International projectionist (Jan-Dec 1937)

FORMULAS FOR VARIOUS PROBLEMS IN PROJECTION OPTICS By CHARLES D. MERCER THE "work" of a lens consists in forcing the wave to assume a different shape or in causing the light to converge toward the focus, instead of, as originally, diverging from the distant source. The amount of this change of form of the wave is a measure of the "power" of the lens and depends on the curvature of the lens surfaces and the lens material. The stronger the curvature of the lens surfaces, the more will they alter the shape of the wave-front, the nearer, as a rule, will the focus lie to the lens. In order to properly characterize a lens, it is necessary to look about for a measure of its strength or power. The focal point has, in general, a real existence and can actually be located by placing a screen so as to receive upon it the image of a distant object; the size of this image, is, as we shall now see, the true measure of the power of the lens. It is apparent from the foregoing that the less curved the lens surfaces are, the farther away from the lens will the image lie and the larger will this image be. The power of a lens is often, but sometimes erroneously, measured by the distance of the focus from the lens, or as we may say, by the back focal distance (CF in Fig. 1). If we compare two lenses of different forms (Fig. 1 and 2), but so constructed as to give images having the same power, we note that the "back focal distance" is entirely different in the two cases, showing at once that this distance is not the true measure of power. The true measure of power is found by measuring the size of the image and the apparent size of the object and finding the ratio mentioned previously. If we divide unity — that is, one — by Figure 1 Figure 2 this ratio, we have what is known as the true or equivalent focal length; this is, as we have seen, not at all the same as the "focal distance", often called "back focus". The power of lenses and lens systems may be recorded either in inches or diopters. When the inch system is employed, the focal length of the lens system is taken in inches; an "eight-inch lens," therefore, is a lens which has its sharpest focus at 8 inches. In the dioptric system, a one-diopter lens is a standard lens which will bring rays of light to a focus from a distance of one meter, or 40 inches. Conversion to Diopters To convert the inch focal length of a lens or a lens system into diopters, we divide the number of inches into 40. A lens with an 8-inch focal length is a 5-diopter lens by the simple process of 40: 8=5. Convex lenses (which are thick in the center and thin on the edges) cause rays of light passing through them to converge at a very definite point, as shown in Fig. 3. Concave lenses (which are thin in the center and thick on the edges) cause rays of light passing through them to become divergent, as in Fig. 4. The lens formulas which find the most frequent application in projection work are those which are employed for the following purposes: A. To determine the position of screen and type of lens to employ in a given optical system, use the formula: 1 11 object dist. image dist. f Given: An object 6 inches distant which it is desired to bring to a focus, we substitute: 111 6 Image dist. 3 for we know from mirrors that if the object distance is more than twice the focal length of the lens, the image will converge at the principal focus. Hence, solving the above: 111 6 v 3 or, v + 6 = 2v; or v = 6 = distance of image from lens. B. To determine the size of the image when the object size and the object and image distance are known, use the formula: Image dist. Size of image Object dist. Size of object Given: An object 4 inches high which Figure 3 Figure 4 is 8 inches from the lens, the image distance being 24 inches, we substitute: 24 Image size 8 4 Hence, solving the above: 2 I = 24, or I = 12 C. To determine the focal length of a lens when the curvature and index are given, use the formula : focal length = radius 1st. side X radius 2nd. side radius 1st. side + radius 2nd. side (index-1) Given : A bi-convex lens, one surface having a radius of curvature of 10 inches, the other having a radius of 8 inches, index 1.50, we substitute: 10 X 8 = F (10 + 8) (1.50 — 1) 80 80 F = = — = 8.88 + inches 18 X .50 9 D. To determine the equivalent focus of two lenses in combination when the focus of each lens is given, use the formula: E. F. = the reciprocal of the sum of the individual focus of each lens. Given: A 10-inch lens combined with a 40-inch lens, we substitute: [22]