On a hot May afternoon a bus runs along a straight highway through the Thar desert near Jaisalmer. Far ahead, the road seems to be covered with a sheet of shining water. The few cars on it appear to float on their own reflections. As the bus drives on, the water keeps its distance and never gets wet. The road is bone dry.
A ray through many layers
The last lesson sent a ray through a stack of parallel slabs. At each boundary Snell's law gave \(\mu_1\sin r_1 = \mu_2\sin r_2\). Because the boundaries are parallel, the angle of refraction at one is the angle of incidence at the next. So the product \(\mu\sin\theta\) is carried unchanged from layer to layer. Here \(\theta\) is measured from the normal to the layers, which is the vertical for flat horizontal layers.
When a ray passes through any number of flat, parallel layers, \(\mu\sin\theta\) keeps the same value in every layer. Here \(\theta\) is the angle between the ray and the normal to the layers. The angle in a layer depends only on that layer's \(\mu\) and on the constant.
Going down into denser layers, \(\mu\) grows, so \(\sin\theta\) must shrink: the ray turns towards the vertical. A ray entering from air at \(60°\) to the vertical has \(K = \sin 60° = 0.866\). In layers of \(\mu = \sqrt{1.5}\) and then \(\sqrt3\), \(\sin\theta = 0.866/\mu\) gives \(45°\) and then \(30°\).
Where the ray turns back
Now send a ray upward into thinner and thinner layers. Now \(\mu\) falls, so \(\sin\theta\) must grow, and the ray bends away from the vertical. It flattens out. When it reaches a layer where \(\mu\) equals \(K\), it would need \(\sin\theta = 1\): the ray runs horizontally. It cannot enter any layer with \(\mu < K\), because that would need \(\sin\theta > 1\). So it is totally reflected there and comes back down.
In the figure above, a start at \(40°\) gives \(K = 1.5\sin 40° = 0.96\), less than every \(\mu\) in the stack, and the ray leaves through the top. A start at \(45°\) gives \(K = 1.06\). The ray crosses the layer of \(1.10\) nearly flat, at \(74.6°\), and turns back under the layer of \(1.00\). The steeper the start, the lower the turning point.
A ray starts in a layer of \(\mu_0 = \sqrt2\) at \(45°\) to the vertical and moves upward into layers of smaller and smaller \(\mu\). In which layer does it become horizontal? A second medium has \(\mu(y) = \mu_0\sqrt{1 - y/a}\) for \(0 \le y < a\), and a ray starts at \(y = 0\) at \(30°\) to the vertical. At what height does it turn back?
First medium: \(K = \sqrt2\sin 45° = 1\). The ray is horizontal where \(\mu = K = 1\). It cannot go higher, so it turns back down there.
Second medium: \(K = \mu_0\sin 30° = \mu_0/2\). Horizontal where \(\mu_0\sqrt{1 - y/a} = \mu_0/2\), so \(1 - y/a = \tfrac14\) and \(y = \tfrac34 a\).
Above that height the ray cannot go; it curves over and comes back down.
A ray enters a stack of horizontal layers from air at \(60°\) to the vertical. What angle, in degrees, does it make with the vertical in a layer of \(\mu = \sqrt3\) far below, whatever the layers in between?
The mirage
Now the road in Rajasthan. The sun heats the tar, and the tar heats the air just above it. Hot air is less dense than cool air, and its refractive index is a little lower. So just above the road, \(\mu\) grows with height: the air is a stack of layers whose index falls as you go down.
Follow the ray in Fig 4.1. Coming down, it passes into layers of lower \(\mu\), so it bends away from the vertical. The upward ray in the figure of layers did the same. It flattens out just above the road and turns up again. The traveller's eye traces it straight back and sees a patch of bright sky lying on the road ahead. Bright sky seen below the horizon looks like water, and hot air rising through it makes it shimmer.
How low must the ray come? Take \(\mu = 1.0000\) at the road and \(1.0003\) at the eye. A ray that leaves the eye at an angle \(\alpha\) below the horizontal makes the angle \(90° - \alpha\) with the vertical. It just grazes the road if \(1.0003\sin(90° - \alpha) = 1.0000\sin 90°\), that is \(\cos\alpha = 1/1.0003\). For a small angle \(\cos\alpha \approx 1 - \alpha^2/2\), so \(\alpha^2 \approx 0.0006\) and \(\alpha \approx 0.024\,\text{rad}\), about \(1.4°\).
Near a hot road \(\mu = 1.0000\) at the ground and \(1.0003\) at the eye, \(1.6\,\text{m}\) up. Only rays that leave the eye flatter than the grazing ray turn back and show the sky. Find the grazing angle \(\alpha\), and estimate how far ahead the "water" begins.
Measure angles from the vertical. At the eye the ray is at \(90° - \alpha\); at the road it grazes, at \(90°\). Apply \(\mu\sin\theta = K\).
\(1.0003\cos\alpha = 1.0000\), so \(\cos\alpha = 0.99970\). With \(\cos\alpha \approx 1 - \alpha^2/2\): \(\alpha^2/2 = 0.00030\).
\(\alpha^2 = 0.0006\), so \(\alpha = 0.0245\,\text{rad} = 1.40°\). A line from the eye at this angle meets the road \(1.6 / 0.0245 \approx 65\,\text{m}\) ahead.
The mirage begins about \(65\,\text{m}\) ahead and lies beyond. Rays steeper than \(1.4°\) reach the road itself, so nearby you see only dry tar. As you drive, the boundary moves with you.
Over a cold sea the air near the water is cooler and denser than the air above, so \(\mu\) falls with height. Rays from a distant ship curve downward and reach the eye from above, and the ship seems to hang above the horizon. Sailors call this looming. It is the mirage turned upside down.
Why is the "water" of a road mirage always seen far ahead and never at your feet?
Explain a mirage on a hot road in three sentences. Say what happens to the refractive index of the air, how a ray from the sky bends, and why the road looks wet.
The air near the hot road is hotter and less dense than the air above, so its refractive index is lower. A ray from the sky coming down at a small angle passes into ever lower refractive index, bends away from the vertical and is turned back upward just above the road. The eye traces that ray straight back to the road, so it sees an image of the bright sky there, which looks like a pool of water.
A smooth change bends the ray smoothly
Real air does not come in steps. Its index changes smoothly with height, like a stack of infinitely thin layers. In such a graded medium, the ray does not bend at corners. It curves all the time, and \(\mu\sin\theta\) stays constant along the whole curve. Which way does it curve?
\[ \mu(y)\sin\theta(y) = K \]
Why. the invariant of the layers holds at every height \(y\), with \(\theta\) measured from the \(y\) axis.
\[ \dfrac{d\mu}{dy}\sin\theta + \mu\cos\theta\,\dfrac{d\theta}{dy} = 0 \]
Why. differentiate with respect to \(y\) by the product rule; the right side is a constant.
\[ \dfrac{d\theta}{dy} = -\dfrac{\sin\theta}{\mu\cos\theta}\,\dfrac{d\mu}{dy} = -\dfrac{\tan\theta}{\mu}\,\dfrac{d\mu}{dy} \]
Read it. solve for \(d\theta/dy\).
When \(\mu\) increases with \(y\), \(d\theta/dy\) is negative. As the ray climbs, its angle with the \(y\) axis shrinks, so it turns towards the direction in which \(\mu\) grows. A ray always bends towards the region of higher refractive index. In the mirage the index is higher above, so the ray curves upward.
This part goes beyond the NCERT and is asked in JEE Advanced. It uses the invariant \(\mu\sin\theta = K\) and one integration. The idea is plain; only the algebra is new.
Let a medium fill the region \(y > 0\), with \(\mu(y) = \mu_0\sqrt{1 + y/a}\), where \(a\) is a length. A ray enters at the origin travelling along the \(x\) axis, at \(90°\) to the \(y\) axis. Along the path, \(\mu\sin\theta = \mu_0\sin 90° = \mu_0\). As the ray climbs into higher \(\mu\), \(\sin\theta\) must fall, so the ray turns steadily upward.
\[ \sin\theta = \dfrac{\mu_0}{\mu} = \dfrac{1}{\sqrt{1 + y/a}} \]
Why. the invariant \(\mu\sin\theta = \mu_0\), with the given \(\mu(y)\).
\[ \cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{\dfrac{y/a}{1 + y/a}}, \qquad \tan\theta = \sqrt{\dfrac{a}{y}} \]
Why. \(\tan\theta = \sin\theta/\cos\theta\); the factor \(\sqrt{1 + y/a}\) cancels.
\[ \dfrac{dx}{dy} = \tan\theta = \sqrt{\dfrac{a}{y}} \]
Why. \(\theta\) is the angle with the \(y\) axis, so along the ray \(dx\) is \(\tan\theta\) times \(dy\).
\[ x = \int_0^y \sqrt{a}\,y^{-1/2}\,dy = 2\sqrt{ay} \]
Why. integrate from the origin, where \(x = 0\) and \(y = 0\).
\[ y = \dfrac{x^2}{4a} \]
Read it. square both sides: \(x^2 = 4ay\). The path is a parabola.
With \(a = 1\,\text{m}\), the ray in Fig 4.2 reaches \(y = 16/4 = 4\,\text{m}\) at \(x = 4\,\text{m}\). At \(x = 2\,\text{m}\), its slope is \(dy/dx = x/2a = 1\), so it makes \(45°\) with the \(x\) axis. The check from the other formula agrees: there \(y = 1\,\text{m}\), and \(\tan\theta = \sqrt{a/y} = 1\) gives \(45°\) with the \(y\) axis too.
In the medium \(\mu(y) = \mu_0\sqrt{1 + y/a}\) with \(a = 0.5\,\text{m}\), a ray enters at the origin along the \(x\) axis. How high, in metres, is it when \(x = 2\,\text{m}\)?
A ray travels in a medium whose refractive index changes smoothly with height \(y\). Which statements are true?
The Earth's air is densest near the ground and thins with height, so its refractive index falls upward. Light from the Sun, just below the horizon, enters the air at a slant and bends down, following the curve of the Earth. The Sun is seen about half a degree higher than it really is. The Earth turns half a degree in about two minutes. So we see the Sun rise about two minutes early, and set about two minutes late. The same bending squashes the setting Sun into an oval.
- Measuring \(\theta\) from the layers instead of from their normal.The invariant is \(\mu\sin\theta\) with \(\theta\) measured from the normal to the layers, the vertical for horizontal layers.
- Thinking that the mirage is a reflection from a layer of water, or from the road.The road is dry. The ray is bent gradually by hot air of lower \(\mu\) and turned back upward; the "water" is an image of the sky.
- Expecting a ray in a graded medium to bend towards the lower refractive index.A ray always bends towards higher \(\mu\): \(d\theta/dy\) has the opposite sign to \(d\mu/dy\).
- Forgetting that a ray turns back where \(\mu\) falls to \(K\).Where \(\mu = K\) the ray is horizontal. It cannot enter a region with \(\mu < K\).