1 1 1 ─── + ───── = ─── 240 _v_ 3 from which the values of _v_ are 3-1/13 inches and 3-1/26 inches respectively, a difference of only 1/26 of an inch. If _u_ were 6 inches, _v_ would also be 6 inches; if u were infinity, _v_ would be 3 inches. The movement of the image, when an object is brought towards it from a great distance, would in the mirror under notice be from 3 inches away to 6 inches away, and at distances of 10 feet and upwards would only differ in position by small fractions of an inch. _Example II._—To find the area of the image of a circular opening, 1 foot in diameter, formed by a mirror of 6 inches radius distant from the opening (_a_) 10 feet; (_b_) 20 feet. Since _d_ _u_ ──────── = ───── ; then, from the results of _d_{1}_ _v_ Example 12 120 ──────── = ──────── at 10 feet distance, _d_{1}_ (3-1/13) and 12 240 ─────── = ──────── at 20 feet. _d_{1}_ (3-1/26) Hence the linear dimensions, _i.e._ the diameters of the circular images, will be 0·308 and 0·152 inch respectively; and the areas 0·074 and 0·0182 square inch. These areas are to each other practically as 4 : 1. That is, the area of the image decreases in size directly as the _square_ of the distance of the object; the squares of the distances being 100 and 400, or as 1 : 4; whereas the areas of the images are as 4 : 1. _Example III._—To find, for a 6-inch mirror, and a junction of 1/10th of an inch in diameter, the greatest distance at which the mirror may be placed from an opening 1 foot in diameter, so as to give an image not less than the junction. From Example I it is evident that at any distance exceeding 20 feet the position of the image will only be a minute and negligible fraction over 3 inches; hence _v_ may be taken as 3. Applying values in the formula _d_ _u_ ─────── = ───── and _d_{1}_ _v_ taking _d_{1}_ as equal to the diameter of the junction, = 0·1 inch, 12 _u_ ─── = ──── , and _u_ = 360 inches, or 30 feet. 0·1 3