Plate X [E. &.H.Spitta,Photos.

" Upper Picture, View of the Saas Grat from Saas Fee. Photographed with Ross 3-inch portable symmetrical F/64 yellow screen. Edward's iso-medium plate, exposure three seconds. Hour 8.30 a.m. Mountain about three miles distant. The portion included in the telephoto view is that immediately under the two asterisks.

"Lower Picture, Telephoto - The Dom from other side of Saas Valley. Dallmeyer 2B patent portrait lens and high power tele-attachment. Camera extension from back of negative attachment 20 inches, F/16 on portrait lens. Edward's iso-medium plate with yellow screen. Hour 9 a.m. Exposure 12 seconds. Mountain 4 1/2 miles distant." (Dr. E. Spitta's description.) always shorter than that of the positive lens, hence we may represent f1 as being a multiple of m (greater than one) of f2; m=f1/f2.........(5) or f1 = mf2.........(6)

Let l1 and l2 represent the positive and negative lenses respectively, Fig. 42, and d the interval. If we plot f1, the focal point of the positive lens to the left of the lens Ll and f2, the focal point of the negative lens to the right of the lens l2, it can be shown that*:

(1) The front focal point of the whole system f1 is distant from the front focal point of the positive lens f1 tof1 = mf;

(2) The back focal point of the whole system f2 is distant from the back focal point of the negative lens f2 to f2 = F/m. (See Notes.)

Hence the distance between the front focal point of the whole system and the front lens:

F1L1 = MF + f1 and the distance between the back focal point of the whole system and the negative lens :

F2 L2 =F/m - f2

In the case of an ordinary positive lens of focal lengthf, we found that the position of its focal points were each equal to f on either side of the lens respectively ; again, for a given " magnification "of -: calling the distance of the object from the lens 0, and the distance of the image from the lens i, calling o the distance of the object from the positive element and I the distance of the image from the negative element,

0 = nf + f; i = 1/n f + f.

For a similar "magnification" with the Telephotographic lens,

* Czapski, " Theorie der Optischen Instrumente nach Abbe"

O = nF + mF + f1.......(7)

I=I/nf + I/mF- f2.......(8)

To compare these results, let f = f ; hence n f = 0 - f ; and I/nF= i - F If we neglect the separation between positive and negative lenses,

0 = 0 - F + mf +fl ;

I = i - F +I/m f -f2; or, written in another form:-

0 = 0 + (f (m - I) +f1)......(9)

I = i -(f(i -I/m)-f2)......(10)

From the last two equations* it is evident that using a Telephoto-graphic lens in place of an ordinary positive lens of identical focal length:

(1) The distance of the object must be greater than with an ordinary positive lens for the same "magnification" - an advantage in perspective.

(2) The distance of the negative lens to the plane of the image (or camera extension) is smaller - an advantage in mechanical means.

* D. P. Rudolph, " Gebrauchsanleitung fur Tele-Objective," May 1896.

From the quantities contained in the brackets in equations (9) and (10), it is evident that the greater we make m the more pronounced will the advantages of the Telephotographic lens become with respect to convenience of usage, and in respect also of the better perspective given by a greater distance between lens and object for a given "magnification."

Note. - Equations (7) and (8) must be remembered and applied in conjunction with the general formula f = f1f2/d (which determines the focal length of the compound system) to find the necessary distance of object from the front lens of the system for a given magnification, and to ascertain the requisite camera extension.

This method, although elegant in theory, necessitates further calculation to arrive at the intensity of the equivalent or compound lens for the time being. This may readily be done by making the diaphragm apertures of definite measurement, in contradistinction to "relative apertures," by dividing the focal length of the lens (for a given interval d) by the aperture used.

When the object is very distant, or as we say at infinity, it is evident that no advantage is derived from the property of the Telephotographic lens shown in equation (9), although the advantage derived from the condition in equation (10) always exists. Referring to equation (3) when n is small, or of the same order as m, then we find the increased distance between object and lens a great advantage, as will be seen when dealing with the subject of portraiture; but when n is great, then m becomes insignificant in comparison, and no advantage is practically attained.

Having ascertained the position of the two focal points f1f2, let us proceed to find the position of the principal points P1 P2, Fig. 42.

It is evident that their distance apart, p1 to P2 = F1 to F2 - 2 F. 64

The whole distance

F1 to f2 = m f + 2 f1 + d - 2f2 + 1/m f;

= f12/d + 2f1+d - 2f1+ f22/d;

=1/d(f1-f2+d)2-(2f1f2/d); but 2 F = 2f1f2/d,

Hence p1 to p2 = 1/d(f1 - f2 + d)2..........(11).

P1 and p2 are equidistant from Fl and f2, as in any positive system, and thus their positions and width are determined.

Equation (11) shows us that when the interval d is small, the principal points are widely separated, so that with the Telephotographic lens we cannot consider them as coinciding.

Let us now see how to determine the position and size of the image of an object.

Plot the two principal points of the lens p1 p2 (which here lie outside the lens) and their corresponding focal points f1 f2 (see Fig. 42 for convenience of following the disposition of the cardinal points). Let A b be an object n times the focal length of the combination distant from the focal point F1 thus bf1 = nf. The ray ap1, parallel to the axis, passes from p2 through the focal point f2 in the direction p2f2a'; the ray an1 proceeding to the front principal point n1 proceeds from the second principal point n2 in a direction parallel to a n1, or in the direction n2 a', meeting the first ray in a', a' is then the image of a; and a' b' the image of a b. As b f1 = n f, we know from our previous study of a positive lens, that b' f2 = 1/n f. Again the size of ae:a'b; :: n : 1.

Let us now give a numerical example comparing a Telephotographic lens with an ordinary positive system of the same, focal length. (See (2) and (3) Fig. 42.)

Let the focal length of the ordinary positive lens be 24 inches.

Let the focal lengths of positive and negative lenses of the Tele-photographic system be 6 and 3 inches respectively; - = 2, or m here = 2. To make this equivalent to the 24-inch lens, the interval d must be § of an inch : 6x3 =24

3 24.

Suppose we want to produce an image \ of the size of the object

AB.

In both cases b F1 = 4 x 24 = 96, and similarly f2b' = 24/4 = 6 inches,

4 and therefore a b : a' b' : : 4 : 1 in each case.

In the case of the Telephotographic lens the distance of the object

(n) (m) (f1) from the lens must be 4 x 24 + 2 x 24 + 6 = 150 inches ; but with the ordinary lens only

(n)

(4 + 1) 24 = 120 inches.

Again for camera extension with the Telephotographic lens

(f2) 24/4+ 24/2 + 3 = 15",

(n) (m) but with the ordinary lens

24 + 24/4 = 30". (n)

The convenience of the shorter camera extension is evident and very marked.

The greater distance between lens and object is of great importance from the point of view of perspective, for the reader must bear in mind that the position and magnitude of the image of an object derived from the four cardinal points, although accurate, does not trace the course of those rays which in reality produce the image. In the example above it is evident that the ordinary positive lens, being nearer to the object, includes it under a greater angle than does the Telephotographic lens at a greater distance. When the distance of the object is very great the advantage of course ceases. It may be stated very approximately that: The distance of the object from the lens determines the perspective. We shall refer to this more fully later.

## Notes On A

If two lenses, f1 , f2, of positive and negative focal length respectively, are separated by a given distance a between their principal or nodal points, the focal length of the combination

F = f1xf2/f2+a - f1

If we make d = f1-f2 - a=o, a = f1 - fs; or when the lenses are separated by a distance equal to the difference of their focal lengths d =o and the focal length of the combination is infinite.

I f d = o, or is any interval greater than the difference of the focal lengths, and equal to or less than f1, the formula takes the form

F=f1xf2/d

When a = f1, d =f1 - f2 -f1= -f2, or the interval d is equal to the focal length of the negative lens; the condition when the combination has its shortest focal length; f =f1 .

If f1 is some multiple of f2, we may say fi = mf2 where m is the multiple. Substituting in : f =f1f2/d; it is evident that

F=1/m(f1/d)2; 67 and also f = m(f22/d).

The distance of the front focal point of the combination from the front focal point of the positive lens = (f12/d) and

Similarly the distance of the back focal point of the combination from the back focal point of the negative lens = (f22/d) and and hence:

Distance of front focal point from positive lens = m f +f1

,, ,, back ,, ,, ,, negative lens = 1/mF - f2