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How much more does blue light bounce around than red light?

Research question: How does the inverse fourth-power law of Rayleigh scattering quantitatively explain why blue light is scattered more than red light?

The records show that Rayleigh scattering intensity scales with the inverse fourth power of wavelength, so halving the wavelength raises re-emission intensity by a factor of 16 2. Blue light's wavelength is about 65% that of red light, which gives roughly six times more scattering for blue than red 4. The solar spectrum varies by only about 50% across the visible range, so it cannot offset this wavelength dependence 6. The records do not derive the law itself; they only attribute the blue sky to scattering by particles much smaller than the wavelength 18. The newest dated source in these facts is from June 2006, and it notes that the λ−4 law does not always hold in some waveguides 3.

Checked 2026-10-14. Only documented and supported facts are used.
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Sources

[1] supported
Strutt explained the blue color of the sky by the scattering of light by small particles, which is called Rayleigh scattering. The document does not give the inverse fourth-power law.
"Strutt erklärte die blaue Farbe des Himmels anhand der Streuung des Lichtes an kleinen Partikeln ( Rayleigh-Streuung )."
John Strutt, 3. Baron Rayleigh
[2] supported
Because Rayleigh scattering intensity scales with the inverse fourth power of wavelength, halving the wavelength increases the re-emission intensity by a factor of 16.
"as light gets halved in wavelength, its re-emission intensity ("scattering," if you will) increases by a factor of 16."
Why Is the Sky Blue?
[3] supported
The document states that the λ−4 wavelength scaling of Rayleigh-scattering loss is a standard law that does not always hold, since in small-core waveguides the wavelength dependence can substantially deviate from it.
"the wavelength dependence of losses induced by Rayleigh scattering can substantially deviate from the λ −4 scaling law"
Non λ−4 wavelength dependence of rayleigh-scattering loss in waveguides (2006)
[4] supported
Blue light's wavelength is about 65% that of red light, so blue light is scattered roughly six times more than red light.
"In fact, the wavelength of blue light is about 65% that of red light, so blue light is indeed scattered about six times more than red light"
Why Is the Sky Blue?
[5] supported
In the limiting case of an ultrathin waveguiding wire surrounded by air, the lower bound of Rayleigh-scattering losses asymptotically scales as exp(−κλ²) rather than as λ−4.
"asymptotically scales as exp(−κλ 2 ) with radiation wavelength λ, with κ being a wavelength-independent constant"
Non λ−4 wavelength dependence of rayleigh-scattering loss in waveguides (2006)
[6] supported
The intensity of the solar spectrum varies by only about 50% across the visible region, so it cannot offset the wavelength dependence of scattering.
"The intensity of the solar spectrum, as one can see from Figure 1, varies only by about 50% through the visible region of the spectrum."
Why Is the Sky Blue?
[7] supported
Violet light actually scatters more than blue light, but the sky looks blue because the eye is less sensitive to violet and sunlight contains less violet than blue.
"Actually, violet light scatters more than blue light."
Why Is the Sky Blue?
[8] supported
When particle size is much smaller than the light's wavelength, Mie scattering reduces to Rayleigh scattering, which is why the sky is blue.
"Thus, in the limit of a << λ, Mie scattering becomes Rayleigh scattering, and the sky is blue!"
Why Is the Sky Blue?