| Date | November 2018 | Marks available | 2 | Reference code | 18N.3.SL.TZ0.10 |
| Level | Standard level | Paper | Paper 3 | Time zone | 0 - no time zone |
| Command term | Calculate | Question number | 10 | Adapted from | N/A |
Question
An optic fibre consists of a glass core of refractive index 1.52 surrounded by cladding of refractive index n. The critical angle at the glass–cladding boundary is 84°.
The diagram shows the longest and shortest paths that a ray can follow inside the fibre.
For the longest path the rays are incident at the core–cladding boundary at an angle just slightly greater than the critical angle. The optic fibre has a length of 12 km.
Calculate n.
The refractive indices of the glass and cladding are only slightly different. Suggest why this is desirable.
Show that the longest path is 66 m longer than the shortest path.
Determine the time delay between the arrival of signals created by the extra distance in (b)(i).
Suggest whether this fibre could be used to transmit information at a frequency of 100 MHz.
Markscheme
«» n1 = 1.52 × sin 84.0° ✔
n1 = 1.51 ✔
to have a critical angle close to 90° ✔
so only rays parallel to the axis are transmitted ✔
to reduce waveguide/modal dispersion ✔
long path is ✔
= 12066 «m» ✔
«so 66 m longer»
speed of light in core is «m s−1» ✔
time delay is «s» ✔
no, period of signal is 1 × 10−8 «s» which is smaller than the time delay/OWTTE ✔
Examiners report
Syllabus sections
- 17N.3.HL.TZ0.15b: Outline how the combination of core and cladding reduces the overall dispersion in the optic...
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18M.3.SL.TZ1.9c.i:
Draw on the axes an output signal to illustrate the effect of waveguide dispersion.
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18M.3.SL.TZ2.10b.ii:
An amplifier can increase the power of the signal by 12 dB. Determine the minimum number of amplifiers required.
- 18N.3.SL.TZ0.10a.ii: The refractive indices of the glass and cladding are only slightly different. Suggest why...
- 18N.3.SL.TZ0.10b.iii: Suggest whether this fibre could be used to transmit information at a frequency of 100 MHz.
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18M.3.SL.TZ2.10b.i:
Calculate the maximum attenuation allowed for the signal.
- 19M.3.SL.TZ2.12bii: An input signal to the fibre consists of wavelengths that range from 1299 nm to 1301 nm. The...
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18M.3.SL.TZ2.10c:
In many places clad optic fibres are replacing copper cables. State one example of how fibre optic technology has impacted society.
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16N.3.SL.TZ0.14a:
State the main physical difference between step-index and graded-index fibres.
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17M.3.SL.TZ1.8a.i:
State two advantages of optic fibres over coaxial cables for these transmissions.
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17M.3.SL.TZ1.8a.ii:
Suggest why infrared radiation rather than visible light is used in these transmissions.
- 16N.3.SL.TZ0.14b: Explain why graded-index fibres help reduce waveguide dispersion.
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18M.3.SL.TZ2.10a:
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.
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18M.3.SL.TZ2.10b.iii:
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 λA and ray B has a wavelength of λB. Discuss the effect that the difference in wavelength has on the rays as they pass along the fibre.
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18N.3.SL.TZ0.10b.i:
Show that the longest path is 66 m longer than the shortest path.
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19M.3.SL.TZ2.12biii:
Explain the shape of the signal you sketched in (b)(ii).
- 19M.3.SL.TZ2.12biv: A signal consists of a series of pulses. Outline how the length of the fibre optic cable...
- 19M.3.SL.TZ2.12a: Outline the differences between step-index and graded-index optic fibres.
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18M.3.SL.TZ1.9a:
Calculate the critical angle at the core−cladding boundary.
- 19N.3.SL.TZ0.9a: Describe why a higher data transfer rate is possible in optic fibres than in twisted pair...
- 19N.3.SL.TZ0.9b(i): State one cause of attenuation in the optic fibre.
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17N.3.HL.TZ0.15a:
Calculate the maximum angle β for light to travel through the fibre.
Refractive index of core = 1.50
Refractive index of cladding = 1.48 -
17M.3.SL.TZ2.10b.i:
Identify the features of the output signal that indicate the presence of attenuation and dispersion.
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19N.3.SL.TZ0.9b(ii):
Determine the distance at which the signal must be amplified.
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18N.3.SL.TZ0.10b.ii:
Determine the time delay between the arrival of signals created by the extra distance in (b)(i).
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17M.3.SL.TZ1.8b:
A signal with an input power of 15 mW is transmitted along an optic fibre which has an attenuation per unit length of 0.30 dBkm–1. The power at the receiver is 2.4 mW.
Calculate the length of the fibre.
- 20N.3.SL.TZ0.13: A single pulse of light enters an optic fibre which contains small impurities that scatter...
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17M.3.SL.TZ2.10a:
The diagram shows a ray of light in air that enters the core of an optic fibre.
The ray makes an angle A with the normal at the air–core boundary. The refractive index of the core is 1.52 and that of the cladding is 1.48.
Determine the largest angle A for which the light ray will stay within the core of the fibre.
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18M.3.SL.TZ1.9b:
The use of optical fibres has led to a revolution in communications across the globe. Outline two advantages of optical fibres over electrical conductors for the purpose of data transfer.
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18M.3.SL.TZ1.9c.ii:
Calculate the power of the output signal after the signal has travelled a distance of 3.40 km in the fibre.
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18M.3.SL.TZ1.9c.iii:
Explain how the use of a graded-index fibre will improve the performance of this fibre optic system.
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17M.3.SL.TZ1.8c:
State and explain why it is an advantage for the core of an optic fibre to be extremely thin.
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17M.3.SL.TZ2.10b.ii:
The length of the optic fibre is 5.1 km. The input power of the signal is 320 mW. The output power is 77 mW. Calculate the attenuation per unit length of the fibre in dBkm–1.
