SPHERICAL ABERRATION 111 



Now find the point Q on the axis such that 



CK = /?' 



CK /z QC 

 By construction p~r = -, =7^' 



i.e. the sides of the triangles PCK, CKQ about the common angle C are proportional. 



:. by Euclid vi. 6, the triangles are equiangular, viz. CKQ=KPC and 

 KQC=CKP. 



sinCKP sinCKP PC /*' _ 

 shTCKQ = snTKPC = CK => =h ' 

 where h is the relative index of refraction of the second medium to the first medium. 



But if P be a source of light, 



sin CKP =sin < =h sin $' h sin CKQ, 

 where <= angle of incidence, and </>' angle of emergence. 



.'. KR must be the refracted emergent ray. 



Now the position of K is arbitrary ; therefore all the light diverging from P, 

 however great the angle U, must after refraction appear to come from Q. (The 

 angle U, or CPK, represents of course merely one-half the cone of light that falls 

 on the lens.) 



This then satisfies the first condition of aplanatism ; in other words, the spherical 

 aberration is corrected. 



With an oil-immersion lens, seeing that the oil and the glass have practically the 

 same index of refraction, the object may for all practical purposes be regarded as 

 in the glass, which is exactly the condition required to satisfy the sine condition. 



For the second condition, the correction of coma: 



From centre C at distances CP and CQ draw arcs PB and QB' (fig. 107). Join 

 CBB'. Then CBB' may be regarded as the axis of the lens, and the image of B will 

 be formed at B'. 



Regard PB as the object o, and QB' as the image i. 



i QB' QC //CK 

 o~PB~PC~ PC "//sinCKP 

 p. sin KPC fi sin U 



~ fj.' sin KQC ~ jtx' sin U' ' 

 that is, 



index of refraction of first medium x sin of hah* the 



dimensions of the image _ _ angle of the rays diverging from the object 

 dimensions of the object ~ index of refraction of final medium x sin of hah* the " 



angle of convergence of the rays forming the image 

 and this is the one and necessary test for aplanatism. 



