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### ANSWERS

10-1: The resolution is best close up in the foreground and diminishes progressively with increasing distance of view (towards the top of the photo). Thus smaller objects can be discerned near the bottom of the picture but these same objects (no change in size) would likely not be distinguishable in the hills farther away across the water. This same simple principle holds as we look out over a landscape: a house nearby would be large, with observable details, but would be small and generalized at a distance, say, of 5 miles. BACK

10-2: 1 inch = 100,000 divided by 12 or 8333.3 feet. In miles that would be 8333/5280 = 1.5783. BACK

10-3: The lake can be found in the last image, an aerial photo printed at the scale of 1:141,000. Look for it, it shows as a dark roundish dotlike feature. A 1:4000 scale image yields 1 inch = 333.33 ft (approximately 100 yards or the length of a football field) = 0.063 miles. On the Interstate (large white road), individual cars are visible. A typical auto is about 15 feet (180 inches) in length. The resolution therefore is at least as good as about 5 yards (meters) - probably even approaches 2 yards but nothing of that size is evident. BACK

10-4: Look at the scale expressed this way: 1:30000 = 1/30000, which might be read as 1 inch per 30000 inches. The denominator is the clue. Small denominators are large scale; large denominators denote small scale. After all, 1/10 is a larger fraction than 1/1000. BACK

10-5: After loading the film, you first would check the film's ASA value and, if your camera is so equipped, turn the timer to a position coincident with, or close to, that value. Then, you would turn (rotate) the lens along its thread, causing its focal length to change, until the scene is in focus for the close-up of the person (done visually through a sighting window and/or by reading distance numbers on the lens barrel). Now, adjust the F/Stop, probably to something like F/11 or F/16. Next, turn the timer knob to some exposure time setting, perhaps around 1/250th of a second. If you have a light meter, look to see that the needle is between two guide marks that indicate near-optimum settings. If the moving needle does not fall between these marks, adjust either F/Stop or exposure time until the needle assumes the right position. You are ready to snap the picture. Then, repeat the process for the distant scene, changing the focal length to infinity. You probably will need to also adjust F/Stop and/or exposure time, since the amount of light from a distant scene will differ. At sunset, the F/Stop will probably be set to F/2 or thereabouts and the time to something around 1/8th to 1/30th of a second, both to allow more light to enter (opening up the lens aperture and keeping it open longer); these settings, too, need to be determined experimentally. (The settings used in both time of day cases are probably reasonably close to adequate even if you don't have a light meter.) BACK

10-6: The rule might read like this: an object of some primary color sends its light onto the negative; the layer that is activated will be the complementary color composed of the other two primary colors; when a print is made, the layers activated are the other two complementary colors. For green, then, the negative will be red + blue = magenta; the positive will be cyan + yellow. Try this now for a red object. BACK

10-7: RF = photographic distance/map distance = 9"/63,360" = 1/7040 or 1:7040. Note: the 63,360 is just 5280 feet x 12 inches/foot. BACK

10-8: 1.75 inches x 1/39.36 inches/meter = 0.04 meters. 0.05/1108 = 1/27700 BACK

10-9: 2.2" x 20000/1 = 44000. 6.83/44000 = 1/6442. BACK

10-10: Scale = f/H* = 0.5/(5000 - 1000) = 1/8000.BACK

10-11: s = f/H*; H* = f/s = 0.152 m/1/2000 = 304 meters. (6 inches x 2.54cm/inch = 15.2 cm = 0.152 m). Note: the photo size does not enter the solution, the actual size could be much larger, with the area of the photo occupying only a fraction of the print area, or smaller, in which case only part of the scene would occupy a 9 inch print. Thus, the scale remains independent of the size of the print. BACK

10-12: First convert feet to meters: 12000 ft x 1/3.28 ft/m = 3658. Then convert 9 inches to millimeters: 9 inches x 25.4 mm/in = 229 mm. Then Rg = 3658/15 x 229 = 1.065 line-pairs/meter. BACK

10-13: Rs = H/f x Rg. Thus: 6000/120 x 4 = 12.5 line-pairs/mm. BACK