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Let ABH be a parallelepiped of crystal, and let the top surface AEHF be a perfect rhombus, the obtuse angles of which are equally divided by the straight line EF, and the acute angles by the straight line AH perpendicular to FE. The section which we have hitherto considered is that which passes through the lines EF, EB, and which at the same time cuts the plane AEHF at right angles.
Let the line N be the length of the travel of light in air during the time in which, within the crystal, it makes, from the centre C, the spheroid QCqgM. Then having drawn CO perpendicular to the ray CR and situate in the plane through CR and AH, let there be adjusted, across the angle ACO, the straight line OK equal to N and perpendicular to CO, and let it meet the straight line AH at K. Supposing consequently that CL is perpendicular to the surface of the crystal AEHF, and that CM is the refraction of the ray which falls perpendicularly on this same surface, let there be drawn a plane through the line CM and through KCH, making in the spheroid the semi-ellipse QMq, which will be given, since the angle MCL is given of value 6 degrees 40 minutes.
About the centre C, which I suppose to be in the intersection of AH and FE, let there be imagined a hemi-spheroid QGqgM, such as the light would form in spreading in the crystal, and let its section by the plane AEHF form the Ellipse QGqg, the major diameter of which Qq, which is in the line AH, will necessarily be one of the major diameters of the spheroid; because the axis of the spheroid being in the plane through FEB, to which QC is perpendicular, it follows that QC is also perpendicular to the axis of the spheroid, and consequently QCq one of its major diameters.
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