United States or Slovakia ? Vote for the TOP Country of the Week !


If then through the point K one now draws KS parallel to Gg, which will be parallel also to QX, the tangent to the Ellipse QGq at Q; and if one conceives a plane passing through KS and touching the spheroid, the point of contact will necessarily be in the Ellipse QMq, because this plane through KS, as well as the plane which touches the spheroid at the point M, are parallel to QX, the tangent of the spheroid: for this consequence will be demonstrated at the end of this Treatise.

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.

In the time then that the line Oo arrives at the surface of the crystal at Kk, all the points of the wave COoc will have arrived at the rectangle Kc along lines parallel to OK; and from the points of their incidences there will originate, beyond that, in the crystal partial hemi-spheroids, similar to the hemi-spheroid QMq, and similarly disposed.