HW4 Solution

1) a) Source should have area <= area of core and divergence angle <= 2 alpha_m
So maximum diameter = 8 mu m
Divergence angle = 2 alpha_m => NA_{source} = 0.18

b) lambda = 1.55 mu m,  w_o = 8/2 (mu m) = 4 ( mu m )
theta = 2 lambda_o / {pi w_o} = 2 1.55 / {pi 4 } = 0.2467 (rad) = 14.13^o
Check alpha_m: alpha_m = sin^{-1} (NA) = 10.37^o > 14.13^o/2
Yes, it is optimum.

c) If the source area = core area, the distance should be zero since beam will expand with distance. In sub-optimal situation, the distance should not be larger z_o = pi w_o^2/lambda = pi 4^2/1.55 = 32.43 (mu m)

d) No, this small beam size will lead to increase in theta such that theta > 2 alpha_m.
Check: theta ={2 times 1.55}/ { pi times 4/2} = 2 times 14.13^o >= 2 times 10.37^o
Even if you can make the distance between source and fiber to zero, light will leak since it is not totally internal reflection the cladding and core interface inside the fiber!

2) a = 25 mu m, n_1 = 1.485, n_2=1.465, lambda = 850 nm, lambda_0 = 1360 nm, S_0 = 0.09 {ps}/{km-nm^2}, Delta lambda = 2nm
a) V = {2 pi a}/ lambda NA = {2 pi times 25}/ 0.85 sqrt{1.485^2 - 1.465^2} = 44.89
b) Since V > 2.405, this fiber operates at multi-mode regime. Number of modes can be found as
M = Int ( V^2 / 2 )= 1007
c) To have single mode operation for 1900nm to 2100nm, we need to have cut-off at the short wavelength and long wavelength will satisfy the single mode condition automatically. Therefore, lambda_c = 1900 nm
2.405 = {2 pi a} / lambda_c NA => NA = 2.405 lambda_c / {2 pi a} =2.909 times 10^{-2}
=> sqrt{n_1^2 - n_2^2} = 2.909 times 10^{-2} => n_1^2 = n_2^2 + ( 2.909 times 10^{-2})^2 => n_1 = 1.46529

3) a = 30 mu m, n_2 = 1.46, D_{modal} = 10 {ns} / km, L = 10km, n_1 approx n_{1g} and n_1 approx n_2
D_{modal}= n_1 Delta / c => Delta = c D_{modal}/ n_2 ={3 times 10^8 times 10^{-8} times 10^{-3}} / 1.46 => Delta = 2.055 times 10^{-3}
NA approx n sqrt {2 Delta} = 1.46 times sqrt {2 times 2.069 times 10^{-3}} = 0.093594
Also NA = n_a sin alpha_m where n_a is the refractive index outside the fiber, i.e. n_a = 1 for air outside the fiber.
=> alpha_m = 5.37^o
Using textbook's formula, R_b = 1 /{2 Delta tau} = 1 / {2 times 10 times 20} ({Gb}/s) = 5 ({Mb}/s)
Using conservative estimate, R_b = 1 /{4 Delta tau} = 2.5 ({Mb}/s)

4) a = 4 mu m, n_1 = 1.467, n_2 = 1.464, lambda = 1.3 mu m
a) V = {2 pi 4} / 1.3 sqrt{1.467^2 - 1.464^2} = 1.813
b) V < 2.405 => single mode regime
c) Single mode of 800nm to 1100nm => lambda_c = 800 nm
2.405 = {2 pi a} / lambda_c NA => a = {2.405 lambda_c} / {2 pi NA} = {2.405 times 0.8} / {2 pi sqrt{1.467^2 - 1.464^2}}
=> a = 3.266 ( mu m ), i.e. diameter <= 6.532 (mu m)
Notes: Core radius needs to be reduced for shorter lambda_c.

5) a= 5 mu m, Delta = 0.002. Take typical n = 1.45 for glass. Single mode cutoff wavelength lambda_c occurs at V=2.405.
2.405 = {2 pi a} / lambda_c NA approx 2.405 = {2 pi a} / lambda_c n sqrt{2 Delta}
lambda_c = {2 pi a NA} / {2.405} = { 2 pi a n sqrt{2 Delta}} / 2.405
lambda_c approx {2 pi times 5 times 1.45 times sqrt{2 times 0.002}} / 2.405 = 1.1979 ( mu m)

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