Sarahisme Posted October 7, 2005 Share Posted October 7, 2005 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? Link to comment Share on other sites More sharing options...
Sarahisme Posted October 9, 2005 Author Share Posted October 9, 2005 ??? Link to comment Share on other sites More sharing options...
swansont Posted October 9, 2005 Share Posted October 9, 2005 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? Link to comment Share on other sites More sharing options...
Sarahisme Posted October 10, 2005 Author Share Posted October 10, 2005 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 Link to comment Share on other sites More sharing options...
Sarahisme Posted October 10, 2005 Author Share Posted October 10, 2005 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)? Link to comment Share on other sites More sharing options...
Sarahisme Posted October 10, 2005 Author Share Posted October 10, 2005 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)?? Link to comment Share on other sites More sharing options...
Sarahisme Posted October 10, 2005 Author Share Posted October 10, 2005 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 © ?? Link to comment Share on other sites More sharing options...
swansont Posted October 10, 2005 Share Posted October 10, 2005 Looks much better Link to comment Share on other sites More sharing options...
Sarahisme Posted October 10, 2005 Author Share Posted October 10, 2005 so yep i think it all looks correct i gather you do too Thanks swansont! Link to comment Share on other sites More sharing options...
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