Reflection

How can we determine this refraction quantitatively?

To answer this, we must first explain reflection.

Suppose a ball is struck from A towards B. At B it meets the surface of the ground CBE, which prevents it from going further and causes it to be deflected.

• The ground is perfectly flat and hard.
• The ball always travels at a constant speed, both when it descends and rebounds upwards.

Where will it go?

The force that causes the speed of the ball is different from the force that causes the direction of the ball. That direction is determined by the position of the racket.

This directional force shows that this ball can be diverted by the encounter with the earth CBE.

Thus, CBE can change the direction that it had towards B without changing the force that gave its speed, since these are two different things.

This is why it does not have to stop at B before bouncing off to F as many of our Philosophers do. If it was stopped at B then there is nothing at B which can cause it to start again.

The determination to move towards a certain side can, as well as movement and generally any other kind of quantity, be divided between all the parts of which it can be imagined to be composed.

A ball that moves from A to B is composed of two others:

1. A force which makes it fall from the line AF to the line CE
2. A force which makes it go from the left AC to the right FE

These together lead it to B along the straight line AB.

The encounter with the earth can only prevent one of these two determinations, and not the other in any way.

It prevents the one that made the ball fall from AF to CE because it occupies all the space that is below CE.

But why would it prevent the other, which made it advance towards the right hand, since it is in no way opposed to it in that sense?

Let us draw a circle with center B, which passes through point A.

In as much time as it will have taken to move from A to B, it must infallibly return from B to some point on the circumference of this circle, since all the points that are as distant from this B as A is, are found in this circumference, and we suppose the movement of this ball to be always equally fast.

To know precisely to which of all the points on this circumference it must return, let us draw 3 straight lines AC, HB, and FE perpendicular to CE, in such a way that there is neither more nor less distance between AC and HB than between HB and FE.

Let’s say that in the same amount of time it took for the ball to advance towards the right side, from point A on line AC to point B on line HB, it must also advance from line HB to some point on line FE.

Because all the points on line FE are equally distant from HB in that sense, just like those on line AC, and it is equally determined to advance towards that side as it was before.

However, it is impossible for it to arrive at any point on the line FE at the same time as reaching any point on the circumference of the circle AFD, except at point D or F. This is because these are the only two points where they intersect each other.

Thus, if the Earth prevents it from passing towards D, it must inevitably go towards F.

This is how reflection occurs at an angle always equal to what is called the angle of incidence.

Fig 7

For example, if a ray coming from point A falls on point B on the surface of the flat mirror CBE, it reflects towards F in such a way that the angle of reflection FBE is neither greater nor smaller than the angle of incidence ABC.

A ball, propelled from A towards B, encounters at point B, not the surface of the earth, but a delicate and thin web CBE. This web is so weak that the ball has the force to break it and pass right through, only losing half of its velocity.