same vanishing points
as the object itself.

[Illustration: Fig. 289.]

Let us suppose _R_ (Fig. 289) to be standing on the water or reflecting
plane. To find its reflection make square [R] equal to the original
square _R_. Complete the reversed cube by drawing its other sides, &c.
It is evident that this lower cube is the reflection of the one above
it, although it differs in one respect, for whereas in figure _R_ the
top of the cube is seen, in its reflection [R] it is hidden, &c. In
figure A of a semicircular arch we see the underneath portion of the
arch reflected in the water, but we do not see it in the actual object.
However, these things are obvious. Note that the reflected line must be
equal in length to the actual one, or the reflection of a square would
not be a square, nor that of a semicircle a semicircle. The apparent
lengthening of reflections in water is owing to the surface being broken
by wavelets, which, leaping up near to us, catch some of the image of
the tree, or whatever it is, that it is reflected.

[Illustration: Fig. 290.]

In this view of an arch (Fig. 290) note that the reflection is obtained
by dropping perpendiculars from certain points on the arch, 1, 0, 2,
&c., to the surface of the reflecting plane, and then measuring the same
lengths downwards to corresponding points, 1, 0, 2, &c., in the
reflection.

CLXV

ANGLES OF REFLECTION

In Fig. 291 we take a side view of the reflected object in order to show
that at whatever angle the visual ray strikes the reflecting surface it
is reflected from it at the same angle.

[Illustration: Fig. 291.]

We have seen that the reflected line must be equal to the original line,
therefore _mB_ must equal _Ma_. They are also at right angles to _MN_,
the plane of reflection. We will now draw the visual ray passing from
_E_, the eye, to _B_, which is the reflection of _A_; and just
underneath it passes through _MN_ at _O_, which is the point where the
visual ray strikes the reflecting surface. Draw _OA_. This line
represents the ray reflected from it. We have now two triangles, _OAm_
and _OmB_, which are right-angled triangles and equal, therefore angle
_a_ equals angle _b_. But angle _b_ equals angle _c_. Therefore angle
_EcM_ equals angle _Aam_, and the angle at which the ray strikes the
reflecting plane is equal to the angle at which it is reflected from it.

CLXVI

REFLECTIONS OF OBJECTS AT DIFFERENT DISTANCES

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