Download to read offline
Fiber Optics Engineering physics-BBT .pdf
- 2. What is anOptical Fibre? • Optical Fiber are long, thin cylindrical system of glass (silica) or clear plastic about the diameter of a hair. • The fiber which are used for optical communication are wave guides made of transparent dielectrics. • The function of optical fiber is to guide visible and infra red light over long distances. Optical Fibre
- 3. Components of OpticalFibre Core: central tube of very thin size made of optically transparent dielectric medium and carries the light from transmitter to receiver. The core diameter can vary from about 5𝜇𝑚 to 10𝜇𝑚. Optical Fibre Cladding: outer optical material that surrounds the core having refractive index lower than core. It helps to keep the light within the core throughout the phenomenon of total internal reflection. Buffer Coating: plastic coating that protects the fibre made of silicon rubber. The typical diameter of fibre after coating is 250𝜇𝑚 to 300𝜇𝑚.
- 4. Working Principal ofOptical Fibre Total Internal reflection • In the optical fibre the rays undergo repeated total internal reflection until it emerges out of the other end of the fibre even if the fibre is bent. Optical Fibre
- 5. Classification of OpticalFibre On the basis of number of modes of propagation, the optical fibres are classified into two types. (i) Single Mode (Monomode) Optical Fibre (SMF) (ii) Multimode Optical Fibre (MMF) Optical Fibre 5 Single Mode Fiber
- 6. (i) Single Mode(Monomode) Optical Fibre (SMF) • Single Mode Optical Fibre is that which is designed for the transmission of a single ray or mode of light as a carrier. • It uses laser light source. • It has very small core and carries only one beam of light. • It is used for long distance signal transmission. • This fibre is used in telecom and CATV networks. Optical Fibre
- 7. (ii) Multimode OpticalFibre (MMF) • The optical cable that carries signals through different paths. • Multi-mode optical fibre cable is about 10 times thicker than single mode cable. • Multi-mode fibre cable can send information only over relatively short distance. • They are used to link computer networks together. Optical Fibre
- 8. Comparison: Multi-mode v/sSingle mode Optical Fibre Multi-mode Single mode • Low cost source (LED) • Low cost connectors. • Lower installation cost • Higher fiber cost • Lower system cost • Higher loss • Lower bandwidth • Distance up to 2 km • High cost source (Laser) • High cost connectors. • Higher installation cost • Lower fiber cost • Higher system cost • Lower loss • Higher bandwidth • Distance to 60 km+
- 9. On the basisof refractive index, there are two types of optical fibres (i) Step index and (ii) Graded index (i) Step Index Optical Fibre • In this type refractive index of core refractive index (𝜇1) is uniform, cladding refractive index (𝜇2) is uniform and core refractive index is greater than cladding refractive index. • The rays entering at different angles travel different paths and emerge out the fibre at different time along different directions. • An input pulse widen as it travels along the fibre. Optical Fibre
- 10. (ii) Graded index •In graded index optical fibre, the core has non uniform refractive index which increases from axis towards cladding, core refractive index is constant and core refractive index is greater than cladding refractive index. • The ray entering is continuously bent and travels a periodic path along the axis. • The rays entering at different angles follow different paths with same period in space and time. Thus there is a periodic self-focusing of the rays. The pulse dispersion is less than that of in step index fiber. Optical Fibre
- 11. Note: The multi-modefibre (MMF) can either be step index or graded index whereas the single mode fibre (SMF) is usually a step index type. Optical Fibre
- 12.
- 13. Acceptance Angle • Theacceptance angle of optical fiber is defined as the maximum value of angle of incidence at the entrance end of the fiber at which the angle of incidence at the core-cladding interface is equal to critical angle of the core. • The angle of incidence at launching end with axis is i show that the angle of incidence at core-cladding interface is equal to critical angle C then i is acceptance angle. Optical Fibre
- 14. • The refractiveindex of core w.r to surrounding of refractive index 𝜇𝑂 is 𝜇1 𝜇𝑜 = 𝑠𝑖𝑛𝑖 𝑠𝑖𝑛𝜃 …(1) • From ∆𝑂𝑃𝑁, 𝜃 = 90 − 𝐶 and 𝑠𝑖𝑛𝜃 = sin 90 − 𝐶 = 𝑐𝑜𝑠𝐶 • Equation 1 becomes, 𝜇1 𝜇𝑜 = 𝑠𝑖𝑛𝑖 𝑠𝑖𝑛𝜃 or, 𝑐𝑜𝑠𝐶 = 𝜇𝑜 𝜇1 𝑠𝑖𝑛𝑖 …(2) Optical Fibre AcceptanceAngle (Contd…)
- 15. • The refractiveindex of cladding w.r. to core is 𝜇2 𝜇1 = 𝑠𝑖𝑛𝐶 𝑠𝑖𝑛90 = 𝑠𝑖𝑛𝐶 or, sinC = 𝜇2 𝜇1 …(3) • Squaring and adding Eq. 2 and Eq. 3 𝜇2 2 𝜇1 2 + 𝜇𝑜 2 𝜇1 2 𝑠𝑖𝑛2 𝑖 = 1 or, 𝑠𝑖𝑛𝑖 = ± 𝜇1 2−𝜇2 2 𝜇𝑜 2 ….(4) • The acceptance angle, 𝑖 = 𝑠𝑖𝑛−1 𝜇1 2−𝜇2 2 𝜇𝑜 2 …(5) • If the fiber is in air, 𝜇𝑂 = 1, the acceptance angle 𝑖 = 𝑠𝑖𝑛−1 𝜇1 2 − 𝜇2 2 …(6) Optical Fibre
- 16. Numerical Aperture (NA) •Numerical Aperture is a figure of merit that is used to describe the light gathering or light collecting capacity of an optical fibre. • The larger the magnitude of NA, the greater the amount of light accepted by the fibre from external source. • Mathematically, NAis sine of the acceptance angle. 𝑁𝐴 = 𝑠𝑖𝑛𝑖 = 𝜇1 2 − 𝜇2 2 …(7) • Now, 𝜇1 2 − 𝜇2 2 = 𝜇1 − 𝜇2 𝜇1 + 𝜇2 • Since 𝜇1 ≈ 𝜇2, 𝜇1 2 − 𝜇2 2 = 2𝜇1 𝜇1 − 𝜇2 = 2𝜇1 2 𝜇1−𝜇2 𝜇1 = 2𝜇1 2 ∆ • Where ∆ = 𝜇1−𝜇2 𝜇1 is fractional refractive index change. • Thus numerical aperture is 𝑁𝐴 = 𝜇1 2∆ …(8) Optical Fibre
- 17. 1. A glassclad fibre is made with core glass of refractive index is1.5 and cladding is doped to give a fractional index difference of 0.005. Find (i) the cladding index, (ii) the critical internal reflection angle, (iii) the external critical acceptance angle, (iv) numerical aperture and (v) acceptance angle. Solution: Given: core refractive index 𝜇1 = 1.5, Fractional index difference ∆= 0.005 Optical Fibre
- 18. Optical Fibre Optical Fibre 1.A glass clad fibre is made with core glass of refractive index is1.5 and cladding is doped to give a fractional index difference of 0.005. Find (i) the cladding index, (ii) the critical internal reflection angle, (iii) the external critical acceptance angle, (iv) numerical aperture and (v) acceptance angle.
- 19. Optical Fibre Optical Fibre Alight ray passes from passes in an optical fiber with refractive index of core 1.5 and that of cladding 1.48. What are the values of critical angle, fractional refractive index change, acceptance angle and the numerical aperture ?
- 20. Normalized frequency orV-Number • The normalized frequency (or V-Number) determines how many modes a fibre can support. It is a dimensionless quantity. • The V-number is related to the fibre’s cutoff wavelength. • The V- number which is often used in the context of step index fibre. • It is defined as 𝑉 = 2𝜋𝑎 𝜆 𝑁𝐴 = 2𝜋𝑎 𝜆 𝜇1 2Δ where λ is the vacuum wavelength, a is the radius of fibre core, and NA is numerical aperture. • When the number of modes in the fiber is very large, and sum up, then the number of modes that propagate inside the optical fiber is approximately given 𝑁 = 𝑉2 2 • For a single mode fiber, V < 2.4 Optical Fibre Optical Fibre
- 21. Fibre Loss • 𝑃𝑖is input power in the fiber then output power 𝑃𝑂 at a distance L is given by 𝑃𝑂 = 𝑃𝑖𝑒−𝛼𝐿 …(1) • 𝛼 is attenuation coefficient. • 𝛼 = 1 𝐿 𝑙𝑛 𝑃𝑖 𝑃𝑜 in per km. • For simplicity in calculating signal attenuation in fibre, the attenuation coefficient is expressed in dB/km given by 𝛼 = 10 𝐿 𝑙𝑜𝑔10 𝑃𝑖 𝑃𝑜 …(2) Optical Fibre Optical Fibre • As the light travels along a fibre, its power decreases exponentially with distance.
- 22. Dispersion • The amplitudeof signal is reduced due to losses. • In transmission of signals, it is expected that the signal observed at the other end unchanged in shape and width. • In some cases pulse at the output is wider than the pulse at input i.e. pulse gets distorted. • The distortion of pulse is due to dispersion effect. • The dispersion is determined using 𝐷 = 𝜇1 𝐶 ∆ 1−∆ in terms of nano second per km. • This term is also called inter modal factor. Where c is speed of light. • In case of graded index fibre, time delay due to intermodal dispersion is 𝐷 = 𝜇1∆2 8𝑐 Optical Fibre Optical Fibre
- 23. Applications of OpticalFiber (i) Communication • Optical fiber is mostly used in communication. • It is use in Wi-Fi router, landline phone and server- connector. • A single optical fibre can carry over 3,000,000 full duplex voice calls or 90,000 TV channels. So it is use in broad bandwidth. (ii) Millitary • Optical fibre is used to make military equipments and weapons. • It is also use to make antenna to communicate in far areas. Optical Fibre
- 24. (ii) Sensor • Mostof the sensors are made from optical fiber. • Optical fibers also use to make detectors i.e. metal detector. (iii) Medical Used as light guides, imaging tools and also as lasers for surgeries (iv) Other applications Optical fibre is used to make lamps, decorative applications, art, toys, microscope, and outer body of the devices. Optical Fibre Optical Fibre
- 25. 2. Find thecore diameter necessary for single mode operation at 450nm in step index fibre with 𝜇1 = 1.48 and 𝜇2 = 1.47. what is the numerical aperture and maximum acceptance angle of this fibre? Given V-number = 2.405. Solution We have V-number 𝑉 = 𝜋𝑑 𝜆 𝜇1 2 − 𝜇2 2 or, 𝑑 = 𝑉𝜆 𝜋 1 𝜇1 2−𝜇2 2 = 2.405×450×10−9 𝜋 1 1.482−1.472 = 2𝜇𝑚 Numerical Aperture 𝑁𝐴 = 𝜇1 2 − 𝜇2 2 = 1.482 − 1.472 = 0.1717 Acceptance angle 𝑖 = 𝑠𝑖𝑛−1 𝑁𝐴 = 𝑠𝑖𝑛−1 0.1717 = 9.88𝑜 Optical Fibre Optical Fibre
- 26. 3. A stepindex fibre is made with a core of index 1.52, a diameter of 29𝜇m, fractional difference index of 0.007. it is operated at a wavelength of 1.3𝜇m. Find (i) the fibre V-number and (ii) number of modes the fibre will operate. Optical Fibre
- 27. 4. Calculate themaximum value of ∆ in the case of single mode fibre of core diameter 10𝜇𝑚 and core refractive index 1.5. The fibre is coupled to a light source of wavelength 1.3 𝜇m. Also calculate the refractive index of cladding and acceptance angle. The maximum V-number = 2.405. Solution V- number of the fibre, 𝑉 = 𝜋𝑑 𝜆 𝑁𝐴 = 𝜋𝑑 𝜆 𝜇1 2Δ Δ = 1 2 𝑉𝜆 𝜋𝑑𝜇1 2 = 1 2 2.405 × 1.3 × 10−6 𝜋 × 10 × 10−6 × 1.5 2 = 0.0022 Since Δ = 𝜇1−𝜇2 𝜇1 𝜇2 = 𝜇1 1 − ∆ = 1.5 1 − 0.0022 = 1.496 Acceptance angle, 𝑖 = 𝑠𝑖𝑛−1(𝑁𝐴) 𝑖 = 𝑠𝑖𝑛−1 𝜇1 2Δ = 𝑠𝑖𝑛−1 1.5 × 2 × 0.0022 = 5.71𝑜 Optical Fibre Optical Fibre
- 28. 5. A certainoptical fibre has an attenuation of 3.5dB/km at 850nm. If 0.5mW of optical power is initially launched with optical fibre, calculate the power level after 4km. Solution Attenuation, 𝛼 = 10 𝐿 𝑙𝑜𝑔 𝑃𝑖 𝑃𝑜 or, 𝑃𝑜 = 𝑃𝑖 𝑎𝑛𝑡𝑖𝑙𝑜𝑔 𝛼 𝐿 10 or, 𝑃𝑜 = 0.5 𝑎𝑛𝑡𝑖𝑙𝑜𝑔 3.5×4 10 = 19.9 𝜇𝑊 Optical Fibre Optical Fibre
- 29. 5. A 15kmoptical fibre link uses fibre with a loss of 1.5dB/km. The fibre is joined every km with connectors, which give attenuation of 0.8dB/km each. Find the minimum optical power which must be launched with the fiber to maintain a mean optical power level of 0.3𝜇𝑊 at the detector. Connector loss for 15 km = 0.8 × 15 = 12𝑑𝐵 Fiber loss for 15 km = 1.5 × 15 = 22.5𝑑𝐵 Total loss, α = 12 + 22.5 = 34.5𝑑𝐵 Optical Fibre
- 30. 6. Compute thedispersion per km of length and total dispersion in a 10km length of single index fibre if refractive index of core is 1.558 and ∆= 0.026 Optical Fibre Optical Fibre
- 31. 7. A stepindex multimode fiber has core index of 1.5 and cladding index of 1.498. Calculate (i) intermodal factor for the cable and (ii) total dispersion in 18 km length. Solution we have ∆= 𝜇1−𝜇2 𝜇1 = 1.5−1.498 1.5 = 0.0013 Intermodal factor, 𝐷 = 𝜇1 𝑐 ∆ 1−∆ = 1.5 3×108 0.0013 1−0.0013 = 6.5 𝑛𝑠/𝑘𝑚 Total dispersion in 18 km = 18× 6.5 = 117 ns Optical Fibre Optical Fibre