filter stacking, part deux

[phillipmarcovallentin]

<p>I'm confused:</p>

<p>An "expert", with large library of books on photo technique, now claims, that one cannot simply add filters, but that they have to be MULTIPLIED:</p>

<p>Example:</p>

<p>an ND8 (3stops) and a CP-L (just under 2 stops), would equal 3 x 2 = 6 stops,<br>

and not 3 + 2 = 5 stops.</p>

<p>Even people from B+W haven't heard about this !</p>

<p>Any input are MOSTLY welcome.</p>

<p>PS: the "multiply" rule, should be mentioned in one of Phil Malpas textbooks,<br>

but of cause this could be just a printing error ?</p>

 

[joshuasigar]

<p>Hmm, I never pay attention but I would think it'll be 5. But why don't you just get two filters and check how the exposure changes between one and two stacked filters?</p>

[jim_momary]

<p>An electrical analogy to filters are resistors. Put a two ohm resistor in series with a 3 ohm resistor and you wind up with a five ohm resistor. It's additive, not multiplicative.<br>

Jim M</p>

[randall ellis]

<p>It's all down to terminology. Multiplying applies to the <em>filter factor</em> , and adding applies to the <em>exposure compensation</em> of a filter. If you use a filter with a filter factor of 2.0 (an exposure compensation of 1 stop) and you want to add another that has a filter factor of 4.0 (an exposure compensation of 2 stops) you would then have to <em>multiply</em> the <em>filter factors</em> (2.0 * 4.0 = 8.0) to get the correct adjustment. An 8.0 <em>filter factor </em> equates to an <em>exposure compensation</em> of 3 stops, which is the same as what you get if you add the exposure compensation for each (1 + 2). It's just an older method of calculating how to adjust for filters that is not used much anymore. It also does not help that many people think of the exposure compensation (number of stops to adjust) as the filter factor, which, unless I've gotten more senile that I thought, is not technically the correct term...</p>

<p>- Randy</p>

[phillipmarcovallentin]

<p>@ Randal:<br>

NOW I know where he got it wrong !<br>

Thanks for clearing u things ;-)</p>

[Alan Marcus]

<p >Filter Factors:</p>

<p >Most books and teachers are grounded in exposure determination using a hand-held light meter. Additionally early cameras with built-in meters handled the meter as an add-on, it was not coupled to the camera. Only in modern times did the camera and meter become integrated. Nowadays the meter reads through-the-lens. </p>

<p > </p>

<p >Now through-the-lens metering get its data after the light has transversed the camera’s optical system. This is advantages because the light absorbing properties of mounted filters are taken into account. This is true even if you stack filters. So you need not concern yourself with filter factors with a through the lens metering system. The main pitfall will be, light meters are calibrated to mimic the response of film or digital chip but they have slightly different sensitivities to some colors i.e. a mismatch can occur with colored filters. Meaning it is wise to bracket important shots. </p>

<p > </p>

<p >If the meter is independent (hand-held) we must manually apply a filter factor.</p>

<p >Several ways to apply:</p>

<ol type="1">

<li >Multiply exposure time by filter factor. Example: filter factor is 4. Filter in mounted on an enlarger. Unfiltered the exposure time is 10 seconds. Correction is 10 x 4 = 40 seconds. Without filter mounted, camera exposure time is 1/60 second. 1/60 x 4 = 1/15 sec.</li>

<li >Divide ISO by filter factor. Media is 200 ISO. A filter with a filter factor of 4 is mounted. 200 ÷ 4 = 50. Reset meter ISO to 50. This modification automatically applies the filter factor when a hand-held meter is employed.</li>

<li >Count on you fingers using the power of 2 number set. Thus 2 – 4 – 8 – 16 – 32 -64. If the filter factor is 4, that’s two fingers. Compensation is 2 f/stops. If the filter factor is 16, that’s 4 fingers thus 4 f/stops.</li>

</ol>

<p > </p>

<p >Stacking filters:</p>

<p >Consider a filter factor of 2 = 1/f stop. This is true because 1/f stop change halves the amount of light that will transverse the optical system. Assume 1000 photons stream through the optical system in one second, no filter. Now mount 1 filter with a filter with a factor of 2, now 500 photons pass. Mount (stack) a second filter with a filter factor of 2. Now 250 photons are allowed to go by. Mount (stack) a third filter with a filter factor of 2. Now 125 photons will be allowed to go by. We have stacked three so the stopping power is 2 x 2 x 2 = 8 that’s 3 f/stops. We multiply filter factors together when stacking to calculate the total. </p>