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Date November 2014 Marks available 2 Reference code 14N.3.SL.TZ0.19
Level Standard level Paper Paper 3 Time zone Time zone 0
Command term Determine Question number 19 Adapted from N/A

Question

This question is about optic fibres.

An optic fibre consists of a thin glass fibre surrounded by a cladding material. The refractive index of the glass is 1.62.

Calculate the critical angle for this optic fibre.

[1]
a.i.

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

N14/4/PHYSI/SP3/ENG/TZ0/19.1.ii

[[N/A]]
a.ii.

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 \(l\).

[2]
b.

Markscheme

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

a.i.

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.

a.ii.

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

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

b.

Examiners report

(a)(i) was well answered.

a.i.

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

a.ii.

Most failed to determine the attenuation in (b).

b.

Syllabus sections

Option C: Imaging » Option C: Imaging (Core topics) » C.3 – Fibre optics
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