DETERMINATION OF THE PATH OF RAYS. 



11 



drawn through the vertex. The plane is, therefore, intersected by 

 the refracted ray (produced backwards) in a point whose distance 

 V from the axis is, within a very small quantity which we may dis- 

 regard, equal to W. The equation for the refracted ray becomes, 

 therefore, 



Q' 



+ W. (2) 



y 



n 



It is evident that the ratio to the second vertex N f may be 

 expressed in a similar way. For this purpose the term & has 

 only to be replaced by another, &*, which determines the point 

 of incidence uon the second refractin. 'Oiirfacc, or, more rceiool 



p 



upon the tangential plane drawn through N r . 

 fore, 



We obtain, there- 



y 





+ b*. 



(3) 



By equating the expressions on the right-hand side of the 

 equations (2) and (3) we get 



6* = #> + 4 N*-N). (4) 



71 



In order, still, to determine the unknown quantity /3' by means of 

 /3 we must draw through M (Fig. 6) a perpendicular which cuts 

 the refracted ray at Q and the prolongation of the incident ray at 



FIG. 6. 



Q. Let the angles which this makes with P Q and P Q' be <f> 

 and </>', and the angles of incidence and refraction a and a. In 

 the triangles P Q M and P Q' M", since P J/ = r, the trigono- 

 metrical ratios are 



MQ = 

 From this results 



sin <f> 



sin 



sin a sin <f> 

 sin a sin 0' ' 



