focal plane of the 
telescope from the no deviation position, the angle being sub- 
tended at the centre of the object-glass. Therefore | 
u(e cos i—f sin 2) — (e cos wz —f sin wi) 
+(esin wi+fcos wi)AQ,;=pr.. . (3) 
Suppose now we have also light of a wave-length slightly 
different from 2X. Then let us suppose that the echelon 
remains in the same position. This new radiation will produce 
an image given by A@. Thus we have 
(w+dy)(ecosi—f sin 7) —(e cos wi—f sin pt) 
+(esin witfcos uw2)AQ,=p(ry+dr). . (4) 
Subtracting (3) from (4), 
du(ecosi—f sin 2) + (esin we +f cos ui)d0,=pdr, . (5) 
where d0,=A@,—A6@,, i.e. is the angular distance between 
the two images. 
Since z is small this becomes 
“ae | 
ag, = POE .) == 
Let the next order image of the radiation X be given by 
Aé,. Then 
p.(e cos 2—f sin i) — (e cos wi—f sin pt) 
+(esin wi+fcos uz)A@;=(p+1)rA. . (7) 
Subtracting (3) from (7), 
(e sin witfcos pr)d@,=rA, . . . - (8) 
where d@,= A@;— AO, = angular distance between two orders. 
lf 2 be small we obtain 

Si x . 
if 
The value of p is ane a Ce ee “. Therefore 
w—i)edr 
dO, fate ae dts 

dO, 
eae aus. 
This is , the usual formula for. evaluating an angular dis- 
tance in Angstrom units. 
