Mirages: Light Bent by Hot and Cold Air

Set the air temperature, how hot or cold the ground is, how the air warms or cools above it, your eye height, and a striped pole some distance away. The page traces lines of sight through the air and shows them side on, and shows what your eye would see: the pole, the sky, the road, and the sky lying in the road.

Worked example — Mirages: Light Bent by Hot and Cold Air

Air temperature at the ground and at your eye	50.0 °C and 30.0 °C
The mirage ray is longer by	4.80 mm

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Mirages: Light Bent by Hot and Cold Air

A mirage is an image formed by light that has curved through air whose refractive index changes with height. Along a ray in horizontal layers n cos θ stays the same (θ above level; on the curved Earth, with m = n + z/R in place of n), so a line of sight leaving the eye at depression α is turned back up before the ground whenever m(H) cos α > m(0). Where m is least at the ground, as over a road hotter than the air, the grazing angle α_c = arccos(m(0)/m(H)) ≈ √(2Δm), with Δm = m(H) − m(0), separates lines of sight that meet the ground from those that turn back up before it, and the eye, extrapolating straight, puts the start of the mirage at H / tan α_c.

What each symbol means

n: refractive index of the air; n − 1 = 7.86×10⁻⁴ p/(273 + t) for dry air, p in kPa and t in °C
α_c: the grazing line of sight's angle below level
m: the index with the Earth's curvature folded in, n + z/R
k: how fast a level ray curves, as a fraction of the Earth's curvature
γ: how fast the air warms with height, in K per metre; negative when it cools

Assumes: Dry air, horizontally uniform layers, and NIST's simple formula for the refractive index, which NIST states for 0 to 35 °C; above that, at a road surface, the page uses its dry-air term, which is the density of an ideal gas, without that guarantee. A real hot surface layer is turbulent, which is why a real mirage shimmers.

The puddle has an edge, and it drives with you 🖖

Press Hot road, from a car. Air at 30 °C over a road at 50 °C, and the refractive index at your eye is 1.62×10⁻⁵ higher than at the road surface: hot air is thinner, and light travels a little faster in it. A line of sight that leaves your eye shallow enough is bent back up by that difference before it reaches the road, and comes back from the sky. The steepest one that makes it grazes the road at 0.3284° below level; anything steeper meets the tarmac. Your eye assumes light goes straight, so it places that grazing direction on the road 209.3 m ahead, and everything beyond looks like sky lying on the road: water. Andrew T. Young's pages on mirages, describing a photograph of a desert mirage, say the same: the mirage begins at a sharp boundary where the line of sight meets the ground at the critical angle. Nothing in that sentence depends on where you are, so the edge drives with you. Press Play and watch the pole come towards you: its inverted image shrinks and is gone once the pole is closer than 235.3 m, where the grazing ray really turns, while the edge stays 209.3 m ahead.

The mirage's light takes the slower path, and not even a locally quickest one 🖖

Two rays from the eye reach the point 1.00 m up the pole: the direct one, nearly straight, and the mirage ray, which dips to 35.7 mm above the road and comes back up. Both are the rays Fermat's principle allows, paths where a small change makes no first-order change in the travel time. But the mirage ray is 4.80 mm longer, and although it spends part of its way in faster air it arrives 12.57 ps later. The results table goes one step further. Bend the mirage ray 1.2 cm up or down in the middle, keeping its ends fixed, and its optical path gets shorter both ways, by 5.17 µm and by 5.53 µm, while a two-humped bend makes it longer: it sits at a saddle, not at a minimum. The direct ray gets longer whichever way it is bent. The inverted image is carried by light whose path is stationary without being least, and the same thing happens to the action of a moving body, as the least-action page shows.

Warm air over a cold sea bends light round the Earth 🖖

Normally the air thins upwards mostly because the pressure falls, and a level ray curves downwards at about a fifth of the Earth's curvature: the Frozen sea, ordinary air preset gives 0.219, so from 14 m the horizon is 15.11 km away instead of the 13.36 km of straight lines. Warm the air with height and the ray bends harder. In air at −20 °C, warming by 0.0921 K per metre makes a level ray curve exactly as the Earth does; the page needs 0.1129 at 0 °C and 0.1295 at 15 °C, in line with the 0.114 °C per metre Young quotes from Alfred Wegener. Press Frozen sea, warm air above: at 0.15 K per metre the ray curves 1.435 times as fast as the Earth, and a 30 m cliff 50 km away, which a moment ago lay below the horizon, is seen upright a few arcminutes above level. Barents' crew, wintering on Novaya Zemlya, saw the Sun on 24 January 1597 while its centre was, by van der Werf and colleagues' reconstruction, still 5° 26′ below the horizon, two weeks before it was due.

Problem solved in full

How hot must the road be for the puddle to start 100 m ahead? 4 steps
A driver's eye is 1.2 m above the road and the air is at 30 °C. The page puts the puddle 209.3 m ahead on a 50 °C road. How hot would the road surface have to be for it to start only 100 m ahead? Press Hot road, from a car to check the first step.

Open this problem in the tool
1. From the page, at the opening state:
  
  $α_{c} = 0.328 4^{\circ}, \frac{H}{tan α _{c}} = \frac{1 . 2}{tan 0 . 328 4 ^{\circ}} = 209.3 m$
2. Turn it round. For the edge to be 100 m ahead, the grazing line of sight must leave the eye at
  
  $α = arctan \frac{1 . 2}{100} = 0.687 5^{\circ}$
3. That line of sight grazes the road, so the refractive index at the road surface must satisfy m(0) = m(H) cos α. At the eye, in air at 30 °C, m(H) = 1.000263003:
  
  $m (0) = m (H) cos α = 1.000263003 \times 0.999928008 \Rightarrow n_{0} - 1 = 1.9099 \times 1 0^{- 4}$
4. Solve NIST's formula for the temperature that gives that refractivity, with the pressure at the road:
  
  $t_{0} = \frac{7 . 86 \times 1 0 ^{- 4} \times 101 . 325}{1 . 9099 \times 1 0 ^{- 4}} - 273 = 144.0^{\circ} C$
Answer

About 144 °C, and building a road at 144 °C into the page's air puts the edge back at 100.0 m. That is hotter than boiling water, and far outside the temperature range NIST vouches for its formula. The edge moves in slowly: 290.1 m for a road 10 °C above the air, 209.3 m for 20 °C, 152.9 m for 40 °C. The puddle stays far ahead because Δn is so small: 1.62×10⁻⁵ for a road 20 °C above the air.

$t_{0} \approx 144^{\circ} C, n_{H} - n_{0} = 7.18 \times 1 0^{- 5} (50^{\circ} C : 1.62 \times 1 0^{- 5})$

Learning path

Least time, least action

References (3)

The refractive index of air: the Ciddor and Edlén equations and, in Appendix B, the shop-floor formula this page uses with the humidity term at zero, with the range of temperatures it is stated for: J. A. Stone and J. H. Zimmerman, "Index of Refraction of Air", Engineering Metrology Toolbox, National Institute of Standards and Technology, Appendix B "Simple Shop-floor Formula for Refractive Index of Air".
How mirages form: the hot-road mirage starting at the line of sight that meets the ground at the critical angle; why rays curve; the gradient at which a level ray follows the Earth; and ducts: A. T. Young, "An Introduction to Mirages" and "Calculating Ray Bending", web pages in the series "Mirages and Green Flashes", San Diego State University.
The Novaya Zemlya effect on Barents' third voyage: the Sun seen on 24 January 1597 while its centre was 5° 26′ below the horizon, reconstructed by ray tracing through a temperature inversion: S. Y. van der Werf, G. P. Können, W. H. Lehn, F. Steenhuisen and W. P. S. Davidson, "Gerrit de Veer's true and perfect description of the Novaya Zemlya effect, 24–27 January 1597", Applied Optics 42(3), 379–389 (2003).

By the 42Math team · Reviewed against published sources. Corrections welcome.