Question

If 1/ 2 3 4 P M Pclad = where Pclad is the optical power flows in the cladding of a fiber, P is the total optical power, and M is the number of modes propagates in the fiber, determine among two signals, one with λ = 820 nm and the other with λ = 1550 nm which one travels longer distance in the fiber. The two signals were launched at the same angle and assume n1 = 1.48, n2 = 1.46, and the core radius is 25 µm.

Answer 1

Solution:

Pclad :P Criven data 3MY Pelad - optical power flows in the cladding of a fobre. P: Total optical power M: number of modes propagated. lg = 8 20x10-9 m. for signal - I 12= 1550x104 for signal - ni = 1.48 n2 come Madius (a) = 25x10-6 m 1. 46 If we have M = y² where Mi no. of modes 2 3 7.89 then Pclad a 4 P Pelad P We can write this expression as Pc ead 1.89 1 Р 2x (0,²-n2²) 12 V 2ka Via nin ²/2 for dis 820nm, we have - 2T X 25x10-6 (1.48² - 1.462 g/ 820 x 10-9 v 3 21 x 25 820 Xlo X 0.242 46.45 t = 24.575 For signal with X2 = 1550nm, we have V 2KX 25 (1.48² ). 46 ²/2 0.82 V 12.4.5724 Now 820 nm 3 1.8856 v Pelad for t pi Pread P. 1= 820nm so 1.885 24.574 0.0406 And Pulad for t= 1550 nm 3 1.88 56 P 46.451 so Pread = 0:07.67 Р i = 1550nm it will have Signal with Lesser number of modes will travel fastt_longer distance, as So greater light power in Hence, Signal_with_1 = 8200m will travel farther t value higher cladding in the fibre.