of the Interference of Homogeneous Light, 85 



To find the positions of the secondary images of the lu- 

 minous point after the passage of the small pencils through 

 the prism, we will suppose one of the primary images at o, 

 fig. 2 ; then after refraction at the first surface the pencil will 

 diverge as from another pointy, such that r being put for the 

 distance o r, and r 1 for the distance p r, we have 



, cos 2 f 



r* ~rp r-7-, 



cos 2 1 



i being the angle of incidence on the first surface, i f that 

 of refraction, anil //, the refractive index of the glass. (This 

 and several other equations which I shall have to introduce 

 being demonstrated in Mr. Coddington's excellent later trea- 

 tise on Optics, I shall here use them without further explana- 

 tion.) Then t being put for the thickness of the glass passed 

 through, we have for r or q s this equation : 



x . , . cos 2 i j cos 2 i 



r x = h J + t) «-r ; = r + t 



\L COS* V jU, COS 3 if 



Now, of the two interfering pencils we may take the axis 

 of one, or the line perpendicular to the line joining the two 

 primary images of the luminous point, as passing through the 

 very angle of the prism where t = ; and hence r\ = r r 



Then the incidence of the axes of the pencils being that of 

 minimum deviation, and 2 a being put for the distance a b, and 

 i for the refracting angle of the prism, we have 

 t = 2 x distance c d x tan \ i 



j i- . l cos ** ^ cos i' 

 cd = distance ao x r = 2a r . 



cos i cos i 



the angle ere being equal to the angle of incidence, and the 

 angle red equal to the angle of refraction. Hence 



cos i' cos f sin i 1 , 



t — 4a - tan \i = 4 a r- tan v = 4 a r , and 



cos i cos i cos i 



, i sin i' cos 2 i ■ cos i. sin i' , . 



r = r + 4>a r- • j-, = r + 4a 5-^- = «'s m 



COS 2 |U, cos- z' • ^ cos 2 1* 



the figure. 



Then, drawing sf perpendicular to be produced, we have 

 the angle csf = 1, and $/* = 2 a. Hence 



the distance cf — 2a tan i, 



and g 6' being drawn perpendicular to b'c and as, we have 



the distance ga' = (r, + 2a tan 1) - A- -f- 4a cos * S1 £A 



From these equations we find the positions of the secondary 

 images of the luminous point, a! and b'. 



