passage, and thence deduced the constant ratio of the Sines. His
demonstration, which may be seen in his printed works, and in the
volume of letters of Mr. Des Cartes, is very long; wherefore I give
here another which is simpler and easier.
[Illustration]
Let KF be the plane surface; A the point in the medium which the light
traverses more easily, as the air; C the point in the other which is
more difficult to penetrate, as water. And suppose that a ray has come
from A, by B, to C, having been refracted at B according to the law
demonstrated a little before; that is to say that, having drawn PBQ,
which cuts the plane at right angles, let the sine of the angle ABP
have to the sine of the angle CBQ the same ratio as the velocity of
light in the medium where A is to the velocity of light in the medium
where C is. It is to be shown that the time of passage of light along
AB and BC taken together, is the shortest that can be. Let us assume
that it may have come by other lines, and, in the first place, along
AF, FC, so that the point of refraction F may be further from B than
the point A; and let AO be a line perpendicular to AB, and FO parallel
to AB; BH perpendicular to FO, and FG to BC.
Since then the angle HBF is equal to PBA, and the angle BFG equal to
QBC, it follows that the sine of the angle HBF will also have the same
ratio to the sine of BFG, as the velocity of light in the medium A is
to its velocity in the medium C. But these sines are the straight
lines HF, BG, if we take BF as the semi-diameter of a circle. Then
these lines HF, BG, will bear to one another the said ratio of the
velocities. And, therefore, the time of the light along HF, supposing
that the ray had been OF, would be equal to the time along BG in the
interior of the medium C. But the time along AB is equal to the time
along OH; therefore the time along OF is equal to the time along AB,
BG. Again the time along FC is greater than that along GC; then the
time along OFC will be longer than that along ABC. But AF is longer
than OF, then the time along AFC will by just so much more exceed the
time along ABC.
Now let us assume that the ray has come from A to C along AK, KC; the
point of refraction K being nearer to A than the point B is; and let
CN be the perpendicular upon BC, KN parallel to BC: BM perpendicular
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