Let the curvature of the surface be the same, but medium different; glass, for instance, having an index of refraction equal to ; then F" will be equal to 3 x 5 = 15 millimetres, and F' will equal 2 x 5 10 millimetres. That is to say, the ratio between the two focal distances is equal to the index of refraction of the refractive medium, and their real lengths are multiples of the radius of curvature. The same proportion exists necessarily between G' and G":

=

values of F, G", F" and G', these formulæ become

We have thus eliminated the expressions n', n" and r, which greatly simplifies the calculations. Provided we know the first and second focal distances, we can easily find the spot where is formed the image of a point situated at a distance f' in front of, or f" behind the refracting surface, or at g' in front of, or g" to the rear of its centre of curvature. We deduce, in fact, from the formulæ 7, for these different lengths, the following formulæ :

Let us call the distance from the object to the first focus (f-F') =l', the distance from the image to the second focus (ƒ” — F")=l".

F'f"

In the formula ƒ =, let us subtract F' from each member of the equation, and we shall obtain

The plus sign is given to f' and g' as well as to f" and g", as long as the object and the image are on opposite sides of the surface.

If, on the contrary, the image and object be both on the same side of the surface, f' and g', or f" and g" take the minus or negative sign. This may happen when the luminous point is removed from the surface beyond infinity, i.e., when the rays coming from this point are neither divergent nor parallel, but convergent. This is not an impossible case. Although no luminous point ever emits convergent rays, still the rays coming from it may be rendered convergent, either by causing them to pass through a convex lens, or by reflecting them from a concave mirror. The object L' is then situated at the point where the convergent rays would be focused if prolonged, i.c., behind the surface, on the same side of it as the image (Fig. 11).

In order to find the place of the point L", we shall have to introduce, with the minus sign, the value of f' in the formula, for f". Let us take an example:

and suppose that the rays which strike the surface converge toward a point L' situated 250 millimetres behind the surface; f' is then negative and equals - 250 millimetres :

L' is therefore situated 18-8 millimetres behind the surface, i.e., nearer to it than is the principal focus. This is conceivable: parallel rays are united at the focus, in p", and rays which are already convergent before reaching the surface must be united nearer the surface than the parallel ones, and all the nearer according as they are more convergent.

Let us invert the case, and make L" the object, situated in the second medium. It follows from the law of reciprocity that, in this case, rays emanating from L" diverge as if they came from a point I' situated behind the surface. L' is, then, the image of L", but this image is virtual, i.e., it has no real existence, since the rays emanating from L' are nowhere focused, and we have obtained L' only by supposing these rays to be prolonged backward. We shall see, however, that these divergent rays do not always pass on to become lost beyond infinity. Let us suppose, for instance, an eye which looks toward this convex surface and which is capable of focusing divergent rays upon its retina. It would collect the rays coming from L", and this point would appear to it as if situated at L'.

Supposing now the incident rays to be less and less convergent, L' and L" will recede more and more from the surface, L' more rapidly than L", until L' shall have reached infinity, when the incident rays are parallel and are focused at ". The image L" is then at the principal focus, 20 millimetres behind the surface, in our example.

When the luminous point L', the object, is within infinity, i.e., at a finite distance in front of the surface, L" should be beyond 4", behind the surface; hence f' and ƒ" are positive (Fig. 8).

Let us take an example. L' being 250 millimetres in front of the surface, and F" and F', equal respectively to 20 and 15 millimetres, as in the preceding case, we shall have—

Hence L" is situated 21.2 millimetres behind the surface. If there be a luminous point at this distance, in the second medium, it will produce its image 250 millimetres in front of the surface.

The nearer the object L' approaches to the surface, the farther its image L" recedes from it. At length, when L' has reached the anterior focus ('), the rays are no longer united, but are, as we have seen, parallel to each other after having passed through the surface (compare Fig. 14).

If the luminous point be brought still nearer the surface, and be situated between it and the anterior focus, the rays must diverge after their refraction. Supposing that the first focal distance remain 15 millimetres, and that L' be 10 millimetres in front of the surface, we shall obtain for f" a negative value:

which indicates that L" is situated on the same side of the surface as L'. In other words, that the rays given off from L' diverge, after their

passage through the surface, as if they came from a point situated 40 millimetres in front of it (compare Fig. 15).

When L' is on the surface itself, the image and object of the point coincide.

IMAGE FORMED BY A SINGLE REFRACTING SURFACE.

We have seen that a luminous point L', situated at a distance f' in front of a spherical surface, on its axis, i.e., upon the line which passes through this point and the centre of curvature of the surface, forms its image L" at the distance f" from the latter.

Another luminous point A' (Fig. 12), situated at the same distance from the surface, ought to act in exactly the same way as L'. It will likewise form its image upon the axis passing through this point and the centre of curvature C of the surface, and at the same distance behind the latter as the rays emanating from L'-since L' and A' are equidistant from the surface. 1

Hence we have only to draw a straight line from A' through C, and to mark upon it the length f", from the point where this line cuts. the surface (or g' from C), in order to find the point A", which will be the image of A'.

The name optic axis or principal axis is generally given to the axis which, like L'L" in our example, is perpendicular to the refracting surface, when the latter is regarded as plane. The axes corresponding to points not situated upon the principal axis are then called secondary axes. These secondary axes have the same significance, for points not situated upon the principal axis, that the latter has for the points which it connects.

1 We suppose the portion of the surface which produces the image of A'B' to be small enough to be considered parallel to A'B'.

Since the same thing takes place in the case of B' as occurs in the case of A', the image of the former will be formed at B".

The same thing is repeated, too, for all points situated between A' and B'; all form their images between A" and B" upon the axes corresponding to them. All the points united form, on the one hand, the linear object A' B', and, on the other, its linear image B" A".

All lines situated in a plane perpendicular to the axis L'L", and, passing through L', will thus form their images in a plane perpendicular to the axis, and passing through the point L".

Hence a plane object, perpendicular to the optic axis, and situated at the distance f" in front of the surface, forms its image at the distance f" behind the surface. This image is real when it is situated on the opposite side of the surface, because it is then formed by the union of rays emanating from every point of the object.

We have, indeed, drawn but one ray for each luminous point, inasmuch as, being already known from the calculation, a single ray was sufficient; but in reality each luminous point gives off an infinite number an entire cone-of rays, which are all united at the point which corresponds to them.-(See Fig. 13 for the three points A', L' and B'.)

at the

The image is inverted relatively to the object. What is above in the object, is below in the image; what is right in the object, is at the left in the image, and vice versa. It is, moreover, geometrically similar to the object. A'B' and A" B" are in fact corresponding sides of two similar triangles, whose angles are equal. Hence, in order to find the size of the image, we write