## 614. To Measure Focal Length

To Measure Focal Length. The practical application of this method is varied. Say that we want to find out what is the focal length of our lens. We focus an object carefully at actual size; then measure accurately the distance between object and focusing screen and divide by four. The quotient is the focal length of the lens. Thus, if we find that the distance from object to focusing screen is 40 inches, the focus of the lens is 10 inches. On the other hand, if we are to photograph an object actual size, what camera extension is required with a 10 inch focus lens? Object and image are equidistant from the lens, each two focal lengths away; consequently we must have camera extension sufficient to give us a distance of 20 inches from center of lens to plate. Or we want to know what distance is required between lens and object to give us an image of a certain size. For instance, we want to know at what distance a 12 inch focus lens will give us a 4 inch image of a person 6 feet tall. Size of object is 72 inches, size of image 4 inches; ratio of image to object (scale of image we will call it) therefore is 1-18. Consequently the object must be 18 focal lengths beyond f1, and 19 focal lengths beyond the center (c) of the lens - 19 times 12 inches, or 19 feet. In the same manner the distance required with a 20 inch focus lens is found to be 19 times 20 inches, or 380 inches, or 31 feet 8 inches.

615. Conversely we want to know the focal length of lens required to give us a 4 inch figure of a person 6 feet tall at a distance of 20 feet. Scale of image, as before, is 1-18, which means that the object must be 18 focal lengths away from f1 and 19 focal lengths from the center (c) of the lens. In other words, the distance of 20 feet or 240 inches must be equal to 19 times the focus of the lens required, or the focus of the lens must be 1-19 of 240 inches. Divide 240 by 19; the quotient is 12 12-19, and we therefore require a lens of 12 12-19 (12 2-3) inches focus. If the distance is 15 feet, the 15 feet or 180 inches must be equal to 19 times the focus of the lens. Divide 180 by 19 and the quotient is 9 9-19; the required focal length is 9 9-19 inches, or just about 9 1/2 inches.

616. Again, what size image will a 10 inch focus lens make of an object 6 feet tall at a distance of 30 feet? Distance, 30 feet or 360 inches equals 36 times the focus of the lens. Distance between f1 and object is therefore 35 focal lengths; scale of image is 1-35, which means that the size of the image is 1-35 of 72 inches, or about 2 inches.

617. For group work in a short gallery the choice of lens has often proven both difficult and puzzling. Say that we require a lens for 8x10 groups in a gallery where we can get an operating distance of 18 feet. The width (or length, if we so wish to term it) of the group is of course, limited by the width of the gallery, so that we can without much difficulty determine how wide will be the widest group which we can arrange there. Let the maximum width of group be 15 feet. On an 8x10 plate we could not very well make it over 9 inches. Size of object then is 15 times 12 inches or 180 inches; size of image 9 inches; scale of image 1-20. Distance from lens to group, 18 feet or 216 inches, must therefore equal 21 focal lengths, or 21 times the focus of the lens required. Dividing 216 by 21, we get the quotient of 10 2-7 - and we know that the lens must have a focus of 10 2-7 inches. It then becomes a question of finding a lens which with this focal length combines sufficient covering power and speed for our purpose - that is, one which will cover the plate satisfactorily with a relative opening large enough to give satisfactory speed.