Figure – Total Internal Reflection 
Hence at critical angle θ1= θc and θ2= 90o . Using Snell‘s law: n1 sin θ1 = n2 sin θ2 n1 sin θc = n2 sin90o sin θc = n2 / n1 θc = sin1 (n2 / n1) 
According to Snell’s law at point A n0 Sin θ0 = n1 Sin θ1 Sin θ0= (n1 / n0) Sin θ1 ………(1) According to Snell’s law at point B n1 Sin(90 θ1) = n2 Sin90 ………(2) n1 Cosθ1 = n2 as (Sin90=1) Cosθ1 = n2 /n1 Sinθ1 = (1Cos2 θ1)1/2 Sinθ1= (1 (n2 /n1)2)1/2 Sinθ1= ( n12 n22 )1/2/ n1 ………(3) We know Sin θ0= (n1 / n0) Sin θ1 from equation (1) Substitute the value of Sinθ1 from equation (3) Sinθ0= (n1 / n0) *( n12 n22 )1/2/ n1 On simplification Sinθ0= ( n12 n22 )1/2/ n0 θ0=Sin1 ( n12 n22 )1/2/ n0 Acceptance Angle is θ0=Sin1 ( n12 n22 )1/2/ n0 ………(4) 
Figure 
NA = Sin(acceptance angle) NA = Sin {Sin1 (( n12 n22 )1/2/ n0)} from equation (4) NA = (( n12 n22 )1/2/ n0) ………(5) If the refractive index of the air medium is unity i.e. n0=1 put in (5) NA = ( n12 n22 )1/2 ………(6) Fractional change in refractive index ∆= (n1 n2)/ n1 n1∆ = (n1 n2) ………(7) from equation (6), we have NA = {( n1 n2 )( n1+n2 )}1/2 NA = { n1∆ (n1+n2 )}1/2 as n1∆ = (n1 n2) by Eq(7) NA = { n1∆ 2n1}1/2 n1 ≈ n2, so n1+n2 =2n1 NA = n1{2∆}1/2 
Structure: Core diameter : 510μm Cladding diameter : Generally around 125μm Protective layer : 250 to 1000μm Numerical aperture : 0.08 to 0.10 Bandwidth : More than 50MHz km.

Structure: Core diameter : 50350μm Cladding diameter : 125μm  500μm Protective layer : 250 to 1100μm Numerical aperture : 0.12 to 0.5 Bandwidth : Less than 50MHz km. 
Figure: 23 
Figure 
Figure

Figure: Schematic of an optical fiber

Figure: (optical fiber with core, cladding and total internally reflected ray) 
Figure: Total Internal Reflection
Figure: Optical Fibre 
Figure: Representation of Optical Fibre

Numerical aperture (NA)

(a) we know that n0 = nair = 1.0003. The numerical aperture is therefore NA = n0 sin θ0max = (1.0003) sin (8.50o) = 0.148. (b) The index of refraction of the cladding can be found from the numerical aperture: n12  n22 = NA2. (c) We know that that the n0 = nwater = 1.33. Since the numerical aperture is a property of the fiber and only depends upon n1 and n2, it will not change when the medium outside the fiber changes. The cutoff angle, however, will have to change if the numerical aperture is to be unaffected by a change in n0: NA = 0.148.

a) the critical angle θc at the core  cladding interface.
b) the numerical aperture N.A. of the optical fiber
c) the angle of acceptance αmax of the the optical fiber system.
A12)
a) θc = sin1 (n2 / n1) = sin1 (1.45 / 1.46) = 83.29 ° 
a) angle of refraction β at the outside  core interface.
b) angle θ
c) and explain why this light ray will be reflected at the core  cladding interface and hence guided along the fiber.
A13)
a) Angle β is found using Snell's law at the outside  core interface as follows 
A14) An optical fibre is characterized by one more important parameter, known as Vnumber which is more generally called normalized frequency of the fibre. It is given by the relation
V – number determines how many modes a fiber can support, It is given by, V = NA Where d is the diameter of the core, l is the wavelength of light used NA is the numerical aperture of the fibre. V = Or V = If V ≤ 2.405, then the fibre is single mode fibre (SMF) If V > 2.405, then the fibre is multimode fibre (MMF)

d = 2r = 2 x 50 x 106 m V = NA = = =94.72= normalized frequency Total number of guided mode = M = V2/2 = 4486.

Given :z=8km P(0) = 120 μW P(z) = 3 μW
a) Overall attenuation is given by,
αp (dB/km ) = 10. log = 16.02
b) Overall attenuation for 10 km, Attenuation per km = αp (dB/km ) = = = 2.00 dB/km Attenuation in 10 km link = 2.00 x 10 = 20 dB In 10 km link there will be 9 splices at 1 km interval. Each splices introducing attenuation of 1 dB. Total attenuation = 20 dB + 9 dB = 29 dB 