Physics · Optics · Year 1

Scattering of light

Why is the sky blue and the sunset red?

The situation

In Tashkent on a summer day the sky is a clear light blue, and in the evening, when the Sun sets, the horizon turns bright red. The clouds over the mountains are always pure white. In a teahouse, tea poured from one teapot looks yellowish in a small bowl and reddish-brown in a glass. Air, cloud and tea all receive the same white sunlight. So why is one light blue, one white and one red?

Topic

Scattering and absorption of light. Why the sky is blue, why the sunset is red, why tea darkens in colour

Key ideas

Rayleigh's law
Mie scattering
Beer–Lambert law

Formulas

I_{s c a t} \propto \frac{I_{0}}{λ^{4}}

Rayleigh's law (particle ≪ λ): blue (450 nm) light is scattered ${(700 / 450)}^{4} \approx 5.9$ times more strongly than red (700 nm).

I = I_{0} e^{- α x}

Bouguer's law: light is attenuated exponentially in a medium; $α$ is the sum of the absorption and scattering coefficients.

A = \lg \frac{I_{0}}{I} = ε c l

The Beer–Lambert law: the absorbance of a solution is proportional to the concentration $c$ and the layer thickness $l$ .

I = I_{0} e^{- τ X}, X \approx \frac{1}{\sin h}

Attenuation in the atmosphere: $τ$ is the optical depth, $X$ the air mass (1 with the Sun overhead, ≈ 38 at the horizon).

P = \frac{\sin^{2} γ}{1 + \cos^{2} γ}

The degree of polarisation of Rayleigh-scattered light; $γ$ is the scattering angle, fully polarised at $γ = 90 °$ .

Theory

How does scattering occur?

The electric field of the light makes the electrons in a molecule oscillate — the molecule becomes a tiny dipole antenna and radiates light in all directions at that frequency. The power of dipole radiation is proportional to $ω^{4}$ , that is, to $1 / λ^{4}$ — this is Rayleigh's law. In a perfectly uniform medium the waves from neighbouring molecules would cancel each other out sideways; in air, however, the density constantly fluctuates because of thermal motion (Smoluchowski–Einstein theory) — the brightness of the sky comes from exactly these random irregularities.

Why is the sky light blue and the sunset red?

The blue part of sunlight is scattered ≈ 5.9 times more strongly than the red, so the sky away from the Sun glows with scattered blue light too (violet is weaker in sunlight and the eye is less sensitive to it). When the Sun approaches the horizon, its light travels 10–38 times farther through the atmosphere: almost all the blue is scattered out of the direct beam — the Sun and the clouds it lights turn red. In the scene the sky is drawn in each direction by exactly this calculation; after sunset the Earth's shadow covers the sky from the east.

Mie scattering: cloud, fog, milk

If the particle size is close to $λ$ or larger ( $x = 2 π r / λ ≳ 1$ ), Rayleigh's law breaks down: scattering depends only weakly on wavelength and is directed mainly forwards (Mie scattering, 1908). Cloud droplets (≈ 5–10 µm) scatter all colours equally — a cloud is white, fog is grey. In optically turbid media (milky water, smoke, dust) small particles scatter the blue and turn the transmitted light red — the “sunset” in the aquarium. Model: single scattering in the tank; the particles have a size distribution, the scattering efficiency is joined smoothly between the Rayleigh and van de Hulst formulas; the colour of the light is drawn more saturated so that it is easy to see.

Polarisation of scattered light

An oscillating dipole does not radiate along its own axis. That is why light scattered at 90° to the Sun is almost completely polarised: $P = \sin^{2} γ / (1 + \cos^{2} γ)$ . In the real sky $P$ reaches at most ≈ 70–80 %: multiple scattering, aerosols and light reflected from the ground reduce the polarisation (on the map the yellow region fades when dust is increased). Bees find their way by seeing the polarisation of the sky, and photographers darken the sky with a polarising filter; in astronomy polarimetry gives information about particles and magnetic fields. Other laws of polarisation — the laws of Malus and Brewster, and birefringence — are in the “Polarisation of light” and “Crystal optics” experiments.

The Beer–Lambert law and colorimetry

In a thin layer $d x$ the light decreases in proportion to its own intensity: $d I = - α I d x$ , so $I = I_{0} e^{- α x}$ . In a solution the absorption coefficient is proportional to the concentration, so the absorbance is $A = \lg (I_{0} / I) = ε c l$ . $ε$ depends on frequency — this is the absorption spectrum: tea absorbs violet and blue light strongly and red weakly, so a thick layer looks reddish-brown. A colorimeter has its filter chosen in the region the solution absorbs strongly; the most accurate measurement is at $A \approx 0.2$ – $0.8$ . A graph of $A (c)$ is drawn from standards, and the concentration of an unknown sample is read from it. Absorption and refractive index are linked: the Kramers–Kronig relations express $n (ω)$ through the absorption at all frequencies — anomalous dispersion is observed near an absorption line. Model: for tea the $ε (λ)$ values are approximate (educational).

Observing through the atmosphere

Light from a star or the Sun is attenuated in the atmosphere by the law $I = I_{0} e^{- τ X}$ ; astronomers write this in stellar magnitudes: $m = m_{0} + k X$ . If you measure at different air masses and plot $\ln I$ against $X$ , you get a straight line (the Langley method): its slope is $- τ$ , and its value at $X = 0$ is $\ln I_{0}$ , that is, the intensity outside the atmosphere. Because blue light is attenuated more, stars near the horizon look reddened — the $B - V$ colour index in the UBV system increases. Air turbulence causes the twinkling of stars and blurring of the image; that is why observatories are built in places with clear, calm air (astroclimate). At the Maidanak Observatory in Uzbekistan the average width of a star image is ≈ 0.7″ — one of the best figures in the world. When measuring faint objects, the signal-to-noise ratio is proportional to the square root of the number of photons $N$ : $S / N \propto \sqrt{N}$ .

Did you know?

The blue of the sky was explained in 1871 by the English physicist John Strutt (later Lord Rayleigh): particles much smaller than the wavelength scatter light in inverse proportion to the fourth power of the wavelength. In 1899 he showed that the air molecules themselves are enough for this.
On Mars the daytime sky is yellowish-brown and the sunset looks light blue — NASA rovers have photographed this. The reason is that the fine dust particles in the Martian air scatter light differently.

Inside the experiment

The experiment starts from a real-life situation: you predict the result, change the controls, run the experiment and draw a conclusion. Inside are missions, theory, a quiz and a guided tour in which the camera moves and explains the key points.

More experiments in this section

Interference and diffraction — Grade 11
Polarisation of light — Grade 11
Laser physics and holography — Grade 11
Luminous flux and illuminance — Grade 11
Infrared, UV and X-rays — Grade 11
Crystal optics — Year 1
Interferometers and LIGO — Year 1
Resolving power — Year 1
Nonlinear optics: the green laser — Year 2