let 4 represent the length 
of a vibration of light, it 
is evident we shall have 
An—dn=(2m—1)s. Sub- 
tracting this equation from 
the equation Sd=SA we 
have Sd+dn—An=Sn— 
X 
An=SA-— (2m—1) 5 eee 
The second member of this : 
equation being constant whatever be the distance, AP ; par 
jectory of the fringe forms one branch of an hyper oes 
iareat AandS. To obtain the equation o this } J Pn=th 
in a form convenient for discussion let SA=2C, AP rie eae 
and then seek the values of 2A and 2B, the axes of the 
ri 
We first take 2A=Sn—An=20—(2m—1)5, whence, 
. ; 37, 
A*=0(0-(@m—1)) neglecting the term which — 
i 
