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Old Oct 25, 2002, 3:22 PM   #1
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Default infrared filter

Does anyone know if doubling to two weak infrared filters will result in one strong one, or not? I have a deep red filter now, and can't afford a new expensive one, but can buy a cheaper, similar one.; would that work?
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Old Oct 28, 2002, 2:50 AM   #2
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I think you're confusing ND (neutral density--grey) filters, which can be added together, with IR. ND filters allow a certain percentage of the light that enters them to pass through, so they have a multiplicative effect. In English, that means if you combine an ND filter that lets 25% of the light that hits it through with a 50% ND filter, you'll wind up with 12.5% (0.25*0.50=0.125) of the light through.

IR filters (good ones, anyway), block all visible light with wavelengths less than about 750 nanometers. Actually, the point at which they start letting light through is determined by the filter...a Hoya R72 has 50% transmissivity at about 720_nm, the R90 at about 900_nm. Check out this link for the skinny:


So, if you look at the 89b, you can see it begins transmitting light right around 675_nm and starts really picking up around 690_nm. It lets about half the light through at the 50%T point of around 722_nm. Since the visible spectrum ranges from about 400_nm (violet) to 700_nm (red), that filter will pass a tiny bit of visible red light. To picture what would happen with two of those filters stacked, just let y=x^2, where in the graph I've linked it's y=x. That means it'd be roughly equivalent to an IR filter with 50%T of about 735_nm...not much change.

I was initially confused by the Heliopan on that graph...I saw the curve going the other way and wondered what kind of IR filter blocks most of the IR spectrum and lets in clearly visible light. Then I did a bit more reading--it turns out that's not an IR filter, it's an IR *cut* filter. In other words, IR light apparently flips photocells when you're taking normal pictures--if you want to get the image of what your eye is actually seeing, you have to shield the CCD from both UV and IR light, and a UV filter in combination with that Heliopan will limit light passing into the camera to that which is visible.

At any rate, sorry to be the bearer of bad news. To significantly move the transmissivity spectrum, you'll have to get a new filter designed for the purpose. Check it out, though...the next one up isn't that expensive, I think. I've learned that the Kodak Wrattans go in this order (from lightest to darkest): 89b -> 88a -> 87 -> 87b -> 87c. Chances are, though, with a digital camera you're probably not going to see much of anything with the 87b or 87c--you'll have to test it in the store, and probably at a very long exposure with the aperture wide open, and it'll still come out way underexposed. That's my guess...I haven't tried it.

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