Physical Analysis

The problem describes the phenomenon of a mirage, where the refractive index of air varies with temperature near the ground. The ray of light from a distant object (like the sky) travels through layers of varying optical density, bends, and eventually grazes the ground layer before curving upwards to the observer. This creates the illusion of a puddle.

We are given the refractive index dependency on temperature:

$$ \mu = 1 + \frac{a}{T} $$

The condition for the observer to see the mirage is that the light ray must become horizontal (total internal reflection or grazing incidence limit) at the lowest point of its trajectory, which is the ground layer.

Application of Snell’s Law in Gradient Medium

Let the temperature at the observer’s height $h$ be $T_h = T_0$.
Let the temperature at the ground layer (height $y=0$) be $T_g = T_0 + \Delta T$.

The refractive indices at these points are:

$$ \mu_h = 1 + \frac{a}{T_0} $$ $$ \mu_g = 1 + \frac{a}{T_0 + \Delta T} $$

For a stratified atmosphere where $\mu$ changes only with height $y$, the generalized Snell’s law states that $\mu(y) \cos \theta(y) = \text{constant}$, where $\theta$ is the angle the ray makes with the horizontal.

At the turning point (the ground), the ray travels horizontally, so $\theta_g = 0^\circ$. At the observer’s eye, let the angle of depression be $\alpha$. Thus:

$$ \mu_h \cos \alpha = \mu_g \cos(0^\circ) $$ $$ \mu_h \cos \alpha = \mu_g $$

Approximating the Angle

Since the changes in refractive index are very small (because $a$ is small and temperatures are around 300 K), the angle $\alpha$ is very small. We can use the approximation $\cos \alpha \approx 1 – \frac{\alpha^2}{2}$.

$$ \mu_h \left( 1 – \frac{\alpha^2}{2} \right) = \mu_g $$ $$ 1 – \frac{\alpha^2}{2} = \frac{\mu_g}{\mu_h} $$ $$ \frac{\alpha^2}{2} = 1 – \frac{\mu_g}{\mu_h} = \frac{\mu_h – \mu_g}{\mu_h} $$

Substituting the expressions for $\mu$:

$$ \mu_h – \mu_g = \left( 1 + \frac{a}{T_0} \right) – \left( 1 + \frac{a}{T_0 + \Delta T} \right) $$ $$ \mu_h – \mu_g = a \left( \frac{1}{T_0} – \frac{1}{T_0 + \Delta T} \right) = a \left( \frac{T_0 + \Delta T – T_0}{T_0(T_0 + \Delta T)} \right) = \frac{a \Delta T}{T_0(T_0 + \Delta T)} $$

Since $\mu_h \approx 1$ in the denominator term, we get:

$$ \frac{\alpha^2}{2} \approx \frac{a \Delta T}{T_0(T_0 + \Delta T)} $$ $$ \alpha = \sqrt{\frac{2a \Delta T}{T_0(T_0 + \Delta T)}} $$

Calculating the Horizontal Distance

For small angles of incidence, the horizontal distance $x$ at which the ray strikes the ground (or appears to originate from) is related to the observer’s height $h$ and the angle of depression $\alpha$ by the geometry of the tangent:

$$ \alpha \approx \tan \alpha = \frac{h}{x} \implies x = \frac{h}{\alpha} $$

Substituting the expression for $\alpha$ derived above:

$$ x = h \sqrt{\frac{T_0(T_0 + \Delta T)}{2a \Delta T}} $$

Numerical Calculation

Substituting the given values:
$h = 1.75 \text{ m}$
$T_0 = 300 \text{ K}$
$\Delta T = 20 \text{ K}$
$a = 1/6 \text{ K}$ (at normal pressure)

$$ x \approx 1.75 \sqrt{\frac{300(300 + 20)}{2(1/6)(20)}} $$ $$ x \approx 1.75 \sqrt{\frac{300 \times 320}{(20/3)}} $$ $$ x \approx 1.75 \sqrt{\frac{96000 \times 3}{20}} $$ $$ x \approx 1.75 \sqrt{14400} $$ $$ x \approx 1.75 \times 120 $$ $$ x = 210 \text{ m} $$