on Mr. Potter's Experiment 011 Interference. 163 



of that colour have reached by equal (or rather equivalent) 

 paths, and therefore is the place of a bright bar for that colour. 

 Lastly, as the interfering pencils for different colours are made 

 to deviate unequally by the prism, the centres of the mixture, 

 for different colours, will not occupy the same place. 



From the last consideration it follows at once that the 

 centre of the mixture will not be the place of the centre of the 

 fringes, inasmuch as similar bars of the different colours are 

 not united there. To find the place where they are united 

 we must consider that the centre of the red mixture, being 

 the least deviated, has made the smallest progress towards the 

 thick side of the prism, but that its bars are the broadest : the 

 centre of the yellow mixture has made the greatest progress, 

 but its bars are the narrowest. A number n of bars can there- 

 fore be found such that the linear deviation of the red centre 

 -\-n x breadth of a red bar = linear deviation of yellow centre 

 + MX breadth of a yellow bar : (if the equality is not exact, 

 n may be chosen so that the second side is the greater, but so 

 that on putting n+ 1 for n the first side will be the greater). 

 The nth red bar and the nth yellow bar, thus determined, will 

 coincide; and their place will be the true centre of the fringes. 

 It is plain that when the linear dispersion produced by the 

 prism is small, that is, when the eye-glass is very near the 

 prism, the centre of the fringes will not sensibly differ from 

 the centre of the mixture for any of the colours ; but that 

 when the linear dispersion is great, or the eye-glass far from 

 the prism, the centre of the fringes will be far from the centre 

 of any of the mixtures. 



Algebraically we may express it thus : — If d be the linear 

 deviation of the centre of the mixture for one standard colour, 

 d + Srfthat for any other colour, b the breadth of a bar for 

 the standard colour, b + $b that for any other colour, then the 

 distances of the nth bars from the origin of d will be d+nb 

 and d+hd+ n (b + $b) respectively; and these will be equal 



if hd+nlb — 0, whence n = r-j-; which makes the di- 

 ll O 



h 7\ rJ 



stance of the place of coincidence d ~y-. As Id and I b 



have different signs, this is greater than d; or the place of 

 coincidence is removed from the origin of d in the direction 

 of deviation, that is, towards the thick side of the prism. And 



as, upon receding from the prism, $d increases, while yr- is 



not altered, it follows that the place of coincidence or centre of 

 the fringes advances further and further towards the thick side 



Y2 



