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Neutral density filter question.

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pbromaghin

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I have an odd-ball neutral density filter. It's a Kodak Series VI and it says it is a 2.5. Looking around I see info on 2.4 and 2.6, but no 2.5.

Does that number mean it is a number 2.5 and so about 8 stop, or is it a factor of 2.5 and so a little more than 1 stop?
 
I have an odd-ball neutral density filter. It's a Kodak Series VI and it says it is a 2.5. Looking around I see info on 2.4 and 2.6, but no 2.5.

Does that number mean it is a number 2.5 and so about 8 stop, or is it a factor of 2.5 and so a little more than 1 stop?

You have it. Why don't you look through it? -1 stop doesn't look at all like -8 stops.
 
You have it. Why don't you look through it? -1 stop doesn't look at all like -8 stops.

Hmm. Good question. It's the only neutral density filter I have. I've never dealt with them before and so have no basis from which to judge.

I plan to be using iso 400 film in an old folder with a fastest shutter of 1/250. In the past a red filter has worked with B&W to get the exposure down to what the shutter can handle. I found this in the drawer and thought it would work for color film.
 
Or just meter through it.

Another good suggestion. It's a Series VI which is quite a bit smaller diameter than any lens I have for a metered camera.
 
Hmm won't fit a camera with a meter...

How about shining a light on a table with a beam that you can cover with the filter... Meter the light before and after.
 
Depending on the filter maker’s policy, ND filters can be marked in:

1. Density units

2. Stops of light absorbed

3. Time multiplication factor

I have examples of each type in my possession.

For example, 0.40 ND = 1.5 stop = 2.5X time factor. Each value describes the same filter expressed in different units.

I suspect that your “2.5” filter is time factor 2.5X.
 
Now that I have seen the filter in good light, I feel really stupid for even starting this thread. It says very clearly on the edge of the filter - 2.5x multiplication factor - meaning a little over 1 stop. Thanks to all of you for your help.
 
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