Bauhaus-Universität Weimar

Helmholtz's treatise on physiological optics. Volume 1. Edited by James P. C. Southall. Translated from the 3rd German edition
Helmholtz, Hermann von
§9. Laws of Optical Imagery 
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 


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