middle of the glass be given, and the point F at which one desires the
rays to be all perfectly reunited, whatever be the first refraction
occurring at the surface AK.
I say that for this the sole requirement is that the outline BDK which
constitutes the other surface shall be such that the path of the
light from the point L to the surface AK, and from thence to the
surface BDK, and from thence to the point F, shall be traversed
everywhere in equal times, and in each case in a time equal to that
which the light employs, to pass along the straight line LF of which
the part AB is within the glass.
[Illustration]
Let LG be a ray falling on the arc AK. Its refraction GV will be given
by means of the tangent which will be drawn at the point G. Now in GV
the point D must be found such that FD together with 3/2 of DG and the
straight line GL, may be equal to FB together with 3/2 of BA and the
straight line AL; which, as is clear, make up a given length. Or
rather, by deducting from each the length of LG, which is also given,
it will merely be needful to adjust FD up to the straight line VG in
such a way that FD together with 3/2 of DG is equal to a given
straight line, which is a quite easy plane problem: and the point D
will be one of those through which the curve BDK ought to pass. And
similarly, having drawn another ray LM, and found its refraction MO,
the point N will be found in this line, and so on as many times as one
desires.
To demonstrate the effect of the curve, let there be described about
the centre L the circular arc AH, cutting LG at H; and about the
centre F the arc BP; and in AB let AS be taken equal to 2/3 of HG; and
SE equal to GD. Then considering AH as a wave of light emanating from
the point L, it is certain that during the time in which its piece H
arrives at G the piece A will have advanced within the transparent
body only along AS; for I suppose, as above, the proportion of the
refraction to be as 3 to 2. Now we know that the piece of wave which
is incident on G, advances thence along the line GD, since GV is the
refraction of the ray LG. Then during the time that this piece of wave
has taken from G to D, the other piece which was at S has reached E,
since GD, SE are equal. But while the latter will advance from E to B,
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