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Homework answers / question archive / How significant is the broadening of a nsec pulse laser in the optical fibers? Their fiber was quartz, 2
How significant is the broadening of a nsec pulse laser in the optical fibers? Their fiber was quartz, 2.5 m long, 600 um in diameter, with numerical aperture in air of 0.22. Let's assume you sent an idealized pulse of 337 nm light (from the N2 laser) with zero temporal width into the fiber. Calculate the minimum and maximum flight times using ray approximations and thereby report the broadening due to the fiber geometry alone. Is this significant for nsec pulses?
Aperture a = 0.22 in air means that the angle A between the axis of the fiber and the ray most inclined from the axis IN AIR has sin A = a
The precise value of the index of refraction of quartz at wavelength 337 nm, should be found in your textbook. I shall take it n = 1.5 for the example here, but if you find a more precise value in your books, use the more precise value.
From the Snell's law, the angle B between the axis of the fiber and the ray most inclined from the axis IN FIBER has
sin B = sin A / n = a/n
A ray following the axis of the fiber exactly travels with speed
v0 = c/n
where the speed of light in vacuum is c = 3e8 m/s.
On the other hand, a ray traveling at angle B to the axis (bouncing from the walls with internal reflection) travels with speed
v1 = v0*cos B
The times needed to travel the length L = 2.5 m of the fiber for the two rays are
t0 = L/v0 = nL/c
and
t1 = L/v1 = nL/(c cos B) = (nL/c) * n/sqrt(n^2-a^2),
respectively.
These are the minimum and maximum times requested in the assignment.
The temporal pulse broadening is the difference between these two times:
Delta t = t1-t0 = (nL/c) * [ n/sqrt(n^2-a^2) - 1 ]
= 1.5 * 2.5 m / (3.e8 m/s) *( 1.5/sqrt(2.25 - 0.0484) - 1) = 1.4e-10 s = 0.14 ns
If the original pulse lasted 1 ns, it will be broadened by 14%.
