BetterPhoto Q&A
Category: New Answers

setting exp with 89B ir fliter

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June 09, 2009

The 89a designation is from the filter catalog of Frederic Wratten (1840-1926) English filter manufacturer and cofounder of the photo supply house Wratten and Wainwright sold to Kodak 1906 with the agreement that the filters shall retain their catalog numbers. This filter blocks visible light and passes IR.

Filters subtract some light energy. Exposure compensation is required. Filter Factors to the rescue.

If you know the filter factor you can apply compensation.

How to use:
1. Multiply the filter factor times the shutter speed. Example: assume the shutter speed without filter is 1/125 second. We mount a filter with a filter factor of 8. We calculate a revised shutter speed thus 1/125 x 8 = 1/15. Most folks have problems multiply a fraction by a whole number nevertheless this is a viable method.

2. Divide the ISO by the filter factor. Example film speed is 400 ISO to be used with a filter that has a factor of 8. Thus 400 ÷ 8 = 50. We reset the meter’s ISO to 50 and allow the metering system to calculate the revised exposure.

3. Since each f/stop is a 2x change in light energy, we count on our fingers in powers of 2 thus: 2 – 4- 8 – 16 -32. If we are using a filter with a factor of 8, we count and 8 lands on the third finger. We now know we must open of the camera’s aperture 3 f/stops.

Now what you want to know is : The Wratten 98a has a filter factor of 16 that means we open up 4 f/stops. Be advised that IR photography involves some trial-and-error testing. You may find the filter factor of 16 is only an approximation thus you should shoot a bracketed exposure test.

Lots of luck!
Alan Marcus
alanmaxinemarcus@att.net

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June 09, 2009

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June 09, 2009

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June 09, 2009