# Path length of a light ray

1. Aug 7, 2010

### roam

1. The problem statement, all variables and given/known data

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 [Broken]

For the aforementioned fibre, what is the total path length of a ray within the fibre that enters the fibre at the acceptance angle?

3. 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?

Last edited by a moderator: May 4, 2017
2. Aug 7, 2010

### ehild

1 μm is not 106 m.

ehild

3. Aug 8, 2010

### roam

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...

4. Aug 8, 2010

### ehild

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

Last edited: Aug 8, 2010