1)discuss advantages of optical fibers over conventional copper cables.ANS: Advantageous Of Optical Fibers Communication: 1. Information bandwidth is more. 2. Optical fibers are small in size and light weighted. 3. Optical fibers are more immune to ambient electrical noise, electromagnetic interference. 4. Cross talk and internal noise are eliminated in optical fibers. 5. There is no risk of short circuit in optical fibers. 6. Optical fibers can be used for wide range of temperature. 7. A single fiber can be used to send many signals of different wavelengths using Wavelengths Division Multiplexing (WDM). 8. Optical fibers are generally glass which is made up of sand and hence they are cheaper thancopper cables. 9. Optical fibers are having less transmission loss and hence less number of repeaters are used. 10. Optical fibers are more reliable and easy to maintain.

Disadvantageous Of Optical Fibers Communication:

1. Attenuation offered by the optical fibers depends upon the material by which it is made.2. Complex electronic circuitry is required at transmitter and receiver.3. The coupling of optical fibers is difficult.4. Skilled labors are required to maintain the optical fiber communication.5. Separated power supply is required for electronic.

2)explain i)cut off wavelength,ii)mode field diameter.ANS:i)cut-off wavelength:-the cut-off wavelength of single mode fiber separates the single mode from the multimode region.At v>2.405,single mode operation occur for step index fiber.only lp01 mode is available at this wavelength.The waveguide dispersion is shown in the below fig.

A 2m length of fiber is used to determine the cut-off wavelength,it comes a single 14cm radius loop.A light having line width not more than 10nm is installed into fiber,in order to excite both the modes lp01 and lp11 uniformly.

The logarithmic ratio,The largest wavelength at which the higher order lp11 mode power relative to the fundamental lp01 power is reduced to 0.1db is known as effective cut-off wavelength.The resultant total dispersions for the modes shown in fig(1)is shown in fig(2)

ii)mode field diameter:-it is a primary parameter of single mode fiber.it is obtained from the mode field distribution of the fundamental mode.The figure shows the distribution of light in a single mode fiber.

Unlike multimode fibers,in single mode fiber,all the light that propagates through the fiber is not carried in the core.it depends on thevvalue.if vis larger than the percentage of optical power at the core is large.The properties like splice loss,cut-off wavelength,bending loss can be illustrated by using MFD.In order to find the MFD for field intensity must be calculated as,MFD=2W0,

Where, 2W0=spot sizeTo avoid complexity,E(r)can be taken as,E(r)=

3) a)briefly explain bending & scattering losses.ANS: 1. Bending LossBending loss is further classified into,(i) Macro bending loss-and(ii) Micro bending loss.(i) Macro bending LossThe light travels in fiber due to occurrence of total internal reflection inside the fiber at theinterface of core and cladding. However the light beam forms a critical angle with the fiber'scentral axis at the fiber face. When the fiber is bend and the light beam travelling through fiberstrikes at the boundary o f core at an angle greater than critical angle then the beam fails toachieve total internal reflection. Hence this beam is lost through the cladding.

Micro bending LossMicro bending loss is caused by micro-deformations of the fiber axis. The beam which travels atthe critical propagation angle before incident on micro-deformations will change the angle ofpropagation after being reflected by the imperfection of fiber and hence the condition for totalinternal reflection is lost and the beam escapes from the fiber through cladding.

2. Scattering LossA light beam propagating through the fiber core at critical angle or less will change its directionafter hitting on an obstacle in the core region. The obstacle can be any particle in core that mayhave diffused inside the core at the time of manufacturing when the light beam hits the particle it get scattered and due to this total internal reflection is not achieved hence, the beam is lostthrough the cladding.

b)explain about splicing technique.ANS: Splicing: A permanent joint formed between two individual optical fibers in the field is known assplicing. The fiber splicing is used to establish optical fiber links, where smaller fiber lengths areneeded to be joined and where there is no requirement for repeated connection anddisconnection.Splicing can be divided into two broad categories depending on the splicing techniqueutilized. These are fusion-splicing, mechanical or welding splicing..Fusion SplicingFusion splicing of single fibers involves the heating of the two prepared fiber ends to theirfusing point with sufficient axial pressure between the two optical fibers. It is essential that thestripped fiber ends are adequately positioned and clamped with the aid of inspection microscope.The most widely used heating technique is an electric arc. This technique offers advantageof consistent, easily controlled heat with adaptability for use under field conditions.The welding of 2 fibers can be shown as illustrated in the following figure

The figure shows basic arc fusion process, which involves the rounding of the fiber endswith a low energy discharge before pressing the fibers together and fusing with a stronger arc.This technique is called perfusion. It removes the requirement for fiber end preparation. Ithas been utilized with multimode fibers giving average splice losses of 0.09 dB.adhesive splicing:A common method involves the use of ah accurately produced rigidalignment tube into which the prepared fiber ends are permanently, bonded. This snug tubesplice may utilize a glass or ceramic capillary with an inner diameter just large enough to acceptthe optical fibers. Transparent adhesive (e.g. epoxy resin) is injected through a transverse bore indie capillary to give mechanical sealing and index matching of the splice. However, in general,snug tube splices exhibit problems with capillary tolerance requirements.

A mechanical splicing-technique which avoids the critical tolerance requirements of the snugtube splice is shown in figure 4.3.This loose tube splice uses an over sized square section metaltube which easily accepts the prepared fiber ends. Transparent adhesive is first insulated in thetube followed by the fibers. The splice is self aligning when the fibers are curved in the sameplane, forcing the fiber ends simultaneously into the same corner of tube.

Other common splicing techniques involve the use of grooves to secure the fibers to be jointed.A simple method utilized a V-groove into which the two prepared fiber ends are pressed. The Vgroovesplice which is shown in figure gives alignment of the prepared fiber ends throughinsertion in the groove. The splice is made permanent by securing the fibers in the V-groove withepoxy resin (i.e., transparent adhesive).

Multiple splices:Multiple simultaneous fusions splicing of an array of fibers in a ribbon cable has beendemonstrated for both multimode and single mode fibers. In both cases a five fiber ribbon wasprepared by scoring and breaking prior to pressing the fiber ends on to a contact plate to avoiddifficulties, with varying gaps between the fibers to be fused

4)explain in detail about step index fiber.

ANS:There are two fiber types(i) Step index fiber(ii) Graded index fiber.(i) Step Index FiberStep index fiber is further divided in two types,1. Single mode step index fiber2. Multi mode step index fiber.Single mode step index fiber is shown below,

The typical dimension of core is 8 to 12 m and cladding is 125 m.In step index fiber, the refractive index of the core is uniform and at the cladding boundary, itundergoes a step change.In single mode step index fiber, there is only one mode of propagation. The multimode stepindex fiber is shown below,

In multimode step index fiber, hundreds of modes are present.The typical dimension of core is 50 to 200 m and cladding is 125 to 400 m. Multimode fiberhas several advantages, which includes, the transmitting the light directly in to fiber using LED.

5)briefly explain fiber material.ANS: Most of the fibers are made up of glass consisting of either Silica (SiO2) or .Silicate. Highloss glass fibers are used for short-transmission distances and low-loss glass fibers are used for long distance applications. Plastic fibers are less used because of their higher attenuation thanglass fibers.glass fiber:-Glass fiber is a mixture of selenides, sulfides and metal oxides. It can be classified into,Glass Fibers.Glass is made of pure SiO2 which refractive index 1.458 at 850 nm. The refractive index of SiO2can be increased (or) decreased by adding various oxides are known as dopant.The oxides GeO2 or P2O3 increases the refractive index and B2O3 decreases the refractive indexof SiO2 The various combinations are,(i) GeO2 SiO2 Core; SiO2 cladding(ii) P2 O3 SiO2 Core; SiO2 cladding(iii) SiO2 Core; B2O3, - SiO2 cladding(iv) GeO2- B2O3- SiO2, Core; B2O3 - SiO2 cladding.From above, the refractive index of core is maximum compared to the cladding.(1) Halide Glass FibersA halide glass fiber contains fluorine, chlorine, bromine and iodine. The most commonHalide glass fiber is heavy "metal fluoride glass". It uses ZrF4 as a major component. Thisfluoride glass is known by the name ZBLAN. since it is constituents are ZrF4, BaF2, LaF3 A1F3,and NaFThe percentages of these elements to form ZBLAN fluoride glass is shown as follows,Materials Molecular percent:

MaterialsMolecular percentage

ZrF4BaF2LaF3A1F3NaF54%20%4.5%3.5%18%

These materials add up to make the core of a glass fiber. By replacing ZrF4 by HaF4, the lowerrefractive index glass is obtained.The intrinsic losses of these glasses is 0.01 to 0.001 dB/km(2) Active Glass FibersActive glass fibers are formed by adding erbium and neodymium to the glass fibers. The abovematerial performs amplification and attenuation. (3) Chalgenide Glass FibersChalgenide glass fibers are discovered in order to make use of the nonlinear properties ofglass fibers.It contains either "S", "Se" or "Te", because theyare highly nonlinear and it also contains one element from Cl, "Br, Cd,Ba orSi.The mostly used chalgenide glass is AS2-S3, AS40S58Se2 is used to make the core and AS2S3 isused to make the cladding material of the glass fiber. The insertion loss is around 1 dB/m.Plastic Optical FibersPlastic optical fibers are the fibers which are made up of plastic material. The core of thisfiber is made up of Polymethylmethacrylate (PMMA) or Perflourmated Polymer (PFP).Plastic optical fibers offer more attenuation than glass fiber and is used for short distanceapplications. These fibers are tough and durable due to the presence of plastic material. Themodulus of this plastic material is two orders of magnitude lower than that of silica and even a 1mm diameter graded index plastic optical fiber can be installed in conventional fiber cableroutes. The diameter of the core of these fibers is 10-20 times larger than that of glass fiberwhich reduces the connector losses without sacrificing coupling efficiencies. So we can useinexpensive connectors, splices and transceivers made up of plastic injection-moldingtechnology.Graded index plastic optical fiber is in great demand in customer premises to deliver high-speedservices due to its high bandwidth

6)explain about dispersion.

7) a)write a short note on LED sructure.ANS: High radiance is obtained by restricting the emission to a small active region within thedevice. A well is etched in a substrate (GaAs) to avoid the heavy absorption of the emitterradiation and to accommodate the fiber. These structures have a low thermal impedance in theactive region and hence radiance emission into the fiber. Double hetero structures are used to getincreased efficiency and less optical absorption. The structure of a high radiance etched well DHS(Double Hetero structure) surface emitter which is also known as burrus type LED is as shown infigure (4.5).

This structure emits light in band of 0.8 to 0.9 um wavelength. The plane of the active lightemitting region is made perpendicular to the fiber axis. The fiber is cemented in a well matchedthrough the substrate of the fiber so that maximum emitted light is coupled to the fiber. Due tolarge band gap conjoining area, the internal absorption is less and the reflection coefficient at theback crystal face is high, hence forward radiance is good. The active area in circle is of 50m indiameter and up to 2.5m thick. The emission from this active area is isotropic with 120 halfpower beam width is used for practical purpose. Isotropic pattern from a surface emitter islambertian pattern.The source is equally bright when viewed from any direction but power diminishes as coswhere $ is the angle between viewing direction and to the normal to the surface. Power is downto 50%, when = 60, so that the total half power beam width is 120. The power coupled into amultimode step index fiber may be estimated from the relationship.PC=(1-r)ARD (NA) 2..(1)Where, PC = Power coupled into fiberr = Fresnel reflection coefficientA = Emission area of sourceRD= Radiance of the sourceNA = Numerical aperturePower coupled into the fiber depends on(i) Distance and alignment between emission area and the fiber.(ii) Medium between the emitting area and the fiber.(iii) Emission pattern of SLEDAddition of Epoxy resin in the etched well reduces the refractive index mismatch and increasesthe external power efficiency of the device. Hence the power coupled in the double heterostructure surface emitters are more than Pc(optical power) that is given by equation (1), Forgraded index fiber-direct coupling requires the source diameter of about one half the fiber corediameter. b)derive relationship b/w photon density & injection current density for a laser diode optical source.

8)a multimedia step index fier has a relative refractine index difference of 1% and a core refractive index of 1.5.the number of modes operating at a wavelength of 1.3micro meter is 1500.determine the diameter of fiber core.

9)explain the difference b/w LED&ILD.ANS: Parameter1)principle of operation2)o/p beam3)spectral width4)data rate5)coupling efficiency6)insertion distance7)cost8)required current9)wavelength available10)application LEDSpontaneous emissionNon-coherentBroas spectrumLowLowSmallerLow50-100ma0.66-1.65micro meterModerate velocity,Low data rate ILDStimulated emissionCoherentVery narrowVery highVery ighBroaderHighHouse hold current0.78-1.65micro meterLong distance,high data rate

10)a multimode step index fiber with a core diameter of 90 micro meter and a relative refractive index difference is 1.5% and is operating at a wavelength of 0.85 micrometer.if the core refractive index is 1.5.estimate normalised frequency for the fiber and the no.of aided modes.