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B A silly question about why the sky is blue

  1. Jan 10, 2019 #1
    I know that 'the sky is blue because higher frequency blue wavelengths are scattered more than red'.

    What has always confused me is that when I imagine this, I imagine the blue wavelengths bouncing around between atoms in the atomosphete. Whilst red light interacts less, so travels through relatively unimpeded. To me that feels like the red light should be what reaches our eyes as these frequencies can make it through the atmosphere without being scattered.

    Obviously I'm wrong. But can someone explain in laymens terms why this picture is incorrect, and what I should be picturing? Thanks.
  2. jcsd
  3. Jan 10, 2019 #2


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    And you're right, and this is exactly what happens when you look at the source of light under high scattering conditions - i.e. at the sun at sunset or sunrise.

    Now, imagine somebody else standing some distance away and also looking at the sunset. They will also see a red sun. The blue light scattered from their line of sight towards the star will be bouncing around instead of hitting their eyes, and some of it will enter your eyes instead. You'll see it as coming from random directions that are not the direction towards the sun, i.e. the blue sky.
  4. Jan 11, 2019 at 7:33 AM #3


    Staff: Mentor

    Sometimes a picture is worth 1000 words.

  5. Jan 11, 2019 at 8:25 AM #4


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    As others have eloquently pointed out, you're right.

    Your question is about the colour of the sky, not the light source.The sky is all scattered light.
    If not for scattered light, the sky would be transparent, and we would see black.
  6. Jan 11, 2019 at 3:37 PM #5


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    I've never liked any of the explanations of this question.
    After some googling, and maths, I now I know why.


    From what I've gathered, only 7% of the light from the sun is scattered.

    wiki; "When the Sun is at the zenith in a cloudless sky, with 1361 W/m^2 above the atmosphere, direct sunlight is about 1050 W/m^2, and total insolation about 1120 W/m^2. This implies that under these conditions the diffuse radiation is only about 70 W/m2 out of the original 1361 W/m^2."

    Hence, 93% of the "direct" light isn't scattered.

    100% of the "not looking directly at the sun" sky is a result of that 7% scattered light.

    The sun only covers 0.002% of the sky.
    The scattered light comes from 99.998% of the sky.

    Which, from my last equation, yielded that 350,000% of the light is scattered.
    At which point, I gave up, and made lunch.

    ps. I now like the "It's because of 'Rayleigh scattering' " explanation quite a bit more.
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