Let this point of contact be at I, then making KC, QC, DC proportionals, draw DI parallel to CM; also join CI.
I say that CI will be the required refraction of the ray RC.

This will be manifest if, in considering CO, which is perpendicular to the ray RC, as a portion of the wave of light, we can demonstrate that the continuation of its piece C will be found in the crystal at I, when O has arrived at K.
38.

Now as in the Chapter on Reflexion, in demonstrating that the incident and reflected rays are always in the same plane perpendicular to the reflecting surface, we considered the breadth of the wave of light, so, similarly, we must here consider the breadth of the wave CO in the diameter G_g_.

Taking then the breadth C_c_ on the side toward the angle E, let the parallelogram CO_oc_ be taken as a portion of a wave, and let us complete the parallelograms CK_kc_, CI_ic_, Kl_ik_, OK_ko_.

In the time then that the line O_o_ arrives at the surface of the crystal at K_k_, all the points of the wave CO_oc_ will have arrived at the rectangle K_c_ 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 QM_q_, and similarly disposed.