How Is Path Length Calculated for Light in an Optical Fiber?

[roam]

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.

$$sin (14.5) = \frac{50 \times 10^6}{x}$$

x= a very huge number!

But the answer must be 3.305 m! What did I do wrong?

 

[ehild]

1 μm is not 10⁶ m.

ehild

 

[roam]

1 μm is not 10⁶ m.

ehild

Oh, I meant 10^-6. But why is it that I end up with 1.9 x 10^-4? This is not the right answer...

 

[ehild]

Oh, I meant 10^-6. But why is it that I end up with 1.9 x 10^-4? This is not the right answer...

There are a lot of reflections in that 3.2 m long fibre. The light goes through the whole length of the fibre, but trawels along a zigzag path which is longer than the fibre length. ehild