194 The Rev. S. Haucurton on the Equilibrium and 
é, 0’ are the angles of incidence and refraction, we shall have, when z= 0, 
¢=¢,=¢’'. Substituting therefore the results obtained above in the equations 
of condition, »/ = 1", f’ =f”, we shall deduce finally, 
a ue 
n/(7—7,) < iN 
Jo 
Wa ibe 
. nN ? 
or, writing n’’ = ky’, and substituting their values for n,n’, A, r’, 
a — 7’, 
Ro, Suds Sa (61) 
whence we obtain 
aav(1 +o) 21, (1x 2 I 
and finally, for the two unknown vibrations, 
| 2tand’ 
T. tand’ 4 Ktand” 
To determine the actual position and magnitude of the second transversal, we 
have evidently, for a single wave, 
/ 
tané’ — xtan@ 
Lap, == 406 ay, a Ted 
tané’ + ktané 
(62) 
ie nsing 
yf Sl SSS 
Te x 
27 lsind 
Fo h — Te aS 3 
whence we obtain 
L 
2 je wT. 
ge ge VP +R =n.;sing, 
i. e. the second transversal lies in the plane «, z, or plane of incidence, and makes 
with the axis of x an angle equal to the angle of incidence. 
2nd. Incident vibration transversal, and in the plane of incidence. In this 
case all the motion will continue in the plane of incidence; we shall, therefore, 
have 7 = 0; and reasoning similar to that used in the former case will show 
that € and ¢ are independent of y. The differential equations of motion will 
therefore become 
