how to find the new value of apurture and shutter
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The best way to work with f/numbers is to memorize the f-number set. The number set for the f/stops: 1 – 1.4 – 2 – 2.8 – 4 – 5.6 – 8 – 11 – 16 – 22 – 32 Each number going right is its neighbor on the left multiplied by 1.4. Each number going left is its neighbor on right divided by 1.4 (or multiplied by 0.7).
Your exposure for this given situation is f/11 @ 1/60 of a second. Your desire is to move the aperture to f/22. Inspect the f/number set above. Note f/22 is two (2) f/stops away from f/11. Going right is smaller aperture diameters (more closed down – less light will be allowed to enter the camera). Now each f/stop represents a doubling or halving of light energy that is allowed to transverse the lens. Your desire is to reduce the aperture two (2) f/stops. To compensate, you must slow the shutter the equivalent of two f/stops. This action will allow more time for the light to play on the image sensor during the exposure. In other words, slowing the shutter allows more time for the exposing light to accumulate; thus you are compensating for the aperture size reduction by allowing more time for the exposing light to collect on the image sensor.
To accomplish this -- if you slow the shutter from 1/60 of a second to 1/30 of a second, this action doubles the time the light can play on the sensor. However, you desire a two f/stop change, so you must do this again. The final and correct shutter speed will be 1/15 of a second. This is a 4X or two stop change.
Mathematically, to make a two stop change to the shutter, remember each change of the shutter is the equivalent of a 2X change. You need to do this twice for a 2 f/stop change; thus the change in shutter speed is 4X. To accomplish, you multiply the original shutter speed by 4. Thus 1/60 X 4/1 = 4/60. This fraction can be reduced 4/60 = 2/30 = 1/15. Nobody said this stuff is easy!.
Your exposure for this given situation is f/11 @ 1/60 of a second. Your desire is to move the aperture to f/22. Inspect the f/number set above. Note f/22 is two (2) f/stops away from f/11. Going right is smaller aperture diameters (more closed down – less light will be allowed to enter the camera). Now each f/stop represents a doubling or halving of light energy that is allowed to transverse the lens. Your desire is to reduce the aperture two (2) f/stops. To compensate, you must slow the shutter the equivalent of two f/stops. This action will allow more time for the light to play on the image sensor during the exposure. In other words, slowing the shutter allows more time for the exposing light to accumulate; thus you are compensating for the aperture size reduction by allowing more time for the exposing light to collect on the image sensor.
To accomplish this -- if you slow the shutter from 1/60 of a second to 1/30 of a second, this action doubles the time the light can play on the sensor. However, you desire a two f/stop change, so you must do this again. The final and correct shutter speed will be 1/15 of a second. This is a 4X or two stop change.
Mathematically, to make a two stop change to the shutter, remember each change of the shutter is the equivalent of a 2X change. You need to do this twice for a 2 f/stop change; thus the change in shutter speed is 4X. To accomplish, you multiply the original shutter speed by 4. Thus 1/60 X 4/1 = 4/60. This fraction can be reduced 4/60 = 2/30 = 1/15. Nobody said this stuff is easy!.
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