Water Droplets

Let AFD be a drop of water, whose half diameter CD or AB I divide into as many equal parts as I want to calculate rays, in order to attribute as much light to some as to others.

Then I consider one of these rays in particular, for example EF, which instead of going straight towards G, turns towards K, and reflects from K towards N, and from there goes towards the eye P

or else reflects again from N towards Q, and from there turns towards the eye R.

having drawn CI at right angles to FK, I know from what has been said in Optics that AE or HF, and CI, have between them the proportion by which the refraction of water is measured.

So that if HF contains 8000 parts, such that AB contains 10000, CI will contain about 5984, because the Refraction of water is slightly greater than three to four, and for the most precisely that I have been able to measure it, it is like 187 to 250.

Thus having the two lines HF and CI, I easily know the two arcs, FG which is 73 degrees and 44 minutes, and FK which is 106. 30.

Then removing twice the arc FK, from the arc FG added to 180 degrees, I have 40. 44. for the quantity of the angle ONP, because I suppose ON parallel to EF.

Removing these 40. 44 from FK, I have 65. 46 for the angle SQR, because I also pose SQ parallel to EF.

And calculating in the same way all the other rays parallel to EF, which pass through the divisions of the diameter AB, I compose the following table:

In this table that there are many more rays that make the angle ONP of about 40 degrees, than there are that make it less; or SQR of about 54, than there are that make it greater.

Then in order to make it even more precise I do this table:

I see here that the largest angle ONP can be 41 degrees 30 minutes.

The smallest SQR is 51.54, to which adding or subtracting about 17 minutes for the half diameter of the sun, I have 41.47 for the largest half diameter of the inner rainbow, and 51.37 for the smallest of the outer one.

When water is hot, its refraction is slightly less than when it is cold, which can change something in this calculation.

However, this cannot increase the half diameter of the inner rainbow by more than one or two degrees at most. And then, that of the outer one will be almost twice as small.

This demonstrates that the refraction of water cannot be much less or greater than I suppose.

For if it were only a little greater, it would make the half diameter of the inner rainbow less than 41 degrees, instead of which, by common belief, it is given 45 degrees;

If we suppose it small enough to make it really 45, we will find that that of the outer one will not be much more than 45 either, instead of which it appears to the eye to be much larger than that of the inner one.

Maurolycus was the first to have determined one of 45 degrees. He determines the other to be about 56.

This shows the little faith that should be added to observations that are not accompanied by true reason.

Red is outside in the inner rainbow and inside in the outer one.

This is based on the same cause for which it appears towards F, rather than towards H, through the MNP prism.

If the eyes is on white linen FGH looking at this crystal, he will see red towards its thicker part MP, and blue towards N. This is because the red-tinted ray that goes to F comes from C, the part of the sun most advanced towards MP.

This means that the thicker center of the raindrops are:

• outside with respect to the colored points that form the inner rainbow.
  • This causes red to appear there outside.
• inside with respect to the colored points that form the outer rainbow.
  • This causes the red to appear there inside.

The only difficulty are the irregularities that occur there.

Examples are when:

• the arc is not exactly round
• the arc’s center is not in the straight line that passes through the eye and the sun.

These can happen if the winds change the shape of the raindrops.

This makes a notable difference in the angle under which the colors should appear.

We have also seen sometimes, so I am told, a rainbow so overturned that its horns were turned upwards, as is represented here in FF.

This is caused by the reflection of the sun’s rays falling on the water of the sea or lake.

As if coming from the part of the sky SS, they fall on the water DAE.

From there, they reflect towards the rain CF. The eye B will see the arc FF, whose center is at point C, so that CB being extended to A, and AS passing through the center of the sun, the angles SAD and BAE are equal, and the angle CBF is about 42 degrees.

This requires there to be:

• no wind at all to disturb the water’s surface towards E

and perhaps with that there is some cloud, like G, which prevents the sunlight going in a straight line towards the rain, do not erase the one that this water E sends there. From which it comes that it happens only rarely.

In addition, the eye can be in such a position relative to the Sun and the rain that the lower part that completes the circle of the rainbow will be seen, without seeing the upper part: and thus it will be taken for an overturned arc: notwithstanding that it will not be seen towards the sky, but towards the water, or towards the earth.

I have also been told that a third rainbow has sometimes been seen above the two ordinary ones; but which was much weaker, and about as far from the second as the second was from the first. Which I do not judge could have happened, unless there were hailstones, very round, and very transparent, mixed with the rain, in which the refraction being notably greater than in water, the outer rainbow should have been much larger there, and thus appear above the other. And for the inner one which by the same reason should have been smaller than the inner one of the rain, it may be that it will not have been noticed, because of the great luster of this one: or that their ends having joined, they will not have been counted both for one.

But for one whose colors will have been arranged differently than usual.

And this reminds me of an invention to make signs appear in the sky, which could cause great admiration to those who would not know the reasons. I suppose you already know how to show the rainbow by means of a fountain. As if the water that comes out of the small holes ABC, jumping high enough, spreads in the air in all directions towards R, and the sun is towards Z, so that ZEM being a straight line the angle MER can be about 42 degrees, the eye E will not fail to see the iris towards R, quite similar to the one that appears in the sky. To which it must now be added that there are oils, spirits of wine, and other liquids, in which refraction is significantly greater or smaller than in ordinary water, and which are no less clear and transparent. So that one could arrange several fountains in order; in which there are various of these liquids, one would see by their means a whole large part of the sky full of the colors of the iris: namely by making the liquids, whose refraction would be the greatest, be the closest to the spectators; and that they do not rise so high that they prevent the view of those that would be behind. Then because by closing part of the holes ABC one can make disappear such part of the iris RR as one wants, without removing the others, it is easy to understand that in the same way, opening and closing the holes of these various fountains at the right time, we can make what appears colored have the figure of a cross, or a column, or some other such thing, which gives rise to admiration. But I admit that it would take skill and expense, in order to proportion these fountains, and to make the liquids jump so high that these figures could be seen from afar by a whole people, without the artifice being discovered.