By treating the lens in this manner we do not necessarily know the separation of the principal points or planes of the two components, and for distant objects the knowledge of the distance between the principal points and planes of the whole system is immaterial. For extreme accuracy in the results, we ought to know the position of the principal point or plane of the negative lens, as it should be from this point that we set the camera extensions. In practice the mechanical centre of the thickness of the negative lens is very nearly the position of its principal point, and negligible errors only are likely to creep in.

It is perhaps needless to say that the separation of the two elements determines the plane of the final image, and we have to adjust the separation of the two lenses mechanically until the image is well defined upon the screen in the position we have assigned to it, in order to bring-about the required magnification.

Again, by considering the final image as a magnification of the image produced by the positive lens alone, we see, at once, from first principles, how to determine the comparative illumination of the two images, which decides the relative photographic exposures.

Suppose we know the correct exposure to give with the positive lens alone, and we magnify this image by means of the negative lens a certain number of times (linear), the surface covered is the square of the linear dimension ; hence if t is the correct time of exposure to give for the positive lens alone, the correct exposure for a given magnification (linear) with the Telephotographic illumination is : t, the exposure = m2 t............(16)

Rule. - To find the correct exposure with the Telephotographic lens : Multiply the time of exposure necessary for the positive lens alone by the square of the magnification.

This result is identical with the following :

Rule. - To find the intensity of the Telephotographic lens : Divide the intensity of the positive lens by the magnification.

We shall refer to intensity in the next chapter, but may point out here that to find the relative exposures of two lenses we have to square the denominators of the fractions expressing them. So that if the intensity of the positive lens alone I =a/f, where a is the aperture, and f1 the focal length, the intensity of the Telephotographic combination for a magnification of M times = a/Mf or the relative exposures are as : f12 : (m/f12) or f12 : f2. And this is the same thing as saying that with a given aperture, or definite size of diaphragm of the lens, the relative exposure for the positive lens alone and the Telephotographic combination are as the squares of their focal lengths.

If we examine Fig. 43, we shall see that only a comparatively small part of the entire image formed by the positive lens is taken up and magnified by the negative lens; the "drawing" of this small area is identical with that of the final image, but of different scale. It is evident that the scale is dependent upon the magnification employed. The magnification is due to a multiple of the focal length of the positive lens, produced at a distance which is only a (nearly equal) multiple of the (shorter) focal length of the negative lens ; hence it is evident that the correct viewing distance of the image cannot be from the distance it is removed from the lens, but a considerably greater distance! We shall examine the correct position in the next chapter.

Let us now investigate, more particularly, the employment of the Telephotographic lens in forming images of near objects, an application of great practical interest and importance to photographers.

We have seen that when we require to produce a certain " magnification" with an ordinary positive lens, the following relations exist:

Distance of object = (n + 1)f1;

,, „ image = n + 1/nf1

We have now only to decide how many times we wish to increase the size of this image to obtain the desired magnification, for we know that the size of the image formed by the positive element is 1/n, the actual size of the object. Let m represent the magnification given by negative lens, as before, and - the entire "magnification" we wish to bring about between the object and final image by the whole system.

1/N = 1/nx M........(l7)

Example. - Suppose the focal length of the positive element f1= 10 inches, that of the negative lens f2=4 inches, and the object is 80 inches distant, and we wish to reproduce it in half natural size.

The lens f1 produces an image \ the size of the object; we require the final image to be but 1/2 or 1/N = 1/2 ; hence we must magnify the image given by f1 ,3 1/2 times, or make m = 3 1/2.

1/2=1/7x 2/7=1/2 applying the formula:

E = f2 (M - I)

= 4 (3 1/2 -1) = 10, or we require a camera extension of 10 inches to arrive at the final magnification wanted.

Let us now determine the true focal length f of the Telephoto-graphic system that has produced the result.

It is obvious that one of the focal points F1 or the position of its focus for parallel rays, must be somewhere between the negative lens and the screen.

If we call its distance from the negative lens f l2=x, f s must be

- ; calling E=the camera extension as before, we have the two following relations: x = e - F/N ; f = m x + f1 hence f = m (e - F/N) + fi ; or f(m/N + 1) = mE + f1; f = mE + f1........(18)

(m/N + 1)

In the example given m = 10/4= 2 1/2 - = 2 1/2,f1 = 10, e = 10, and 1/N =1/2.

F (2 1/2/2 + 1) = 2 1/2 x 10 + 10;

1 5/4 F = 35; F = 15 5/9

The knowledge of the true focal length is necessary for determining the intensity of the lens, and therefore equation (18) must be remembered and applied. It will be noted that the result obtained is identical with that given by (8).