Problem 1 Consider a fibre with $n_1=1.48, n_2=1.46$ and core radius $a=30 \mu \mathrm{m}$.

(a) Show that all rays making an angle $\theta\lt9.43^{\circ}$ with the axis will be guided through the fibre.

(b) Assume $\theta=3.43^{\circ}$ and calculate the number of total internal reflections that will occur in propagating through one $\mathrm{km}$ length of the fiber.

Solution:

(a) At the core-cladding surface the critical angle is given by,

$ \phi_c=\sin ^{-1}\left(\frac{n_2}{n_1}\right)=\sin ^{-1}\left(\frac{1.46}{1.48}\right)=80.57^{\circ} $

Therefore, the refracted ray at the entrance makes an angle of $90^{\circ}-\phi_c=9.43^{\circ}$ with the axis.

We know that if a ray makes an angle more than the critical angle at the core-cladding surface, the total internal reflection is possible.

So in our case for $\phi\lt9.43^{\circ}$, the total internal reflection is there in the fiber.

(b) We know that the total no. of internal reflection through a fiber of length L is,

$ n=\frac{L \tan \theta}{a}-1\\ $

Given,

$a=30 \mu \mathrm{m}$

$ =30 \times 10^{-6} \mathrm{~m} $

$ =30 \times 10^{-9} \mathrm{~km} $

$\theta=3.43^{\circ}$

$ n=\frac{1 \times \tan 3.43}{30 \times 10^{-9}}-1 \approx 2 \times 10^6-1 \approx 2 \times 10^6 $