 Christopher A. Walrath |
Physics of the Flash-Lit Photograph Thought that might get your attention. I know how to use a flash manually. I know that if you have a flash guide number of, say 88, that a subject 22 feet away would require an aperture of f/4 and a subject 4 feet away would require an aperture of f/22 and so forth. I know how a flash works and how to use it. WHY? Why does a larger aperture not capture the flash-elevated values in nearer subjects as well as further ones? Why do guide numbers work the way they do (other than, that's just the way it is, now bug off)? I need a gobbledygookish tutorial on the physics of the flash-lit photograph. HELP!!!!!!!!!
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- Gregory LaGrange Contact Gregory LaGrange Gregory LaGrange's Gallery |
If you're wondering why a large aperture, flash, and close subjects you find it difficult to to get things properly exposed even when they're just a small distance away from each other, than you'll find the answer in how light behaves relative to the distance it travels. The farther it has to go, the rate of the fall-off decreases. Just think of any light you have and the difference in exposure you'll get at 15ft versus 30ft. But with sunlight, you'll get the same exposure at one spot that you would if you stood 5 miles away. So with close things that are only 1ft away, if there's another thing that's merely another ft behind it, if the front object is exposed right, then that second object that's only a foot behind it would a whole f/stop off. And a large aperture brings in more ambient light anyway, so with a distant subject you'll getting more of an area where the light doesn't change in exposure very much. That's what I understood you question to be.
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Jon Close |
Excluding lasers, light does not stay at the same intensity at all distances from its source. It spreads out and dissipates very quickly. The level of light intensity decreases with the inverse square of distance from its source. That fits neatly with f/stops. 1.4x farther distance is 1 stop lower light intensity (1/2 power), 2x distance is 2 stops (1/4 power) etc. Rather than give the maximum light output for a speedlight in lumens or watts/seconds, it is converted to a guidenumber to simplify exposure calculation. At the flash's maximum output level, the guidenumber is the maximum distance for which a subject will be properly exposed at ISO 100 and f/1 lens. Stop down the lens, then less light gets to the film/sensor, so the maximum flash distance is reduced. One stop smaller aperture = 1 "stop" less flash range (divide by 1.4). Increasing the film/sensor sensitivity (higher ISO) means proper exposure is achieved with less light, so the flash range is extended. One stop greater ISO = 1 "stop" (1.4x) greater flash range. The complete guidenumber equation is: Ex) GN of 43m, ISO 400, and f/5.6 gives maximum flash range of 43 x sqrt(400/100) ÷ 5.6 = 42x2/5.6 = 15m The formula can be rearranged so that you can determine the aperture for a given subject distance:
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Alan N. Marcus |
Hi Christopher, Why do guide numbers work the way they do? Consider a point source in space. It can be seen from any direction. We can draw these using lines that resemble spokes radiating out from the central point of a wheel. The Latin word for spoke is radius; thus we say the light radiates. Light falls off with distance. When the light originates from a point source it performs according to the law-of-the-inverse square. Consider a slide projector with a square slide projecting a picture on the wall. Assume the projector is 1 foot from the wall and the image on the wall measures 1 foot by 1 foot. The area of the illumined frame is 1 times 1 or 1 sq. ft. We measure the projected image brightness and find it to be 1000 foot candles. Now we move the projector back to the 2 foot mark and re-focus. The image on the wall enlarges. It now measures 2 feet by 2 feet. Its area is 4 sq. ft. The lamp inside the projector did not change brilliance, however the projected image is dimmer. Actually it’s now only 25% as bright as it was at the 1 foot distance mark. A meter reading reveals the brightness to be 250 foot candles. This is so because the same light energy is now required to cover 4 sq. ft. (i.e. the total quantity of light remains the same but the area it must play upon has increased) hence the light on the wall is more dilute. We can use the law of the inverse square to calculate brightness for any distance. We write the distance as a fraction. At the two foot mark we write 2/1; we invert 1/2 At the four foot mark we write 4/1; we invert 1/4 Since light intensity is defined as the quantity of light per unit area, we substitute lumen (Latin for light) for the energy covering one sq. ft. originating from a standard candle situated 1 foot from the target. To find your own guide number: Example: Best frame is f/11 – distance is 10 feet. Thus 11 x 10 = 110. Alan Marcus (marginal technical gobbledygook)
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Alan N. Marcus |
It’s time again for me to don the dunce cap. Once again an error. Wrong! Corrected: The idea was to explain how drastically light falls off with distance. This same law of the inverse square is applicable to sound, radiant heat, magnetism and gravity. Alan Marcus (marginal dunce)
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Christopher A. Walrath |
Hi all. More to learn. More to chew on. Better to become. Thank you very much. Chris
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