We want to couple the energy of a metal vapour laser beam of wavelength (lambda

Question

We want to couple the energy of a metal vapour laser beam of wavelength (lambda = 628 nm) into a fiber optic cable. The beam of the laser has a cross-sectional area of 10 mm^2 and all the energy from the beam has to be coupled into a diameter of 2.5 mu m (fiber core). A lens is used to couple the laser beam into the fiber core. What should be the focal length f of the lens to be able to couple the beam into the fiber core? What is the ratio f/D to just couple the beam into the fiber core? Calculate the power density in the fiber core in part a) if the laser power is 1W. Good commercial fibers can withstand a power density ranging between 10^7 - 10^W/cm^2. Is your answer within the specifications given for commercial fibers? Power density (P) = Power/C.S.A., where C.S.A. stands for cross- sectional area. What is the average power density if the laser is pulsed and the pulse width T is equal to one half the period T? Averaged power density () = Averaged power/C.SA = /C.S.A What is the average power density if the laser is pulsed and the pulse width T is equal to one third the period T? What is the critical angle if n_1 = 1.50 and n_2 = 1.49? Below what value of lambda does the fiber in part f) start being a multi-mode fiber?

Explanation / Answer

(a) For a Gaussian laser beam, the required beam diameter 'D' (Laser beam diameter) incident upon focusing lens of focal length 'f' to produce a focused spot of diameter (core diameter) 'w' is

D = 4f/( w)

  f =  Dw/4

Given = 628 nm = 628 x 10-9 m,

Laser beam cross sectional area = r2 = 10 mm2 r2 = 3.18 mm2   r = 1.785 mm

D = Laser beam diameter = 2r = 2 x 1.785 = 3.57 mm = 3.57 x 10-3 m

focused spot of diameter (fiber core diameter) w = 2.5 m = 2.5 x 10-6 m

since, f =  Dw/4 f = ( 3.57 x 10-3 x x 2.5 x 10-6 ) / (4 x 628 x 10-9) = 0.0112 m = 11.2 cm

(b)   f/D ratio

f/D = 0.0112 / 3.57 x 10-3 = 3.14

(c) Power density in the fiber core from part (a)

Power of the laser beam = 1 W

focused spot of diameter (fiber core diameter) w = 2.5 m = 2.5 x 10-6 m

radius of the core R = w/2 = 1.25 x 10-6 m

Cross sectional area of the fiber core = R2 = 3.14 x (1.25 x 10-6)2 = 4.91 x 10-12 m2

Power density = Power/C.S.A. = 1 W / 4.91 x 10-12 m2 = 2.04 x 1011 W/m2 = 2.04 x  107 W/cm2

Thus it is well within the range of power density of (107 - 109 ) commercial fibers.

(d) average power of pulsed laser = Pavg = E/ T

if t is half the time period T, t = T/2

Thus, Pavg = E/ (T/2) = 2 (E/T)

Avg. Power density = Avg. Power/C.S.A.

Power increases by two times, thus, power density increases by two times.

(e) t = T/3, Power increases by three times, thus, power density increases by three times.

(f) Critical angle c = sin-1(n2/n1) = sin-1(1.49/1.5) = 83.38 degrees

(g) Cutoff Wavelength c = [2·R ·sqrt ((n1)2– (n2)2)]  / Vc

R is the fiber core radius = R = w/2 = 1.25 x 10-6 m (given)

n1 is the index of refraction of the core

n2 is the index of refraction of the cladding

c is the cutoff wavelength

Vc is the cutoff V number (waveguide parameter), equals 2.4045

The cutoff wavelength is the minimum wavelength in which a particular fiber still acts as a single mode fiber. Below this wavelength it acts as multi mode fiber and above it is single mode fiber.

Cutoff Wavelength c = [2·R ·sqrt ((n1)2– (n2)2)]  / Vc

= [2 x 1.25 x 10-6 x sqrt ((1.5)2– (1.49)2)]  / 2.4045 = 564.5 x 10-9 = 564.5 nm

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