254 Geometrical Propositions 
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plusreflected at the second surface, then minus reflected at the first surface, and 
finally emerging from the second surface in a direction parallel to that of incidence. 
Its path within the crystal is PpM. 
These examples indicate the general method of expressing the path of a ray. 
40. Suppose light to be moving in the same direction and with the same velocity 
along two proximate parallel rays, so that it is at the point A in one ray when it is at 
the point B in the other; and through the points 4 and B conceive two planes per- 
pendicular to the common direction of the rays. These planes are either coincident, 
or maintain a constant distance. In the first case, the rays are said to be in complete 
accordance. In the second case, the constant distance between the planes is called 
the interval between the portions of light composing the rays, or the interval between 
the waves that move along the rays. 
We proceed to find the lengths of these intervals in the case of rays emerging 
parallel to each other, at either side of the crystal that we have been hitherto con- 
sidering. 
41. Let the tangent planes at P, M, p, m, intersect the plane of the figure 
(Fig. 8) inthe right lines PP, MI, pp,, mm,, which of course are tangents to the 
section of the surface of refraction represented in the figure ; let a perpendicular at O 
to the face of the crystal cut these tangents in the points P, 11, p,, m,; and let 
the lines OP”, OM", Op", Om", respectively parallel to PP, MM, pp,, mm,, cut 
the line S#s in the points P”, 1”, p”, m’. 
The length of the path which a ray P describes within the crystal, is equal to the 
thickness © of the crystal divided by the cosine of the angle P’ OP, which the path 
of the ray makes with a perpendicular to the faces of the crystal; and the velocity of 
: Os dere é 
P isequalto V x op (37) dividing therefore the length of the path by the velocity, 
By iy: - ; @x OP’ 
we find that the time in which a ray P crosses the crystal is equal to Fx 08x Cos POP 
: : ORS 
But as OP’ is perpendicular to the tangent plane at P, we have Gos POR, 
=OP =PP". Therefore the time is equal to A ask 
which rays M, p,m, pass from one surface of the crystal to the other, are equal to 
OxMM” expp” Oxmm’ ; 
x08? Vx08? Pos” respectively. 
42. Now suppose the path of a ray P to be be projected perpendicularly on a right 
line having any proposed direction in space. Through O conceive a right line OL 
parallel to the proposed direction, and meeting in L the tangent plane at P. ‘The 
length of the projection is equal to the length of the path multiplied by the cosine of — 
the angle P’OL which the ray P makes with OL ; that is, the projection is equal to 
Cos POL | , 
Similarly, the times in 
But because OP’ is perpendicular to the tangent plane at P, we have 
Cos P'OP, ' 1 2p 
5 F OF : OP' OP Cos POL PP" 
Cos P'OL= OL” and 08 Oe OP,= PP” ; therefore Cos POP, = OL Henee 
the projection is equal to on 
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