Helmholtz's treatise on physiological optics. Volume 1. Edited by James P. C. Southall. Translated from the 3rd German edition Hermann von Helmholtz lit39649
50.] §9. Laws of Optical Imagery 65 meets the surface at c and is refracted there into the second medium. The first question is, what is its new path? By the law of refraction given above, it must remain in the plane of incidence determined by the incident ray and the normal at the point of incidence. The radius of a sphere being everywhere perpendicular to its surface, the incidence normal in this case is simply the prolongation of the radius ac, and the plane of incidence is the plane containing the lines pc and ad. The line pq also lies in this plane, since it contains the points p and a that are on pq. The refracted ray, therefore, must intersect the straight line pa in some point q. Where is this point with reference to the point &amp;? Should the ray happen to be parallel to pa, the intersection q may be regarded as at an infinite distance. Its position depends on the fact that n&gt; sin (pcd) = n'&gt;sin(qca)......(1) where n&gt; and denote the indices of refraction of the first and second media, respectively. By the law of sines in trigonometry, we derive the following relations from the triangles ape and aqc: sin (pea) _ ap sin (epa) ac sin(qca) aq sin(cqa) ac Dividing the second of these equations by the first, and noting that the sine of the angle pea is equal to the sine of its supplement pcd, we obtain : sin(pcd) sin(cqa) _ap sin(qca) sin(cpa) aq Equation (1) may be written: sin(pcd) n&gt;' sin(qca) n&gt; and from the triangle peq we have : sin(cqa) cp sin(cpa) cq Combining the last three equations, we.find: n"-cp ap n&gt;‘Cq aq (2)