Of Light and Colours,

or Halo’s, is because there is but one par^ticular Point N in all the Part of the Drop

between

Then CFN = CXN + F.NX, but FNX CNG— CNF or CFN; therefore CFN =CXN-(-CNX — CFN; that is, aCFN—CNX = CXN.Or 80° 24—59° 23'—21° oi'rCXNi therefore2CXN = SXR = 40° 02', which is the Measure ofthe Angle that the incident and emerging Rays, which,are the least refrangible, contain with each other.

23. If instead of the Ratio 108 to 8r, yve take thatof 109 to 81, we shall find the Values of y' 3RR and\/ P—R? such as will give the Angle of IncidenceBCN, or the Arch BN — 58° 40', and the AngleS X R ~ 40° 17', which will be the Cafe for the mostrefrangible, or extreme Violet Rays.

24. If the Ray be twice reflected, viz. at F and G,as in the Production of the exterior Bow, and emergesatH in the Direction HA intersecting the incident RayS N in Y ; then we may find the Angle AY 8, whichthose Rays contain with each other, thus. ProduceA H, till it meets G X produced in R ; then in the Tri-angle HGR, the external Angle HGX = HRG-(-G H R. But because of equal Angles of Reflection atF and G, it is G H R = F G X ; therefore H G X —FGX = HGF = HRG = 2CGF or CNF. And(in Art. 22.) we had S X R = 4CNF — 2 C N X jtherefore in the Triangle YXR we have the two inter-nal Angles R -fX = 6 CNF —?2CNX — the ex-ternal Angle at Y, viz. AY N,

25. In this Case to find the Angles of Incidence and

Refraction, we have y/ 8 R R : \/ P—R? :: Radius: the Co-Sine of the Angle of Incidence ; whence thesaid Angle of Incidence will be found 71° 50' -CNX.And as 198 : 81 Sine of 71° 50' : Sine o f 45° 27' —C N F the Angle of Refraction ; therefore 45° 27" X 6— 2 X 71050' = X29° 02' = AY N, and therefore itsComplement AY S = 50° 58' the Angle required, forthe least refrangible Rays.

B b 4 26. But