Knowing that F=15 5/9", the distance between object and image to bring about the magnification of 1/2 must be (or would be thought to be)

4 1/2 f or 70 inches ; but (neglecting the separation of positive and negative lenses) the whole distance = 80" (distance of object from positive lens) + 10" (camera extension from negative lens = E) or 90 inches, hence the separation of the principal points of the whole system is 90-70, or as much as 20 inches!

If we required an ordinary positive lens-system to give the same "magnification" (1/2), at the same distance from the object, its focal length would have to be -, or 27 inches nearly, and

3 the camera extension 40 inches.

The photographer will see at once the advantage gained by using the Telephotographic lens, giving the same size of image at the same distance as the ordinary lens of longer focal length: (1) in respect of rapidity, if equal effective apertures are employed, or (2) in respect of greater "depth of focus," if both lenses are used at the same intensity, the ratio of the effective aperture to focal length being the same in each case.

Equation (18) shows us that when n is very great, the above advantages of the Telephotographic lens no longer hold good, for f (m/N+1) then becomes f (m/∞ + 1), or is equal to f, and the equation becomes as before: f=m e+f1 (See Equation (15).)

Let us now proceed to find the distance that must intervene between the object and the Telephotographic lens for a given "magnification" of - when its focal length is known.  Plate XIII

Taken from the end of Boscombe Pier, which is 600 yards distant from the small house on the cliff. The stratification of the cliff immediately below the house is hardly discernible when photographed with an ordinary lens. Astigmatic F/6-, 10-in. focal length was used. (By the Author.) Plate XIV

Taken from the same standpoint as Plate XIII. by a telepho lens composed of an 8 1/4-in. c. de v. lens, combined with a 2-in. negative lens. The stratification is here clearly visible. Camera extension 20 in. {By the Author).

In the first place we must determine the distance of the focal point from the negative lens in order that it may correspond to the focal length f chosen, and to this add - for the given magnification, thus determining the entire camera extension e that will be required.

Now we must determine the magnification given to the image formed by the positive lens f1, to bring about the final relation between object and image, as from the size of the image formed by f1 we at once find the necessary distance for the object to be placed : as before

M =E/f2 +I, and

I/N = M/n, or n = MN..........(19) and the distance of the object

= f1(n + 1) . . cp(2)

Example : Take the case of the Telephotographic lens illustrated in Fig. 43 where the focal length is 24 inches. To find the necessary distance between object and lens for a magnification of 1/5.

Here e = 9 +24/5 = 13 4/5; m= 13 4/5 + 1 = 5 3/5;

1/N = 1/5 or N = 5 n = 28/5 x 5 = 28, or the distance of object=6 (28 + 1) = 174 inches.

The principal points are here separated by a distance of 174-(6 x 24 +15) = 15 inches.

For an ordinary positive system of the same focal length Distance of object =f1 (n + 1) = 144 inches. „ image = f1 (N+1) = 28 4/5.

N

In other words the ordinary lens must be placed in the position where the refraction caused by the Telephotographic lens appears to originate. (See Fig. 43.)

We thus confirm again our conclusion that a greater distance must intervene between object and lens with a Telephotographic lens than with an ordinary positive system, both having the same focal length, in order to produce the same size of image, or the same magnification.

If both lenses have the same intensity, or ratio of effective aperture to focal length, the Telephotographic lens has the advantage (1) in giving better perspective, and (2) in greater "depth of focus."

It must be remembered, however, that under these particular circumstances the Telephotographic lens will require a longer exposure than the ordinary lens in the proportion of the square of their distances from the object; in our particular case as (174)2: (144)2, or very nearly as 3 : 2.

Very little consideration will show the reason for this. When images of the same size are produced by either type of lens worked at the same intensity and at the same distance, the relative exposures must be the same; but here, although the size of the image is the same in both instances, the distance the light has to travel from the object before reaching the Telephotographic lens is greater, and the comparative illumination of the two images obeys the "law of inverse squares."

The reader will find more examples illustrating the use of the lens for photographing near objects in Chapter VII (Practical Applications Of The Telephotographic Lens). on the "Application of the Lens."

To sum up, this method (B) seems to offer the following advantages to the photographer:

(1) It only extends the already familiar principle of producing images of objects in a given scale.

(2) The magnification of the original scale carries with it direct information on the question of correct exposure.

(3) The determination of the "equivalent" lens of the system is arrived at more readily than in the method (A), when it is necessary to find it. In (A) it is always necessary, and involves a knowledge of the "interval" between the component lenses, as well as their focal lengths. In the method (B) the knowledge of the "equivalent" lens is only really necessary in the case of photographing near objects.

(4) Fewer preliminary considerations are required in arranging the scale for the final image. The distance between object and lens decides the scale given by the positive lens, and the subsequent magnification necessary is immediately known and attained.

(5) Less mental effort or calculation is required for equally accurate results.