imagining a spheroidal wave about the centre C, I knew, because I have
just explained it, that its axis must be in the same plane, the half
of which axis I have marked CS in the next figure: and seeking by
calculation (which will be given with others at the end of this
discourse) the value of the angle CGS, I found it 45 degrees 20
minutes.
[Illustration]
27. To know from this the form of this spheroid, that is to say the
proportion of the semi-diameters CS, CP, of its elliptical section,
which are perpendicular to one another, I considered that the point M
where the ellipse is touched by the straight line FH, parallel to CG,
ought to be so situated that CM makes with the perpendicular CL an
angle of 6 degrees 40 minutes; since, this being so, this ellipse
satisfies what has been said about the refraction of the ray
perpendicular to the surface CG, which is inclined to the
perpendicular CL by the same angle. This, then, being thus disposed,
and taking CM at 100,000 parts, I found by the calculation which will
be given at the end, the semi-major diameter CP to be 105,032, and the
semi-axis CS to be 93,410, the ratio of which numbers is very nearly 9
to 8; so that the spheroid was of the kind which resembles a
compressed sphere, being generated by the revolution of an ellipse
about its smaller diameter. I found also the value of CG the
semi-diameter parallel to the tangent ML to be 98,779.
[Illustration]
28. Now passing to the investigation of the refractions which
obliquely incident rays must undergo, according to our hypothesis of
spheroidal waves, I saw that these refractions depended on the ratio
between the velocity of movement of the light outside the crystal in
the ether, and that within the crystal. For supposing, for example,
this proportion to be such that while the light in the crystal forms
the spheroid GSP, as I have just said, it forms outside a sphere the
semi-diameter of which is equal to the line N which will be determined
hereafter, the following is the way of finding the refraction of the
incident rays. Let there be such a ray RC falling upon the surface
CK. Make CO perpendicular to RC, and across the angle KCO adjust OK,
equal to N and perpendicular to CO; then draw KI, which touches the
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