Calculate the critical angle for this optic fibre.

The diagram shows a straight optic fibre. Sketch the passage of a ray of light through the fibre.

The input power to the fibre is 150 mW. The attenuation per unit length of the glass fibre is ${\text{12.0 dB}}\,{\text{k}}{{\text{m}}^{ - 1}}$. When the light has travelled a distance $l$ its power has fallen to 3.00 mW, at which point amplification of the signal is required. Determine $l$.

$\left( {\sin C = \frac{1}{n} = \frac{1}{{1.62}} \Rightarrow C = } \right){\text{ }}38(.1)^\circ$ or 0.665 rad;

rays with an angle of greater than ${38^ \circ }$ shown with total internal reflection;

normal drawn onto diagram for at least one point;

pairs of angles of incidence and reflection the same;

Judge by eye.

attenuation/${\text{dB}} = \left( {10\log \frac{{{I_1}}}{{{I_2}}} = } \right){\text{ }}10\log \frac{{3{\text{ mW}}}}{{150{\text{ mW}}}}$$\,\,\,$or$\,\,\,$$- 17.0{\text{ dB}}$;

${\text{length}} = \left( {\frac{{{\text{attenuation}}}}{{{\text{attenuation per unit length}}}}{\text{ = }}\frac{{ - 17.0}}{{12}} = } \right){\text{ 1.42 km}}$;

(a)(i) was well answered.

In (a)(ii), many candidates were not careful to obey the law of reflection.

Most failed to determine the attenuation in (b).