8 ELEMENTARY PRINCIPLES OF MICROSCOPICAL OPTICS 



(Problem) V. 3 : 



1-5 x sin 45 1-5x707 1-0605 



1-6 



1-6 



I 



= 663; 



'=41^ (found by table). 



As a final instance. Suppose the ray to be travelling in the 

 opposite direction, so that G F is the incident ray and B F G, or 

 <^/ 41l ? be gi ven? the media being the same as in the last case, 

 //=l-6 and /*=l-5, find </>, or the angle of the refracted ray. 



(Problem) V. 4 : 



Sin 0= 



u! sin 



1-6 sin 41V 1-6 x -663 



1-5 



1-5 



0=45 (found by table). 



The importance of the prism in practical optics is well known. 



Its geometrical form in per- 

 spective and in section is shown 

 in fig. 4. 



By means of the above pro- 

 blems and their solutions we 

 are now able to trace the diver- 

 gence of a ray through a prism. 

 In fig. 5 let ABC repre- 

 sent a prism of very dense nint 

 glass whose absolute refractive 

 indices /x/ for red light is 1*7, 

 and ft" for blue light is 175. 

 Let the refracting angle B A C 

 of the prism =50, and let the 

 angle of incidence of a ray of 

 white light I) E=45 =0 in 

 air, /x= 1 . The dotted lines show 

 the normals. Then by (problem) 

 v. 1 for red light we have for 

 the angle of refraction 0'. 



FIG. 4. The geometrical form of the prism. 

 (From the ' Forces of Nature.') 



sn = 



404; 



/A sin 1 sin 45 707 

 ~~f~ ~TT~ = 1 T 7 = 

 0'= 241 (found by table). 



And for Uue, light : 



,,_/Ksin_0__l sin 45_7p7_ 

 fj. ff 1-75 ~~175~~ 



0"=23f (found by table). 



Now, for the red ray draw E F (fig. 5), 24^ to the normal, and 

 let it meet the other side of the prism A C in F. At F draw 

 another normal. 



On the scale of our diagram it is not possible to draw two lines 

 E F, one for the red ray and the other for the blue, for they are too 

 close together, their angular divergence being only j. But by 



