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roam
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Homework Statement
A glass optical fibre of length L = 3.2 m is in a medium of glycerine with a refractive index n0 = 1.47 . The fibre has a core of refractive index, n1 = 1.58 and diameter, d = 100μm surrounded by a thin cladding of refractive index, n2 = 1.53. The end of the fibre is cut square (see diagram).
[PLAIN]http://img69.imageshack.us/img69/4976/imagenz.gif
For the aforementioned fibre, what is the total path length of a ray within the fibre that enters the fibre at the acceptance angle?
The Attempt at a Solution
Here are the stuff that I have already calculated:
* The minimum angle, θ2, for total internal reflection at the core/cladding boundary: 75.5 degrees.
* The angle, θ1, to the axis of the fibre that corresponds to the minimum angle, θ2: 14.5 degrees.
* The acceptance angle θ0: 15.6 degrees.
So, now to find the path length of a ray within the fibre that enters the fibre at the acceptance angle I tried using trigonometry:
Since d=100.0 μm = 100 x 106 m. We have d/2 = 50 x 106.
[tex]sin (14.5) = \frac{50 \times 10^6}{x}[/tex]
x= a very huge number!
But the answer must be 3.305 m! What did I do wrong?
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