NUMERICAL APERTURE. 159 



cale, and represent the passage from air into glass and from glass 

 into air.) The wave advancing along A O must have got to some 

 point on this semi-circle by the time the wave along B D has 

 reached C : from C draw a tangent, C E, to the semi-circle, touching 

 it in E join O E. Then C E represents the tangent to the advanc- 

 ing waves in the new medium, for take any other wave direction in 

 the beam, represented by K L, and meeting M N in L ; draw L R 

 parallel to O E, meeting EC in R, and L Q parallel to O D, 

 meeting L C in Q, and L S parallel to C E, meeting O E in S. 

 (This is only drawn in Fig. 2, a, to avoid confusion of lines.) 

 Then 



O L: LC :: O S : LR 1 T. ,. , ,„ 



\ Euclid VI., 4 

 OL:LC::KL:QC/ 



. • . O S : L R : : K L : Q C. 



OS + LR:LR::KL + QC:QC. 



OE:LR::DC:QC. 



LR:QC::OE:DC. 



Or, the distance L R is the right amount of advance in the new 

 medium for the wave along K L, whilst the wave along B C is 

 covering the distance Q C ; that is to say that C E represents the 

 tangent to the advancing waves in the new medium. 



Through O draw G H perpendicular to ]\I N. Then the 

 angle A O G is called the angle of incidence, and the angle 

 HOE the angle of refraction. 



Let angle A O G = /, 

 and angle H O E = ;- 



Then, since A O D is a right angle, and G O C is a right angle, 



take away the common angle GOD, then / = angle U O C. Also, 



since angle HOC is a right angle, and angle E O C + angle 



O C E are equal to a right angle ; take away the common angle 



C O E J therefore, r =■ angle O C E. 



..DC . . , O E 



JNow, — -— = sm /, and -— -^ = sm ;- 



DC sin / 



O E sin r 



D C 



But — — - is the ratio expressing the speed of propagation of 



light in the two mediums, and is constant ; there is, therefore, an 



