KLEMENTAKY PRINCIPLES <>F MK'KnsCOPICAL OPTICS 



(Problem) V. :! : 



u sin <& 1-5 < sin 45 I'.") x '7<>7 1-0605 



rj * | _ rt /> o 



>>in a = 7 = . :; 7; = ^5 ^'oOo ; 



p' I'D I'D 1'6 



/ = 41J j (found by table). 



A> a iinal instance. Suppose the ray to lie travelling in the 

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

 / = 41.V. be given, the media being the same as in the last case. 

 //=]<) and p.= \-~). iind 0. or the angle of the refracted ray. 



(Problem) V. 4 : 



_/_sin 0' 1-6 sin 4H _l-6x"663 1-0608 



' 1 1 1 C' ^^ ~ " -, 'z ^ ~ ~ ^ ^ ~ */'_'/ 



II 'I l**^l l -r l 



W. 1 *J L fj JL <J 



= 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. 



l>y means of the above pro- 

 blems and their solutions we 

 are now able to trace the diver- 

 gence of a i'<i i/ tltrouyh a jirixnt. 



In fig. 5 let ABC repre 

 sent a prism of very dense Hint 

 glass whose absolute refractive 

 indices p! for red light is 1'7. 

 and p." for blue light is 1'7.">. 

 Let the refracting angle B A < ' 

 of the prism =50, and let the 

 angle of incidence of a ray of 

 white light 1) E = 45 =0 in 

 air. //,= 1 . The dotted lines show 



< t. The geometrical form of the prism, the normals. Then by (problem) 

 From the 'Forces of Nature.') y _ j f()r ml j^ ^^ J 



the angle of refraction 0'. 



,_fjL sin Isin 4") 707 



p! "17 ' \'T~ : 41<i ' 



7 == 241 (f oun ,i by t .,i,i,.). 



And for blue light: 



p sin 1 sin 45 '707 



P" 1-75 = l-7^ = 



0' / =23| (found by tabl,.). 



Now, for the red ray draw K V (fig. :,). i>4^ to the normal, and 

 let it meet the other side of the prism A f' in V. At V dra\\ 

 another normal. 



On the scale of onr diagram it is not pos>ible to draw two lines 



F, one for the red ray and the other tbr the bine, for the} are too 



close together, their angular divergence being only : , io . ' Hut by 



