An optic fibre of refractive index 1.4475 is surrounded by air. The critical angle for the core – air boundary interface is 44°. Suggest, with a calculation, why the use of cladding with refractive index 1.4444 improves the performance of the optic fibre.

Calculate the maximum attenuation allowed for the signal.

An amplifier can increase the power of the signal by 12 dB. Determine the minimum number of amplifiers required.

The graph shows the variation with wavelength of the refractive index of the glass from which the optic fibre is made.

Two light rays enter the fibre at the same instant along the axes. Ray A has a wavelength of λ_A and ray B has a wavelength of λ_B. Discuss the effect that the difference in wavelength has on the rays as they pass along the fibre.

In many places clad optic fibres are replacing copper cables. State one example of how fibre optic technology has impacted society.

sin c = \(\frac{{1.4444}}{{1.4475}}\) or sin c = 0.9978

critical angle = 86.2«°»

with cladding only rays travelling nearly parallel to fibre axis are transmitted

OR

pulse broadening/dispersion will be reduced

OWTTE

[3 marks]

attenuation = «10 log\(\frac{I}{{{I_0}}}\)» = 10 log\(\frac{{2.0 \times {{10}^{ - 6}}}}{{400 \times {{10}^{ - 6}}}}\)

attenuation = «–»23 «dB»

Accept 10 log\(\frac{{400}}{{2.0}}\) for first marking point

[2 marks]

185 × 0.200 = 37 loss over length of cable

«\(\frac{{37 - 23}}{{12}}\) = 1.17» so two amplifiers are sufficient

[2 marks]

mention of material dispersion

mention that rays become separated in time

OR

mention that ray A travels slower/arrives later than ray B

[2 marks]

high bandwidth/data transfer rates

low distortion/Low noise/Faithful reproduction

high security

fast «fibre» broadband/internet

high quality optical audio

medical endoscopy

Allow any other verifiable sensible advantage

[1 mark]