Atmospheric Scattering (Rayleigh) Calculator

Rayleigh scattering is the scattering of electromagnetic radiation, including visible light, by particles much smaller than its wavelength, such as individual nitrogen and oxygen molecules in the atmosphere. Its defining feature is that scattering intensity increases sharply as wavelength decreases, following an inverse fourth-power relationship.

Sunlight contains all visible wavelengths, but Rayleigh scattering redirects shorter, blue wavelengths far more strongly than longer, red wavelengths, because intensity scales as 1/wavelength^4. That scattered blue light reaches your eyes from every direction across the sky, which is why the sky appears blue rather than the sun's actual white color.

Scattering intensity I is proportional to 1/lambda^4, where lambda is the wavelength of light. This calculator normalizes that relationship to a 550nm green reference wavelength: I_relative(lambda) = (550/lambda)^4, so a value greater than 1 means that wavelength scatters more strongly than 550nm green light.

Comparing a typical blue wavelength (450nm) to a typical red wavelength (650nm) using the 1/lambda^4 law gives a ratio of (650/450)^4, approximately 4.4 times more scattering for blue than for red, which is the core reason the sky looks blue rather than red.

At sunset or sunrise, sunlight travels through a much longer, more oblique path through the atmosphere to reach an observer. Rayleigh scattering removes most of the blue light along that long path before it arrives, leaving mostly the less-scattered red and orange wavelengths, which is why the sun and sky near the horizon take on warm colors at those times.

Violet light does scatter even more strongly than blue under the 1/lambda^4 law. The sky still looks blue rather than violet mainly because the sun's spectrum contains less violet light than blue to begin with, and the human eye's cone cells are considerably less sensitive to violet wavelengths than to blue.

No. Rayleigh scattering specifically applies to particles much smaller than the wavelength of light, such as individual air molecules. Larger particles like water droplets, dust, and aerosols scatter light through Mie scattering instead, which is far less wavelength-dependent, which is why clouds and fog appear white or gray rather than blue.

Mars has a much thinner atmosphere composed mostly of carbon dioxide, and its atmosphere carries significant suspended dust. The dust particles scatter light through Mie-like scattering rather than pure Rayleigh scattering, and combined with the thin atmosphere, this produces Mars's characteristic butterscotch or reddish sky color rather than a blue one.

No, they are different mechanisms. Ocean water appears blue mainly because water molecules themselves weakly absorb red light more than blue light as light travels through the water, combined with some scattering, rather than the atmospheric Rayleigh scattering mechanism responsible for sky color.

At high cruising altitude, an aircraft is above much of the denser lower atmosphere, so there are fewer air molecules along the line of sight to scatter sunlight, especially when looking away from the sun. With less scattered light reaching the eye, the sky appears darker, deeper blue, or even near-black at very high altitudes, compared to the paler blue seen from the ground.

The underlying 1/lambda^4 relationship holds for any electromagnetic wavelength, so the calculator will still compute a mathematically valid relative intensity outside 380 to 750nm. However, wavelengths outside that range fall outside human visible perception, so a note appears reminding you the result describes physical scattering strength, not a color you could actually see.