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⁄d1 = u⁄v;
then from the results of Example I,
12⁄d1 = 120⁄(3-1⁄13) at 10 feet distance, and
12⁄d1 = 240⁄(3-1⁄26) at 20 feet.
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.