filters - colour absorption question

pellicle · Sep 2, 2009

Hi

pardon me, but I'm not sure where's the best place to ask this, so I thought I'd start here.

I normally print black and white, and I have multigrade filters for that, folks who have experience in colour printing will no doubt understand the filters better than me, so here I am.

If I was wanting to absorb more of the green and some of the red spectra from a light source what colour filter would I use? I know that a blue appearing filter will remove more of the red and green, but what colour will a filter which removes some of the green as well appear?

Thanks

Nick Zentena · Sep 2, 2009

For printing colour? The colour of the filter removes the same colour. So more green in the filter means more green removed.

Or do you mean something else?

pellicle · Sep 2, 2009

Nick Zentena said:
For printing colour? The colour of the filter removes the same colour. So more green in the filter means more green removed.

Or do you mean something else?

nope, I didn't mean something else. That's what I meant .. so a filter which removes some blue, and some of the green would look what colour?

keithwms · Sep 2, 2009

I don't think it's very useful to think of 'removing' colour. It's easier to think in terms of what is transmitted. A filter has a certain colour because that colour is what is transmitted. A red filter transmits red... and blocks the other colours.

So... a filter will simply change the relative weights of the colours. If you use a filter (say, red) then the light transmitted through that filter will be more red weighted: there will be less green, less blue... relative to red. If you want more blue and green transmitted then you could use an orange or yellow filter.

You could also do a multiple exposure: do one shorter exposure with all the colours (no filter) and then another exposure with a filter for the one you want to dominate.... that would allow you to tune the relative exposures very finely with time.

P.S. my post is at odds with Nick's assertion that a green filter will block green; not sure what Nick means, let's see...

Nick Zentena · Sep 2, 2009

If your colour print is too green adding green to your filter will remove the green. With normal RA-4 printing. I guess this is subtractive?

keithwms · Sep 2, 2009

Oh you are talking about RA4.

Bob-D659 · Sep 2, 2009

For RA4 printing:
Too green - subtract Magenta
Too Cyan - subtract Magenta and Yellow in equal amounts
Too Blue - subtract Yellow
Too Yellow - add Yellow
Too Red - Add Magenta and Yellow in equal amounts
Too Magenta - add Magenta

Mr Bill · Sep 2, 2009

If I was wanting to absorb more of the green and some of the red spectra from a light source what colour filter would I use? I know that a blue appearing filter will remove more of the red and green, but what colour will a filter which removes some of the green as well appear?

Pellicle, from the simple viewpoint that we often take as photographers, we consider white light to be a mix of some reddish, some greenish, and some bluish light. From that viewpoint, and considering your desire to remove (filter out) both some green and some red light, you would want to let the third color, blue, to come through unaffected. since mainly blue light is coming through, if you look through that filter, toward a white light), the filter appears bluish. But it sounds as if you already know this part.

Now, if you want to remove, for example, two units of green and only one unit of red (this is what you are asking, I think), here is one way to do it: first, use a blue filter with enough strength to remove both one unit of green and one unit of red. Now it remains only to remove one more unit of green. It turns out that such a filter, a green absorber, exists under the name of magenta. So your final filter pack would consist of the combination of blue filter and magenta filter. And if you hold it up to the light, and look through it, it has the appearance of a magenta-ish blue, or a bluish magenta, whatever you wish to call it.

Now, if you didn't know what magenta was, I'd describe it as a filter which blocks green, allowing red and blue to come through unscathed. So you could now see that my previous filter pack, blue plus magenta, is roughly equivalent to a double blue filter plus red filter, having a slightly reddish-blue appearance. Making any sense?

If you try to understand filters any deeper than these simple ideas, I think you need to look at spectral data. But for color printing, and using filters with only partial absorption, the ideas of red, green, and blue, combined with their complements of cyan, magenta, and yellow, work perfectly.

With regard to color printing, a rule I learned long, long time ago, which seems virtually infallible, is simply this, "Always do the wrong thing." If your print is too green, it seems like the wrong thing is to add green filtration. But do it anyway, it will work. Except, we don't normally use green filter, so you do the equivalent thing, which is to REMOVE magenta filtration. (I said it was equivalent, but actually it changes the exposure a bit).

A bit more comlicated example, per Bob-...'s "Too Cyan..." example: If your print is too cyan, you want to do the wrong thing, which is to add more cyan. However, conventionally, we don't adjust cyan; you want to do the equivalent thing, which is to remove red. However, we don't usually handle red filters, either, so you jump to the equivalent of removing both magenta and yellow. Once you work with this a bit, these equivalents will become second nature. In the meantime, you might want to consult that diagram, with three overlapping circles, labeled red, green, and blue. The little section opposite red is cyan, opposite green is magenta, and opposite blue is yellow.

Did I make this too long and complicated? If so, just remember, "always do the wrong thing."

tim k · Sep 2, 2009

Wow, kinda makes black and white sound pretty attractive.

Mr Bill · Sep 2, 2009

Wow, kinda makes black and white sound pretty attractive.

hehe...but think about it Tim, with B&W, you have to do everything right. But for color, just use my method, to always do the wrong thing. If you know zilch about color, almost for sure you'll do things wrong, right? So in no time at all, you'll be making great color prints!

In truth, though, it's not as difficult as it seems. Once you work through that color diagram a few times, things fall into place.

pellicle · Sep 3, 2009

Folks

thanks for all the kind answers. I will digest soon, but I have a busy time with work for next 7 days so I will not be able to experiment and perhaps ask further questions till after than.

I just get confused with additive and subtractive colour models.... but:

If your print is too cyan, you want to do the wrong thing, which is to add more cyan.

is exactly what was throwing me I think ... once I developed an 'adage' "that when it doesn't seem to fit its because everything is opposite to what I'd expect"

this especially holds true in my work (computers)

dhroane · Sep 25, 2009

A bit more comlicated example, per Bob-...'s "Too Cyan..." example: If your print is too cyan, you want to do the wrong thing, which is to add more cyan. However, conventionally, we don't adjust cyan; you want to do the equivalent thing, which is to remove red. However, we don't usually handle red filters, either, so you jump to the equivalent of removing both magenta and yellow. Once you work with this a bit, these equivalents will become second nature. In the meantime, you might want to consult that diagram, with three overlapping circles, labeled red, green, and blue. The little section opposite red is cyan, opposite green is magenta, and opposite blue is yellow.

Hello, Mr. Bill. I have two sets of questions.

1)What does "filtering" connote? In other words, does the name of a colored filter designate the color that is being blocked or the color that is being allowed to pass through?

2)In the portion of your entry which I quote above, you mention that to block red light, you instead do the equivalent thing which would be to block magenta and yellow. To block magenta and yellow would assume that these two colors "mix" to create the red you are trying to block. Doesn't this assumption contradict the very fact that, in the additive system where light is concerned, magenta and yellow are secondary colors, which when combined, cannot form red (a primary additive), but rather form white?

Thank you.

Mr Bill · Sep 25, 2009

Hi dhroane, let me see if I can explain a bit better.

1)What does "filtering" connote?

Normally, we name a filter by what color it passes. For example, if we look at a white light, then raise a filter in front of our eyes and it appears to be red, we would call this a red filter. And of course, it looks red because only (or mainly) red light is passing through to our eyes.

In the case of "light" we cant's see, such as UV (ultraviolet) or IR (infrared), the naming is not always clear. So rather than say that something is an IR filter, it's better to call it either an IR-pass or an IR-blocking filter.

To block magenta and yellow would assume that these two colors "mix" to create the red you are trying to block.

Well, as I said in my first post, in order to truly understand, I think it is necessary to look at the full spectral data. The following explanation, using words, is going to sound complicated and hard to follow, but it should work. ps; no one really needs to know this, it's only for intellectual curiousity, I think, or to make you feel more confident in the rules of "doing the wrong thing."

So in keeping with the simple concept, where white light is made up of reddish, greenish, and bluish light, here's as simple as I think I can exlain it. The primary filters, red, green, and blue, will pass only one color (their own), while blocking the other two. The secondary filters, cyan, magenta, and yellow are considered to pass two colors, while blocking only one. For example, a magenta filter will block green, but passes both red and blue (red and blue light, mixed, appear to us as magenta).

So, for your question about blocking magenta and yellow: to block magenta, as you now know, actually means to block both red and blue, while passing the third color, green. in a similar manner, to block yellow means to block both red and green, while passing blue. Clearly these two things are at odds with each other; the one passes only green, while the other passes only blue. If you combine both filters, it seems that nothing should come through. And this IS the case with perfect filters, where they completely pass or block. However, color printing filters are not in the class of all or nothing; they block certain percentages of light. For example a 30 cc yellow filter should block about 50% of the blue light, while allowing virtually ALL of the red and green to come through.

So lets redo the "blocking magenta and yellow" thing with 30CC filters. When we start with white light (100% red, 100% green, and 100% blue), then try to block magenta, this calls for a green filter. So after passing through our 30CC green filter, we'll find this: 50% of red, 100% of green, and 50% of blue light. The next step, blocking yellow light, calls for a blue filter. So applying a 30CC blue filter to the now greenish light, here's what happens: blue light will pass through unaffected, but only 50% of the existing red and green light gets through. Consequently we'll end up with: 25% of red, 50% of green, and 50% of blue. So the final result is that green and blue have equal strengths, while the red has been weakened to half of them.

The net effect is, that red light has been relatively weakened, giving the visual appearance of greenish-blue light, which we might also call cyan. So you can see that things DID behave correctly, our initial intent was to (relatively) block red light, and this was the actual result, although we did it by trying to block magenta and yellow light.

Once you are satisfied that things work, it is much easier to simply draw the little color diagram, either the triangle thing or the overlapping circles, and then follow the rule of "always do the wrong thing."

Mr Bill · Sep 25, 2009

Doesn't this assumption contradict the very fact that, in the additive system where light is concerned, magenta and yellow are secondary colors, which when combined, cannot form red (a primary additive), but rather form white?

Now that I have tried to confuse everyone with the previous post, I see that I didn't read your questions completely.

In your example, I think that actually the result WILL BE reddish light, not white. Here's how: you want to add (not filter) lights, using magenta light and yellow light. In terms of red, green, and blue, the magenta light contains red and blue (it is missing green light). Similarly, the yellow light contains red and green (it is missing blue light). So combining these, we get (red and blue) plus (red and green), in other words, a double portion of red light, plus a single portion of green light and a single portion of blue light. so the overall result should have a distinctly reddish appearance. Of course, it is not a pure red light - it has been contaminated with some green and some blue, but reddish nonetheless.

I hope this is some help, rather than just confusing.

dhroane · Sep 27, 2009

Mr Bill said:
Mr Bill said:

Hi dhroane, let me see if I can explain a bit better.

Normally, we name a filter by what color it passes. For example, if we look at a white light, then raise a filter in front of our eyes and it appears to be red, we would call this a red filter. And of course, it looks red because only (or mainly) red light is passing through to our eyes.

In the case of "light" we cant's see, such as UV (ultraviolet) or IR (infrared), the naming is not always clear. So rather than say that something is an IR filter, it's better to call it either an IR-pass or an IR-blocking filter.

Click to expand...

Thank you so much, Mr. Bill! I appreciate the clarity!

So, for your question about blocking magenta and yellow: to block magenta, as you now know, actually means to block both red and blue, while passing the third color, green. in a similar manner, to block yellow means to block both red and green, while passing blue. Clearly these two things are at odds with each other; the one passes only green, while the other passes only blue. If you combine both filters, it seems that nothing should come through. And this IS the case with perfect filters, where they completely pass or block. However, color printing filters are not in the class of all or nothing; they block certain percentages of light. For example a 30 cc yellow filter should block about 50% of the blue light, while allowing virtually ALL of the red and green to come through.

Click to expand...

Thank you again, Mr. Bill! Permitting a percentage of the light which the filter is designed to block, to pass through anyways, allows for the combinatory effects gained my using multiple filters.

I really appreciate the time you spent providing me with thoughtful, and most significantly, CLEAR explanations!

You would make a fantastic teacher (if you aren't one already)!

So lets redo the "blocking magenta and yellow" thing with 30CC filters. When we start with white light (100% red, 100% green, and 100% blue), then try to block magenta, this calls for a green filter. So after passing through our 30CC green filter, we'll find this: 50% of red, 100% of green, and 50% of blue light. The next step, blocking yellow light, calls for a blue filter. So applying a 30CC blue filter to the now greenish light, here's what happens: blue light will pass through unaffected, but only 50% of the existing red and green light gets through. Consequently we'll end up with: 25% of red, 50% of green, and 50% of blue. So the final result is that green and blue have equal strengths, while the red has been weakened to half of them.

Click to expand...

Now your explanation has caused me to return to the issue which prompted all of this—the tamping down of a print that is "too cyan" by adding, of all things, more cyan!

Correct me if I am wrong, but the nomenclature for filters is what can be the source of confusion. Though we are trying to "block" red with the use of green and blue filters, each one, nevertheless, still allows enough red light to pass through to effectively neutralize at least a percentage of the cyan that would appear in the final print.

But it would still seem that the effect of whatever amount of red light allowed to pass would be overpowered by the additional cyan you have added with your use of the blue and green filters. This begs the question, why go through all of this? Why not devise a red filter, which, too, can be modified in its effects, if necessary, via the partial transmission of the other primaries?

Am I confusing things too much?

dhroane · Sep 27, 2009

Mr Bill said:
Mr Bill said:

Now that I have tried to confuse everyone with the previous post, I see that I didn't read your questions completely.

In your example, I think that actually the result WILL BE reddish light, not white. Here's how: you want to add (not filter) lights, using magenta light and yellow light. In terms of red, green, and blue, the magenta light contains red and blue (it is missing green light). Similarly, the yellow light contains red and green (it is missing blue light). So combining these, we get (red and blue) plus (red and green), in other words, a double portion of red light, plus a single portion of green light and a single portion of blue light. so the overall result should have a distinctly reddish appearance. Of course, it is not a pure red light - it has been contaminated with some green and some blue, but reddish nonetheless.

I hope this is some help, rather than just confusing.

Click to expand...

This does clarify things. As a painter, I made the mistake of confusing the results that would be gained in a subtractive system of light, for what we were really talking about, which was what would occur in an ADDITIVE system of light.

Again, Mr. Bill...thank you.

Mr Bill · Sep 28, 2009

You would make a fantastic teacher (if you aren't one already)!

Ha ha, thanks for the compliment. No, I'm not a teacher; I'm just a lifelong photo enthusiast who never lost his interest in it. Although I work in some aspect of the photo industry (end user, not manufacurer), I'm here on APUG just as a hobbyist. By the way, you are quite welcome for any help I've given. And I appreciate your feedback, often it's not clear if anyone has benefitted from advice.

Though we are trying to "block" red with the use of green and blue filters, each one, nevertheless, still allows enough red light to pass through to effectively neutralize at least a percentage of the cyan that would appear in the final print.

It might seem so, but if you work through the numbers, like before, the result turns out to be effectively the same as having used a cyan filter. I say effectively, because the cyan does not throw away so much of the light, so less exposure correction is needed.

Here's roughly how it goes, comparing a 30cc cyan filter vs the combination of 30cc green plus 30cc blue: Note that the 30cc designation means these filters will pass virtually all of their own color, but block about 50% of everything else.

For 30 cc Cyan: this will block 50% of the red light, and pass 100% of both green and blue light. Result, compared to original 100% of each color: red = 50%, green = 100%, and blue = 100%. So the relative result is that green and blue have equal strength, and the red is reduced to 1/2 of theirs.

For 30 cc green plus 30cc blue: green will block 50% each of the red and blue light, passing 100% of green light. Applied to this light is a blue filter which will block 50% each of the red and green light, passing 100% of blue light. The combined result is: red = 50% x 50% = 25%, green = 100% x 50% = 50%, and blue = 50% x 100% = 50%. So the relative color result is the same as the previous example, where green and blue light have equal strength, and the red light is 1/2 of that strength. However, in this case, all of the actual light transmission percentages are only 1/2 of the previous example, consequently we have to double the exposure time.

Am I confusing things too much?

Nah! You have to develop some comfort, or confidence, that these ideas DO work properly, whether by doing it repetitively, or figuring it out, or perhaps someone you trust says that it is so. Otherwise, whenever something goes wrong, you'll always be wondering if the method is wrong, or if you simply made a mistake.

dhroane · Oct 4, 2009

For 30 cc green plus 30cc blue: green will block 50% each of the red and blue light, passing 100% of green light. Applied to this light is a blue filter which will block 50% each of the red and green light, passing 100% of blue light. The combined result is: red = 50% x 50% = 25%, green = 100% x 50% = 50%, and blue = 50% x 100% = 50%. So the relative color result is the same as the previous example, where green and blue light have equal strength, and the red light is 1/2 of that strength. However, in this case, all of the actual light transmission percentages are only 1/2 of the previous example, consequently we have to double the exposure time.

Ahhh, I didn't think of that. Thank you, Mr. Bill.

But, Mr. Bill, I STILL do not understand why, if a print is "too cyan" in hue, you would "correct it" by adding more cyan AT ALL whether that be with the use of a cyan filter or rather with a combination of blue and green filters? With either method, it appears you are still adding proportionately more cyan than red to a print that the artist has already deemed to be "too cyan."

I know that in your industry you do not handle red filters. Pardon my asking, but why this complication at all? It seems unnecessary given that you could just devise a red filter and be done with it. I don't mean to sound irreverent of a craft, which I, on the contrary, respect very much. But as an admitted layperson to photography, this decision not to use red filters sounds completely arbitrary and self-handicapping.

Thank you, Mr. Bill.

Q.G. · Oct 4, 2009

dhroane said:
But, Mr. Bill, I STILL do not understand why, if a print is "too cyan" in hue, you would "correct it" by adding more cyan AT ALL whether that be with the use of a cyan filter or rather with a combination of blue and green filters? With either method, it appears you are still adding proportionately more cyan than red to a print that the artist has already deemed to be "too cyan."

Think of it as a negative process the result of which, the output, will be the opposite of the input.
As in black and white, the print gets darker where the negative is lighter, the print will get more cyan where there is less of it in the negative.

So a print that is too cyan needs more cyan to become less cyan.
Just like a black and white print that is too light needs more light to become less light.

When going from positive to positive (scene to print, or slide to print), the opposite is true, and you simply need to remove what there is too much of.
So if the print is too cyan, you need to use a 'minus cyan' filter in front of the camera lens to get a correct print without having to use filters to correct for it at the printing stage.

mattmoy_2000 · Oct 6, 2009

If you're printing negatives and the print comes out too XXX, then by adding colour XXX filters (or -YYY and -ZZZ) in with the negative, you're removing XXX from the print, because you're adding it (proportionally) to the light.

In this example XXX, YYY and ZZZ are three arbitrary primary colours. Of course, if you're printing in Cibachrome, then if the print comes out too XXX then you add -XXX filters in.

nickandre · Oct 6, 2009

Cyan cancels Red, Magenta cancels Green, and Yellow Cancels blue.

Q.G. · Oct 6, 2009

And red cancels cyan, green cancels magenta, and blue cancels yellow.

dhroane · Oct 11, 2009

Thank you, QG and Mattmoy 2000!

filters - colour absorption question

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