2. Fibre Optic Telecommunications (a) Optical fibres can carry light over very l
ID: 1770214 • Letter: 2
Question
2. Fibre Optic Telecommunications (a) Optical fibres can carry light over very long distances Loss in transmission sys- tems (also known as attenuation) is measured in Decibels per unit length (usually per km), with the attenuation coefficient a measured in dB/km. The signal carried by a medium is 10% of its original strength after 2 km. (i) What is the attenuation in dB/km of the carrier medium? (i) What is the signal strength after z = 1 km and after z 100 km? (5 marks) (b) Fibre Bragg Gratings (FBG) are used as narrowband filters in telecommunications. What property of the fibre core is modified and what is the most common way to fabricate FBGs in optical fibres? State the equation for the Bragg condition and use diagrams to support your answer and explanation. (5 marks) (c) Referring to the lefthand figure below, what does TIR stand for and how does it work? Make sure that you define the Critical Angle 0c in terms of the refractive indices for the two optical media, by stating the formula and defining the terms. Use diagrams to support your answer. (5 marks) Optical fibre Low High NA Light Ray Silica cladding Polymer jacket n2 n. Ge doped Silica core 1% higher index (d) For a given optical fibre the number of transverse modes that may propagate along the fibre is a function of the fibre geometry and the propagating wavelength. Since the effect of the geometry scales with wavelength, i is therefore convenient to introduce a normalized quantity called the Normalized Frequency or V-parameter (V). Single moded behavior (i.e. only one propagating mode) exists when V 2.405 the fibre may support multiple optical modes. Assume that a fibre (righthand figure above) has a core diameter of 9 microns (um) and a cladding refractive index n1-1.5 (i) Calculate the maximum refractive index value of the fibre core (n2) allowed for single moded behavior (ie. V2.405) at = 1.5 m (ii) How many modes (M) will this same fibre support at = 500 nm?Explanation / Answer
2) a) Formula: I/I0 = e –alpha*z
Where, I = Intensity after travelling Z km, I0 =Initial intensity, alpha = attenuation
Substituting all the values in above equation,
(10/100) = e –alpha*2km => (100/10) = e alpha*2km
Taking Logarithm both side , ln(10) = alpha* (2km) => alpha = ln(10)/2km
alpha = (2.3025)/2km = 1.1525 (dB/km)
Submitting in the formula: I/I0 = e –alpha*z => I1/I0 = e –1.1525*1km
In this case, I100 = ?, alpha =1.1525, Z1= 100 km
Submitting in the formula: I/I0 = e –alpha*z => I100 = I0 / 1.12834E+50
=> I100 = 8.8626E-51* I0 dB
(b) To support narrow band filter, diameter of core should be made small
Bragg’scondition: 2*d*sin = n*wavelength ; d= width between slits, n = order, =angle of diffraction
c) TIR : Total Internal Relation:
By Snell’s law, n1/n2 = sin 1/ sin 2
Here, 2 > 900 then whole light reflects in the core with out loss of energy
when light travels from denser to rarer media, critical angle is the angle of incidence at which the light ray moves parallel to the boundary of separation of both media . Here, 2 = 900
Snell’s law become, n1/n2 = sin 1/ sin 90 => sin-1(n1/n2) = 1 =Angle of incidence =Critical angle
(d) (i)Given, ncore =1.5, ncladd = ?, r= 4.5 micro meter, wavelength =1.5 micro meter, V=2.405
V= 2 * (pi) * NA * r / wavelength
NA = Numerical Aperture= [(ncore)2- (ncladd)2]0.5
a = fiber radius =
lambda = wavelength =
Substituting all the values, we get
Substituting in the above equation,
We get, V =16.39
Therefore No. of modes =V/2.405 =6.81 ~7 modes are possible in the given fibre
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