Crystal and have become double; for it is quite otherwise for rays in
all other planes which intersect the Crystal, as we shall see
afterwards.
11. I recognized at first by these experiments and by some others that
of the two refractions which the ray suffers in this Crystal, there is
one which follows the ordinary rules; and it is this to which the rays
KL and OQ belong. This is why I have distinguished this ordinary
refraction from the other; and having measured it by exact
observation, I found that its proportion, considered as to the Sines
of the angles which the incident and refracted rays make with the
perpendicular, was very precisely that of 5 to 3, as was found also by
Mr. Bartholinus, and consequently much greater than that of Rock
Crystal, or of glass, which is nearly 3 to 2.
[Illustration]
12. The mode of making these observations exactly is as follows. Upon
a leaf of paper fixed on a thoroughly flat table there is traced a
black line AB, and two others, CED and KML, which cut it at right
angles and are more or less distant from one another according as it
is desired to examine a ray that is more or less oblique. Then place
the Crystal upon the intersection E so that the line AB concurs with
that which bisects the obtuse angle of the lower surface, or with some
line parallel to it. Then by placing the eye directly above the line
AB it will appear single only; and one will see that the portion
viewed through the Crystal and the portions which appear outside it,
meet together in a straight line: but the line CD will appear double,
and one can distinguish the image which is due to regular refraction
by the circumstance that when one views it with both eyes it seems
raised up more than the other, or again by the circumstance that, when
the Crystal is turned around on the paper, this image remains
stationary, whereas the other image shifts and moves entirely around.
Afterwards let the eye be placed at I (remaining always in the plane
perpendicular through AB) so that it views the image which is formed
by regular refraction of the line CD making a straight line with the
remainder of that line which is outside the Crystal. And then, marking
on the surface of the Crystal the point H where the intersection E
appears, this point will be directly above E. Then draw back the eye
towards O, keeping always in the plane perpendicular through AB, so
48