Solved Problems Unit 1

 

Unit-1

Q1)A wave is specified by y=8cos2π(2t-0.8z)where y is expressed in micrometer

and the propaation constant is iven by !m -".#nd the a)amp$it!de b)wave$enth c)an!$ar fre%!ency d) disp$acement at t=0 and z=&!m.' Sol-we now that the  general form is given by-

y = (amplitude) cos( t - kz) = A cos [2 ( t - z/ )].

Therefore (a) amplitude = 8 µm. (b) wavelength: 1/λ = .8 µm-1 so that λ = 1.!" µm (#) ω = 2πν = !π(!) = $π (d) %t t =  and & = $ µm we have y = 8 #os '!π(-.8 µm-1)($ µm) y= 8 #os '!π(-.!) = !.$*! Q2)*iht trave$in in air stries a $ass p$ate at an an$e of ++ 0!pon striin the

$ass part of the beam is re,ected and part of it is refracted. f the refracted and re,ected ray mae an an$e of 0 0 with each other what is refractive index of the $ass' what is the critica$ an$e for this $ass' Sol-

%##ording to +nell,s lawn1cos 1 = n2cos 2 where n1 = 1 θ1 = ° and θ! = ° - ° = "*° ∴ n! =#os °/#os "*= 1."$ The #riti#al angle is found from nglass sinφglass = nair sinφair  with φair = ° and nair = 1.

∴ φ#riti#al = ar#sin (sin inverse) φ= ar#sin(1/ nglass) =ar#sin(1/1."$) = $.".($.$!)

Q3)% point sour#e of light 1! #m below the surfa#e of a large body of water(refra#tive =1.)what is radius of the largest #ir#le #ir#le on the water surfa#e through whi#h the the light #an emerge0 Sol-largest radius #an be a#hieve only when light is at #riti#al so from +nell,s law  n1 sin 1 = n2 sin c n1 sin θ1=1 hen n! = 1. then θ# = $8.*"°  tan θ# =r /1! #m

whi#h yields r = 1.* #mwhi#h will be largest radius. Q)% $"-$"- prism is immersed in al#ohol(refra#tive inde2 =1.$").what is the minimum refra#tive inde2 the prism must have if a ray in#ident normally on one of the short fa#es is to be totally refle#ted at the long fa#e of the prism 0 Sol- 3sing +nell,s law  nglass sin θ# = nal#ohol sin ° where θ# = $"° we havenglass =1.$"/sin $"°= !." Q!)4al#ulate the numeri#al aperture of a step inde2 fiber having refra#tive inde2 n1=1.$8 and n!=1.$.what is the ma2imum entran#e angle θma2 for this fiber if the outer medium is air0 Sol- we 5now that "A=(n12-n22)1/2 +ubstituting value of n1 and n! we have 6%= .!$! #$ma% =

a&csin ("A/n)  = ar#sin(.!$!/1.) = 1$°

Q') % step inde2 multimode fiber with a 6% of .! support appro2imately 1 modes at an 8"nm wavelength a)what is the diameter of its #ore0  b)how many modes does the fiber support at 1!nm0 #)how many modes does the fiber support at 1""nm0

+ole 5now that =22a2(n12-n22)/ here (n1!-n!!)=6%!=.! 7=1 avelength=8"nm en#e a #an be #al#ulated by the above e9uation a= .!"µm *e kno* t+at , = 2a therefore  =." µm (b) using the above e9uation and substituting a=.!"um e have 7=$1$ at the wavelength 1!nm (#) %t 1"" nm 7 = 

Q)% manufa#turer wishes to ma5e a sili#a-#ore step inde2 fiber with ;=*" and a numeri#al aperture 6%=. to be used at 8!nm.if n1=1.$"8 what should the #ore si&e and #ladding inde2 be 0

Sol e 5now that "A=(n12-n22)1/2

en#e n!=(n1!-6%!)1/! +ubstituting n1=1.$"8  6%=. e have n!by above e9uation as =1.$!* e 5now that a=;(wavelength )/!<(6%) hen#e putting ;=*"  6%=.wavelength =8!nm #ore si&e #an be #al#ulated.

Q)a #ertain opti#al fiber has an attenuation of .db/5m at 11nm and .db/5m at 1"" nm. suppose the following two opti#al signal are laun#hed simultaneously into the fiber: an opti#al  power of 1"u at 11 nm and opti#al power of 1u at 1""nm.what are the power level in u of these two signal at a)85m b)!m0 Sol-+in#e the attenuations are given in d>/5m first we need to find the power levels in d>m for  1 µ and 1" µ. These are respe#tively ?(1 µ) = 1 log (1 µ/1. m) = 1 log (.1) = - 1. d>m

?(1" µ) = 1 log (1" µ/1. m) = 1 log (.1") = - 8.!$ d>m (a) %t 8 5m we have the following power levels: ?11(8 5m) = - 8.! d>m @ (. d>/5m)(8 5m) = - 1. d>m = " µ ?1""(8 5m) = - 1. d>m @ (. d>/5m)(8 5m) = - 1!.$ d>m = "*." µ (b) %t ! 5m we have the following power levels: ?1(! 5m) = - 8.! d>m @ (. d>/5m)(! 5m) = - !.! d>m = ."" µ ?1""(! 5m) = - 1. d>m @ (. d>/5m)(! 5m) = - 1. d>m = !".1 µ

Q)%n opti#al signal at a spe#ifi# wavelength has lost ""A of its power after traversing *. m of fiber.what is the attenuation in d>/5m of this fiber0 Sol- we 5now that 0(d/km)=1#/z lo(p(#)/p(z))

a##ording to problem we have ?out = .$" ?in (as ""A is lost)

α = (1/* 5m) log (1/.$") = ." d>/5m

Q1#)% #ontinuous $ 5m long opti#al fiber lin5 has a loss of .$ db/m a) hat is the minimum opti#al power level that must be laun#hed into the fiber to maintain an opti#al power level of !.u at the re#eiving end0  b) hat is the re9uired input power if the fiber has a loss of .db/5m0 Sol- a)first of all #onverting power in dbm we will have output power = -!*dbm

en#e we will have -!*B(.$)($)=1 log(?in()/1m) Therefore ?in from the above e9uation is given by =*.$!u  b)+imilarly e have -!*B(.)($)=1log(?in()/1m)= - en#e ?in="1.18*u.