3. Total Internal Reflection When a ray of light travels from a denser to a rarer medium such that the angle of incidence is greater than the critical angle, the ray reflects back into the medium. This is called total internal reflection.
4. Optical Fibres Consists of a very thin glass core surrounded by a material of slightly lower refractive index called cladding The thin fibre can be bent without breaking and a ray of light can be sent down the fibre’s core Total internal reflection takes place at the boundary of the core and the cladding
5. Acceptance angle The maximum angle of incidence that a ray can make that will result in total internal reflection is called acceptance angle
6. PROBLEM The refractive index of the core of an optical fibre is 1.50 and that of the cladding is 1.40.Calculate the acceptance angle of the fibre. Ans:330 The refractive index of the core of an optical fibre is 1.50 and the critical angle of the core- cladding boundary is 750.Calculate the refractive index of the cladding. Ans: 1.45
7. Material Dispersion Light of different wavelengths have different refractive index and hence come out of the fibre at different times. This is called material dispersion
9. Modal dispersion Rays that undergo many internal reflections are said to follow a high order mode paths Rays undergoing fewer reflections follow low order mode paths Set of rays having same wavelength reach the end at different times due to different paths taken. This is called modal dispersion
10. Monomode & Multimode fibre In multimode fibres, the core has a diameter of about 100μm and the cladding is about 20μm thick The rays passing through multimode fibres undergo material as well as modal dispersion In monomode fibres, the core has a diameter of about 8 -10μm and the cladding is about 125μm thick. Rays follow just one path eliminating modal dispersion
11. Step Index fibre The refractive index of core is constant The refractive index of cladding is constant The refractive index of cladding is slightly lower than that of core
12. Graded index fibre Refractive index of core decreases smoothly from the centre to the outer edge Refractive index of cladding is constant
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14. ATTENUATION Attenuation in an optic fibre is caused by the impurities of the glass core. The amount of attenuation depends on the wavelength of the light being transmitted. Power loss in decibels is defined as Power loss = 10log (Pfinal / Pinitial) in dB Thus a power loss of 16 decibels means that the initial power of, say,8.0mW has been reduced to 0.2mW
15. PROBLEM An amplifier amplifies an incoming signal of power 0.34mW to a signal of power 2.2mW.Calculate the power gain of the amplifier in decibels. Ans: 8.1dB A signal of power 12mW is input to a cable of specific attenuation 4.0 dB/km. Calculate the power of the signal after it has travelled 6.0km in the cable. Ans: 0.048mW
17. Attenuation & Wavelength The specific attenuation ( power loss in dB per unit length ) actually depends on the wavelength of the radiation travelling along the optic fibre The graph shows minima at 1310nm and 1550nm, which implies that these are desirable wavelengths for optimal transmission These are infra red wavelengths
18. DETECTION The light that enters an optic fibre travels down the length of the fibre and the arrival of light is registered by a photodiode In the absence of any light, falling on the photodiode, the current is zero When light of a specific wavelength falls on the photodiode, a current flows. The magnitude of the current is proportional to the intensity of light
20. NOISE Source of noise in a cable: Random motion of electrons which creates additional electric fields contaminating the signal. This increases with temperature. Lightning Charged particles emitted by the sun during intense solar activity
21. NOISE IN OPTICAL FIBRES Main source is the dark current of the photodiode. This is the small current that flows even when the photodiode is dark Signal to noise ratio (SNR) is defined as SNR = 10log Psignal / Pnoise
22. PROBLEM The minimum SNR considered acceptable for a certain signal is 30dB.If the power of the noise is 2.0mW, calculate the least acceptable signal power. The SNR in a certain signal is 10dB.the signal passes through an amplifier of gain 6.0dB.What will be the signal to noise ratio after amplification?
