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Why is white paper white and not black

  1. May 1, 2013 #1
    Why is white paper "white" and not black

    I understand this may seem a silly question but I really would like to know.

    As far as I understand it (may be wrong) additive colours (visible wavelengths) when combined equally produce white. Subtractive colours, when all the colours are equally combined produce black.

    Additive colours are produced by a light source. Subtractive colours are produced by a non-light source (e.g. reflection, filter, lens, etc).

    So, if I look at a piece of white paper I am seeing all the colours (with none absorbed)combined so why does it appear white and not black, as all the colours are being reflected equally.
     
    Last edited: May 1, 2013
  2. jcsd
  3. May 1, 2013 #2
    I'll put it another way.

    If direct source wavelengths between (very approx. 400nm – 750nm) are reflected with some omissions, why is the result produced as an additive, “plus” the omissions and not as a subtractive series.

    Or – why does a non complete series of direct source wavelengths produce the same result as non direct source series.

    Or – why does white paper look white when (as reflected, and so subtractive, wavelengths) is should appear black.

    It was a genuine question
     
  4. May 1, 2013 #3

    Borek

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    When you look at the paper light that you see has its source on the paper surface and carries information about this surface properties (plus some information about properties of the original light source, but to make things easier let's assume it was a just a Sun).
     
  5. May 1, 2013 #4

    jedishrfu

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    The subtraction occurs because the reflecting surface absorbs selected frequencies of lights and as a result they are not part of the reflected light (hence they are subtracted from it). So white paper looks white because it reflects all the frequencies of light that you can see.

    Here's some info about how varying amounts of ink affect the reflected color:

    http://en.wikipedia.org/wiki/Subtractive_colors
     
    Last edited: May 1, 2013
  6. May 1, 2013 #5
    So the light is reflecting from the paper and as all the wavelength are reflected equally and it looks white??

    But if this is the case and I were to add to red ink to the paper this would absorb more wavelengths outside of (very approx) 680nm to 720nm so this would be a subtraction from the spectrum and the effect would be a red view.

    The paper’s colour can’t be both additive and subtractive at the same time.

    Assuming that the paper isn’t a light source it should (as the light is being reflected from the paper) be black. But of course it isn’t. I don’t understand why.
     
  7. May 1, 2013 #6

    Bandersnatch

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    Think of it this way:
    "White" is the whole spectrum of visible light. All wavelengths between ~360-720(or something like that) nanometres must reach your eye to produce white colour response in your brain.

    When you look at a source of light, it has to emit all these wavelengths to be percieved as white. You can look at two separate sources producing one-half of the visible spectrum each, and since the total spectrum adds up to the whole visible spectrum, you see white.
    You can have as many sources as you want, and as long as the sum of them nets you all of the visible spectrum, you see white. Hence this is an additive effect.
    With RGB displays you've got three such sources. If only two of them are emitting, say red an blue, then you see all parts of the spectrum but green, netting you the colour magenta.

    When light, let's say white to make it easier, gets reflected from a surface, parts of the spectrum may get absorbed, depending on the properties of the surface.
    If a part gets absorbed, then whatever remains reaches your eye, and you no longer see the whole spectrum.
    To see the colour magenta, as in the previous example, you need the surface that absorbs the green wavelengths out of white incident light.

    What we call "white surfaces" reflect all wavelengths equally, so you see whatever colour the source is.

    Conversely, "black" absorbs all wavlenghts, so it doesn't matter what colour the source is.

    You can have a source that shines red an green, and a surface that absorbs green, and as a result you'd see red only(the surface subtracts green from the incident spectrum).

    If you've got a pure green source, and a surface that absorbs just the green wavelengths, you'll see black.

    And so on.
     
  8. May 1, 2013 #7
    I think I have it.
    The paper is white as all the light is reflecting equally, so the colour I view is in reality additive - being that I am seeing light.

    If I add red ink this will “subtract” some wavelengths and the light I see will be red - so the cause is subtractive but the effect is additive.

    So, if I view the red light through a coloured lens the result would be subtractive (and so get darker – or more toward black)

    Is that right ???
     
  9. May 1, 2013 #8

    Bandersnatch

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    Pretty much, yeah.
     
  10. May 1, 2013 #9
    You’re a genius.

    I have been wondering about this all day.

    Thank you
     
  11. May 1, 2013 #10

    Dale

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    The papers color is subtractive. It subtracts nothing. If you shine white light on it then nothing is subtracted and you see white. If you shine green light on it then nothing is subtracted and you see green. Etc. If you shine no light on it then it subtracts nothing and you see nothing.
     
  12. May 1, 2013 #11

    sophiecentaur

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    If you cover a sheet of paper with tiny dots of two different colours, which do not overlap, and view from a distance or through a diffuser you will get additive colour mixing when illuminated by white light. If you use three 'good' primary colours for the spots, you will not get a full brightness 'white' of course because the three pigments absorb a lot of light and the reflectivity of the three has to be just right (as well as the hue / chrominance / colour). But you can get a good grey, which a camera with an accurate analysis (matching the eye's analysis) will end up giving you equal values of RG and B signals. Right chrominance but wrong luminance.

    The white paper can be regarded as a load of fully reflective spots - giving white, natch.
    I guess a system of colour printing could be devised which uses this process but afaik, it is not possible to get coloured dots small enough for good resolution but which will not overlap (bleed) one another and cause subtractive mixing at the edges. Hence, ink jet printers accept this and do it subtractively, with mixtures of inks on each spot.
     
  13. May 1, 2013 #12
    Right let’s see if I have this.

    Light is always additive. When all the colours are added you get white
    Non light source objects (with a very few exception) are always subtractive. When all the colours are subtracted you get black.

    So if I shine a light onto a white object all the different colours will reflect back and merge into white.

    If I shine white light (e.g. sunlight) on a red object all the colours, other than red, will be absorbed (subtracted) and the red section of the white light will be reflect back as red light.

    As light is always additive if I shine a white light onto a red object and also a green object and (somehow) focus the two reflections together, the result will be yellow. In this case the object’s surfaces are subtractive – but of course the light reflected is additive.

    If I shine a white light onto a yellow object the light reflected is yellow. If I view this through magenta coloured lens the object will appear red. In this case the pigment in the lens is subtractive.

    I hope that’s correct.
     
  14. May 1, 2013 #13

    Bandersnatch

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    It is. Good job.
     
  15. May 1, 2013 #14

    sophiecentaur

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    This is ok as far as it goes but I'm not sure that you are making enough distinction between the wavelengths and amplitudes of light in the spectrum of the light from a source and human perception of the 'colour', which is very subjective and involves just three very broad analysis curves. As I have said many times, the eye is not a spectrometer. There are an infinity of combinations of spectra that will produce the sensation of a particular colour. Actual spectral colours (single wavelengths) are very rarely seen, in fact (even a rainbow is mostly white light from the rest of the sky). Most colours we see consist of a complicated spectrum. Laser light is one massive exception to that statement but we did not evolve to look at laser light!
    The 'colour triangle' is a very crude description of reality because it makes no distinction between analysis and synthesis of colours. The spectrum of light from a source consist of a distribution of wavelengths (not 'colours) and it is interpreted by the three sets of sensors in the eye (plus the brain) as a simple 'colour'.
    Talking in terms of "adding colours" is really not a precise enough description. The (RGB) primaries used in additive systems like colour TV displays are each specially chosen to have a narrow band of wavelengths and to be as bright as possible. This is to allow the widest possible gamut of accurately displayable perceived colours and brightnesses in the projected picture.

    The so called primaries that are used in subtractive mixing are far less easily specified. They basically correspond to -R (Cyan), -G (Magenta) and -B (Yellow). Look at the colour of inks that your colour inkjet printer uses. Manufacturers all have their own 'best' combinations. Subtractive mixing is such a minefield that most good quality printing uses more than just those three basic inks / dyes and even has to use special pigments for producing the 'special' colours that everyone knows - like the Coca Cola Red. Artists in oils have to use a whole range of tubes of colours because mixing two other pigments to get a 'strong' colour will also make it very dark. Cinema film is really really bad colourimetrically - but we accept it and even get to like it.
     
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