Optical Fiber 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 = n1 on the axis to n = n2 (with 1

Question

Optical Fiber

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 = [latex]n_{1}[/latex]on the axis to n = [latex]n_{2}[/latex] (with 1 < [latex]n_{2}[/latex] < [latex]n_{1}[/latex]) at a distance a from the axis, according to the formula

[latex]n=n(x)=n_{1}\;{\sqrt{1-\alpha^{2}x^{2}}}[/latex]

where x is the distance from the core axis and α is a constant. The core is surrounded by a cladding made of a material with constant refraction index [latex]n_{2}[/latex], Outside the fiber is air, of refractive index [latex]n_{0}[/latex] .

Let Oz be the axis of the fiber, with O the center of the fiber end. Give [latex]n_{0}[/latex] = 1. 000; [latex]n_{1}[/latex] = 1. 500; [latex]n_{2}[/latex] = 1. 460, a = 25 μm.

(1) A monochromatic light ray enters the fiber at point 0 under an
incident angle [latex] heta_{i}[/latex], the incident plane being the plane xOz.

(a) Show that at each point on the trajectory of the light in the fiber,
the refractive index n and the angle θ between the light ray and the Oz axis satisfy the relationship ncos θ = C where C is a constant. Find the expression for C in terms of [latex]n_{1}[/latex]and [latex] heta_{i}[/latex] •

(b) Use the result found in (1) (a) and the trigonometric relation [latex]\cos heta=(1+ an^{2} heta)^{-\frac{1}{2}}[/latex] , where[latex] an heta={\frac{\mathrm{d}x}{\mathrm{d}z}}=x^{\prime}[/latex]is the slope of the tangent to the trajectory at point (x, z), derive an equation for x'. Find the full expression for α in terms of [latex]n_{1}[/latex] , [latex]n_{2}[/latex] and a. By differentiating the two sides of this equation versus z, find the equation for the second derivative x".

(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.

(d) Sketch one full period of the trajectories of the light rays entering the fiber under two different incident angles [latex] heta_{i}[/latex].

(2) Light propagates in the optical fiber.

(a) Find the maximum incident angle [latex] heta_{im}[/latex] under which the light ray still can propagate inside the core of the fiber.

(b) Determine the expression for coordinate z of the crossing points of a light ray with Oz axis for [latex] heta_{i}[/latex]≠ 0.

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

(a) Determine the time r it takes the light to travel from point 0 to the first crossing point with Oz for incident angle [latex] heta_{i}[/latex]≠ 0 and [latex] heta_{i}[/latex] ≤ [latex] heta_{im}[/latex] .

The ratio of the coordinate z of the first crossing point and τ is called the propagation speed of the light signal along the fiber. Assume that this speed varies monotonously with [latex] heta_{i}[/latex].

Find this speed (called [latex]v_{m}[/latex]) for [latex] heta_{i}[/latex]= [latex] heta_{im}[/latex].

Find also the propagation speed (called [latex]v_{0}[/latex]) of the light along the axis Oz.

Compare the two speeds.

(b) The light beam bearing the signals is a converging beam entering the fiber at 0 under different incident angles [latex] heta_{i}[/latex] with 0 ≤ [latex] heta_{i}[/latex] ≤ [latex] heta_{iM}[/latex] Calculate the highest repetition frequency f of the signal pulses, so that at a distance z = 1000 m two consecutive pulses are still separated (that is, the pulses do not overlap).

Attention
1. The wave properties of the light are not considered in this problem.
2. Neglect any chromatic dispersion in the fiber.
3. The speed of light in vacuum is c = 2. 998 X [latex]10^{8}[/latex] m/ s.
4. You may use the following formulae:

The length of a small arc element ds in the xOz plane is

[latex]\mathrm{d}s=\mathrm{d}z{\sqrt{1+\left({\frac{\mathrm{d}x}{\mathrm{d}z}} ight)^{2}}}\,;[/latex]

[latex]\int {\frac{\mathrm{d}x}{\sqrt{a^{2}-b^{2}x^{2}}}}={\frac{1}{b}}\mathrm{Arcsin}\ {\frac{b x}{a}};[/latex]

[latex]\int {\frac{x^{2}\mathrm{d}x}{\sqrt{a^{2}-b^{2}x^{2}}}}=-\displaystyle{\frac{x\;{\sqrt{a^{2}-b^{2}x^{2}}}}{2b^{2}}}+\displaystyle{\frac{a^{2}\mathrm{Arcsin}\;{\frac{b x}{a}}}{2b^{3}}};[/latex]

Arcsin 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 = arcsin x, sin y = x.

Accepted Answer

(1) (a) At both sides of the point 0 (outside and inside the fiber), according to Snell law, we have

[latex]n_{\mathrm{o}}\sin heta_{i}=n_{1}\sin heta_{1}\,,[/latex] (1)

where [latex]θ_1 [/latex] is the value of angle θ 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

[latex]n\sin i=(n+\mathrm{d}n).\sin(i+\mathrm{d}i)\,,[/latex] (2)

where i is the angle between the light trajectory and x direction. Because θ+ [latex]i={\frac{\pi}{2}},[/latex]

[latex]n\mathrm{cos}\, heta=\,(n+\mathrm{d}n)\,·\mathrm{cos}(\, heta+\mathrm{d} heta).[/latex] (3)

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

[latex]n\mathrm{cos}\, heta=n_{1}\;\sqrt{1-\alpha^{2}x^{2}}\,\mathrm{cos}\, heta=n_{1}\mathrm{cos}\, heta_{1}.[/latex] (4)

Because

[latex]\cos heta_{1}=\sqrt{1-\sin^{2} heta_{1}}=\sqrt{1-\frac{\sin^{2} heta_{i}}{n_{1}^{2}}}\,,[/latex] (5)

we have

[latex]n\cos heta=n_{1}\cos heta_{1}=n_{1}\sqrt{1-\frac{\sin^{2} heta_{i}}{n_{1}^{2}}}\,=\,\sqrt{n_{1}^{2}-\sin^{2} heta_{i}}.[/latex]

Then

[latex]n\mathrm{cos}\, heta=C=\,\sqrt{n_{1}^{2}-\sin^{2} heta_{i}}.[/latex] (6)

(1) (b) Becaus [latex]{\frac{\mathrm{d}x}{\mathrm{d}z}}=x^{\prime}= an heta,[/latex] from (6) we have:

[latex]n_{1}\;\sqrt{1-\alpha^{2}x^{2}}\cos heta=n_{1}\;\sqrt{1-\alpha^{2}x^{2}}\,(1+ an^{2}\! heta)^{-\frac{1}{2}}=C.[/latex] (7)

Squaring the two sides, we obtain

[latex](1-\alpha^{2}x^{2})\,(1+ an^{2}\! heta)^{-1}={\frac{C^{2}}{n_{1}^{2}}}[/latex]

and

[latex]1+x^{\prime\,2}=(1-\alpha^{2}x^{2})\,\frac{n_{\mathrm{1}}^{2}}{C^{2}}.[/latex] (8)

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

[latex]x^{\prime\prime}+{\frac{\alpha^{2}n_{1}^{2}}{C^{2}}}x=0.[/latex] (9)

Because [latex]n=n_{1}\ {\sqrt{1-\alpha^{2}x^{2}}}[/latex] and

[latex]n=n_{1}\;\mathrm{at}\,x=0\,,[/latex]

[latex]n=n_{2}\,,\,\mathrm{at}\,x=a,[/latex]

we get

[latex]\alpha=\frac{\sqrt{n_{1}^{2}-n_{2}^{2}}}{a·n_{1}}.[/latex]

Finally, we get the equation for x":

[latex]x^{\prime\prime}+{\frac{n_{1}^{2}-n_{2}^{2}}{a^{2}\,(n_{1}^{2}-\sin^{2} heta_{i})}}· x=0.[/latex] (10)

(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

[latex]x=x_{0}\sin( ho z+q)[/latex] (11)

with

[latex]{p}=\frac{1}{a}\,\sqrt{\frac{n_{1}^{2}-n_{2}^{2}}{n_{1}^{2}-\sin^{2} heta_{i}}}.[/latex]

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, [latex]x^{′}={\frac{\mathrm{d}x}{\mathrm{d}z}}= an heta_{1}[/latex],then

[latex]x_{o}={\frac{ an heta_{\mathrm{1}}}{p}}[/latex]

[latex]={\frac{a\cdot\sin heta_{i}}{\sqrt{n_{1}^{2}-n_{2}^{2}}}}.[/latex] (12)

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

[latex]x={\frac{a\sin heta_{i}}{\sqrt{n_{1}^{2}-n_{2}^{2}}}}\ ·\,\sin\Biggl({\sqrt{\frac{n_{1}^{2}-n_{2}^{2}}{n_{1}^{2}-\sin^{2} heta_{i}}}}\·\,{\frac{z}{a}}\Biggr).[/latex] (13)

(d) Here is a sketch of the trajectories of two rays entering the fiber at 0, under different incident angles.

(2) (a) The condition for the light to propagate along the fiber is that [latex] x_{0}[/latex] ≤ a. This means

[latex]\frac{a\sin heta_{i}}{\sqrt{n_{1}^{2}-n_{2}^{2}}}\leq a[/latex]

or

[latex]\sin heta_{i}\leq\sqrt{n_{1}^{2}-n_{2}^{2}}\,.[/latex] (14)

Thus the incident angle [latex] heta_{i}[/latex] must not exceed [latex] heta_{im}[/latex] , with

[latex]\sin heta_{\mathrm{im}}=\,\sqrt{n_{1}^{2}-n_{2}^{2}}\,=\,0.344[/latex] (14a)

or

[latex] heta_{i}\leq heta_{i\mathrm{m}}=\arcsin(\,{\sqrt{n_{1}^{2}-n_{2}^{2}}}\,)\,=\,{\mathrm{arcsin}}\,0.344[/latex]

[latex]=0.351\,{\mathrm{rad}}=20.13^{\circ}.[/latex]

(b) The crossing points of the light beam with Oz axis must satisfy the condition pz = kπ, with k - an integer. The z coordinates of these points are

[latex]z={\frac{k\pi}{p}}=k\pi a{\sqrt{\frac{n_{1}^{2}-\sin^{2}{ heta}_{i}}{n_{1}^{2}-n_{2}^{2}}}}[/latex] (15)

except for [latex] heta_{i}[/latex] = 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 [latex] au[/latex] it takes the light to propagate from point 0 to its first crossing point with Oz axis. This is twice the time it takes the light to propagate from point O to its position most distant from Oz axis

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

[latex]\mathrm{d}t={\frac{n}{c}}\mathrm{d}s={\frac{n}{c}}\;\;\sqrt{\mathrm{d}x^{2}+\mathrm{d}z^{2}}={\frac{n}{c}}\,\sqrt{1+{\frac{\mathrm{d}z^{2}}{\mathrm{d}x^{2}}}}·\mathrm{d}x[/latex]

[latex]={\frac{n}{c}}{\sqrt{1+\left({\frac{1}{ an heta}} ight)^{2}}}·\mathrm{d}x={\frac{n}{c}}·{\frac{\mathrm{d}x}{\sin heta}}[/latex]

From (6), we have

[latex]\mathrm{d}t={\frac{n_{1}^{2}(1-\alpha^{2}x^{2})}{c·\,{\sqrt{\sin^{2} heta_{i}-n_{1}^{2}\alpha^{2}x^{2}}}}}·\,\mathrm{d}x[/latex]

and

[latex]{\frac{ au}{2}}\int_{\mathrm{o}}^{x_{\mathrm{o}}}\mathrm{d}t=[/latex]

[latex]\frac{n_{1}^{2}}{c}\Bigg(\int_{0}^{x_{0}}\frac{\mathrm{d}x}{\sqrt{\sin^{2} heta_{i}-n_{1}^{2}\alpha^{2}x^{2}}}-[/latex]

[latex]\alpha^{2}\int_{0}^{x_{0}}{\frac{x^{2}\,\mathrm{d}x}{\sqrt{\sin^{2} heta_{i}-n_{1}^{2}\alpha^{2}\,x^{2}}}}\Biggr)[/latex]

[latex]=\frac{n_{1}^{2}}{c}(I_{1} - \alpha^2 I_{2}),[/latex] (16)

where

[latex]I_{1}=\frac{1}{n_{1}\alpha}\mathrm{Arcsin~}\frac{n_{1}\alpha x}{\sin heta_{i}}\biggr|_{0}^{x_{0}}=\frac{\pi a}{2\sqrt{n_{1}^{2}-n_{2}^{2}}}\,,[/latex] (17)

[latex]I_{2}=\frac{-{x}\sqrt{\sin^{2}\! heta_{i}-n_{1}^{2}\alpha^{2}x^{2}}}{2{n^{2}_{1}}{\alpha^{2}}}\biggl|_{0}^{x_{0}}+\frac{\sin^{2}\! heta_{i}\cdot\mathrm{Aresin~}\frac{{ n}_{1}\alpha x}{\sin heta_{i}}}{2{n^{3}_{1}}{\alpha^{3}}}\biggr|_{0}^{x_{0}}[/latex]

[latex]={\frac{\pi\sin^{2} heta_{i}}{4n_{1}^{3}\alpha^{3}}}.[/latex] (18)

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

[latex] au={\frac{\pi a·n_{1}^{2}}{c\ {\sqrt{n_{1}^{2}-n_{2}^{2}}}}}\left(1-{\frac{\sin^{2} heta_{i}}{2n_{1}^{2}}} ight).[/latex] (19)

The propagation speed along the fiber is [latex]v\,=\,{\frac{z}{ au}}\,,[/latex]where z is the r coordinate of the first crossing point, which is determined by (15) for k = 1, Because z and [latex] au[/latex] depend on the incident angle [latex] heta_{i}[/latex], v also depends on [latex] heta_{i}[/latex] .

For [latex] heta_{i}[/latex] = [latex] heta_{im}[/latex] , from (14a) , we get

[latex]v_{m}={\frac{\pi a n_{2}}{\sqrt{n_{1}^{2}-n_{2}^{2}}}}\cdot{\frac{2c\,{\sqrt{n_{1}^{2}-n_{2}^{2}}}}{\pi a n_{1}^{2}}}\Big(1+{\frac{n_{2}^{2}}{n_{1}^{2}}}\Big)^{-1}={\frac{2cn_{2}}{n_{1}^{2}+n_{2}^{2}}}[/latex] (20)

and

[latex]v_{m}={\frac{2 imes2.998 imes10^{8} imes1.460}{1.500^{2}+1.460^{2}}}=1.998 imes10^{8}\,\mathrm{{m/s}}.[/latex] (20a)

The propagation speed of the light along the Oz axis is

[latex]v={\frac{c}{n_{1}}}[/latex] (21)

because the refraction index is [latex]n_{1}[/latex]on the axis of the fiber.

The numerical value is

[latex]v_{0}={\frac{2.998 imes10^{8}}{1.5}}=1.999 imes10^{8}\ \mathrm{{m/s}}.[/latex] (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 [latex] heta_{i}[/latex] = 0 and [latex] heta_{i}[/latex] = [latex] heta_{im}[/latex] arrive to the plane z with a time delay

[latex]\Delta t={\frac{z}{v_{m}}}-{\frac{z}{v_{o}}}={\frac{z}{c}}·{\frac{(n_{1}-n_{2})^{2}}{2n_{2}}}.[/latex] (22)

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

[latex]f_{\mathrm{m}}=(\Delta\iota)^{-1}={\frac{2\cdot c \cdot n_{2}}{z\cdot (n_{1}-n_{2})^{2}}}.[/latex] (23)

If z = 1000 m , then

[latex]f_{\mathrm{m}}={\frac{2 imes 2.998 imes10^{8} imes 1.460}{1000 imes (1.500-1.460)^{2}}}=547.1\ \mathrm{MHz}.[/latex]

Question 5.TC.2: Optical Fiber An optical fiber consists of a cylindrical cor......

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