Съдържание

WoPhO 2011, Теория — Задача 3. Theoretical 3: Mirage

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от WoPhO 2011

Edwin Soeryadjaya Problem. Theoretical 3: Mirage

Условие

The refractive index of the air varies with temperature. Cold air is denser than warm air and has therefore a greater refractive index. Thus a temperature gradient in the atmosphere is always associated with a gradient of the refractive index. Under certain conditions, this gradient of the refractive index could be strong enough so that light rays are bent to produce a displaced image of distant objects. This amazing phenomenon is called a mirage. Mirages can be categorized as ”inferior” and ”superior”. Inferior mirages can be seen on deserts and highways, and superior mirages occur over the sea.

To describe in detail the phenomenon of the mirage, we need to analyze the path of light rays in media with a refractive index gradient.

Two photographs side by side. Left: a lighthouse and buildings on a small island above a blue sea, stretched vertically by a superior mirage. Right: a desert plain with distant mountains, where the lower part of objects near the horizon is reflected as if in water (inferior mirage). — Figure 1: Left: A superior image. Right: An inferior image

1 The path of a light ray and the trajectory of a mass point

Media with a refractive index gradient can be treated as being formed from thin homogeneous layers with different refractive indices. The path of a light ray can be determined then by analyzing refractions of the light ray at interfaces between these thin homogeneous layers. But there exists a more convenient way for determining the path of a light ray in media with a refractive index gradient. Instead of analyzing the propagation of the light, one may study the motion of a mass point that moves along the path of the light ray, driven by a conservative force. The potential energy of the conservative force field depends on the distribution the refractive index. Once the potential energy is established, one may study the motion of the mass point by using well developed tools in classical mechanics, and find the trajectory of a mass point which is also the path of the light ray.

1(a) Refraction of a light ray at the interface between a medium with a refractive index $n_1$ and a medium with a refractive index $n_2$ is shown in Fig. 2a. The angles $i_1$ and $i_2$ obey the Snell’s law

n_1\sin i_1 = n_2\sin i_2 \qquad (1)

The path of a mass point in a conservative force field is illustrated in Fig. 2b. The speed of the mass point is $v_1$ in the region with the potential energy $E_{p1}$ , and $v_2$ in the region with the potential energy $E_{p2}$ . Find expressions for $v_1$ and $v_2$ so that the relation $n_1\sin i_1 = n_2\sin i_2$ hold also for the trajectory of the mass point. You may use an arbitrary constant expressions $v_0$ with the dimension of speed in your expressions. (0.2 points)

Two diagrams. Left (a): a horizontal interface between media n1 (above) and n2 (below) with a dashed normal; an incident ray at angle i1 to the normal is refracted to angle i2. Right (b): the same geometry for a mass point, with speed v1 in the region of potential energy Ep1 above and speed v2 in the region Ep2 below, angles i1 and i2 to the dashed normal. — Figure 2: (a) Left: Refraction of a light ray. (b) Right: Deviation of a mass point in a conservative force field

1(b) Assume that the mass of the mass point is $m$ , and the total energy of the mass point equals to zero. Find the potential energies $E_{p1}$ and $E_{p2}$ in $v_0$ , $n_1$ and $n_2$ . (0.2 points)

1(c) The trajectory of a mass point with a mass $m$ in a conservative force field is the same as the path of a light ray in a medium with a refractive index $n(\vec r)$ which is a function of the position. The total energy of this mass point is zero. Find expressions for the potential energy $E_p(\vec r)$ of the conservative force field and the speed of the mass point $v(\vec r)$ . (0.4 points)

1(d) To describe the motion of a mass point, one expresses the position of the mass point as a function of time: $\vec r(t)$ . To describe the trajectory of a mass point, one needs to express the position of the mass point as a function of the distance $s$ traveled by the mass point from the start point of the trajectory to the current position: $\vec r(s)$ . Derive a differential equation for the trajectory $\vec r(s)$ of a mass point with a mass $m$ and a total energy $E$ that moves in a conservative force field. (0.6 points)

Hint: in a conservative force field, $\vec r(t)$ satisfies Newtons second law:

m\frac{d^2\vec r}{dt^2} = -\vec\nabla E_p\,, \qquad (2)

where

\frac{ds}{dt} = v(\vec r)\,, \qquad (3)

\vec\nabla = \hat x\frac{\partial}{\partial x}+\hat y\frac{\partial}{\partial y}+\hat z\frac{\partial}{\partial z}\,. \qquad (4)

[0,6 т.]

1(e) Derive the light ray equation (a differential equation for the path of a light ray) by using the results obtained in (c) and (d). (0.6 points)

Hint: $\vec\nabla f^2(\vec r) = 2f(\vec r)\vec\nabla f(\vec r)$ [0,6 т.]

2 The inferior mirage

When an inferior mirage appears, an image of a distant object can be seen under the real object. A direct image of that object is seen because some of the light rays enter the eye in a straight line without being refracted. The double images seem to be that of the object and its upside-down reflection in water. For exhausted travelers in the desert it seems like that there is a lake of water in front of them.

An inferior mirage occurs when a strong positive gradient of refractive index is present near the ground. We use the following model to describe the variation of the refractive index of the air with elevation:

n^2(z) = \begin{cases} n_0^2+\alpha z & \text{for } h\ge z\ge 0\\ n_0^2+\alpha h & \text{for } z>h \end{cases} \qquad (5)

with $n_0^2-1\ll 1$ and $\alpha h\ll 1$ .

A man stands at the origin of z (vertical) and x (horizontal) axes; his eye is at height H. A light ray enters his eye from below the horizontal at angle theta. A dashed horizontal line marks height h above the ground. — Figure 3: The geometry for analyzing an inferior mirage

2(a) Find the path of a light ray (i.e. $z$ as a function of $x$ ) that enters in an observer’s eye at an angle $\theta$ (see Fig. 3). The height of the observers eye is $H$ . (1.9 points)

2(b) Due to the inferior mirage, the observer can see an inverted image of the upper part of a camel at a large distance. The parameters for the refractive index of the air are $\alpha = 3.0\times10^{-5}\ \mathrm{m^{-1}}$ , $h = 0.50\ \mathrm{m}$ . The height of the observers eye is $H = 1.5\ \mathrm{m}$ , and the height of the camel is $l = 2.2\ \mathrm{m}$ . Find the minimum values for the distance $D_m$ between the observer and the camel so that the observer still can see the inverted image of the upper part of the camel. You may use the approximation $n_0^2\approx 1$ . (0.7 points)

2(c) As a result of the light refraction in the region with refractive index gradient, the observer cannot see the lower part of the camels legs. Find the height of the lowest point ( $l_m$ ) on the camel at the distance $D_m$ that can still be seen by the observer. (0.7 points)

2(d) Find the distance $d$ between the observer and the imaginary lake of water. (0.3 points)

3 The superior mirage

When a superior mirage occurs, light rays that were originally directed above the line of sight of the observer will reach the observer’s eyes. Thus, an object ordinarily below the horizon will be apparently above the horizon.

A superior mirage occurs when a negative gradient of refractive index is present over a body of water or over large sheets of ice. We use the following model to describe the refractive index of the atmosphere (see Fig. 4):

n^2(r) = \begin{cases} n_0^2\left[1-\beta(r-r_0)\right] & \text{for } b\ge(r-r_0)\ge 0\\ n_0^2\left[1-\beta b\right] & \text{for } (r-r_0)>b \end{cases} \qquad (6)

where $r_0$ is the radius of the earth, $b\ll r_0$ and $\beta b\ll 1$ .

An arc of the Earth's surface (thick line) with centre below; a dashed arc at height b above it bounds the layer. A curved light ray leaves the surface on the right at angle gamma to the vertical (radius r0) and returns to the surface on the left. A radius r at polar angle phi from the dash-dotted vertical axis points to the ray; the word Earth is written inside. — Figure 4: The geometry for analyzing a superior mirage

3(a) Find the path of a light ray (i.e. $r$ as a function of $\phi$ ) within the range $b\ge(r-r_0)\ge 0$ . Use $\gamma$ , the angle between the light ray and the vertical direction at the sea level as the parameter of the path. (1.4 points)

Hint: The trajectory of a mass point with mass $m$ , angular momentum $L$ and total energy $E$ in a conservative force field with

E_p = -\frac{mA}{r}+E_0 \qquad (7)

r = \frac{\rho}{1-\varepsilon\cos\phi}\,, \qquad (8)

where

\rho = \frac{L^2}{m^2A}\,,\qquad \text{and}\qquad \varepsilon = \sqrt{1+\frac{2(E-E_0)L^2}{m^3A^2}}\,. \qquad (9)

We also have

\frac{1}{r}\approx\frac{1}{r_0}-\frac{r-r_0}{r_0^2}\qquad \text{if}\quad |r-r_0|\ll r_0. \qquad (10)

[1,4 т.]

3(b) Find the minimum value of $\beta$ ( $\beta_m$ ) at which the superior mirage occurs. Use the following values for $n_0$ and $r_0$ : $n_0\approx 1$ , $r_0 = 6.4\times10^{6}\ \mathrm{m}$ . (0.8 points)

3(c) Under a certain atmospheric condition, with $b = 100\ \mathrm{m}$ and $\beta = 6.0\times10^{-7}\ \mathrm{m^{-1}}$ , calculate the largest distance $D_M$ at which the surface of the sea can be seen by an observer at an altitude $y = 10\ \mathrm{m}$ ( $y$ is the altitude of the observer’s eye). (1.4 points)

Useful formula: $\cos\phi\approx 1-\frac{1}{2}\phi^2$ for $\phi\ll 1$ . [1,4 т.]

3(d) For comparison, calculate the largest distance $D'_M$ at which the surface of the sea can be seen by an observer at the same altitude $y = 10\ \mathrm{m}$ , when the refractive index of the air is constant. (0.3 points)

3(e) Calculate the angular difference $\Delta\vartheta$ between the apparent horizon when a superior mirage occurs as described in 3(c) and the apparent horizon in a normal day as described in 3(d), seen at the same altitude $y = 10\ \mathrm{m}$ . (0.5 points)

Отговори

1(a) $v_1 = n_1v_0,\quad v_2 = n_2v_0$
1(b) $E_{p1} = -\frac{1}{2}mn_1^2v_0^2,\quad E_{p2} = -\frac{1}{2}mn_2^2v_0^2$
1(c) $v(\vec r) = n(\vec r)v_0,\quad E_p(\vec r) = -\frac{1}{2}mn^2(\vec r)v_0^2$
1(d) $mv\frac{d}{ds}\left(v\frac{d\vec r}{ds}\right) = -\nabla E_p$

The official solution prints the accompanying speed as $v = \sqrt{\frac{E-E_p}{2m}}$ ; from energy conservation it should be $v = \sqrt{\frac{2(E-E_p)}{m}}$ . Kept as printed in the solution text.
1(e) $\frac{d}{ds}\left(n\frac{d\vec r}{ds}\right) = \nabla n$
2(a) Piecewise path: $z = H-x\tan\theta$ for $x<x_1$ ; $z = \frac{\alpha}{4n^2(h)\cos^2\theta}\left(x-\frac{1}{2}(x_1+x_2)\right)^2-\frac{n^2(h)\sin^2\theta}{\alpha}+h$ for $x_2\ge x\ge x_1$ ; $z = (x-x_2)\tan\theta+h$ for $x>x_2$ , with $x_1 = (H-h)\cot\theta$ and $x_2 = (H-h)\cot\theta+\frac{4n^2(h)\cos\theta\sin\theta}{\alpha}$ .
2(b) $D_m = \frac{H+l+2h}{\sqrt{\alpha h}} = 1.21\times10^3\ \mathrm{m}$
2(c) 0.475 m
2(d) $d = H\cot\theta_M\approx\frac{H}{\sqrt{\alpha h}}$ ≈ $386$ m
3(a) $r = \frac{\rho}{1-\varepsilon\cos\phi},\quad \rho = \frac{2}{\beta}\sin^2\gamma,\quad \varepsilon = \sqrt{1-4\beta^{-2}r_0^{-2}(r_0\beta-1)\sin^2\gamma}$
3(b) $\beta_m = \frac{2}{r_0} = 3.1\times10^{-7}\ \mathrm{m^{-1}}$
3(c) $D_M = 2\sqrt{\frac{r_0(b-y)}{\beta r_0-2}}+2\sqrt{\frac{r_0b}{\beta r_0-2}} = 7.3\times10^4\ \mathrm{m}$
3(d) $D'_M\approx\sqrt{2r_0y} = 1.1\times10^4\ \mathrm{m}$
3(e) 0.0069 rad

The official solution also gives $0.40^{\circ}$ .

Решение

Solution

1 The path of a light ray and the trajectory of a mass point

1.1 The component of the conservative force parallel to the interface is null, thus

v_1\sin i_1 = v_2\sin i_2

This relation becomes

n_1\sin i_1 = n_2\sin i_2

v_1 = n_1v_0 \quad \mathit{and} \quad v_2 = n_2v_0

(0.2 point)

1.2 $E_{p1} = 0-\frac{1}{2}mv_1^2 = -\frac{1}{2}mn_1^2v_0^2$ and $E_{p2} = 0-\frac{1}{2}mv_2^2 = -\frac{1}{2}mn_2^2v_0^2$ (0.2 point)

1.3 $v(\vec r) = n(\vec r)v_0$ (0.2 point)

and

$E_p(\vec r) = 0-\frac{1}{2}mv^2(\vec r) = -\frac{1}{2}mn^2(\vec r)v_0^2$ (0.2 point)

1.4 $\frac{d}{dt} = \frac{ds}{dt}\frac{d}{ds} = v\frac{d}{ds}$ , thus $\frac{d^2\vec r}{dt^2} = v\frac{d}{ds}\left(v\frac{d\vec r}{ds}\right)$ . (0.2 point)

The equation for the trajectory is

$mv\frac{d}{ds}\left(v\frac{d\vec r}{ds}\right) = -\nabla E_p$ with $v = \sqrt{\frac{E-E_p}{2m}}$ (0.4 point)

1.5 $mv\frac{d}{ds}\left(v\frac{d\vec r}{ds}\right) = -\nabla E_p$ becomes $mnv_0\frac{d}{ds}\left(nv_0\frac{d\vec r}{ds}\right) = -\nabla\left(-\frac{1}{2}mn^2v_0^2\right)$

(0.2 point)

But $-\nabla\left(-\frac{1}{2}mn^2v_0^2\right) = mv_0^2n\nabla n$ and $mnv_0\frac{d}{ds}\left(nv_0\frac{d\vec r}{ds}\right) = mnv_0^2\frac{d}{ds}\left(n\frac{d\vec r}{ds}\right)$

Therefore

\frac{d}{ds}\left(n\frac{d\vec r}{ds}\right) = \nabla n

(0.4 point)

2 The inferior mirage

2.1 The path of light ray is a straight line for $z>h.$

Path of a light ray for the inferior mirage: from the eye at height H the ray descends at angle theta, crosses the dashed line z = h at x1, bends along a parabola below h and rises again, crossing z = h at x2 at angle theta'. Axes z (vertical) and x (horizontal); heights H and h marked with arrows.

Thus

z = H-x\tan\theta \quad \mathit{for} \quad x<x_1

(0.2 point)

with

x_1 = (H-h)\cot\theta

(0.1 point)

For $x_2\ge x\ge x_1$ , the corresponding potential energy is

E_p(z) = -\frac{1}{2}mv_0^2n_0^2-\frac{1}{2}mv_0^2\alpha z

(0.2 point)

The equation of motion of the corresponding mass point is then

\begin{cases} m\dfrac{d^2z}{dt^2} = \dfrac{1}{2}mv_0^2\alpha\\ m\dfrac{d^2x}{dt^2} = 0 \end{cases}

(0.2 point)

With the initial condition

\left.\frac{dx}{dt}\right|_{t=0} = n(h)v_0\cos\theta \quad \mathit{and} \quad \left.\frac{dz}{dt}\right|_{t=0} = -n(h)v_0\sin\theta

(0.2 point)

So we have

\begin{cases} z = \dfrac{1}{4}v_0^2\alpha t^2-n(h)v_0t\sin\theta+h\\ x = n(h)v_0t\cos\theta+x_1 \end{cases}

(0.2 point)

By eliminating $t$ in above equations, we obtain

z = \frac{\alpha}{4n^2(h)\cos^2\theta}(x-x_1)^2-x\tan\theta+H

(0.2 point)

This expression can also be written as

z = \frac{\alpha}{4n^2(h)\cos^2\theta}\left(x-(H-h)\cot\theta-\frac{2n^2(h)\cos\theta\sin\theta}{\alpha}\right)^2-\frac{n^2(h)\sin^2\theta}{\alpha}+h

This path may reach in the region with $x>x_2$ if

-\frac{n^2(h)\sin^2\theta}{\alpha}+h\ge 0

\sin\theta\le\frac{\sqrt{\alpha h}}{n(h)}

(0.2 point)

The value of $x_2$ can determined by the following condition

h = \frac{\alpha}{4n^2(h)\cos^2\theta}\left(x_2-(H-h)\cot\theta-\frac{2n^2(h)\cos\theta\sin\theta}{\alpha}\right)^2-\frac{n^2(h)\sin^2\theta}{\alpha}+h

We find

x_2 = (H-h)\cot\theta+\frac{4n^2(h)\cos\theta\sin\theta}{\alpha}

(0.1 point)

We also have

\tan\theta' = \left.\frac{dz}{dx}\right|_{x=x_2} = \tan\theta

(0.1 point)

Then we have for $x>x_2$

z = (x-x_2)\tan\theta+h

(0.2 point)

We can express the path of light ray as

z = \begin{cases} H-x\tan\theta & \mathit{for}\quad x<x_1\\ \dfrac{\alpha}{4n^2(h)\cos^2\theta}\left(x-\dfrac{1}{2}(x_1+x_2)\right)^2-\dfrac{n^2(h)\sin^2\theta}{\alpha}+h & \mathit{for}\quad x_2\ge x\ge x_1\\ (x-x_2)\tan\theta+h & \mathit{for}\quad x>x_2 \end{cases}

with

x_1 = (H-h)\cot\theta

and

x_2 = (H-h)\cot\theta+\frac{4n^2(h)\cos\theta\sin\theta}{\alpha}

2.2 Let $x_c$ be the coordinate of the camel. To see the inverted image of the upper part of a camel we must have

l = (x_c-x_2)\tan\theta+h

(0.1 point)

and

\sin\theta\le\frac{\sqrt{\alpha h}}{n(h)}

(0.1 point)

Substituting the expression for $x_2$ we find

x_c = (H+l-2h)\cot\theta+\frac{4n^2(h)\cos\theta\sin\theta}{\alpha}

(0.2 point)

For small $\theta$ , $x_c$ decreases with $\theta$ , so $x_c$ is minim for

\theta = \theta_M = \arcsin\left(\frac{\sqrt{\alpha h}}{n(h)}\right) = 3.87\times10^{-3}\,\mathrm{rad}

Thus

\begin{aligned} D_m &= (H+l-2h)\cot\theta_M+\frac{4n^2(h)\cos\theta_M\sin\theta_M}{\alpha}\\ &\approx\frac{H+l-2h}{\sqrt{\alpha h}}+4\frac{\sqrt{\alpha h}}{\alpha}\\ &= \frac{H+l+2h}{\sqrt{\alpha h}}\\ &= 1.21\times10^3\,\mathrm{m} \end{aligned}

(0.3 point)

2.3 Let $z_c$ be the height of a point on the camel. If the observer can see this point, then we must have

z_c = \frac{\alpha}{4n^2(h)\cos^2\theta}(D_m-x_1)^2-D_m\tan\theta+H

(0.1 point)

with $\theta<\theta_M$ . As $\theta<\theta_M\ll 1,$ so $\cos\theta\approx 1.$ We also have $n(h)\approx 1$ and

\tan\theta = \frac{H-h}{x_1}

(0.1 point)

Let $y = \frac{D_m-x_1}{D_m}$ , we have then

z_c = \frac{\alpha}{4}D_m^2y^2-\frac{H-h}{1-y}+H

To find to minim of $z_c$ we need to calculate the derivative $\frac{dz_c}{dy}$

\frac{dz_c}{dy} = \frac{\alpha}{2}D_m^2y-\frac{H-h}{(1-y)^2}

(0.1 point)

The condition for minim $z_c$ is then

y = \frac{2(H-h)}{D_m^2\alpha(1-y)^2} = \frac{0.0455}{(1-y)^2}

This equation can be solved numerically. Let $y_1 = 0.0455$ and

y_{n+1} = \frac{0.0455}{(1-y_n)^2}

We found

$n$	1	2	3	4	5
$y_n$	0.0455	0.0499	0.0504	0.0505	0.0505

Thus $z_c$ reaches minim at $y_m = 0.0505$

l_m = \frac{\alpha}{4}D_m^2y_m^2-\frac{H-h}{1-y_m}+H = 0.475\,\mathrm{m}

(0.4 point)

2.4 If a refracted light ray enters the observer’s eye at the angle $\theta$ , the observer will consider the ray being reflected from a water surface at a distance equal to $H\cot\theta$ . The minim value of this distance is the distance between the observer and the “lake”. Thus

d = H\cot\theta_M\approx\frac{H}{\sqrt{\alpha h}} = 386\,\mathrm{m}

(0.3 point)

3 The superior mirage

3.1 For $b\ge(r-r_0)\ge 0$

\begin{aligned} E_p(r) &= -\frac{1}{2}mv_0^2n_0^2+\frac{1}{2}mv_0^2n_0^2\beta(r-r_0)\\ &\approx -\frac{mA}{r}+E_0 \end{aligned}

(0.1 point)

with

A = \frac{1}{2}v_0^2n_0^2r_0^2\beta

(0.2 point)

and

E_0 = \frac{mA}{r_0}-\frac{1}{2}mv_0^2n_0^2

(0.2 point)

The angular momentum equals to

L = mv_0n_0r_0\sin\gamma

(0.2 point)

And the total energy $E = 0$ . Thus the path is given by

r = \frac{\rho}{1-\varepsilon\cos\phi}

(0.1 point)

with

\rho = \frac{2}{\beta}\sin^2\gamma

(0.3 point)

and

\varepsilon = \sqrt{1-4\beta^{-2}r_0^{-2}(r_0\beta-1)\sin^2\gamma}

Superior mirage geometry: the Earth's surface (thick arc) with a dashed arc at height b; a curved light ray leaves the surface at angle gamma to the radius r0 and reaches maximum radius r_m on the vertical axis; a radius r at angle phi from the vertical; the word Earth inside.

(0.3 point)

3.2 Let $r_m$ be the maxim value of $r$ . When the superior mirage occurs, we have $r_m\le r_0+b$ . The angular momentum is constant

L = mv_0n_0r_0\sin\gamma = mv_0n(r_m)r_m

(0.2 point)

Thus

n_0r_0\ge n_0r_0\sin\gamma = n(r_m)r_m

That implies

\frac{r_0^2}{r_m^2}\ge 1-\beta(r_m-r_0)

But

\frac{r_0^2}{r_m^2}\approx 1-\frac{2(r_m-r_0)}{r_0}

So we must have

\beta\ge\frac{2}{r_0}

(0.4 point)

Therefore

\beta_m = \frac{2}{r_0} = 3.1\times10^{-7}\,\mathrm{m^{-1}}

(0.2 point)

3.3 The maxim distance $D_M$ is given by the maxim values of $\phi_1$ and $\phi_2$

D_M = r_0(\phi_{1M}+\phi_{2M})

(0.1 point)

Ray from the sea surface (angle gamma to radius r0) rising to maximum radius r_m on the vertical axis and descending to the observer's eye at altitude y; angles phi1 and phi2 at the Earth's centre on either side of the vertical axis.

According to expression for the path of light ray, we have

r_0 = \frac{\rho}{1-\varepsilon\cos\phi_2}

(0.1 point)

r_0+y = \frac{\rho}{1-\varepsilon\cos\phi_1}

(0.1 point)

and

r_m = \frac{\rho}{1-\varepsilon}

(0.1 point)

From above relations, we obtain

\cos\phi_1 = \frac{1}{\varepsilon}\left(1-\frac{\rho}{r_0+y}\right),\quad \cos\phi_2 = \frac{1}{\varepsilon}\left(1-\frac{\rho}{r_0}\right) \quad \mathit{and} \quad 1 = \frac{1}{\varepsilon}\left(1-\frac{\rho}{r_m}\right)

By eliminating $\varepsilon$ we find

\begin{aligned} \cos\phi_2 &= \left(1-\frac{\rho}{r_0}\right)\left(1-\frac{\rho}{r_m}\right)^{-1}\\ &\approx 1-\left(\frac{\rho}{r_0}-\frac{\rho}{r_m}\right)\left(1-\frac{\rho}{r_0}\right)^{-1}\\ &\approx 1-(r_m-r_0)\frac{\rho}{r_0^2}\left(1-\frac{\rho}{r_0}\right)^{-1} \end{aligned}

Thus

\phi_2\approx\sqrt{\frac{2\rho(r_m-r_0)}{r_0(r_0-\rho)}} = \sqrt{\frac{4(r_m-r_0)\sin^2\gamma}{r_0(\beta r_0-2\sin^2\gamma)}}\approx 2\sqrt{\frac{r_m-r_0}{r_0(\beta r_0-2)}}

and

\phi_{2M}\approx 2\sqrt{\frac{b}{r_0(\beta r_0-2)}}

(0.4 point)

Similarly

\phi_1\approx 2\sqrt{\frac{r_m-r_0-y}{r_0(\beta r_0-2)}}

and

\phi_{1M}\approx 2\sqrt{\frac{b-y}{r_0(\beta r_0-2)}}

(0.4 point)

Therefore

\begin{aligned} D_M &= r_0(\phi_{1M}+\phi_{2M})\\ &= 2\sqrt{\frac{r_0(b-y)}{\beta r_0-2}}+2\sqrt{\frac{r_0b}{\beta r_0-2}}\\ &= 7.3\times10^4\,\mathrm{m} \end{aligned}

Constant refractive index case: a straight line of length D'_M from the observer's eye at altitude y is tangent to the Earth's surface (thick arc); radius r0 to the tangent point.

(0.2 point)

3.4

(D'_M)^2+r_0^2 = (r_0+y)^2

Thus

D'_M = \sqrt{(r_0+y)^2-r_0^2}\approx\sqrt{2r_0y} = 1.1\times10^4\,\mathrm{m}

(0.3 point)

Apparent horizon angles: from the observer's eye at altitude y above the Earth's surface, the curved ray of height b and the straight tangent D'_M; angles theta1 and theta2 between the lines at the eye; angle theta2 also at the Earth's centre between the radii; radius r0 marked.

3.5 we have

\Delta\vartheta = \vartheta_1+\vartheta_2

where

\vartheta_2 = \frac{D'_M}{r_0} = \sqrt{\frac{2y}{r_0}}

(0.1 point)

and

(r_0+y)n(r_0+y)\cos\vartheta_1 = (r_0+b)n(r_0+b)

Thus

\cos\vartheta_1 = 1-\frac{1}{2}(\beta r_0-2)\frac{b-y}{r_0}

\vartheta_1 = \sqrt{(\beta r_0-2)\frac{b-y}{r_0}}

(0.3 point)

Therefore

\begin{aligned} \Delta\vartheta &= \sqrt{(\beta r_0-2)\frac{b-y}{r_0}}+\sqrt{\frac{2y}{r_0}}\\ &= 6.9\times10^{-3}\,\mathrm{rad}\\ &= 0.40^{\circ} \end{aligned}

(0.1 point)

Оригинал в Архива: WoPhO_2011_Q13.pdf · официални решения: WoPhO_2011_S13.pdf