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Spectrum flip dependent on background

  1. Feb 17, 2007 #1

    This is probably pretty basic but I've asked a few times in various internet forums over the years and never gotten an answer.

    If you print a black bar on a white packground and a white bar on a black background and view each bar through a prism, held at the same angle, the spectrum flips. I don't have a prism on hand so forget the color order but this is basically what you see at the transitions from black to white.

    black bar/white background

    white bar/black background

    I'm a sculptor by training so have only a rudimentary knowlege of optics but for all appearances, it would seem that red and blue light follow the exact same path through the prism (are refracted the same) depending on whether the background is dark or light. Can anyone explain the cause of this spectrum flip?


  2. jcsd
  3. Feb 18, 2007 #2
    Cute. Does it help you to think of the white bar like the gap between two black bars?
  4. Feb 18, 2007 #3

    Claude Bile

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    The black bar/white background case will act like a two white bars, with a black bar in between them. What you are therefore seeing is most likely the blue end of the spectrum from one white bar and the red end from the other white bar sandwiched together.

  5. Feb 21, 2007 #4
    Thanks for the response Claude.

    In order to double check my old observations, I dug out my prism from storage and looked again, this time with a sheet of paper half black and half white. This eliminates the "sandwich effect" you mentioned above while still showing spectrum flip.

    Holding the prism so the apex points to the right

    For white left/black right you see
    white cyan blue violet black

    For black left/white right you see
    black red orange yellow white

    I'm not claiming anything earthshaking here, I just find it curious how the simple act of rotating a piece of paper 180 degrees can so effectively shift color from the blue end of the spectrum to the red end. For white/black the red end of the spectrum is completely absent. For black/white the blue end is absent.

    View angle doesn't seem to matter as there is no detectable color change with change in viewing angle with either orientation. So it's not just a matter of the blue spectrum or red spectrum not being focused into your eye. They seem to disappear completely.

    The only things I can think of to explain this effect are:
    1. The white is overpowering the hidden end of the spectrum
    2. There is enough light reflecting off the white surface that is exactly i/4 wavelength out of phase to cancel the missing end of the spectrum.

    Option 1 seems plausible except (especialy noticeable in the case of white cyan blue violet black) the line between white and cyan is fairly sharp. I would expect to see yellow, orange, red fade in intensity but that's not what you see. You see white, bam, cyan.

    Option 2 seems plausible except there wouldn't seem to be any logical reason why this 1/4 out of phase light would selectively restrict one end of the spectrum.

    Last edited: Feb 21, 2007
  6. Feb 21, 2007 #5
    Rotation? Phase? Will it help if I draw a picture?
    Code (Text):

      black bar/white background           black bar/white background
    red-----------------------blue      red-----------------------blue

  7. Feb 21, 2007 #6
    Thanks. But there's no need for sarcasm. If you look at my first post, you''ll see an almost identical diagram, so yes, I am aware of this relationship. The question was not what was being observed, but why.

    One half of the spectrum disappears and that disappearance has some logical cause. Angle in the prism, phase cancellation, something That's what I'm trying to find out.

  8. Feb 22, 2007 #7
    Your prism is deflecting all the blue light to the left, and the red light not so much to the left (and colours in between stay that way). If your light source was just a pin point, you'd get a pretty spectrum (since the colours would no longer overlap). Since you're using an extended source, you should instead get overlapping images: a blue image of the source far to the left, overlapping a green image not-quite-so-far, etc, overlapping a red image to the right. However, those bits in the middle (where multiple colours overlap) appear white, so your result is a picture of the original source (still white) except with a blue edge on the left of the source's light areas, and red on every right-edge.
  9. Jul 12, 2007 #8
    | I just find it curious how the simple act of rotating a piece
    | of paper 180 degrees can so effectively shift color from
    | the blue end of the spectrum to the red end.

    what you're describing is the the basis for why goethe rejected newton's theory in 1810 -- goethe believed that the spectrum was a compound phenomena which could be explained by the simpler phenomena of colour arising at light-dark boundaries.

    from his experiments, he arrived at different conclusions than newton. he thought that colour arises from the interaction of light and dark -- he (like you) observed that when a light area borders a dark area -- blue violet results; and when a dark area borders a light area -- yellow-red results.

    goethe observed that with a prism, yellow-red colours arose along the top light-dark boundary, and a blue-violet edge arose along the bottom dark-light boundary, and that the spectrum with green in the middle only arose when when the beam of light was made small enough for these coloured edges to overlap.

    it doesn't fit in with newton's theory (who held that colour was comprised of particles/corpuscles moving at different speeds), or with young and fresnel (who held that colour was a result of different wavelengths). but the fact can nevertheless be demonstrated by experiment.

    for more information, check out the goethe's 'theory of colours' (MIT Press).
    Last edited: Jul 12, 2007
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