these divergences in the direction parallel to the surface of the
crystal.
30. To find the length of the line N, in proportion to CP, CS, CG, it
must be determined by observations of the irregular refraction which
occurs in this section of the crystal; and I find thus that the ratio
of N to GC is just a little less than 8 to 5. And having regard to
some other observations and phenomena of which I shall speak
afterwards, I put N at 156,962 parts, of which the semi-diameter CG is
found to contain 98,779, making this ratio 8 to 5-1/29. Now this
proportion, which there is between the line N and CG, may be called
the Proportion of the Refraction; similarly as in glass that of 3 to
2, as will be manifest when I shall have explained a short process in
the preceding way to find the irregular refractions.
31. Supposing then, in the next figure, as previously, the surface of
the crystal _g_G, the Ellipse GP_g_, and the line N; and CM the
refraction of the perpendicular ray FC, from which it diverges by 6
degrees 40 minutes. Now let there be some other ray RC, the refraction
of which must be found.
About the centre C, with semi-diameter CG, let the circumference _g_RG
be described, cutting the ray RC at R; and let RV be the perpendicular
on CG. Then as the line N is to CG let CV be to CD, and let DI be
drawn parallel to CM, cutting the Ellipse _g_MG at I; then joining CI,
this will be the required refraction of the ray RC. Which is
demonstrated thus.
[Illustration]
Let CO be perpendicular to CR, and across the angle OCG let OK be
adjusted, equal to N and perpendicular to CO, and let there be drawn
the straight line KI, which if it is demonstrated to be a tangent to
the Ellipse at I, it will be evident by the things heretofore
explained that CI is the refraction of the ray RC. Now since the angle
RCO is a right angle, it is easy to see that the right-angled
triangles RCV, KCO, are similar. As then, CK is to KO, so also is RC
to CV. But KO is equal to N, and RC to CG: then as CK is to N so will
CG be to CV. But as N is to CG, so, by construction, is CV to CD. Then
as CK is to CG so is CG to CD. And because DI is parallel to CM, the
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