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I have a set of neutral density filters that are rated as .3, .6, and .9 stops. Are they actually multiples of 0.3, or are they really multiples of 1/3? For example 1/3, 2/3, and 1 stop? If I stack the .3, .6, and the .9 is that 1.8 stops or 2 stops?

If it really is multiples of .3, what's the reasoning for that?

marked as duplicate by mattdm, dpollitt, Philip Kendall, Itai, MikeW Dec 13 '15 at 20:42

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • For the record, it's neither 0.3 stop or ¹⁄₃ stop. See the answers to the question linked above for details. – mattdm Dec 13 '15 at 18:20
  • The accepted answer was completely wrong on the point you are questioning though - it is corrected now. – Mark Fisher Dec 13 '15 at 18:43
  • D'oh! Sorry for the duplicate! I didn't find it when I searched. – user1118321 Dec 13 '15 at 23:27
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It is neither one. There are three different widely accepted systems used to describe the transmissive properties of filters.

The system you have used in your question is called the ND x.x number system and is used by Lee and Tiffen to describe their filters. It is the only system that uses decimals in the notation. The decimal values are based on optical density, not f-stop reduction. An ND 0.3 filter has a one stop reduction in terms of f-stop, as half the light striking the filter is allowed to pass through. An ND 0.6 has a two stop reduction as 1/4 of the light passes through. An ND 0.9 rating is a three stop filter. Each increase of ND 0.3 results in an additional stop of light reduction. So an ND 1.8 is a six stop filter, while an ND 2.0 is 6 2/3 stop filter, and so on. Back before calculators and other digital aids were available this system allowed photographers with external light meters to compute the effect of a filter or several filters combined on exposure using simple addition and subtraction instead of more complex mathematical operations when in the field.

Another system, used by others, is the ND1xx notation. All of the numbers begin with a "1" and include two other digits. The second and third digits express the number of stops of light the filter reduces. An ND 101 filter is a one stop filter, an ND 102 is a two stop filter, and ND 106 is a six stop filter, and so on.

A third system, used by Hoya, B+W, and Cokin, is the ND 1/x (or 1/2^x) system. Each filter is described as the reciprocal of the amount of light allowed to pass through the filter. An ND2 allows one-half the light to pass for a one-stop reduction. An ND4 allows one-fourth the light to pass for a two-stop reduction, an ND64 filter allows 1/64 the light to pass for a six-stop reduction. Note that each increase of one-stop in this system is a power of the number "2".

To see a chart that shows each system and how filters in one system relate to filters using one of the other notations, please see this chart at wikipedia.

-1

Photo scientists in the late 1800’s, like the rest of the scientific community, had no calculators or computers. They mostly relied on the slide rule or log tables. Don’t laugh; giant ships and building etc. were designed with the aid of a slide-rule. Now the slide-rule uses logarithmic notation. The number 2 is written as 10 elevated to the 0.30 power. The number 4 is 10 elevated to 0.60 power. The number 8 is 10 elevated to the 0.90 power. Log in base 10

2 = 10^0.30 (2x change = 1 stop)

4 = 10^0.60 (4x change = 2 stops)

8 = 10^0.90 (8x change – 3 stops)

16 = 10^1.20 (16x change = 4 stops)

32 = 10^1.50 (32x change = 5 stops)

64 = 10^1.80 (64x change = 6 stops)

128 = 10^2.10 (128x change = 7 stops)

256 = 10^2.40 (256x change = 8 stops)

512 = 10^2.70 (512x change = 9 stops)

1024 = 10^3.00 (1024x change = 10 stops)

0.05 = 1/6 f/stop

0.10 = 1/3 f/stop

0.15 = 1/2 f/stop

0.20 = 2/3 f/stop

0.30 = 1 f/stop

0.40 = 1 1/3 f/stop

0.50 = 1 2/3 f/stop

0.60 = 2 f/stop

etc.

Now to compensate for the absorption of a filter, we use a filter factor. This is a multiplier. A filter that attenuates 2 f/stops has a filter factor of 4. We use the filter factor by multiplying the exposure time without filter by the filter factor. Suppose the exposure without filter is f/11 @ 1/100 of a second. We mount a filter with a filter factor of 4 (2 stops attenuation). The revised exposure time is 1/100 x 4/1 = 1/25. In other words, the revised exposure is f/11 @ 1/25 of a second.

Note: To avoid repetition it is customary to drop the 10^ and only write the mantissa of the log (the elevated proton). Thus for a filter factor of 4, instead of writing 10^0.60, we just write .6.

Nobody said this stuff is easy.

  • How, exactly, does this answer the question? How does this information allow someone with no knowledge of ND filter notation to understand the relation between an ND 0.9 filter and an ND8 or ND 103 filter? – Michael C Dec 13 '15 at 19:42
  • 2
    The position of the number 2 on a slide rule is an expression of log(base 10) 2=0.3, NOT 2^0.3=2 (WHICH IS NOT CORRECT). 2^0.3=1.2311444...(base 10). – Michael C Dec 13 '15 at 19:56
  • You are so correct! Thanks' I lost it -- have edited the answer to 10^. A tip of the hat from Alan Marcus – Alan Marcus Dec 13 '15 at 20:33

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