IPhO 1984 · Theory problem 1

Refraction in a layered medium and the desert mirage

Download PDFIPhO 1984

a) Consider a plane-parallel transparent plate, where the refractive index, \(n\), varies with distance, \(z\), from the lower surface (see figure). Show that \(n_A\sin\alpha=n_B\sin\beta\). The notation is that of the figure.

Figure 1

b) Assume that you are standing in a large flat desert. At some distance you see what appears to be a water surface. When you approach the “water” is seems to move away such that the distance to the “water” is always constant. Explain the phenomenon.

c) Compute the temperature of the air close to the ground in b) assuming that your eyes are located 1.60 m above the ground and that the distance to the “water” is 250 m. The refractive index of the air at \(15^\circ\mathrm{C}\) and at normal air pressure (101.3 kPa) is 1.000276. The temperature of the air more than 1 m above the ground is assumed to be constant and equal to \(30^\circ\mathrm{C}\). The atmospheric pressure is assumed to be normal. The refractive index, \(n\), is such that \(n-1\) is proportional to the density of the air. Discuss the accuracy of your result.

Official English text of the problem.

Answers

  • a) The conserved quantity is \(n(z)\sin\theta(z)\), where \(\theta\) is the angle to the normal. Consequently, \[ \boxed{n_A\sin\alpha=n_B\sin\beta}. \]
  • b) This is an inferior mirage: hot, less dense air near the ground bends sky light upward into the observer’s eyes. The apparent water remains at a constant distance because the horizontal temperature distribution and the observer’s eye height remain unchanged.
  • c) Interpreting 250 m as the apparent distance to the mirage’s near edge, \[ \boxed{T_{\mathrm{ground\ air}}\simeq329\,\mathrm K\simeq56^\circ\mathrm C}. \] The small-angle approximation contributes less than \(0.001^\circ\mathrm C\); uncertainty in the apparent edge and departures from ideal atmospheric layering are more significant.

Solution

a)

Divide the plate into many thin horizontal layers, each having an approximately constant refractive index. Let \(n_i\) be the index of layer \(i\), and let \(\theta_i\) be the ray’s angle to the normal there.

Snell’s law at each interface gives \[ n_i\sin\theta_i=n_{i+1}\sin\theta_{i+1}. \] Since all interfaces are parallel, their normals have the same direction. Including the entrance and exit interfaces, \[ n_A\sin\alpha =n_1\sin\theta_1 =\cdots =n_N\sin\theta_N =n_B\sin\beta. \]

Taking the limit of infinitesimally thin layers establishes the result for a continuously varying index: \[ \boxed{n(z)\sin\theta(z)=p=\text{constant}}, \qquad \boxed{n_A\sin\alpha=n_B\sin\beta}. \]

b)

Sunlight heats the desert surface, which heats the air immediately above it. At approximately constant pressure, this warmer air has a lower density and hence a smaller refractive index. Thus the refractive index increases with height through the hot layer.

Consider a ray of sky light travelling downward at a shallow angle to the ground. From part a), \[ \sin\theta(z)=\frac{p}{n(z)}. \] As the ray enters air of smaller refractive index, its angle \(\theta\) to the vertical increases. If it reaches a height where \(n(z)=p\), it becomes horizontal. It then turns upward toward regions of larger refractive index and can reach the observer.

The observer traces the arriving light backward along a straight line and therefore sees sky light apparently coming from below the horizon. This resembles the reflection of the sky in water. In a smooth temperature gradient, the turning occurs through continuous refraction.

If the atmosphere is horizontally uniform, moving horizontally leaves the optical situation unchanged. At a fixed eye height, the angular position of the mirage’s near edge—and therefore its apparent distance—remains constant.

c)

Let \(n_e\) denote the refractive index at eye height and \(n_g\) the minimum index in the hottest air next to the ground. Assume geometrical optics, horizontal stratification, and a temperature that increases toward the ground.

First determine the refractive index as a function of temperature. For air treated as an ideal gas at constant pressure, \[ \rho\propto\frac1T, \] where \(T\) is the absolute temperature. Since \(n-1\propto\rho\), \[ n(T)-1=\frac{A}{T}, \] with \[ A=(1.000276-1)(273.15+15) =0.0795294\,\mathrm K. \]

The eyes are above the 1 m level, so their surrounding air is at \[ T_e=303.15\,\mathrm K. \] Consequently, \[ n_e=1+\frac{0.0795294}{303.15} =1.000262343. \]

Let \(\delta\) be the depression angle of the mirage’s near edge below the horizontal. Interpreting the stated distance as the apparent ground-plane distance, straight backward projection of the viewing direction gives \[ \tan\delta=\frac{h}{D} =\frac{1.60}{250}=0.00640, \] so \[ \delta=0.0063999\,\mathrm{rad}\simeq0.367^\circ. \]

The near edge is determined by the limiting ray that becomes horizontal in the minimum-index air. At eye height its angle to the vertical is \(90^\circ-\delta\); at its turning point that angle is \(90^\circ\). Part a) therefore gives \[ n_e\sin(90^\circ-\delta)=n_g\sin90^\circ, \] or \[ n_g=n_e\cos\delta =\frac{n_e}{\sqrt{1+(h/D)^2}}. \]

Rays with a smaller depression angle turn above the ground and bring sky light to the observer. Rays with a larger depression angle reach the ground.

Numerically, \[ n_g =\frac{1.000262343}{\sqrt{1+(1.60/250)^2}} =1.000241859. \] Thus \[ T_g=\frac{A}{n_g-1} =\frac{0.0795294}{0.000241859} =328.826\,\mathrm K, \] and hence \[ \boxed{T_g\simeq56^\circ\mathrm C}. \]

For the accuracy, the relevant refractive-index decrease is \[ n_e-n_g\simeq2.05\times10^{-5}. \] The angle is sufficiently small that replacing \[ \cos\delta=\frac1{\sqrt{1+(h/D)^2}} \] by \(1-\tfrac12(h/D)^2\) changes the calculated temperature by only about \(0.0009\,\mathrm K\).

The observational and atmospheric assumptions matter more. Differentiating the temperature formula, while holding the reference index and upper-air temperature fixed, gives at the stated values \[ dT_g\simeq55.7\,\mathrm K \left(\frac{dh}{h}-\frac{dD}{D}\right). \] Thus a 1% error in \(h/D\) changes the inferred temperature by approximately \(0.56^\circ\mathrm C\).

A blurred mirage edge, turbulence, uneven heating, and departures from horizontal layering can introduce larger errors. The detailed vertical profile is unnecessary for the ideal limiting-angle calculation, but a distinct edge associated with the hottest air is required. Since measurement uncertainties are not supplied, a numerical error bar is not justified.

The solution was checked against the official solution.