Solved Problems Unit 1
Unit-1 Q1)A wave is specified by y=8cos2(2t-0.8z)where y is expressed in micrometer and the propagation constant is given by um-1.find the a)amplitude b)wavelength c)angular frequency d) displacement at t=0 and z=4um.?Sol-we know that, the general form is given by-y = (amplitude) cos(t - kz) = A cos [2(t - z/)]. Therefore(a) amplitude = 8 m.(b) wavelength: 1/= 0.8 m-1 so that = 1.25 m(c) = 2(2) = 4(d) At t = 0 and z = 4 m we have y = 8 cos [2(-0.8 m-1)(4 m)] y= 8 cos [2(-3.2)] = 2.472Q2)Light traveling in air strikes a glass plate at an angle of 330,upon striking the glass part of the beam is reflected and part of it is refracted. If the refracted and reflected ray make an angle of 900 with each other what is refractive index of the glass? what is the critical angle for this glass?Sol-According to Snell's law-n1cos1 = n2cos 2where n1 = 1, 1 = 33, and 2 = 90- 33= 57n2 =cos 33cos 57= 1.540The critical angle is found fromnglass singlass = nair sinairwith air = 90and nair = 1.0critical = arcsin (sin inverse) arcsin(1/ nglass) =arcsin(1/1.540) = 40.50.(40.49266)Q3)A point source of light 12 cm below the surface of a large body of water(refractive =1.33)what is radius of the largest circle on the water surface through which the light can emerge?Sol-largest radius can be achieve only when light is at critical so from Snell's law n1 sin 1 = n2 sin c n1 sin 1=1 When n2 = 1.33, then c = 48.75 tan c =r12 cmwhich yields r = 13.7 cm,which will be largest radius.Q4)A 45-45-900 prism is immersed in alcohol(refractive index =1.45).what is the minimum refractive index the prism must have if a ray incident normally on one of the short faces is to be totally reflected at the long face of the prism ?Sol- Using Snell's law nglass sin c = nalcohol sin 90where c = 45we have,nglass =1.45/sin 45= 2.05 Q5)Calculate the numerical aperture of a step index fiber having refractive index n1=1.48 and n2=1.46.what is the maximum entrance angle 0,max for this fiber if the outer medium is air?Sol- we know that NA=(n12-n22)1/2 Substituting value of n1 and n2 we have NA= 0.2420,max = arcsin (NA/n) = arcsin(0.242/1.0)= 14 Q6) A step index multimode fiber with a NA of 0.20 support approximately 1000 modes at an 850nm wavelength a)what is the diameter of its core?b)how many modes does the fiber support at 1320nm?c)how many modes does the fiber support at 1550nm?Sol-We know that M=22a2(n12-n22)/Where (n12-n22)=NA2=0.20M=1000Wavelength=850nm Hence a can be calculated by the above equation a30.25mwe know that D = 2atherefore D =60.5 m(b) using the above equation and substituting a=30.25umWe have M=414 at the wavelength 1320nm(c) At 1550 nm, M = 300Q7)A manufacturer wishes to make a silica-core step index fiber with V=75 and a numerical aperture NA=0.30 to be used at 820nm.if n1=1.458 what should the core size and cladding index be ?Sol We know that NA=(n12-n22)1/2 Hence n2=(n12-NA2)1/2 Substituting n1=1.458NA=0.30 We have n2,by above equation as =1.427We know that a=V(wavelength )/2(NA) hence putting V=75 , NA=0.3,wavelength =820nm core size can be calculated.Q8)a certain optical fiber has an attenuation of 0.6db/km at 1310nm and 0.3db/km at 1550 nm. suppose the following two optical signal are launched simultaneously into the fiber: an optical power of 150uW at 1310 nm and optical power of 100uW at 1550nm.what are the power level in uW of these two signal at a)8km b)20Km?Sol-Since the attenuations are given in dB/km, first we need to find the power levels in dBm for100 W and 150 W. These are, respectively,P(100 W) = 10 log (100 W/1.0 mW) = 10 log (0.10) = - 10.0 dBmP(150 W) = 10 log (150 W/1.0 mW) = 10 log (0.15) = - 8.24 dBm(a) At 8 km we have the following power levels:P1310(8 km) = - 8.2 dBm (0.6 dB/km)(8 km) = - 13.0 dBm = 50 WP1550(8 km) = - 10.0 dBm (0.3 dB/km)(8 km) = - 12.4 dBm = 57.5 W(b) At 20 km we have the following power levels:P1300(20 km) = - 8.2 dBm (0.6 dB/km)(20 km) = - 20.2 dBm = 9.55 WP1550(20 km) = - 10.0 dBm (0.3 dB/km)(20 km) = - 16.0 dBm = 25.1 WQ9)An optical signal at a specific wavelength has lost 55% of its power after traversing 7.0 Km of fiber.what is the attenuation in dB/km of this fiber?Sol- we know that (db/km)=10/z log(p(0)/p(z))according to problem we have Pout = 0.45 Pin (as 55% is lost)= (10/7 km) log (1/0.45) = 0.5 dB/kmQ10)A continuous 40 km long optical fiber link has a loss of 0.4 db/Km a) What is the minimum optical power level that must be launched into the fiber to maintain an optical power level of 2.0uW at the receiving end?b) What is the required input power if the fiber has a loss of 0.6db/km? Sol- a)first of all converting power in dbm we will have output power = -27dbmHence we will have -27+(.4)(40)=10 log(Pin(W)/1mW)Therefore Pin from the above equation is given by =79.432uW b)Similarly We have -27+(.6)(40)=10log(Pin(W)/1mW)= -3 Hence Pin=501.187uW.