444 THE OUTGO OF ENEEGY 



Draw a line from the point i of the arrow 

 through k. This line, since it passes through 

 the centre of curvature, will coincide with the 

 perpendicular to the refracting surface, and there- 

 fore will not be refracted. Draw from the point 

 i to the cornea a line parallel to the axis. This 

 parallel ray will be refracted through the poste- 

 rior principal focus c/> 2 . The two rays will unite 

 at their point of intersection, i 2 , which point is 

 the image of i and is its conjugate focus. The 

 arrow ij was. vertical to the axis. Hence its 

 image will also be vertical to the axis. Draw, 

 therefore, from t' 2 a line vertical to the axis. 

 From the end j of the arrow draw a line through 

 k. The intersection of this line with the verti- 

 cal line just drawn will be the image / 2 of the 

 point j. 



Calculation of the Position of the Conjugate 

 Foci. The conjugate foci may be found by the 

 following formulas. Let /i be the conjugate 

 focal distance hi i, and / a be the conjugate focal 

 distance 



For virtual images the formulas become 



F- 



*l-/i 



