ND Filter question.

hepabst

hepabst

Active member
I'm having a brain freeze trying to understand ND Filters. Here is an example: It is a bright sunny day I want to keep a 1/60 shutter speed at ISO 250. I use my hand held meter and it says to use f18. An ND4 will reduce the light by 2stops. So if I open the aperture to f16 and install the ND4 I should still have the correct exposure. Is this correct use of a ND filter?

In this example, if I want to open up lens to f8 I need a ND filter that reduce the light by 10 stops?

I understand the math, but I'm asking this question because I remember reading here, somewhere I can't find it, that ND filters work differently, meaning how much light they reduce, depending on the intensity of the light or something like that.

I'm a bit confused about this. I've tried this out with my D90(still mode) trying understand this but I don't get a consistent result.
 
Your understanding of f-stops is incorrect. The difference between f16 and f18 isn't two stops, it's barely about 1/6 of one stop.

F-stops are numbered 1.4, 2.0, 2.8, 4.0, 5.6, 8, 11, 16, 22, 32, etc.

So f16 with a two stops ND filter would result in f/8.
 
So f-stop is not the same as "stops". What is the math for this? Where can Learn more about this?

EDit: I just noticed the meaning of your example. A ND4(2 stops) is two clicks down on the lens (f16 to f8) do I have it correct?
 
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An f-stop is a "stop". A stop is a measurement of cutting the amount of light in half or doubling the amount of light that gets transmitted.

And yes, two stops would be from f/16 to f/8. Which results in 4x as much light getting let into the lens.
 
the math behind the numbers is actually very simple. the f-stop is the diameter of the aperture, but light doesn't pass through diameters, it passes through areas. the area of a hole is the radius squared times pi. if you square 1^2=1, 1.4^2=2, 2^2=4, 2.8^2=8, 4^2=16, 5.6^2=32, see how it doubles with every stop? that's the definition of a stop as barry said.
 
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