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 &?
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> sin (pcd) = n'>sin(qca)......(1)
where n> 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>'
sin(qca) n>
and from the triangle peq we have :
sin(cqa) cp
sin(cpa) cq
Combining the last three equations, we.find:
n"-cp ap
n>‘Cq aq
(2)