This should be rather easy for these of you who own and use 0.45 ND or 0.75 ND filters so here we go. Just for fun I'm writing Android app that calculates exposure time when certain filter is applied. Times for 1/2 stops adjustment are easy no problem there. However with 1/3 adjustment where calculated exposure time still less then 30 seconds things get little "interesting".

What would be exposure time with 0.45 ND filter while exposure without filter is 1/1000? Would it be 1/400 or 1/320? (I would have try it on my own but I do not have 0.45 or 0.75 filter and I need to make a decision on it as to get Android app to calculate correctly)


1 Answer 1


The ND filter values are just numbers. Round numbers in whatever your favorite units are don't make the calculations different.

ND filters are often rated in "density" values, which are powers of 10. A .45 ND filter therefore attenuates by factor of 10.45 = 2.82. Yes, it really is that easy. Similarly, a ".75" ND filter attenuates by a factor of 10.75 = 5.62. If, for example, a picture is properly exposed without a ND filter at 1/1000 seconds, then to compensate for a .75 ND filter you would need to expose it for 1/(1000 / 5.62) seconds = 1/180 seconds.

To compute the same correction in terms of f-stops, you need to convert the attenuation ratio to a power of 2. Log2(5.62) = 2.5, so a .75 ND filter attenuates by 2.5 f-stops. You can also compute the conversion factor from ND filter density to f-stops directly, since it is just Log2(10) = 3.3. Take the ND filter density value, multiply by 3.3, and that gives you the attenuation factor expressed in f-stops. For example .75 density x 3.3 = 2.5 f-stops.

It's all just very straight forward math, and it doesn't change whether some of the numbers are in round numbers of f-stops or not.

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    \$\begingroup\$ How does 1/(180 / 5.62) seconds = 1/180 seconds? Shouldn't the equation be 1/(1000 / 5.62) seconds = 1/180 seconds (1/177.9 seconds to be exact)? \$\endgroup\$
    – Michael C
    Oct 13, 2013 at 19:53
  • \$\begingroup\$ @Michael: Oops. Looks like I got ahead of myself typing 180 before the answer. Fixed. \$\endgroup\$ Oct 13, 2013 at 22:14
  • \$\begingroup\$ @OlinLathrop sorry for bothering again but I came against a blocker. According to leefilters.com/index.php/camera/bigstopper 2sec with big stopper should result into 32min. However above formula will result into 10 on factor of 3 = 1000, and then 1000 * 2 = 2000sec = 33min. Would you trust application that tells you 33min while help table that came with big stopper is telling you 32min only? \$\endgroup\$
    – peter_budo
    Nov 27, 2013 at 18:04
  • \$\begingroup\$ @peter: Really? You're worried about .044 f-stops? Obviously there is some rounding going on in one of the answers. Do you actually think that the exposure system in your camera, the f-stop setting accuracy of your lens, and the manufacturing tolerance of the filters together get anywhere near .044 f-stops? If so, you need to get out into the real world more. \$\endgroup\$ Nov 28, 2013 at 16:41
  • \$\begingroup\$ @OlinLathrop I guess you misunderstood my request and took it too much personally. My apology I will not bother you. \$\endgroup\$
    – peter_budo
    Nov 28, 2013 at 17:13

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