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APhO 2004, Теория — Задача 2. Optical fiber

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Theoretical Competition-Problem No. 2. Optical fiber

Условие

Perspective sketch of the fiber: an end face drawn as two concentric ellipses with centre O, the axis z to the right and the x axis upward; dashed lines mark the core boundary at distance a from the axis, solid lines the cladding.

Longitudinal section of the fiber in the xOz plane: air with n0=1 to the left of the end face, a ray arriving at O at incident angle θi to the z axis and refracting into the core of index n1; the cladding of index n2 lies above and below the core; axes x (up) and z (right).

An optical fiber consists of a cylindrical core of radius $a$ , made of a transparent material with refraction index varying gradually from the value $n=n_1$ on the axis to $n = n_2$ (with $1<n_2<n_1$ ) at a distance $a$ from the axis, according to the formula

n = n(x) = n_1\sqrt{1-\alpha^2x^2}

where $x$ is the distance from the core axis and $\alpha$ is a constant. The core is surrounded by a cladding made of a material with constant refraction index $n_2$ . Outside the fiber is air, of refractive index $n_0$ .

Let O $z$ be the axis of the fiber, with O - the center of the fiber end.

Given $n_0=1.000$ ; $n_1=1.500$ ; $n_2= 1.460$ , $a = 25\ \mu\mathrm{m}$ .

A monochromatic light ray enters the fiber at point O under an incident angle $\theta_\mathrm{i}$ , the incident plane being the plane $x$ O $z$ .

1.a. Show that at each point on the trajectory of the light in the fiber, the refractive index $n$ and the angle $\theta$ between the light ray and the O $z$ axis satisfy the relationship $n\cos\theta = C$ where $C$ is a constant. Find the expression for $C$ in terms of $n_1$ and $\theta_\mathrm{i}$ . [1.0 points]

1.b. Use the result found in 1.a. and the trigonometric relation $\cos\theta = \left(1+\tan^2\theta\right)^{-\frac{1}{2}}$ , where $\tan\theta = \dfrac{dx}{dz} = x'$ is the slope of the tangent to the trajectory at point ( $x$ , $z$ ), derive an equation for $x'$ . Find the full expression for $\alpha$ in terms of $n_1$ , $n_2$ and $a$ . By differentiating the two sides of this equation versus $z$ , find the equation for the second derivative $x''$ . [1.0 points]

1.c. Find the expression of $x$ as a function of $z$ , that is $x = f(z)$ , which satisfies the above equation. This is the equation of the trajectory of light in the fiber. [1.0 points]

1.d. Sketch one full period of the trajectories of the light rays entering the fiber under two different incident angles $\theta_\mathrm{i}$ . [1.0 points]

Light propagates in the optical fiber.

2.a. Find the maximum incident angle $\theta_\mathrm{iM}$ , under which the light ray still can propagate inside the core of the fiber. [1.5 points]

2.b. Determine the expression for coordinate $z$ of the crossing points of a light ray with O $z$ axis for $\theta_\mathrm{i} \neq 0$ . [1.5 points]

The light is used to transmit signals in the form of very short light pulses (of negligible pulse width).

3.a. Determine the time $\tau$ it takes the light to travel from point O to the first crossing point with O $z$ for incident angle $\theta_\mathrm{i} \neq 0$ and $\theta_\mathrm{i} \le \theta_\mathrm{iM}$ .

The ratio of the coordinate $z$ of the first crossing point and $\tau$ is called the propagation speed of the light signal along the fiber. Assume that this speed varies monotonously with $\theta_i$ .

Find this speed (called $v_\mathrm{M}$ ) for $\theta_\mathrm{i} = \theta_\mathrm{iM}$ .

Find also the propagation speed (called $v_0$ ) of the light along the axis Oz.

Compare the two speeds. [3.25 points]

3.b. The light beam bearing the signals is a converging beam entering the fiber at O under different incident angles $\theta_i$ with $0 \le \theta_i \le \theta_\mathrm{iM}$ . Calculate the highest repetition frequency $f$ of the signal pulses, so that at a distance $z = 1000\ \mathrm{m}$ two consecutive pulses are still separated (that is, the pulses do not overlap). [1.75 points]

Small arc element of a curve: the arc ds drawn as an arrow along the curve, with its horizontal component dz and vertical component dx.

Attention

The wave properties of the light are not considered in this problem.
Neglect any chromatic dispersion in the fiber.
The speed of light in vacuum is c = $2.998\times 10^{8}\ \mathrm{m/s}$
You may use the following formulae:

The length of a small arc element $ds$ in the $x$ O $z$ plane is

ds = dz\sqrt{1+\left(\frac{dx}{dz}\right)^2}

\int\frac{dx}{\sqrt{a^2-b^2x^2}} = \frac{1}{b}\,\mathrm{Arc}\sin\frac{bx}{a}

\int\frac{x^2dx}{\sqrt{a^2-b^2x^2}} = -\frac{x\sqrt{a^2-b^2x^2}}{2b^2} + \frac{a^2\,\mathrm{Arc}\sin\dfrac{bx}{a}}{2b^3}

$\mathrm{Arc}\sin x$ is the inverse function of the sine function. Its value equals the less angle the sine of which is $x$ . In other words, if $y = \mathrm{Arc}\sin x$ then $\sin y = x$ .

Отговори

Решение

Solutions of Problem No. 1

Optical fiber

1.a. At both sides of the point O (outside and inside the fiber), according to Snell law, we have:

n_0\sin\theta_\mathrm{i} = n_1\sin\theta_1 \qquad (1)

where $\theta_1$ is the value of angle $\theta$ at point O inside the fiber.

The light trajectory lays in the $xOz$ plane. Because the refraction index $n$ varies along $x$ direction, we divide $Ox$ axis into small elements $dx$ , so that in each of these elements $n$ can be considered as constant. We have, then:

n\sin i = (n+dn).\sin(i+di) \qquad (2)

where $i$ is the angle between the light trajectory and $x$ direction. Because $\theta + i = \dfrac{\pi}{2}$ , then

n\cos\theta = (n+dn).\cos(\theta+d\theta) \qquad (3)

Thus, at each point of coordinate $x$ on the light trajectory, we have:

n\cos\theta = n_1\sqrt{1-\alpha^2x^2}\,\cos\theta = n_1\cos\theta_1 \qquad (4)

Because

\cos\theta_1 = \sqrt{1-\sin^2\theta_1} = \sqrt{1-\frac{\sin^2\theta_\mathrm{i}}{n_1^2}} \qquad (5)

we have

n\cos\theta = n_1\cos\theta_1 = n_1\sqrt{1-\frac{\sin^2\theta_\mathrm{i}}{n_1^2}} = \sqrt{n_1^2-\sin^2\theta_\mathrm{i}}

Then

n\cos\theta = C = \sqrt{n_1^2-\sin^2\theta_\mathrm{i}} \qquad (6)

1.2. Because $\dfrac{dx}{dz} = x' = \tan\theta$ , from (6) we have:

n_1\sqrt{1-\alpha^2x^2}\,\cos\theta = n_1\sqrt{1-\alpha^2x^2}\left(1+\tan^2\theta\right)^{-\frac{1}{2}} = C \qquad (7)

Squaring the two sides, we obtain:

\left(1-\alpha^2x^2\right)\left(1+\tan^2\theta\right)^{-1} = \frac{C^2}{n_1^2}

and

1+x'^2 = (1-\alpha^2x^2)\frac{n_1^2}{C^2} \qquad (8)

After derivating the two sides of (8) versus $z$ , we get:

x'' + \frac{\alpha^2n_1^2}{C^2}x = 0 \qquad (9)

Because $n = n_1\sqrt{1-\alpha^2x^2}$ and

$n = n_1$ at $x$ =0
$n = n_2$ at $x$ = $a$

we get

\alpha = \frac{\sqrt{n_1^2-n_2^2}}{a.n_1}

Finally, we get the equation for $x''$

x'' + \frac{n_1^2-n_2^2}{a^2(n_1^2-\sin^2\theta_i)}\cdot x = 0 \qquad (10)

1.c. The equation for the light trajectory is obtained by solving (10). This is an equation similar to that for an harmonic oscillation, which solution can be written right away

x = x_0\sin(pz+q) \qquad (11)

with

p = \frac{1}{a}\sqrt{\frac{n_1^2-n_2^2}{n_1^2-\sin^2\theta_\mathrm{i}}}

The parameters $p$ and $q$ are determined from the boundary conditions:

at $z$ =0, $x$ =0, hence $q$ =0
at $z$ =0 inside the fiber, $x' = \dfrac{dx}{dz} = \tan\theta_1$ , then

x_0 = \frac{\tan\theta_1}{p} = \frac{a.\sin\theta_\mathrm{i}}{\sqrt{n_1^2-n_2^2}} \qquad (12)

The equation for the trajectory of the light inside the fiber is:

x = \frac{a\sin\theta_\mathrm{i}}{\sqrt{n_1^2-n_2^2}}.\sin\left(\sqrt{\frac{n_1^2-n_2^2}{n_1^2-\sin^2\theta_\mathrm{i}}}.\frac{z}{a}\right) \qquad (13)

1.d. Here is a sketch of the trajectories of two rays entering the fiber at O, under different incident angles.

Sketch of two sinusoidal ray trajectories starting at O on the fiber axis, entering at two different incident angles; both cross the axis and reach opposite extremes within the core boundaries, which are marked at distance a above and below the axis.

2.a. The condition for the light to propagate along the fiber is that $x_0 \le a$ . This means that:

\frac{a\sin\theta_\mathrm{i}}{\sqrt{n_1^2-n_2^2}} \le a

or:

\sin\theta_\mathrm{i} \le \sqrt{n_1^2-n_2^2} \qquad (14)

Thus the incident angle $\theta_\mathrm{i}$ must not exceed $\theta_\mathrm{iM}$ , with

\sin\theta_\mathrm{iM} = \sqrt{n_1^2-n_2^2} = 0.344 \qquad (14a)

or:

\theta_\mathrm{i} \le \theta_\mathrm{iM} = \mathrm{Arc}\sin\left(\sqrt{n_1^2-n_2^2}\right) = \mathrm{Arc}\sin 0.344 = 0.351\,\mathrm{rad} = 20.13^{\circ}

2.b. The crossing points of the light beam with O $z$ axis must satisfy the condition $pz = k\pi$ , with $k$ - an integer. The $z$ coordinates of these points are:

z = \frac{k\pi}{p} = k\pi a\sqrt{\frac{n_1^2-\sin^2\theta_\mathrm{i}}{n_1^2-n_2^2}} \qquad (15)

except for $\theta_\mathrm{i}$ =0.

3.a. The rays entering the fiber at different incident angles have different trajectories. As a consequence, the propagation speeds of the rays along the fiber should be different.

The light trajectories are sinusoidal as given in (13). Let us calculate the time $\tau$ it takes the light to propagate from point O to its first crossing point with O $z$ axis. This is twice the time it takes the light to propagate from point O to its position most distant from O $z$ axis.

The time required for the light to travel a small segment $ds$ along its trajectory is

dt = \frac{n}{c}ds = \frac{n}{c}\sqrt{dx^2+dz^2} = \frac{n}{c}\sqrt{1+\frac{dz^2}{dx^2}}.dx

Small arc element of the trajectory: the arc ds drawn as an arrow along the curve, with its horizontal component dz and vertical component dx.

= \frac{n}{c}\sqrt{1+\left(\frac{1}{\tan\theta}\right)^2}.dx = \frac{n}{c}\frac{dx}{\sin\theta}

From (6), we have

dt = \frac{n_1^2\left(1-\alpha^2x^2\right)}{c.\sqrt{\sin^2\theta_\mathrm{i}-n_1^2\alpha^2x^2}}.dx

and

\frac{\tau}{2} = \int_0^{x_0}dt = \frac{n_1^2}{c}\left[\int_0^{x_0}\frac{dx}{\sqrt{\sin^2\theta_\mathrm{i}-n_1^2\alpha^2x^2}} - \alpha^2\int_0^{x_0}\frac{x^2dx}{\sqrt{\sin^2\theta_\mathrm{i}-n_1^2\alpha^2x^2}}\right] \qquad (16)

= \frac{n_1^2}{c}\left[I_1-\alpha^2I_2\right]

where

I_1 = \frac{1}{n_1\alpha}\,\mathrm{Arc}\sin\left.\frac{n_1\alpha x}{\sin\theta_\mathrm{i}}\right|_0^{x_0} = \frac{\pi a}{2\sqrt{n_1^2-n_2^2}} \qquad (17)

I_2 = \left.\frac{-x\sqrt{\sin^2\theta_\mathrm{i}-n_1^2\alpha^2x^2}}{2n_1^2\alpha^2}\right|_0^{x_0} + \left.\frac{\sin^2\theta_\mathrm{i}.\mathrm{Arc}\sin\dfrac{n_1\alpha x}{\sin\theta_\mathrm{i}}}{2n_1^3\alpha^3}\right|_0^{x_0} = \frac{\pi\sin^2\theta_\mathrm{i}}{4n_1^3\alpha^3} \qquad (18)

Using (16), (17), (18), we obtain

\tau = \frac{\pi a.n_1^2}{c\sqrt{n_1^2-n_2^2}}\left(1-\frac{\sin^2\theta_\mathrm{i}}{2n_1^2}\right) \qquad (19)

The propagation speed along the fiber is $v = \dfrac{z}{\tau}$ , where $z$ is the coordinate of the first crossing point, which is determined by (15) for $k = 1$ . Because $z$ and $\tau$ depend on the incident angle $\theta_\mathrm{i}$ , $v$ also depends on $\theta_\mathrm{i}$ .

For $\theta_\mathrm{i} = \theta_\mathrm{iM}$ , from (14a), we get

v_\mathrm{M} = \frac{\pi a n_2}{\sqrt{n_1^2-n_2^2}}.\frac{2c\sqrt{n_1^2-n_2^2}}{\pi a n_1^2}\left(1+\frac{n_2^2}{n_1^2}\right)^{-1} = \frac{2cn_2}{n_1^2+n_2^2} \qquad (20)

and

v_\mathrm{M} = \frac{2\times 2.998\times 10^8\times 1.460}{1.500^2+1.460^2} = 1.998\times 10^8\ \mathrm{m/s} \qquad (20a)

The propagation speed of the light along the O $z$ axis is

v = \frac{c}{n_1} \qquad (21)

because the refraction index is n $_1$ on the axis of the fiber.

The numerical value is

v_0 = \frac{2.998\times 10^8}{1.5} = 1.999\times 10^8\ \mathrm{m/s} \qquad (21a)

3.b. If the beam of the light pulses is formed by rays converging at O, then the rays with different incident angles has different propagation speeds. The two rays of incident angles $\theta_i = 0$ and $\theta_i = \theta_\mathrm{iM}$ arrive to the plane $z$ with a time delay

\Delta t = \frac{z}{v_\mathrm{M}} - \frac{z}{v_0} = \frac{z}{c}.\frac{(n_1-n_2)^2}{2n_2} \qquad (22)

This means that a very short light pulse becomes a pulse of finite width $\Delta t$ given by (22) at the plane $z$ . If two consecutive pulses enter the fiber with a delay greater than $\Delta t$ , then at the plane $z$ , they are separated. Hence the repetition frequency of the pulses must not exceed the maximal value:

f_\mathrm{M} = (\Delta t)^{-1} = \frac{2.c.n_2}{z.(n_1-n_2)^2} \qquad (23)

If $z = 1000\ \mathrm{m}$ , then

f_\mathrm{M} = \frac{2\times 2.998\times 10^8\times 1.460}{1000\times(1.500-1.460)^2} = 547.1\mathrm{MHz}

Оригинал в Архива: 2.pdf · официални решения: 2_sol.pdf