optical fibre related problem (i find it hard! :P )

[Sarahisme]

Hey i am having a few problems with this question related to optical fibres. or at least i think i am . if i put up my answers, could someone who has a little bit of free time tell me if they are correct?

 

Thanks

 

Sarah

 

Here is the question:

 

my answers are:

(a) [math] L_2 - L_1 = 68.63 m [/math]

 

(b) 1%

 

© [math] F =( \frac{n_1 - n_2}{n_1 + n_2} )^2 [/math] , where F is the fraction of power that is reflected back.

 

so since [math] n_1 = 1.459 - 0.002(0.5) [/math]

 

and [math] n_2 = air = 1 [/math]

 

the plugging this into to F gives :

 

[math] F = 0.0348 [/math]

 

so therefore the fraction of power reflected from the interface as a percentage of the input power is 3.48 %

 

how was all that?

[Sarahisme]

???

[swansont]

It makes it easier if you show your work. a looks fine. But how is it that a higher fraction of power gets reflected than makes it to the end of the fiber in the first place?

[Sarahisme]

oh i think what i've got means that 3.48% of the 1% of the power that makes it to the end is reflected...right?

 

part (a) is the one i am most unsure about, to do that i just used the fact that

 

n = c/v and v = L/t

[Sarahisme]

ok here is my working,

 

(a) [math] L_1 = 100 \times 10^{3} m [/math]

[math] \lambda_1 = 0.5 \times 10^{-6} m [/math]

[math] \lambda_2 = 1 \times 10^{-6} m [/math]

 

so

[math] n(\lambda_1) = 1.459 - 0.002(0.5) = n_1[/math]

and [math] n(\lambda_2) = 1.459 - 0.002(1) = n_2 [/math]

 

now

 

[math] n_1 = \frac{c}{v_1} \implies v_1 = \frac{c}{n_1} [/math]

 

[math] v_1 = \frac{L_1}{t} \implies t = \frac{L_1}{v_1}[/math]

 

likewise

 

[math] n_2 = \frac{c}{v_2} [/math]

 

[math] v_2 = \frac{L_2}{t} [/math]

 

so [math] v_2 = \frac{c}{n_2} [/math]

 

[math] L_2 = v_2t [/math]

 

so [math] L_2 = \frac{c}{n_2} \frac{L_1}{v_1} = \frac{c}{n_2} \frac{L_1n_1}{c} = \frac{L_1n_1}{n_2} [/math]

 

so the difference in length between the fibres is:

 

[math] |L_1 - L_2| = 100 \times 10^{3} m - \frac{L_1n_1}{n_2} m [/math]

 

plugging in the values gives:

[math] |L_1 - L_2| = 68.63 m [/math]

 

 

so how was that for part (a)?

[Sarahisme]

now for part (b):

 

we are given that :

 

[math] \alpha = -0.2 dB/km[/math]

 

and since [math] \alpha = \frac{10log_{10}(\frac{I_{out}}{I_{in}})}{L}

[/math]

 

we want to find [math] \frac{I_{out}}{I_{in}} [/math]

 

so rearranging gives:

[math] log_{10}(\frac{I_{out}}{I_{in}}) = \frac{L \alpha}{10} [/math]

 

plugging in for the RHS gives:

 

[math] log_{10}(\frac{I_{out}}{I_{in}}) = \frac{(100km) (-0.2dB/km)}{10} = -0.2[/math]

 

so [math] \frac{I_{out}}{I_{in}} = 0.01 [/math]

 

and to get [math] \frac{I_{out}}{I_{in}} [/math] as a percentage, you multiply by 100 so therefore the fraction of power arriving at the end of the 100km fibre as a percentage of the power entering the fibre is 1%

 

 

how was that for part (b)??

[Sarahisme]

now for part ©:

 

using the formula:

[math] F =( \frac{n_1 - n_2}{n_1 + n_2} )^2 [/math] , where F is the fraction of power that is reflected back.

 

we know [math] n_1 \ and \ n_2 [/math] from part (a)

 

so plugging these values in we get:

 

F = 0.0348

 

but only 1% of the input power makes it to the end of the fibre (that is [math] \frac{I_{out}}{I_{in}} ) [/math], so the amount of power reflected from the end of the fibre (as a fraction) is [math] F \frac{I_{out}}{I_{in}} [/math]

 

So the amount of power reflected from the end of the fibre is:

 

[math] F \frac{I_{out}}{I_{in}} = (0.0348)(0.01) = 0.000348 [/math] which as a percentage is 0.0348%

 

now i just relised that there is another bit to part ©, which is that we also have to find the fraction of power that arrives back at the begining of the fibre as a percentage of the input power.

 

so to do this:

 

so we know only 1% of the power leaving one end of the fibre reaches the other end.

 

so the amount of the reflected power that arrives back at the begining as a fraction of the input power is

 

(0.01)(0.000348) = 0.00000348 = 0.000348%

 

so how was that for part © ??

[swansont]

Looks much better

[Sarahisme]

so yep i think it all looks correct i gather you do too

 

Thanks swansont!