| Date | May 2019 | Marks available | 2 | Reference code | 19M.3.SL.TZ1.10 |
| Level | Standard level | Paper | Paper 3 | Time zone | 1 |
| Command term | Calculate | Question number | 10 | Adapted from | N/A |
Question
The diagram shows a light ray incident from air into the core of an optic fibre. The angle of incidence is θ. Values of refractive indices are shown on the diagram.
Calculate the critical angle at the core–cladding boundary.
Show that the maximum value of θ for which total internal reflection will take place at the core–cladding boundary is about 90°.
Comment on your answer to part (a)(ii).
A signal consists of two rays that enter the core at angle of incidence θ = 0 and θ = θmax. Identify a disadvantage of this fibre for transmitting this signal.
Outline the significance of optic fibres in modern communications.
Markscheme
«sin θc = » sin θc = ✔
θc = 51.97° ✔
Award [2] for bald correct answer.
«angle of refraction at air-core boundary is 90°−θc «=90.00° − 51.97° = 38.03°»✔
1.000 × sinθmax =1.620 × sin 38.03° ✔
θmax = 86.41° ✔
«θmax is almost 90° which means that» a ray entering the core almost at any angle will be totally internally reflected/will not escape ✔
rays will follow very different paths in the core ✔
leading to waveguide dispersion/different arrival times/pulse overlap ✔
Reference to 2 of:
secure/encrypted transfer of data ✔
high bandwidth/volume of data transferred ✔
high quality/minimal noise in transmission ✔
free from cross talk ✔
low «specific» attenuation ✔
Examiners report
Optic fibre. The critical angle and also the maximum value of θ were calculated well by better candidates.
Optic fibre. The critical angle and also the maximum value of θ were calculated well by better candidates.
Optic fibre. The critical angle and also the maximum value of θ were calculated well by better candidates. However, the wide acceptance angle was only well identified and commented in iii) by the best candidates. Surprisingly, only a few candidates made numerical mistakes throughout this question, though weaker students were still prone to not be able to correctly determine arcsin values. Working with an appropriate number of significant figures was vital in this question otherwise sin values greater than 1 were sometimes arrived at, thus making it difficult to correctly comment in part iii)
In iv), the waveguide dispersion was well identified and discussed by most the candidates. Many, however, incorrectly referred to speed being slower for the ray that reflects off the boundary.
In b), low attenuation and cross talk (or outside EM interference) was the most popular responses. Many candidates vaguely presented “high speed of communication via optic fibres” without details about data transfer.
